# phy6

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Solutions For University Physics 11th Edition

Chapter 1

1.1: 1 mi × ( 5280 ft mi ) × (12 in. ft ) × ( 2.54 cm in .) × 1 km 105 cm = 1.61 km Although rounded to three figures, this conversion is exact because the given conversion from inches to centimeters defines the inch. 1.2:  1000 cm 3   1in  3 0.473 L ×    1 L  ×  2.54 cm  = 28.9 in .      
3

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1.3: The time required for light to travel any distance in a vacuum is the distance divided by the speed of light; 103 m = 3.33 × 10− 6 s = 3.33 × 103 ns. 8 3.00 × 10 m s 1.4: g  1 kg   100 cm  4 kg 11.3 × 3  1000 g  ×  1 m  = 1.13 × 10 m 3 .    cm    
3

1.5:

(327 in ) × ( 2.54 cm in ) × (1 L 1000 cm ) = 5.36 L.
3 3 3

1.6:

 1000 L   1 gal   128 oz.   1 bottle  1 m3 ×   1 m 3  ×  3.788 L  ×  1 gal  ×  16 oz. .                = 2111.9 bottles ≈ 2112 bottles

The daily consumption must then be bottles  1 yr  bottles 2.11 × 10 3 ×  365.24 da  = 5.78 da .  yr   1.7:

(1450 mi

hr ) × (1.61 km mi ) = 2330 km hr . 2330 km hr × 103 m km × (1 hr 3600 s ) = 648 m s.

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1.8:

180,000

furlongs  1 mile   1 fortnight   1 day  mi ×  8 furlongs  ×  14 day  ×  24 h  = 67 h .      fortnight      

1.9:

15.0

km  1 mi   3.788 L  mi × ×  1 gal  = 35.3 gal .  L  1.609 km   

 mi  1.10: a)  60   hr 

 1h    3600 s    

 5280 ft  ft   1 mi  = 88 s   

m  ft   30.48 cm   1 m  b)  32 2    1ft   100 cm  = 9.8 s 2   s   
3

g   100 cm   c) 1.0   cm 3   1 m  

 1 kg  3 kg   1000 g  = 10 m 3   

1.11: The density is mass per unit volume, so the volume is mass divided by density. V = 60 × 103 g 19.5 g cm 3 = 3077 cm 3 4 Use the formula for the volume of a sphere, V = πr 3 , 3 1/ 3 to calculate r : r = ( 3V 4π ) = 9.0 cm

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1.12:

(3.16 × 10 7 s − π × 10 7 s) (3.16 × 10 7 s) × 100 = 0.58% 10 m = 1.1 × 10 −3 %. 3 890 × 10 m

1.13: a)

b) Since the distance was given as 890 km, the total distance should be 890,000 meters. To report the total distance as 890,010 meters, the distance should be given as 890.01 km. 1.14: a) (12 mm ) × ( 5.98 mm ) = 72 mm 2 (two significant figures). .98 mm b) 512 mm = 0.50 (also two significant figures). c) 36 mm (to the nearest millimeter). d) 6 mm. e) 2.0.

1.15: a) If a meter stick can measure to the nearest millimeter, the error will be about 0.13%. b) If the chemical balance can measure to the nearest milligram, the error will be about 8.3 × 10−3%. c) If a handheld stopwatch (as opposed to electric timing devices) can measure to the nearest tenth of a second, the error will be about 2.8 × 10 −2 %.

1.16: The area is 9.69 ± 0.07 cm2, where the extreme values in the piece’s length and width are used to find the uncertainty in the area. The fractional uncertainty in the cm 2 area is 0..07 cm 2 = 0.72%, and the fractional uncertainties in the length and width are 9 69
0.01 cm 5.10 cm

= 0.20% and

0.01 cm 1.9 cm

= 0.53%.

1.17: a) The average volume is ( 8.50 cm ) 2 ( 0.050 cm ) = 2.8 cm 3 π 4 (two significant figures) and the uncertainty in the volume, found from the extreme values of the diameter and thickness, is about 0.3 cm 3 , and so the volume of a cookie is 2.8 ± 0.3 cm3 . (This method does not use the usual form for progation of errors, which is not addressed in the text. The fractional uncertainty in the thickness is so much greater than the fractional uncertainty in the diameter that the fractional uncertainty in the volume is 10% , reflected in the above answer.) b)
8.50 .05

= 170 ± 20.

1.18: (Number of cars × miles/car.day)/mi/gal = gallons/day (2 × 108 cars × 10000 mi/yr/car × 1 yr/365 days)/(20 mi/gal) = 2.75 × 108 gal/day 1.19: Ten thousand; if it were to contain ten million, each sheet would be on the order of a millionth of an inch thick. 1.20: If it takes about four kernels to fill 1 cm3, a 2-L bottle will hold about 8000 kernels.

1 20

L (50 cm3) per beat, and about

1 4

gallon per

1.28: The moon is about 4 ×108 m = 4 × 1011 mm away. Depending on age, dollar bills can be stacked with about 2-3 per millimeter, so the number of bills in a stack to the moon would be about 1012. The value of these bills would be \$1 trillion (1 terabuck).

1.29:

(9,372,571 km

( Area of USA ) ( Area
2

× 10 m km
6 2

2

) (15.6 cm × 6.7 cm × 1 m

bill ) = number of bills .
2

104 cm 2 = 9 × 1014 bills

)

9 × 1014 bills 2.5 108 inhabitant s = \$3.6 million inhabitant . 1.30:

1.31:

7.8 km, 38  north of east

1.32:

a) b) c) d) 1.33:

11.1 m @ 28.5 m @ 11.1 m @ 28.5 m @

77.6 o 202 o 258o 22 o

144 m, 41 south of west.

1.34:

1.35:

 A; Ax = (12.0 m ) sin 37.0  = 7.2 m, Ay = (12.0 m ) cos 37.0  = 9.6 m.  B; B x = (15.0 m ) cos 40.0  = 11.5 m, B y = −(15.0 m ) sin 40.0  = −9.6 m.  C ; C x = −( 6.0 m ) cos 60.0  = −3.0 m, C y = −( 6.0 m ) sin 60.0  = −5.2 m.
Ay − 1.00 m = = −0.500 AX 2.00 m

1.36:

(a)

tan θ =

θ = tan −1 ( − 0.500 ) = 360  − 26.6  = 333 Ay 1.00 m (b) tan θ = = = 0.500 Ax 2.00 m θ = tan −1 ( 0.500) = 26.6  Ay 1.00 m tan θ = = = −0.500 Ax − 2.00 m

(c )

(d)

θ = tan −1 ( − 0.500) = 180  − 26.6  = 153 Ay − 1.00 m tan θ = = = 0.500 Ax − 2.00 m θ = tan −1 ( 0.500) = 180  + 26.6  = 207 

1.37: Take the +x-direction to be forward and the +y-direction to be upward. Then the second force has components F2 x = F2 cos 32.4 = 433 N and F2 y = F2 sin 32.4 = 275 N. The first force has components F1x = 725 N and F1 y = 0. Fx = F1x + F2 x = 1158 N and Fy = F1 y + F2 y = 275 N The resultant force is 1190 N in the direction 13.4 above the forward direction. 1.38: (The figure is given with the solution to Exercise 1.31). The net northward displacement is (2.6 km) + (3.1 km) sin 45o = 4.8 km, and the net eastward displacement is (4.0 km) + (3.1 km) cos 45o = 6.2 km. The magnitude of the resultant displacement is (4.8 km) 2 + (6.2 km) 2 = 7.8 km, and 48 the direction is arctan ( 6..2 ) = 38o north of east.

 1.39: Using components as a check for any graphical method, the components of B are  Bx = 14.4 m and By = 10.8 m, A has one component, Ax = −12 m .
a) The x - and y - components of the sum are 2.4 m and 10.8 m, for a magnitude  10.8  2 2  of ( 2.4 m ) + (10.8 m ) = 11.1 m, , and an angle of   = 77.6 . 2.4   b) The magnitude and direction of A + B are the same as B + A. c) The x- and y-components of the vector difference are – 26.4 m and − − 10.8 m, for a magnitude of 28.5 m and a direction arctan ( −10..8 ) = 202. Note that 26 4
− 180 must be added to arctan( −10..8 ) = arctan( 10..8 ) = 22 in order to give an angle in the 26 4 26 4 third quadrant.

  ˆ ˆ ˆ d) B − A = 14.4 mi + 10.8 mˆ + 12.0 mi = 26.4 mi + 10.8 mˆ. j j Magnitude =

( 26.4 m ) 2 + (10.8 m ) 2

 10.8   = 28.5 m at and angle of arctan  = 22.2 .  26.4 

1.40: Using Equations (1.8) and (1.9), the magnitude and direction of each of the given vectors is: a) o 180 – 31.2o). b) c) o 360 – 19.2o). (−8.6 cm) 2 + (5.20 cm) 2 = 10.0 cm, arctan

( −58.20 ) .60

= 148.8o (which is

(−9.7 m) 2 + (−2.45 m) 2 = 10.0 m, arctan (7.75 km) 2 + (−2.70 km) 2

( −−29.45 ) .7

= 14o + 180o = 194o. = 340.8o (which is

= 8.21 km, arctan

. ( −.2757 ) 7

1.41:

The total northward displacement is 3.25 km − 1.50 km = 1.75 km, , and the total westward displacement is 4.75 km . The magnitude of the net displacement is

(1.75 km ) 2 + ( 4.75 km ) 2

= 5.06 km. The south and west displacements are the same, so

The direction of the net displacement is 69.80 West of North. 1.42: a) The x- and y-components of the sum are 1.30 cm + 4.10 cm = 5.40 cm, 2.25 cm + (–3.75 cm) = –1.50 cm. b) Using Equations (1-8) and (1-9), (5.4 0 cm) 2 (−1.50 cm) 2 = 5.60 cm, arctan
− ( +15..50 ) 40

= 344.5o ccw.

c) Similarly, 4.10 cm – (1.30 cm) = 2.80 cm, –3.75 cm – (2.25 cm) = –6.00 cm. d) (2.80 cm) 2 + (−6.0 cm) 2 = 6.62 cm, arctan
.00 ( −26.80 )

= 295o (which is 360o – 65o).

  1.43: a) The magnitude of A + B

is

 ( 2.80 cm ) cos 60.0 + (1.90 cm ) cos 60.0 2   + ( 2.80 cm ) sin 60.0 − (1.90 cm ) sin 60.0 

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)

  = 2.48 cm 2 

and the angle is

 ( 2.80 cm ) sin 60.0 − (1.90 cm ) sin 60.0   arctan   ( 2.80 cm ) cos 60.0 + (1.90 cm ) cos 60.0  = 18   

  b) The magnitude of A − B is

 ( 2.80 cm ) cos 60.0 − (1.90 cm ) cos 60.0 2   + ( 2.80 cm ) sin 60.0 + (1.90 cm ) sin 60.0 

(

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)

  = 4.10 cm 2 

and the angle is

 ( 2.80 cm ) sin 60.0 + (1.90 cm ) sin 60.0   arctan   ( 2.80 cm ) cos 60.0 − (1.90 cm ) cos 60.0  = 84        c) B − A = − A − B ; the magnitude is 4.10 cm and the angle is 84  + 180  = 264 .

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1.44:

→

ˆ A = (–12.0 m) i . More precisely,    A = (12.0 m ) cos 180  i + (12.0 m ) sin 180  j .

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 ˆ ˆ B = (18.0 m ) cos 37  i + (18.0 m ) sin 37  ˆ = (14.4 m ) i + (10.8 m ) ˆ j j

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1.45:

 ˆ ˆ A = (12.0 m ) sin 37.0 i + (12.0 m ) cos 37.0 ˆ = ( 7.2 m ) i + ( 9.6 m ) ˆ j j  ˆ  ˆ ˆ B = (15.0 m ) cos 40.0 i − (15.0 m ) sin 40.0 j = (11.5 m ) i − ( 9.6 m ) ˆ j  ˆ ˆ C = −( 6.0 m ) cos 60.0 i − ( 6.0 m ) sin 60.0 ˆ = −( 3.0 m ) i − ( 5.2 m ) ˆ j j

1.46: a) b)

 ˆ ˆ A = ( 3.60 m ) cos 70.0 i + ( 3.60 m ) sin 70.0 ˆ = (1.23 m ) i + ( 3.38 m ) ˆ j j  ˆ  ˆ ˆ B = −( 2.40 m ) cos 30.0 i − ( 2.40 m ) sin 30.0 j = ( − 2.08 m ) i + ( − 1.20 m ) ˆ j    C = ( 3.00 ) A − ( 4.00 ) B ˆ ˆ = ( 3.00 )(1.23 m ) i + ( 3.00 )( 3.38 m ) ˆ − ( 4.00 )( − 2.08 m ) i − ( 4.00 )( − 1.20 m ) ˆ j j ˆ = (12.01 m ) i + (14.94 ) ˆ j

c)

(Note that in adding components, the fourth figure becomes significant.) From Equations (1.8) and (1.9),  14.94 m  2 2  C = (12.01 m ) + (14.94 m ) = 19.17 m, arctan   = 51.2  12.01 m 

1.47: a) b) c) d)

A=

( 4.00) 2 + ( 3.00) 2

= 5.00, B =

( 5.00) 2 + ( 2.00) 2

= 5.39

  ˆ ˆ A − B = ( 4.00 − 3.00 ) i + ( 5.00 − ( − 2.00) ) ˆ = ( − 1.00 ) i + ( 5.00) ˆ j j (1.00) 2 + ( 5.00) 2 = 5.10, arctan 5.00  = 101.3    - 1.00 

ˆ j ˆ 1.48: a) i + ˆ + k = 12 + 12 + 12 = 3 ≠ 1 so it is not a unit vector b)  A=
2 2 Ax + Ay + Az2

 If any component is greater than + 1 or less than –1, A ≥ 1 , so it cannot be a unit  vector. A can have negative components since the minus sign goes away when the component is squared. c)
2

 A =1 a 2 ( 3.0) + a 2 ( 4.0) = 1
2

a 2 25 = 1 a=± 1 = ±0.20 5.0

  ˆ ˆ j B = Bx i + By ˆ. j 1.49: a) Let A = Ax i + Ay ˆ,   ˆ A + B = ( Ax + Bx ) i + ( Ay + By ) ˆ j   ˆ B + A = ( Bx + Ax ) i + ( By + Ay ) ˆ j     Scalar addition is commutative, so A + B = B + A
  A ⋅ B = Ax Bx + Ay B y   B ⋅ A = Bx Ax + By Ay

    Scalar multiplication is commutative, so A ⋅ B = B ⋅ A   ˆ ˆ A × B = ( Ay Bz − Az B y ) i + ( Az Bx − Ax Bz ) ˆ + ( Ax B y − Ay Bx ) k j   ˆ ˆ B × A = ( B y Az − Bz Ay ) i + ( Bz Ax − Bx Az ) ˆ + ( Bx Ay − B y Ax ) k j Comparison of each component in each vector product shows that one is the negative of the other.
b)

1.50: Method 1: ( Pr oduct of magnitudes × cos θ ) BC cos θ = (15 m × 6 m ) cos 80  = 15.6 m 2

AB cos θ = (12 m × 15 m ) cos 93 = −9.4 m 2 AC cos θ = (12 m × 6 m ) cos 187  = −71.5 m 2

Method 2: (Sum of products of components) A ⋅ B = ( 7.22 ) (11.49) + (9.58)(−9.64) = −9.4 m 2 B ⋅ C = (11.49)(−3.0) + (−9.64)(−5.20) = 15.6 m 2 A ⋅ C = (7.22)( −3.0) + (9.58)(−5.20) = −71.5 m 2

1.51: a) From Eq.(1.21),   A ⋅ B = ( 4.00 )( 5.00 ) + ( 3.00 )( − 2.00 ) = 14.00. b) A ⋅ B = AB cos θ , so θ = arccos [ (14.00) ( 5.00 × 5.39 ) ] = arccos( .5195) = 58.7 . 1.52: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to give the angle φ as    A ⋅ B  Ax Bx + Ay B y   . φ = arccos     AB  = arccos AB     In the intermediate calculations given here, the significant figures in the dot products and in the magnitudes of the vectors are suppressed. a)   A ⋅ B = − 22, A = 40, B = 13, and so

 − 22    φ = arccos   40 13  = 165 .   b) c)   60   A ⋅ B = 60, A = 34, B = 136, φ = arccos  = 28 .  34 136    A ⋅ B = 0, φ = 90.

1.53: Use of the right-hand rule to find cross products gives (a) out of the page and b) into the page. 1.54: a) From Eq. (1.22), the magnitude of the cross product is

(12.0 m )(18.0 m ) sin (180 − 37 ) = 130 m 2
The right-hand rule gives the direction as being into the page, or the – z-direction. Using Eq. (1.27), the only non-vanishing component of the cross product is C z = Ax B y = ( − 12 m ) (18.0 m ) sin 37 = −130 m 2 b) The same method used in part (a) can be used, but the relation given in Eq. (1.23) gives the result directly: same magnitude (130 m2), but the opposite direction (+z-direction).

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1.55: In Eq. (1.27), the only non-vanishing component of the cross product is C z = Ax By − Ay Bx = ( 4.00)( − 2.00) − ( 3.00)( 5.00) = −23.00,   ˆ so A × B = −( 23.00) k , and the magnitude of the vector product is 23.00.

  1.56: a) From the right-hand rule, the direction of A× B is into the page (the – z-direction). The magnitude of the vector product is, from Eq. (1.22), AB sin φ = ( 2.80 cm )(1.90 cm ) sin 120 = 4.61 cm 2 . Or, using Eq. (1.27) and noting that the only non-vanishing component is C z = Ax B y − Ay B x = ( 2.80 cm ) cos 60.0  ( − 1.90 cm ) sin 60  = −4.61 cm 2 gives the same result. b) Rather than repeat the calculations, Eq. (1-23) may be used to see that   B × A has magnitude 4.61 cm2 and is in the +z-direction (out of the page). − ( 2.80 cm ) sin 60.0  (1.90 cm ) cos 60.0 

1.57: a) The area of one acre is mile. b)

1 8

1 mi × 80 mi =

1 640

mi 2 , so there are 640 acres to a square

2   (1 acre) ×  1 mi  ×  5280 ft  = 43,560 ft 2  640 acre   1 mi     

2

(all of the above conversions are exact).  7.477 gal  5 (1 acre-foot) = 43,560 ft 3 ×   = 3.26 × 10 gal, 3  1 ft  which is rounded to three significant figures. c) (\$4,950,000 102 acres) × (1acre 43560 ft 2 ) × 10.77 ft 2 m 2 = \$12 m 2 . (\$12 m 2 ) × (2.54 cm in) 2 × (1 m 100 cm) 2 = \$.008 in 2 . \$.008 in 2 × (1in × 7 8 in ) = \$.007 for postage stamp sized parcel.

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1.58: a) b) c)

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1.59: a) To three significant figures, the time for one cycle is 1 = 7.04 ×10 −10 s. 9 1.420 × 10 Hz b)  9 cycles   3600 s  12 cycles 1.420 ×10 ×  = 5.11×10 s   1h  h 

c) Using the conversion from years to seconds given in Appendix F,  3.156 ×10 7 s  9  × 4.600 ×10 9 y = 2.06 ×10 26. 1.42 ×10 Hz ×    1y  

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d) 4.600 × 10 9 y = 4.60 × 10 4 1.00 × 10 5 y , so the clock would be off by 4.60 × 10 4 s. 1.60: Assume a 70-kg person, and the human body is mostly water. Use Appendix D to find the mass of one H2O molecule: 18.015 u × 1.661 × 10–27 kg/u = 2.992 × 10–26 kg/molecule. (70 kg/2.992 × 10–26 kg/molecule) = 2.34 × 1027 molecules. (Assuming carbon to be the most common atom gives 3 × 1027 molecules. 1.61: a) Estimate the volume as that of a sphere of diameter 10 cm: 4 V = πr 3 = 5.2 × 10 −4 m 3 3 Mass is density times volume, and the density of water is 1000 kg m3 , so m = ( 0.98) 1000 kg m3 5.2 × 10− 4 m3 = 0.5 kg

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b) Approximate as a sphere of radius r = 0.25 μm (probably an over estimate) 4 V = πr 3 = 6.5 × 10 −20 m 3 3 m = ( 0.98) 1000 kg m 3 6.5 × 10−20 m3 = 6 × 10 −17 kg = 6 × 10-14 g

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c) Estimate the volume as that of a cylinder of length 1 cm and radius 3 mm: V = πr 2l = 2.8 × 10 −7 m 3 m = ( 0.98) 1000 kg m 3 2.8 × 10 −7 m 3 = 3 × 10 −4 kg = 0.3 g

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1.62: a)

M M , so V = V ρ 0.200 kg x3 = = 2.54 × 10 −5 m 3 3 3 7.86 × 10 kg/m ρ= x = 2.94 × 10 −2 m = 2.94 cm

b)

4 3 πR = 2.54 × 10 −5 m 3 3 R = 1.82 × 10 −2 m = 1.82 cm

1.63: Assume each person sees the dentist twice a year for checkups, for 2 hours. Assume 2 more hours for restorative work. Assuming most dentists work less than 2000 hours per year, this gives 2000 hours 4 hours per patient = 500 patients per dentist. Assuming only half of the people who should go to a dentist do, there should be about 1 dentist per 1000 inhabitants. Note: A dental assistant in an office with more than one treatment room could increase the number of patients seen in a single dental office.  6.0 × 1023 atoms mole (6.0 × 10 kg) ×   14 × 10− 3 kg mole 
24

1.64: a) b)

  = 2.6 × 1050 atoms.  

The number of neutrons is the mass of the neutron star divided by the mass of a neutron:

(2) (2.0 × 10 30 kg) = 2.4 × 10 57 neutrons. (1.7 × 10 −27 kg neutron) c) The average mass of a particle is essentially 2 the mass of either the proton or 3 the neutron, 1.7 × 10 −27 kg. The total number of particles is the total mass divided by this average, and the total mass is the volume times the average density. Denoting the density by ρ (the notation introduced in Chapter 14). 4 π R 3ρ (2π) (1.5 × 1011 m) 3 (1018 kg m 3 ) M = 3 = = 1.2 × 10 79. − 27 2 mave (1.7 × 10 kg ) mp 3 Note the conversion from g/cm3 to kg/m3.

 1.65: Let D be the fourth force.
        A + B + C + D = 0, so D = − A + B + C Ax = + A cos 30.0  = +86.6 N,

(

Ay = + A cos 30.0  = +50.00 N

)

B x = − B sin 30.0  = −40.00 N, B y = + B cos 30.0  = +69.28 N C x = +C cos 53.0  = −24.07 N, C y = −C sin 53.0  = −31.90 N Then D x = −22.53 N,
2 D = D x2 + D y = 90.2 N;

D y = −87.34 N

tan α = Dy / Dx = 87.34 / 22.53

α = 75.54 φ = 180  + α = 256  , counterclockwise from + x - axis
1.66:

R x = Ax + B x = (170 km) sin 68  + (230 km) cos 48  = 311.5 km R y = Ay + B y = (170 km) cos 68  − (230 km) sin 48  = −107.2 km
2 2 R = Rx + R y =

( 311.5 km ) 2 + ( − 107.2 km ) 2
Ry Rx = 107.2 km = 0.344 311.5 km

= 330 km

tanθ R = θ R = 19  south of east

1.67: a)

    b) Algebraically, A = C − B, and so the components of A are

Ax = C x − Bx = ( 6.40 cm ) cos 22.0 − ( 6.40 cm ) cos 63.0  = 3.03 cm

Ay = C y − B y = ( 6.40 cm ) sin 22.0  + ( 6.40 cm ) sin 63.0 = 8.10 cm. c) A=

( 3.03 cm ) 2 + ( 8.10 cm ) 2

 8.10 cm   = 8.65 cm, arctan  = 69.5  3.03 cm 

1.68:a) Rx = Ax + Bx + C x

= (12.0 m ) cos 90 − 37 + (15.00 m ) cos − 40 + ( 6.0 m ) cos 180 + 60 = 15.7 m, and = (12.0 m ) sin 90 − 37 + (15.00 m ) sin − 40 + ( 6.0 m ) sin 180 + 60 = −5.3 m.

(

)

(

)

(

)

Ry = Ay + By + C y

(

)

(

)

(

)

2 2 The magnitude of the resultant is R = Rx + Ry = 16.6 m , and the direction from

−5 3 the positive x-axis is arctan ( 15..7 ) = −18.6 . Keeping extra significant figures in the intermediate calculations gives an angle of − 18.49°, which when considered as a positive counterclockwise angle from the positive x-axis and rounded to the nearest degree is 342 .

b)

S x = −3.00 m − 7.22 m − 11.49 m = −21.71 m;  (−5.14)   θ = arctan   = 13.3  (−21.71)  S = (−21.71 m) 2 + (−5.14 m) 2 = 22.3 m

S y = −5.20 m − ( − 9.64 m ) − 9.58 m = −5.14 m;

1.69:

Take the east direction to be the x - direction and the north direction to be the y - direction. The x- and y-components of the resultant displacement of the first three displacements are then

( − 180 m ) + ( 210 m ) sin 45 + ( 280 m ) sin 30  = 108 m, − ( 210 m ) cos 45 + ( 280 m ) cos 30  = +94.0 m,
keeping an extra significant figure. The magnitude and direction of this net displacement are (108 m ) 2 + ( 94.0 m ) 2 = 144 m, arctan 94 m  = 40.9.    108 m  The fourth displacement must then be 144 m in a direction 40.9 south of west.

1.70:

The third leg must have taken the sailor east a distance

( 5.80 km ) − ( 3.50 km ) cos 45 − ( 2.00 km ) = 1.33 km
and a distance north

( 3.5 km ) sin 45 = ( 2.47 km )
The magnitude of the displacement is (1.33 km) 2 + (2.47 km) 2 = 2.81 km
2 47 and the direction is arctan ( 1..33 ) = 62° north of east, which is 90° − 62° = 28° east of north. A more precise answer will require retaining extra significant figures in the intermediate calculations.

1.71: a)

b) The net east displacement is − ( 2.80 km ) sin 45  + ( 7.40 km ) cos 30  − ( 3.30 km ) cos 22  = 1.37 km, and the net north displacement is − ( 2.80 km ) cos 45  + ( 7.40 km ) sin 30  − ( 3.30 km ) sin 22.0  = 0.48 km, and so the distance traveled is

(1.37 km ) 2 + ( 0.48 km ) 2

= 1.45 km.

1.72: The eastward displacement of Manhattan from Lincoln is

(147 km ) sin 85 + (106 km ) sin 167 + (166 km ) sin 235 = 34.3 km
and the northward displacement is

(147 km ) cos 85 + (106 km ) cos 167 + (166 km ) cos 235 = −185.7 km
(A negative northward displacement is a southward displacement, as indicated in Fig. (1.33). Extra figures have been kept in the intermediate calculations.) a) b) (34.3 km) 2 + (185.7 km) 2 = 189 km The direction from Lincoln to Manhattan, relative to the north, is  34.3 km   arctan   − 185.7 km  = 169.5    and so the direction to fly in order to return to Lincoln is 169.5 + 180 + 349.5 .

20 1.73: a) Angle of first line is θ = tan −1 ( 200 −10 ) = 42. Angle of 210 −

second line is 42 + 30 = 72. Therefore X = 10 + 250 cos 72 = 87 Y = 20 + 250 sin 72 = 258 for a final point of (87,258). b) The computer screen now looks something like this:

The length of the bottom line is

tan −1 ( 258 −−200 ) = 25 below straight left. 210 87

( 210 − 87 ) 2 + ( 200 − 258) 2 = 136

and its direction is

1.74: a)

b)

To use the method of components, let the east direction be the x-direction and the north direction be the y-direction. Then, the explorer’s net xdisplacement is, in units of his step size,

( 40) cos 45 − ( 80) cos 60 = −11.7
and the y-displacement is

( 40) sin 45 + ( 80) sin 60 − 50 = 47.6.
The magnitude and direction of the displacement are  47.6   (−11.7) 2 + (47.6) 2 = 49, arctan   = 104 .  − 11.7  (More precision in the angle is not warranted, as the given measurements are to the nearest degree.) To return to the hut, the explorer must take 49 steps in a direction 104 − 90 = 14 east of south.

 1.75: Let +x be east and +y be north. Let A be the displacement 285 km at 40.0  north  of west and let B be the unknown displacement.
    A + B = R, where R = 115 km, east    B = R− A B x = R x − Ax , B y = R y − Ay Ax = − A cos 40.0  = −218.3 km, Ay = + A sin 40.0  = +183.2 km R x = 115 km, R y = 0 Then B x = 333.3 km, B y = −183.2 km.
2 2 B = B x + B y = 380 km.

tanα = B y Bx = (183.2 km ) ( 333.3 km ) α = 28.8 , south of east 1.76:

(a) (b) (c)

ωpar = ω sin α ωperp = ω cosα ωpar = ω sin α ω= ωpar sin α = 550 N = 960 N sin 35.0 0

1.77:

 B is the force the biceps exerts.  E is the force the elbow exerts.    E + B = R, where R = 132.5 N and is upward. E x = Rx − Bx , E y = R y − B y B x = − B sin 43 = −158.2 N, B y = + B cos 43 = +169.7 N R x = 0, R y = +132.5 N Then E x = +158.2 N, E y = −37.2 N
2 2 E = E x + E y = 160 N;

tanα = E y E x = 37.2 158.2 α = 13 , below horizontal

1.78: (a) Take the beginning of the journey as the origin, with north being the ydirection, east the x-direction, and the z-axis vertical. The first displacement is then ˆ ˆ − 30k , the second is − 15 ˆ , the third is 200i (0.2 km = 200 m) , and the fourth is 100 ˆ . j j Adding the four: ˆ ˆ ˆ ˆ − 30k − 15 ˆ + 200i + 100 ˆ = 200i + 85 ˆ − 30k j j j (b) The total distance traveled is the sum of the distances of the individual segments: 30 + 15 + 200 + 100 = 345 m. The magnitude of the total displacement is:
2 D = D x2 + D y + D z2 = 200 2 + 85 2 + ( − 30) = 219 m 2

 1.79: Let the displacement from your camp to the store be A. A = 240 m, 32  south of east   B is 32  south of west and C is 62  south of west Let + x be east and + y be north    A+ B+C = 0 Ax + B x + C x = 0, so A cos 32  − B cos 48  − C cos 62  = 0 Ay + B y + C y = 0, so − A sin 32  + B sin 48  − C sin 62  = 0 A is known so we have two equations in the two unknowns B and C. Solving gives B = 255 m and C = 70 m. 1.80: Take your tent's position as the origin. The displacement vector for Joe's tent is ˆ ˆ 21 cos 23 i − 21 sin 23 ˆ = 19.33i − 8.205 ˆ. The displacement vector for Karl's tent is j j  ˆ  ˆ ˆ 32 cos 37 i + 32 sin 37 j = 25.56i + 19.26 ˆ . The difference between the two j displacements is:

( (

) ( ) (

) )

(19.33 − 25.56) iˆ + ( − 8.205 − 19.25) ˆ = −6.23iˆ − 27.46 ˆ . j j
The magnitude of this vector is the distance between the two tents: D=

( − 6.23) 2 + ( − 27.46) 2

= 28.2 m

1.81: a) With Az = B z = 0, Eq. (1.22) becomes Ax Bx + Ay B y = ( A cos θ A )( B cos θ B ) + ( A sin θ A )( B sin θ B ) = AB( cos θ Acos θ B + sin θ Asin θ B ) = AB cos( θ A − θ B ) = AB cos φ where the expression for the cosine of the difference between two angles has been used (see Appendix B).

 ˆ b) With Az = Bz = 0, C = C z k and C = C z . From Eq. (1.27),
C = Ax Bx − Ay Bx = ( A cos θ A )( B cos θ B ) − ( A sin θ A )( B cos θ A ) = AB cos θ A sin θ B − sin θ A cos θ B = AB sin ( θ B − θ A ) = AB sin φ The angle between the vectors is 210 − 70 = 140 , and so Eq. (1.18) gives   A ⋅ B = ( 3.60 m )( 2.40 m ) cos 140 = −6.62 m 2 Or, Eq. (1.21) gives   A ⋅ B = Ax Bx + Ay By = ( 3.60 m ) cos 70 ( 2.4 m ) cos 210 + ( 3.6 m ) sin 70 ( 2.4 m ) sin 210 = −6.62 m 2 b) From Eq. (1.22), the magnitude of the cross product is

1.82:

a)

( 3.60 m )( 2.40 m ) sin 140 = 5.55 m 2 ,
and the direction, from the right-hand rule, is out of the page (the   +z-direction). From Eq. (1-30), with the z-components of A and B vanishing, the z-component of the cross product is Ax By − Ay Bx = ( 3.60 m ) cos 70 ( 2.40 m ) sin 210 − ( 3.60 m ) sin 70 ( 2.40 m ) cos 210 = 5.55 m 2

1.83: a) Parallelog ram area = 2 × area of triangle ABC Triangle area = 1 2 ( base )( height ) = 1 2 ( B)( A sin θ ) Parellogram area = BA sin θ b) 90


1.84: With the +x-axis to the right, +y-axis toward the top of the page, and +z-axis out       of the page, A × B x = 87.8 cm 2 , A × B y = 68.9 cm 2 , A × B z = 0.

(

)

(

)

(

)

1.85: a)

A= B=

( 2.00) 2 + ( 3.00) 2 + ( 4.00) 2 = 5.39. ( 3.00) 2 + (1.00) 2 + ( 3.00) 2 = 4.36.
ˆ ˆ = ( − 5.00) i + ( 2.00) ˆ + ( 7.00 ) k j

b)

  ˆ ˆ A − B = ( Ax − Bx ) i + ( Ay − By ) ˆ + ( Az − Bz ) k j

c)

( 5.00) 2 + ( 2.00) 2 + ( 7.00) 2

= 8.83,

  and this will be the magnitude of B − A as well.
1.86: The direction vectors each have magnitude 3 , and their dot product is (1) (1) + (1) (–1) + (1) (–1) = –1, so from Eq. (1-18) the angle between the bonds is arccos − = 3 1 3 = arccos ( − 1 ) = 109 . 3

( )

1.87: The best way to show these results is to use the result of part (a) of Problem 1-65, a restatement of the law of cosines. We know that C 2 = A2 + B 2 + 2 AB cosφ ,   where φ is the angle between A and B .

  a) If C 2 = A 2 + B 2 , cos φ = 0, and the angle between A and B is 90  (the vectors are perpendicular).   b) If C 2 < A2 + B 2 , cosφ < 0, and the angle between A and B is greater than 90  .   c) If C 2 > A 2 + B 2 , cosφ > 0, and the angle between A and B is less than 90  .

1.88: a) This is a statement of the law of cosines, and there are many ways to derive it. The most straightforward way, using vector algebra, is to assume the linearity of the dot product (a point used, but not explicitly mentioned in the text) to show that   the square of the magnitude of the sum A + B is       ( A + B ) ⋅ ( A + B ) = A ⋅ A + A ⋅ B + B ⋅A + B ⋅ B     =A⋅ A+ B⋅B + 2 A⋅B   = A2 + B 2 + 2 A ⋅ B = A 2 + B 2 + 2 AB cos φ Using components, if the vectors make angles θA and θB with the x-axis, the components of the vector sum are A cos θA + B cos θB and A sin θA + B sin θB, and the square of the magnitude is      

( A cos θ A + B cos θB ) 2 + ( A sin θ A + B sin θB ) 2

= A 2 cos 2 θ A + sin 2 θ A + B 2 cos 2θ B + sin 2 θ B = A 2 + B 2 + 2 AB cos( θ A − θ B ) = A 2 + B 2 + 2 AB cos φ where φ = θA – θB is the angle between the vectors. b)

(

)

+ 2 AB( cosθ A cos θ B + sin θ A sin θ B )

(

)

    A geometric consideration shows that the vectors A, B and the sum A + B   must be the sides of an equilateral triangle. The angle between A, and B is 120o, since one vector must shift to add head-to-tail. Using the result of part (a), with A = B, the condition is that A 2 = A2 + A 2 + 2 A 2 cos φ , which solves for 1 = 2 + 2 cos φ, cos φ = − 1 , and φ = 120o. 2 Either method of derivation will have the angle φ replaced by 180o – φ, so the cosine will change sign, and the result is A 2 + B 2 − 2 AB cos φ . Similar to what is done in part (b), when the vector difference has the same magnitude, the angle between the vectors is 60o. Algebraically, φ is obtained from 1 = 2 – 2 cos φ, so cos φ = 1 and φ = 60o. 2

c) d)

1.89: Take the length of a side of the cube to be L, and denote the vectors from a to b, a    to c and a to d as B, C , and D . In terms of unit vectors,    ˆ ˆ j ˆ B = Lk , C = L ˆ+k , j ˆ D = L i + ˆ+k .

(

)

(

)

Using Eq. (1.18),

   B⋅D  L2  = arccos  arccos   ( L) L 3  BD       C ⋅D  2 L2  = arccos  arccos   L 2 L  CD    

(

)

(

)(

  = 54.7 ,     = 35.3. 3  

)

1.90: From Eq. (1.27), the cross product is  ˆ ˆ ˆ  6.00  ˆ − 11.00 k . ˆ (−13.00) i + (6.00) ˆ + (−11.00) k = 13 − (1.00 ) i +  j j 13.00   13.00    The magnitude of the vector in square brackets is 1.93, and so a unit vector in this   direction (which is necessarily perpendicular to both A and B ) is ˆ ˆ  − (1.00) i + (6.00 13.00) ˆ − (11.00 13) k  j  . 1.93   The negative of this vector, ˆ ˆ  (1.00) i − (6.00 13.00) ˆ + (11.00 13) k  j  , 1.93     is also a unit vector perpendicular to A and B .

1.91:

    A and C are perpendicular, so A ⋅ C = 0. AxC x + Ay C y = 0, , which gives

5.0C x − 6.5C y = 0.   B ⋅ C = 15.0, so − 3.5C x + 7.0C y = 15.0 We have two equations in two unknowns C x and C y . Solving gives C x = 8.0 and C y = 6.1.   A × B = AB sin θ   A× B ( − 5.00) 2 + ( 2.00) 2 sin θ = = = 0.5984 ( 3.00)( 3.00) AB θ = sin −1 ( 0.5984 ) = 36.8

1.92:

1.93: a) Using Equations (1.21) and (1.27), and recognizing that the vectors    A, B, and C do not have the same meanings as they do in those equations,    ( A × B ) ⋅ C = (( A B
y z

 ˆ ˆ − Az B y ) i + ( Az B x − Ax B z ) ˆ + ( Ax B y − Ay B x ) k ⋅ C j

)

= Ay B z C x − Az B y C x + Az B x C y − Ax B z C y + Ax B y C z − Ay B x C z . A similar calculation shows that

   A ⋅ B × C = Ax By C z − Ax Bz C y + Ay Bz C x − Ay BxC z + Az BxC y − Az By C x and a comparison of the expressions shows that they are the same.

(

)

b) Although the above expression could be used, the form given allows for ready   compuation of A × B the magnitude is AB sin φ = ( 20.00) sin 37.0 and the direction is, from the right-hand rule, in the +z-direction, and so    ( A × B ) ⋅ C = +( 20.00) sin 37.0 ( 6.00) = +72.2.


1.94: a)

The maximum and minimum areas are (L + l) (W + w) = LW + lW + Lw, (L – l) (W – w) = LW – lW – Lw, where the common terms wl have been omitted. The area and its uncertainty are then WL ± (lW + Lw), so the uncertainty in the area is a = lW + Lw.

b) The fractional uncertainty in the area is a lW + Wl l w = = + A WL L W, the sum of the fractional uncertainties in the length and width. c) The similar calculation to find the uncertainty v in the volume will involve neglecting the terms lwH, lWh and Lwh as well as lwh; the uncertainty in the volume is v = lWH + LwH + LWh, and the fractional uncertainty in the volume is v lWH + LwH + LWh l w h = = + + , V LW H L W H the sum of the fractional uncertainties in the length, width and height.

1.95: The receiver's position is

( + 1.0 + 9.0 − 6.0 + 12.0) iˆ + ( − 5.0 + 11.0 + 4.0 + 18.0) ˆ = (16.0) iˆ + ( 28.0) ˆ. j j
The vector from the quarterback to the receiver is the receiver's position minus the ˆ quarterback's position, or (16.0 ) i + ( 35.0 ) ˆ , a vector with magnitude j = 38.5, given as being in yards. The angle is arctan( 16..0 ) = 24.6 to the 35 0 right of downfield.

(16.0) 2 + ( 35.0) 2

1.96: a)

b)

i) ii) iii)

In AU, In AU, In AU,

(0.3182) 2 + (0.9329) 2

= 0.9857.

(1.3087) 2 + (−.4423) 2 + (−.0414) 2 = 1.3820

(0.3182 − (1.3087)) 2 + (0.9329 − (−.4423)) 2 + (0.0414) 2 = 1.695. c) The angle between the directions from the Earth to the Sun and to Mars is obtained from the dot product. Combining Equations (1-18) and (1.21),  ( −0.3182)(1.3087 − 0.3182) + (−0.9329)(−0.4423 − 0.9329) + (0)   = 54.6°. φ = arccos    (0.9857)(1.695)   d) Mars could not have been visible at midnight, because the Sun-Mars angle is less than 90o.

1.97: a)

The law of cosines (see Problem 1.88) gives the distance as

(138 ly ) 2 + ( 77 ly ) 2 + 2(138 ly )( 77 ly ) cos154.4 

= 76.2 ly,

where the supplement of 25.6 has been used for the angle between the direction vectors. b) Although the law of cosines could be used again, it's far more convenient to use the law of sines (Appendix B), and the angle is given by  sin 25.6   arcsin  138 ly  = 51.5  , 180  − 51.5  = 129  ,  76.2 ly    where the appropriate angle in the second quadrant is used.  ˆ ˆ Define S = Ai + Bˆ + Ck j   ˆ ˆ ˆ ˆ r ⋅ S = ( xi + yˆ + zk ) ⋅ ( Ai + Bˆ + Ck ) j j = Ax + By + Cz    If the points satisfy Ax + By + Cz = 0, then r ⋅ S = 0 and all points r are  perpendicular to S .

1.98:

Chapter 2

2.1: a) During the later 4.75-s interval, the rocket moves a distance 1.00 ×10 3 m − 63 m , and so the magnitude of the average velocity is 1.00 × 10 3 m − 63 m = 197 m s . 4.75 s b)
1.00×10 3 m 5.90 s

= 169 m s

2.2:

a) The magnitude of the average velocity on the return flight is (5150 × 10 3 m) = 4.42 m s . (13.5 da ) (86, 400 s da )

ˆ The direction has been defined to be the –x-direction (−i ). b) Because the bird ends up at the starting point, the average velocity for the round trip is 0. 2.3: Although the distance could be found, the intermediate calculation can be avoided by considering that the time will be inversely proportional to the speed, and the extra time will be  105 km hr  (140 min)   70 km hr − 1 = 70 min.    2.4: The eastward run takes (200 m 5.0 m s) = 40.0 s and the westward run takes (280 m 4.0 m s) = 70.0 s. a) (200 m + 280 m)/(40.0 s + 70.0 s) = 4.4 m s to two significant figures. b) The net displacement is 80 m west, so the average velocity is ˆ (80 m 110.0 s) = 0.73 m s in the –x-direction (−i ). 2.5: In time t the fast runner has traveled 200 m farther than the slow runner: (5.50 m s)t + 200 m = (6.20 m s)t , so t = 286 s . Fast runner has run (6.20 m s)t = 1770 m. Slow runner has run (5.50 m s)t = 1570 m.

2.6:

The s-waves travel slower, so they arrive 33 s after the p-waves. t s = t p + 33 s d = vt → t = d d = + 33 s vs v p d d = + 33 s 3.5 km 6.5 km s s d = 250 km d v

2.7: a) The van will travel 480 m for the first 60 s and 1200 m for the next 60 s, for a total distance of 1680 m in 120 s and an average speed of 14.0 m s . b) The first stage of 240 the journey takes 8.0 mms = 30 s and the second stage of the journey takes (240 m 20 m s) = 12 s , so the time for the 480-m trip is 42 s, for an average speed of 11.4 m s . c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same. 2.8: From the expression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) 20.8 m − 5.60 m 5.60 m − 0 20.8 m −0 = 7.6 m s 4.00 s = 5.2 m s c) 2.00 s = 2.80 m s b) 2.00 s

2.9:

a) At t1 = 0, x1 = 0 , so Eq (2.2) gives x 2 (2.4 m s 2 )(10.0 s) 2 − (0.120 m s 3 )(10.0 s) 3 v av = = = 12.0 m s . t2 (10.0 s)

b) From Eq. (2.3), the instantaneous velocity as a function of time is v x = 2bt − 3ct 2 = (4.80 m s 2 )t − (0.360 m s )t 2 , so i) v x (0) = 0, ii) v x (5.0 s) = ( 4.80 m s )(5.0 s) − (0.360 m s )(5.0 s) 2 = 15.0 m s , and iii) v x (10.0 s) = (4.80 m s )(10.0 s) − (0.360 m s )(10.0 s) 2 = 12.0 m s . c) The car is at rest when v x = 0 . Therefore ( 4.80 m s )t − (0.360 m s )t 2 = 0 . The only time after t = 0 when the car is at rest is t =
4.80 m s 2 0.360 m s 3 2 3 2 3 2 3 3

= 13.3 s

2.10: a) IV: The curve is horizontal; this corresponds to the time when she stops. b) I: This is the time when the curve is most nearly straight and tilted upward (indicating postive velocity). c) V: Here the curve is plainly straight, tilted downward (negative velocity). d) II: The curve has a postive slope that is increasing. e) III: The curve is still tilted upward (positive slope and positive velocity), but becoming less so. 2.11: Time (s) 16 Acceleration (m/s2) 0 0 2 1 4 2 6 2 8 3 10 1.5 12 1.5 14 0

a) The acceleration is not constant, but is approximately constant between the times t = 4 s and t = 8 s.

2.12: The cruising speed of the car is 60 km hr = 16.7 m s . a)

16.7 m s 10 s

= 1.7 m s (to

2

two significant figures). b) 0−16.7sm s = −1.7 m s 2 c) No change in speed, so the acceleration 10 is zero. d) The final speed is the same as the initial speed, so the average acceleration is zero. 2.13: a) The plot of the velocity seems to be the most curved upward near t = 5 s. b) The only negative acceleration (downward-sloping part of the plot) is between t = 30 s and t = 40 s. c) At t = 20 s, the plot is level, and in Exercise 2.12 the car is said to be cruising at constant speed, and so the acceleration is zero. d) The plot is very nearly a straight line, and the acceleration is that found in part (b) of Exercise 2.12, − 1.7 m s 2 . e)

2.14: (a) The displacement vector is:  ˆ ˆ r (t ) = −(5.0 m s)ti + (10.0 m s)tˆ + (7.0 m s)t − (3.0 m s 2 )t 2 k j

(

)

The velocity vector is the time derivative of the displacement vector:  d r (t ) ˆ ˆ = (−5.0 m s)i + (10.0 m s) ˆ + (7.0 m s − 2(3.0 m s 2 )t ) k j dt and the acceleration vector is the time derivative of the velocity vector:  d 2 r (t ) ˆ = −6.0 m s 2 k dt 2 At t = 5.0 s:

 2 ˆ ˆ r (t ) = −(5.0 m s)(5.0 s) i + (10.0 m s)(5.0 s) ˆ + (7.0 m s)(5.0 s) − (−3.0 m s )(25.0 s 2 ) k j ˆ ˆ = (−25.0 m)i + (50.0 m) ˆ − (40.0 m)k j  d r (t ) 2 ˆ ˆ = (−5.0 m s)i + (10.0 m s) ˆ + ((7.0 m s − (6.0 m s )(5.0 s)) k j dt ˆ ˆ = (−5.0 m s)i + (10.0 m s) ˆ − (23.0 m s) k j  d 2 r (t ) 2 ˆ = − 6 .0 m s k 2 dt

(

)

(b) The velocity in both the x- and the y-directions is constant and nonzero; thus the overall velocity can never be zero. (c) The object's acceleration is constant, since t does not appear in the acceleration vector.

2.15: ax =

vx =

dx = 2.00 cm s − (0.125 cm s 2 )t dt

dv x = −0.125 cm s 2 dt a) At t = 0, x = 50.0 cm, v x = 2.00 cm s , a x = −0.125 cm s 2 .

b) Set v x = 0 and solve for t : t = 16.0 s. c) Set x = 50.0 cm and solve for t. This gives t = 0 and t = 32.0 s. The turtle returns to the starting point after 32.0 s. d) Turtle is 10.0 cm from starting point when x = 60.0 cm or x = 40.0 cm. Set x = 60.0 cm and solve for t : t = 6.20 s and t = 25.8 s. At t = 6.20 s, v x = +1.23 cm s . At t = 25.8 s, v x = −1.23 cm s . Set x = 40.0 cm and solve for t : t = 36.4 s (other root to the quadratic equation is negative and hence nonphysical). At t = 36.4 s, v x = −2.55 cm s . e)

2.16: Use of Eq. (2.5), with ∆t = 10 s in all cases, a) b) c)

( ( 5.0 m/s ) − (15.0 m/s ) ) / (10 s ) = −1.0 m/s 2 ( ( − 15.0 m/s ) − ( − 5.0 m/s ) ) / (10 s ) = −1.0 m/s 2 ( ( − 15.0 m/s ) − ( − 15.0 m/s ) ) /(10 s ) = −3.0 m/s 2 .

In all cases, the negative acceleration indicates an acceleration to the left. 2.17: a) Assuming the car comes to rest from 65 mph (29 m/s) in 4 seconds, a x = ( 29 m s − 0) (4 s) = 7.25 m s 2 . b) Since the car is coming to a stop, the acceleration is in the direction opposite to the velocity. If the velocity is in the positive direction, the acceleration is negative; if the velocity is in the negative direction, the acceleration is positive.

2.18: a) The velocity at t = 0 is (3.00 m s ) + (0.100 m s 3 ) (0) = 3.00 m s , and the velocity at t = 5.00 s is (3.00 m s ) + (0.100 m s 3 ) (5.00 s)2 = 5.50 m s , so Eq. (2.4) gives the average acceleration as (5.50 m s) − (3.00 m s) 2 = .50 m s . (5.00 s) b) The instantaneous acceleration is obtained by using Eq. (2.5), ax =
3

dv 3 = 2 βt = (0.2 m s )t. dt

Then, i) at t = 0, ax = (0.2 m s ) (0) = 0, and ii) at t = 5.00 s, ax = (0.2 m s ) (5.00 s) = 1.0 m s .
3 2

2.19: a)

b)

2.20: a) The bumper’s velocity and acceleration are given as functions of time by dx 2 6 vx = = (9.60 m s )t − (0.600 m s )t 5 dt dv 2 6 ax = = (9.60 m s ) − (3.000 m s )t 4 . dt There are two times at which v = 0 (three if negative times are considered), given by t = 0 and t4 = 16 s4. At t = 0, x = 2.17 m and ax = 9.60 m s 2. When t4 = 16 s4, x = (2.17 m) + (4.80 m s 2) b) (16 s 4 ) – (0.100) m s )(16 s4)3/2 = 14.97 m,
6 6

ax = (9.60 m s 2) – (3.000 m s )(16 s4) = –38.4 m s 2.

2.21: a) Equating Equations (2.9) and (2.10) and solving for v0, 2( x − x0 ) 2(70 m) v0 x = − vx = − 15.0 m s = 5.00 m s . t 7.00 s b) The above result for v0x may be used to find v −v 15.0 m s − 5.00 m s 2 ax = x 0x = = 1.43 m s , t 7.00 s or the intermediate calculation can be avoided by combining Eqs. (2.8) and (2.12) to eliminate v0x and solving for ax, x − x   15.0 m s 70.0 m  v 2 a x = 2 x − 2 0  = 2  7.00 s − (7.00 s) 2  = 1.43 m s .  t    t 

2.22: a) The acceleration is found from Eq. (2.13), which v0 x = 0;
4470 m 2 (173 mi hr ) 0.1 mi hr s vx ax = = 1m 2( x − x 0 ) 2( (307 ft ) ( 3.281 ft ) )

(

(

))

2

= 32.0 m s ,

2

where the conversions are from Appendix E. b) The time can be found from the above acceleration, t=
0.4470 m s v x (173 mi hr ) 1 mi hr = = 2.42 s. 2 ax 32.0 m s

(

)

The intermediate calculation may be avoided by using Eq. (2.14), again with v0x = 0, t=
1m 2( (307 ft ( 3.281 ft ) ) 2( x − x 0 ) = = 2.42 s. 4470 m vx (173 mi hr ) 0.1 mi hr s

(

)

2.23: From Eq. (2.13), with v x = 0, a x = x>

2 v0 2 ( x − x0 )

< a max . Taking x0 = 0,

2 v0 x ((105 km hr)(1 m s)(3.6 km hr )) 2 = = 1.70 m. 2a max 2(250 m s 2 )

2.24: In Eq. (2.14), with x – x0 being the length of the runway, and v0x = 0 (the plane starts from rest), v x = 2 x −t x0 = 2 280sm = 70.0 m s . 8 2.25: a) From Eq. (2.13), with v0 x = 0,
2 vx (20 m s) 2 ax = = = 1.67 m s 2 . 2( x − x0 ) 2(120 m)

b) Using Eq. (2.14), t = 2( x − x0 ) v = 2(120 m) (20 m s) = 12 s. c) (12 s)(20 m s) = 240 m.

2.26: a)

x0 < 0, v0x < 0, ax < 0

b)

x0 > 0, v0x < 0, ax > 0

c)

x0 > 0, v0x > 0, ax < 0

2.27: a) speeding up: x − x0 = 1320 ft, v0 x = 0, t = 19.9 s, a x = ? x − x0 = v0 x t + 1 a x t 2 gives a x = 6.67 ft s 2 2 slowing down: x − x0 = 146 ft, v0 x = 88.0 ft s , v x = 0, a x = ?
2 2 v x = v0 x + 2a x ( x − x0 ) gives a x = −26.5 ft s 2 .

b) x − x0 = 1320 ft, v0 x = 0, a x = 6.67 ft s , v x = ?
2 2 v x = v0 x + 2a x ( x − x0 ) gives v x = 133 ft s = 90.5 mph. a x must not be constant.

2

c) v0 x = 88.0 ft s, a x = −26.5 ft s 2 , v x = 0, t = ? v x = v0 x + a x t gives t = 3.32 s.

2.28: a) Interpolating from the graph: At 4.0 s, v = +2.7 cm s (to the right) At 7.0 s, v = −1.3 cm s (to the left) b) a = slope of v - t graph = − 8.06.cms / s = −1.3 cm s 2 which is constant 0 c) ∆x = area under v-t graph

First 4.5 s: ∆x = ARectangle + ATriangle  cm  1  cm  = ( 4 .5 s )  2  + ( 4.5 s )  6  = 22.5 cm  s  2  s  From 0 to 7.5 s:

The distance is the sum of the magnitudes of the areas. 1  cm  1  cm  d = ( 6 s) 8  + ( 1 .5 s )  2  = 25.5 cm 2  s  2  s  d)

2.29: a)

b)

2.30: a)

2.31: a) At t = 3 s the graph is horizontal and the acceleration is 0. From t = 5 s to t = 2 9 s, the acceleration is constant (from the graph) and equal to 45 m s−s20m s = 6.3 m s . From 4 t = 9 s to t = 13 s the acceleration is constant and equal to 2 0− 45 m s = −11.2 m s . 4s b) In the first five seconds, the area under the graph is the area of the rectangle, (20 m)(5 s) = 100 m. Between t = 5 s and t = 9 s, the area under the trapezoid is (1/2)(45 m/s + 20 m/s)(4 s) = 130 m (compare to Eq. (2.14)), and so the total distance in the first 9 s is 230 m. Between t = 9 s and t = 13 s, the area under the triangle is (1 2)(45 m s)(4 s) = 90 m , and so the total distance in the first 13 s is 320 m. 2.32:

2.33: a) The maximum speed will be that after the first 10 min (or 600 s), at which time the speed will be (20.0 m s )(900 s) = 1.8 × 10 4 m s = 18 km s . b) During the first 15 minutes (and also during the last 15 minutes), the ship will travel (1 2)(18 km s)(900 s) = 8100 km , so the distance traveled at non-constant speed is 16,200 km and the fraction of the distance traveled at constant speed is 16,200 km 1− = 0.958, 384,000 km keeping an extra significant figure. km − c) The time spent at constant speed is 384 ,00018 km16 , 200 km = 2.04 × 104 s and the time spent s during both the period of acceleration and deceleration is 900 s, so the total time required for the trip is 2.22 × 10 4 s , about 6.2 hr.
2

2.34: After the initial acceleration, the train has traveled 1 2 (1.60 m s )(14.0 s) 2 = 156.8 m 2 (from Eq. (2.12), with x0 = 0, v0x = 0), and has attained a speed of (1.60 m s )(14.0 s) = 22.4 m s . During the 70-second period when the train moves with constant speed, the train travels ( 22.4 m s ) ( 70 s ) = 1568 m. The distance traveled during deceleration is given by Eq. (2.13), with vx = 0, v0 x = 22.4 m s and ax = −3.50 m s 2 , so the train moves a distance − ( 22 .4 m / s ) x − x 0 = 2( −3.50 m/s ) = 71.68 m. The total distance covered in then 156.8 m + 1568 m + 71.7 m
2 2

2

= 1.8 km. In terms of the initial acceleration a1, the initial acceleration time t1, the time t2 during which the train moves at constant speed and the magnitude a2 of the final acceleration, the total distance xT is given by at  1 2 1 (a1t1 ) 2  a1t1  a1t1 + (a1t1 )t 2 + =  t1 + 2t 2 + 1 1 ,  2 2 | a2 | | a2 |   2   which yields the same result. xT =

2.35: a)

b) From the graph (Fig. (2.35)), the curves for A and B intersect at t = 1 s and t = 3 s. c)

d) From Fig. (2.35), the graphs have the same slope at t = 2 s . e) Car A passes car B when they have the same position and the slope of curve A is greater than that of curve B in Fig. (2.30); this is at t = 3 s. f) Car B passes car A when they have the same position and the slope of curve B is greater than that of curve A; this is at t = 1 s.

2.36: a) The truck’s position as a function of time is given by xT = vTt, with vT being the truck’s constant speed, and the car’s position is given by xC = (1/2) aCt2. Equating the two expressions and dividing by a factor of t (this reflects the fact that the car and the truck are at the same place at t = 0) and solving for t yields 2v 2(20.0 m s) t= T = = 12.5 s 2 aC 3.20 m s and at this time xT = xC = 250 m. b) aCt = (3.20 m/s2)(12.5 s) = 40.0 m/s (See Exercise 2.37 for a discussion of why the car’s speed at this time is twice the truck’s speed.) c)

d)

2.37: a)

The car and the motorcycle have gone the same distance during the same time, so their average speeds are the same. The car's average speed is its constant speed vC, and for constant acceleration from rest, the motorcycle's speed is always twice its average, or 2vC. b) From the above, the motorcyle's speed will be vC after half the time needed to catch the car. For motion from rest with constant acceleration, the distance traveled is proportional to the square of the time, so for half the time one-fourth of the total distance has been covered, or d 4 . 2.38: a) An initial height of 200 m gives a speed of 60 m s when rounded to one significant figure. This is approximately 200 km/hr or approximately 150 mi hr . (Different values of the approximate height will give different answers; the above may be interpreted as slightly better than order of magnitude answers.) b) Personal experience will vary, but speeds on the order of one or two meters per second are reasonable. c) Air resistance may certainly not be neglected. 2.39: a) From Eq. (2.13), with v y = 0 and a y = − g , v0 y = 2 g ( y − y0 ) = 2(9.80 m s )(0.440 m) = 2.94 m s , which is probably too precise for the speed of a flea; rounding down, the speed is about 2.9 m s . b) The time the flea is rising is the above speed divided by g, and the total time is twice this; symbolically, t=2 or about 0.60 s.
2 2.40: Using Eq. (2.13), with downward velocities and accelerations being positive, v y = 2

2 g ( y − y0 ) g

=2

2( y − y 0 ) 2(0.440 m) =2 = 0.599 s, g (9.80 m/s 2 )

(0.8 m s )2 + 2(1.6 m s )(5.0 m) = 16.64 m 2 s 2 (keeping extra significant figures), so vy = 4.1 m s .
2

2.41: a) If the meter stick is in free fall, the distance d is related to the reaction time t by d = (1 2) gt 2 , so t = 2d g . If d is measured in centimeters, the reaction time is t= 2 2 d = d = ( 4.52 ×10 −2 s) d (1 cm) . 2 g 980 cm s

b) Using the above result, (4.52 × 10 −2 s) 17.6 = 0.190 s. 2.42: a) (1 2) gt 2 = (1 2)(9.80 m s 2 )(2.5 s) 2 = 30.6 m. b) gt = (9.80 m s 2 )(2.5 s) = 24.5. m s . c)

2.43: a) Using the method of Example 2.8, the time the ring is in the air is t=
2 v0 y + v0 y − 2 g ( y − y 0 )

g
2

(5.00 m s) + (5.00 m s) − 2(9.80 m s 2 )(−12.0 m) = (9.80 m s 2 ) = 2.156 s, keeping an extra significant figure. The average velocity is then 12.0 ms = 5.57 m s , down. 2.156 As an alternative to using the quadratic formula, the speed of the ring when it hits the 2 2 ground may be obtained from v y = v 0 y − 2 g ( y − y 0 ) , and the average velocity found from y 2 0 y ; this is algebraically identical to the result obtained by the quadratic formula. b) While the ring is in free fall, the average acceleration is the constant acceleration due to gravity, 9.80 m / s 2 down. c) 1 y = y0 + v0 y t − gt 2 2
v +v

1 0 = 12.0 m + (5.00 m s)t − (9.8 m s 2 )t 2 2 Solve this quadratic as in part a) to obtain t = 2.156 s.
2 d) v 2 = v0 y − 2 g ( y − y0 ) = (5.00 m s) 2 − 2(9.8 m s 2 )(−12.0 m) y

v y = 16.1 m s e)

2.44: a) Using ay = –g, v0y = 5.00 m s and y0 = 40.0 m in Eqs. (2.8) and (2.12) gives i) at t = 0.250 s, 2 y = (40.0 m) + (5.00 m s )(0.250 s) – (1/2)(9.80 m s )(0.250 s)2 = 40.9 m, vy = (5.00 m s ) – (9.80 m s )(0.250 s) = 2.55 m s and ii) at t = 1.00 s, y = (40.0 m) + (5.00 m/s)(1.00 s) – (1/2)(9.80 m/s2)(1.00 s)2 = 40.1 m, 2 vy = (5.00 m s ) – (9.80 m s )(1.00 s) = – 4.80 m s . b) Using the result derived in Example 2.8, the time is t= (5.00 m s) + (5.00 m s) 2 − 2(9.80 m s )(0 − 40.0 m) (9.80 m s )
2 2 2

= 3.41 s.

c) Either using the above time in Eq. (2.8) or avoiding the intermediate calculation by using Eq. (2.13),
2 v 2 = v0 y − 2 g ( y − y 0 ) = (5.00 m s) 2 − 2(9.80 m s )(−40.0 m) = 809 m 2 s , y 2 2

vy = 28.4 m s . d) Using vy = 0 in Eq. (2.13) gives y= e)
2 v0 (5.00 m s) 2 + y0 = + 40.0 m = 41.2 m. 2 2g 2(9.80 m s )

2.45: a) v y = v0 y − gt = (−6.00 m s) − (9.80 m s )(2.00 s) = −25.6 m s , so the speed is 25.6 m s . 1 1 2 b) y = v0 y t − gt 2 = (−6.00 m s)(2.00 s) − (9.80 m s )(2.00 s) 2 = −31.6 m, with the 2 2 minus sign indicating that the balloon has indeed fallen. c) 2 2 v y = v 0 y − 2 g ( y 0 − y ) = (6.00 m s) 2 − 2(9.80 m s 2 )(−10.0 m) = 232 m 2 s 2 , so v y = 15.2 m s . 2.46: a) The vertical distance from the initial position is given by y = v0 y t − solving for v0y, v0 y = (−50.0 m) 1 y 1 2 + gt = + (9.80 m s )(5.00 s) = 14.5 m s . t 2 (5.00 s) 2 1 2 gt ; 2

2

2 2 b) The above result could be used in v y = v 0 y − 2 g ( y − y 0 ) , with v = 0, to solve for y – y0 = 10.7 m (this requires retention of two extra significant figures in the calculation 2 for v0y). c) 0 d) 9.8 m s , down.

e) Assume the top of the building is 50 m above the ground for purposes of graphing:

2.47: a) (224 m s) (0.9 s) = 249 m s 2 . b)
2

249 m s 2 9.80 m s 2

= 25.4. c) The most direct way to
2

find the distance is vave t = ((224 m s) 2)(0.9 s) = 101 m. d) (283 m s) (1.40 s) = 202 m s but 40 g = 392 m s , so the figures are not consistent.

2.48: a) From Eq. (2.8), solving for t gives (40.0 m s – 20.0 m s )/9.80 m s = 2.04 s. b) Again from Eq. (2.8), 40.0 m s − (−20.0 m s) 9.80 m s
2

2

= 6.12 s.

c) The displacement will be zero when the ball has returned to its original vertical position, with velocity opposite to the original velocity. From Eq. (2.8), 40 m s − (−40 m s) 9.80 m s
2

= 8.16 s.

(This ignores the t = 0 solution.) 2 d) Again from Eq. (2.8), (40 m s )/(9.80 m s ) = 4.08 s. This is, of course, half the time found in part (c). 2 e) 9.80 m s , down, in all cases. f)

2.49: a) For a given initial upward speed, the height would be inversely proportional to the magnitude of g, and with g one-tenth as large, the height would be ten times higher, or 7.5 m. b) Similarly, if the ball is thrown with the same upward speed, it would go ten times as high, or 180 m. c) The maximum height is determined by the speed when hitting the ground; if this speed is to be the same, the maximum height would be ten times as large, or 20 m.

2.50: a) From Eq. (2.15), the velocity v2 at time t v2 = v1 + = v1 +

∫ αt dt
t1

t

α 2 (t − t12 ) 2 α 2 α 2 = v1 − t1 + t 2 2 3 3 = (5.0 m s ) – (0.6 m s )(1.0 s)2 + (0.6 m s ) t2
= (4.40 m s ) + (0.6 m s ) t2.
3

At t2 = 2.0 s, the velocity is v2 = (4.40 m s ) + (0.6 m s )(2.0 s)2 = 6.80 m/s, or 6.8 m s to two significant figures. b) From Eq. (2.16), the position x2 as a function of time is
3

x 2 = x1 + ∫ tt1 v x dt = (6.0 m) + ∫ tt1 ((4.40 m / s) + (0.6 m / s 3 )t 2 )dt = (6.0 m) + (4.40 m s)(t − t1 ) + At t = 2.0 s, and with t1 = 1.0 s, x = (6.0 m) + (4.40 m s )((2.0 s) – (1.0 s)) + (0.20 m s )((2.0 s)3 – (1.0 s)3) = 11.8 m.
3

( 0 .6 m s ) 3 (t − t13 ). 3

3

c)

2.51: a) From Eqs. (2.17) and (2.18), with v0=0 and x0=0, v x = ∫ ( At − Bt 2 ) dt = A 2 B 2 t − t = (0.75 m s 3 )t 3 − (0.040 m s 4 )t 3 0 2 3 t A B 3 A 3 B 4 x = ∫  t 2 − t  dt = t − t = (0.25 m s 3 )t 3 − (0.010 m s 4 )t 4 . 0 2 3  6 12 
t

b) For the velocity to be a maximum, the acceleration must be zero; this occurs at t=0 A and t = B = 12.5 s . At t=0 the velocity is a minimum, and at t=12.5 s the velocity is v x = (0.75 m s 3 )(12.5 s) 2 − (0.040 m s 4 )(12.5 s) 3 = 39.1 m s.

2.52: a) Slope = a = 0 for t ≥ 1.3 ms b) hmax = Area underv − tgraph ≈ ATriangle + ARectangle 1 cm   (1.3 ms )133  + (2.5 ms − 1.3 ms )(133 cm s) 2 s   ≈ 0.25 cm ≈

c) a = slope of v– t graph a (0.5 ms ) ≈ a (1.0 ms ) ≈ 133 cm s 2 = 1.0 × 105 cm s 1.3 ms a (1.5 ms ) = 0 because the slope is zero.

d) h = area under v– t graph h (0.5 ms ) ≈ ATriangle = h (1.0 ms ) ≈ ATriangle 1  cm  −3 (0.5 ms ) 33  = 8.3 ×10 cm 2 s   1 = (1.0 ms )(100 cm s) = 5.0 × 10 −2 cm 2

1 cm   h (1.5 ms ) ≈ ATriangle + ARectangle = (1.3 ms )133  + (0.2 ms )(1.33) 2 s   = 0.11 cm

2.53: a) The change in speed is the area under the ax versus t curve between vertical lines at t = 2.5 s and t = 7.5 s. This area is (4.00 cm s + 8.00 cm s )(7.5 s − 2.5 s) = 30.0 cm s This acceleration is positive so the change in velocity is positive.
1 2 2 2

b) Slope of vx versus t is positive and increasing with t.

2.54: a) To average 4 mi hr , the total time for the twenty-mile ride must be five hours, so the second ten miles must be covered in 3.75 hours, for an average of 2.7 mi hr . b) To average 12 mi hr , the second ten miles must be covered in 25 minutes and the average speed must be 24 mi hr . c) After the first hour, only ten of the twenty miles have been covered, and 16 mi hr is not possible as the average speed.

2.55: a)

The velocity and acceleration of the particle as functions of time are vx (t ) = (9.00 m s3 )t 2 − (20.00 m s 2 )t + (9.00 m s) a x (t ) = (18.0 m s3 )t − (20.00 m s 2 ). b) The particle is at rest when the velocity is zero; setting v = 0 in the above expression and using the quadratic formula to solve for the time t, t= (20.0 m s3 ) ± (20.0 m s3 ) 2 − 4(9.0 m s 3 )(9.0 m s) 2(9.0 m s 3 )

and the times are 0.63 s and 1.60 s. c) The acceleration is negative at the earlier time and positive at the later time. d) The velocity is instantaneously not changing when the acceleration is zero; solving the above expression for a x (t ) = 0 gives 20.00 m s 2 = 1.11 s. 18.00 m s 3 Note that this time is the numerical average of the times found in part (c). e) The greatest distance is the position of the particle when the velocity is zero and the acceleration is negative; this occurs at 0.63 s, and at that time the particle is at (3.00 m s 3 )(0.63 s) 3 − (10.0 m s 2 )(0.63 s) 2 + (9.00 m s)(0.63 s) = 2.45 m. (In this case, retaining extra significant figures in evaluating the roots of the quadratic equation does not change the answer in the third place.) f) The acceleration is negative at t = 0 and is increasing, so the particle is speeding up at the greatest rate at t = 2.00 s and slowing down at the greatest rate at t = 0. This is a situation where the extreme values of da a function (in the case the acceleration) occur not at times when = 0 but at the dt endpoints of the given range.

2.56: a) 25.0 m = 1.25 m s . 20.0 s 25 m b) 15 s = 1.67 m s . c) Her net displacement is zero, so the average velocity has zero magnitude. d) 50.0 m = 1.43 m s . Note that the answer to part (d) is the harmonic mean, not the 35.0 s arithmetic mean, of the answers to parts (a) and (b). (See Exercise 2.5). 2.57: Denote the times, speeds and lengths of the two parts of the trip as t1 and t2, v1 and v2, and l1 and l2. a) The average speed for the whole trip is l1 + l 2 l1 + l 2 (76 km) + (34 km) = = 76 km = 82 km h, 34 km t1 + t 2 (l1 v1 ) + (l 2 v 2 ) 88 km h + 72 km h

(

) (

)

or 82.3 km/h, keeping an extra significant figure. b) Assuming nearly straight-line motion (a common feature of Nebraska highways), the total distance traveled is l1–l2 and v ave = l1 − l 2 (76 km) − (34 km) = 76 km = 31 km h. 34 km t1 + t 2 88 km h + 72 km h

(

) (

)

( 31.4 km hr to three significant figures.) 2.58: a) The space per vehicle is the speed divided by the frequency with which the cars pass a given point; 96 km h = 40 m vehicle . 2400 vehicles h An average vehicle is given to be 4.5 m long, so the average spacing is 40.0 m – 4.6 m = 35.4 m. b) An average spacing of 9.2 m gives a space per vehicle of 13.8 m, and the traffic flow rate is 96000 m h = 6960 vehicle h . 13.8 m vehicle

2.59: (a) Denote the time for the acceleration (4.0 s) as t1 and the time spent running at constant speed (5.1 s) as t2. The constant speed is then at1, where a is the unknown acceleration. The total l is then given in terms of a, t1 and t2 by l= and solving for a gives a= l (1 2)t + t1t 2
2 1

1 2 at1 + at1t 2 , 2

=

(100 m) = 3 .5 m s 2 . 2 (1 2)(4.0 s) + (4.0 s)(5.1 s)

(b) During the 5.1 s interval, the runner is not accelerating, so a = 0. (c) ∆v ∆t = [(3.5 m s 2 )(4 s)] (9.1 s) = 1.54 m s 2 . (d) The runner was moving at constant velocity for the last 5.1 s. 2.60: a) Simple subtraction and division gives average speeds during the 2-second intervals as 5.6, 7.2 and 8.8 m s . b) The average speed increased by 1.6 m s during each 2-second interval, so the acceleration is 0.8 m s . c) From Eq. (2.13), with v0 = 0, v = 2(0.8 m s )(14.4 m) = 4.8 m s . Or, recognizing that for constant acceleration the average speed of 5.6 m/s is the speed one 2 second after passing the 14.4-m mark, 5.6 m s – (0.8 m s )(1.0 s) = 4.8 m s . d) With both the acceleration and the speed at the 14.4-m known, either Eq. (2.8) or Eq. (2.12) gives the time as 6.0 s. 2 e) From Eq. (2.12), x – x0 = (4.8 m s )(1.0 s) + 1 (0.8 m s )(1.0 s)2 = 5.2 m. This 2 is also the average velocity (1/2)(5.6 m s + 4.8 m s ) times the time interval of 1.0 s. 2.61: If the driver steps on the gas, the car will travel (20 m s)(3.0 s) + (1 2)(2.3 m s 2 )(3.0 s) 2 = 70.4 m. If the brake is applied, the car will travel (20 m s)(3.0 s) + (1 2)(−3.8 m s 2 )(3.0 s) 2 = 42.9 m, so the driver should apply the brake.
2 2

2.62: a)

d = ct = (3.0 × 108 = 9.5 × 1015 m

 365 1 d  24 h  3600 s  m 4 )(1 y)  1 y  1d  1h     s    

b) c) d) e)

m )(10 −9 s) = 0.30 m s 11 d 1.5 ×10 m t= = = 500 s = 8.33 min c 3.0 × 108 m s d = ct = (3.0 ×108 t= t= d 2(3.84 × 108 m) = = 2 .6 s c 3.0 ×108 m s 3 × 10 9 mi d = = 16,100 s = 4.5 h c 186,000 mi s

2.63: a) v = 2πRE t = 464 m s b) v = 2πr t = 2.99 × 104 m s (r is the radius of the earth's orbit) c) Let c be the speed of light, then in one second light travels a distance c(1.00 s). The number of times around the earth to which this corresponds is c(1.00 s) 2πRE = 7.48

2.64: Taking the start of the race as the origin, runner A's speed at the end of 30 m can be found from:
2 2 vA = v0A + 2a A ( x − x0 ) = 0 + 2(1.6 m s 2 )(30 m) = 96 m 2 s 2

vA = 96 m 2 s 2 = 9.80 m s A’s time to cover the first 30 m is thus: t= and A’s total time for the race is: vA − v0A 9.80 m s = = 6.13 s 2 aA 1 .6 m s

(350 − 30) m = 38.8 s 9.80 m s B’s speed at the end of 30 m is found from: 6.13 s +
2 2 vB = v0B + 2aB ( x − x0 ) = 0 + 2(2.0 m s )(30 m) = 120 m 2 s 2 2

vB = 120 m 2 s 2 = 10.95 m s B’s time for the first 30 m is thus v −v 10.95 m s t = B 0B = = 5.48 s 2 aB 2 .0 m s and B's total time for the race is: (350 − 30) m 5.48 s + = 34.7 s 10.95 m s B can thus nap for 38.8 − 34.7 = 4.1 s and still finish at the same time as A. 2.65: For the first 5.0 s of the motion, v 0 x = 0, t = 5.0 s. v x = v 0 x + a x t gives v x = a x (5.0 s). This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s: v0 x = a x (5.0 s), t = 5.0 s, x − x0 = 150 m. x − x0 = v 0 x t + 1 a x t 2 gives 150 m = (25 s 2 )a x + (12.5 s 2 )a x and a x = 4.0 m s 2 2 Use this ax and consider the first 5.0 s of the motion: x − x0 = v0 x t + 1 a x t 2 = 0 + 1 (4.0 m s 2 )(5.0 s) 2 = 50.0 m. 2 2

2.66: a) The simplest way to do this is to go to a frame in which the freight train (which moves with constant velocity) is stationary. Then, the passenger train has an initial relative velocity of vrel,0 = 10 m s . This relative speed would be decreased to
, zero after the relative separation had decreased to vrelrel0 = + 500 m. Since this is larger in 2a magnitude than the original relative separation of 200 m, there will be a collision. b) The time at which the relative separation goes to zero (i.e., the collision time) is found by solving a quadratic (see Problems 2.35 & 2.36 or Example 2.8). The time is given by 1 2 t = vrel ,0 − vrel , 0 + 2axrel ,0 a
2

(

)

= (10 s 2 m)(10 m s − 100 m 2 s − 40 m 2 s ) = (100 s) (1 − 0.6 . Substitution of this time into Eq. (2.12), with x0 = 0, yields 538 m as the distance the passenger train moves before the collision. 2.67: The total distance you cover is 1.20 m + 0.90 m = 2.10 m and the time available is 1.20 m 1.50 m s = 0.80 s . Solving Eq. (2.12) for ax, ax = 2 ( x − x0 ) − v0 xt (2.10 m) − (0.80 m s)(0.80 s) 2 =2 = 4.56 m s . 2 2 t (0.80 s)

2

2

(

)

2.68: One convenient way to do the problem is to do part (b) first; the time spent 20 m accelerating from rest to the maximum speed is 2.5 m ss2 = 80 s. At this time, the officer is x1 = v12 ( 20 m s) 2 = = 80.0 m. 2a 2( 2.5 m s 2 )

This could also be found from (1 / 2)a1t12 , where t1 is the time found for the acceleration. At this time the car has moved (15 m s )(8.0 s) = 120 m, so the officer is 40 m behind the car. a) The remaining distance to be covered is 300 m – x1 and the average speed is (1/2)(v1 + v2) = 17.5 m s , so the time needed to slow down is 360 m − 80 m = 16.0 s, 17.5 m s and the total time is 24.0 s. c) The officer slows from 20 m s to 15 m s in 16.0 s (the time found in part (a)), so the acceleration is –0.31 m s . d), e)
2

2.69: a) xT = (1 2)aT t 2 , and with xT = 40.0 m , solving for the time gives t=
2 ( 40.0 m) (2.10 m s)

= 6.17 s

b) The car has moved a distance 1 2 aC 3.40 m s 2 aCt = x1 = 40.0 m = 64.8 m, 2 aT 2.10 m s 2 and so the truck was initially 24.8 m in front of the car. c) The speeds are aT t = 13 m s and aC t = 21 m s . d)

2.70: The position of the cars as functions of time (taking x1 = 0 at t = 0) are x1 = 1 2 at , 2 x 2 = D − v0 t.

The cars collide when x1 = x2; setting the expressions equal yields a quadratic in t, 1 2 at + v0 t − D = 0, 2 the solutions to which are t= 1 a

(

2 v0 + 2aD − v0 ,

)

t=

1 2 − v0 + 2aD − v0 . a

(

)

The second of these times is negative and does not represent the physical situation. b) c) v1 = at =

(

2 v0 + 2aD − v0

)

2.71: a) Travelling at 20 m s , Juan is x1 = 37 m − (20 m s)(0.80 s) = 21 m from the spreader when the brakes are applied, and the magnitude of the acceleration will be 2 1 a = 2vx1 . Travelling at 25 m s , Juan is x2 = 37 m − (25 m s)(0.80 s) = 17 m from the spreader, and the speed of the car (and Juan) at the collision is obtained from  v2  x   17 m  2 2 2 2 v x = v 0 x − 2a x x 2 = v0 x − 2 1  x 2 = v0 x − v12  2  = (25 m s) 2 − (20 m s) 2   21 m   x   2x     1  1 2 2 = 301 m s and so v x = 17.4 m s . b) The time is the reaction time plus the magnitude of the change in speed (v 0 −v) divided by the magnitude of the acceleration, or v −v 25 m s − 17.4 m s t flash = t reaction + 2 0 2 x1 = (0.80 s) + 2 (21 m) = 1.60 s. v0 (20 m s) 2 2.72: a) There are many ways to find the result using extensive algebra, but the most straightforward way is to note that between the time the truck first passes the police car and the time the police car catches up to the truck, both the truck and the car have travelled the same distance in the same time, and hence have the same average velocity over that time. Since the truck had initial speed 3 v p and the average speed is vp, the 2 truck’s final speed must be
1 2

vp .

2.73: a) The most direct way to find the time is to consider that the truck and the car are initially moving at the same speed, and the time of the acceleration must be that which gives a difference between the truck's position and the car's position as 24 m + 21m + 26 m + 4.5 m = 75.5 m, or t = 2(75.5 m) (0.600 m s 2 ) = 15.9 s. b) v0 xt + (1 2)axt 2 = (20.0 m s)(15.9 s) + (1 2)(0.600 m s 2 )(15.9 s) 2 = 394 m. c) v0 x + a xt = (20.0 m s) + (0.600 m s 2 )(15.9 s) = 29.5 m s.

2.74: a) From Eq. (2.17), x(t) = αt – β t 3 = (4.00 m s )t – (0.667 m s )t3. From Eq. 3
3

(2.5), the acceleration is a(t) = –2βt = (– 4.00 m s )t.
3

b) The velocity is zero at t = ±

α β

(a = 0 at t = 0, but this is an inflection point, not an

extreme). The extreme values of x are then  α β x = ±α −  β 3 

α3 β3
1

The positive value is then

3  =±2 α .  3 β 

2  (4.00 m s) 3  2 2 2  x=   2.00 m s 3  = 3 32 m = 3.77 m. 3  2.75: a) The particle's velocity and position as functions of time are v x (t ) = v 0 x + ∫ ((−2.00 m s ) + (3.00 m s )t ) dt
2 3 0 t

 3.00 m s 3  2 t , = v 0 x − ( 2.00 m s )t +    2  
2

x(t ) = ∫ v x (t )dt = v0 x t − (1.00 m s 2 )t 2 + (0.50 m s 3 )t 3
0

t

= t (v0 x − (1.00 m s 2 )t + (0.50 m s 3 )t 2 ), where x0 has been set to 0. Then, x(0) = 0, and to have x(4 s) = 0, v 0 x − (1.00 m s 2 )(4.00 s) + (0.50 m s 3 )(4.00s) 2 = 0, which is solved for v0 x = −4.0 m s. b) v x (4 s) = 12.0 m s. 2.76: The time needed for the egg to fall is t= 2∆h = 9 2(46.0 m − 1.80 m) (9.80 m s )
2

= 3.00 s,

and so the professor should be a distance vyt = (1.20 m s )(3.00 s) = 3.60 m.

2.77: Let t1 be the fall for the watermelon, and t2 be the travel time for the sound to return. The total time is T = t1 + t 2 = 2.5 s . Let y be the height of the building, then, y = 1 gt12 and y = vs t 2. There are three equations and three unknowns. Eliminate t2, solve 2 for t1, and use the result to find y. A quadratic results: at 2 + bt + c = 0, then t =
−b ± b 2 − 4 ac 2a
2

1 2

gt12 + vs t1 − vsT = 0. If

.

Here, t = t1 , a = 1 2 g = 4.9 m s , b = vs = 340 m s , and c = −vs T = −(340 m s)(2.5 s) = −850 m

Then upon substituting these values into the quadratic formula, t1 = t1 = − (340 m s) ± (340 m s) 2 − 4(4.9 m s )(−850 m) 2 ( 4 .9 m s )
2 2

− ( 340 m s ) ± ( 363.7 m s ) 2 4.9 m/s 2

(

)

= 2.42 s . The other solution, –71.8 s has no real physical meaning.

Then, y = 1 gt12 = 1 (9.8 m s 2 )(2.42 s) 2 = 28.6 m . Check: (28.6 m) (340 m s) = .08 s, the 2 2 time for the sound to return. 2.78: The elevators to the observation deck of the Sears Tower in Chicago move from the ground floor to the 103rd floor observation deck in about 70 s. Estimating a single )( floor to be about 3.5 m (11.5 ft), the average speed of the elevator is ( 103703s.5 m ) = 5.15 m s . Estimating that the elevator must come to rest in the space of one floor, the acceleration 2 2 . m 2 is about 0 −2( 5315m ) s ) = −3.80 m s . ( .5

2.79: a) v = 2 gh = 2(9.80 m s 2 )(21.3 m) = 20.4 m s ; the announcer is mistaken. b) The required speed would be v0 = v 2 + 2 g ( y − y0 ) = (25 m s) + 2(9.80 m s 2 )(−21.3 m) = 14.4 m s, which is not possible for a leaping diver.
2

2.80: If the speed of the flowerpot at the top of the window is v0, height h of the window is h = vave t = v0t + (1 2) gt 2 , or v0 = h − (1 2) gt. t

The distance l from the roof to the top of the window is then 2 v 2 ((1.90 m) /(0.420 s) − (1 2)(9.80 m s )(0.420 s))2 l= 0 = = 0.310 m. 2 2g 2(9.80 m s ) An alternative but more complicated algebraic method is to note that t is the difference between the times taken to fall the heights l + h and h, so that t= 2(l + h) − g 2l , g gt 2 2 + l = l + h.

Squaring the second expression allows cancelation of the l terms, (1 2) gt 2 + 2 gt 2 l 2 = h, which is solved for 1 h  l=  − (1 2) gt  , 2g  t  which is the same as the previous expression.
2

= 2( (59.00 m ss) 2 ) = 1.27 m above the window, so .80 m the greatest height is 13.27 m or 13.3 m to the given precision. 2.81: a) The football will go an additional
v2 2g
2

b) The time needed to reach this height is

2(13.3 m) (9.80 m s 2 ) = 1.65 s.

2.82: a)

2.83: a) From Eq. (2.14), with v0=0, v y = 2a y ( y − y0 ) = 2(45.0 m s 2 )(0.640 m) = 7.59 m s.

b) The height above the release point is also found from Eq. (2.14), with v0 y = 7.59 m s, v y = 0 and a y = − g , (7.59 m s) 2 h= = = 2.94 m 2 g 2(9.80 m s 2 ) (Note that this is also (64.0 cm)
2 v0 y

(

45 m s 2 g

) .The height above the ground is then 5.14 m.

c) See Problems 2.46 & 2.48 or Example 2.8: The shot moves a total distance 2.20 m –1.83 m = 0.37 m, and the time is (7.59 m s) + (7.59 m s) 2 + 2(9.80 m s 2 ) (0.37 m) = 1.60 s. (9.80 m s 2 )

2.84: a) In 3.0 seconds the teacher falls a distance 1 2 1 gt = (9.8 m s 2 )(9.0 s 2 ) = 44.1 m 2 2 To reach her ears after 3.0 s, the sound must therefore have traveled a total distance of h + (h − 44.1) m = 2h − 44.1 m ,where h is the height of the cliff. Given 340 m/s for the speed of sound: 2h − 44.1 m = (340 m s)(3.0 s) = 1020 m , which gives h = 532 m or 530 m to the given precision. y=
2 b) We can use v 2 = v0 y + 2 g ( y − y 0 ) to find the teacher's final velocity. This gives y 2 v y = 2(9.8 m s 2 )(532 m) = 10427 m 2 s 2 and v y = 102 m s .

2.85: a) Let +y be upward. At ceiling, v y = 0, y − y 0 = 3.0 m, a y = −9.80 m s 2 . Solve for v0y.
2 2 v y = v 0 y + 2a y ( y − y 0 ) gives v 0 y = 7.7 m s .

b) v y = v0 y + a y t with the information from part (a) gives t = 0.78 s . c) Let the first ball travel downward a distance d in time t. It starts from its maximum height, so v0 y = 0. y − y 0 = v0 y t + 1 a y t 2 gives d = (4.9 m s 2 )t 2 2 The second ball has v0 y = 1 (7.7 m s) = 5.1 m s . In time t it must travel upward 3 3.0 m − d to be at the same place as the first ball. y − y 0 = v0 y t + 1 a y t 2 gives 3.0 m − d = (5.1m s)t − (4.9 m s 2 ) t 2 . 2 We have two equations in two unknowns, d and t. Solving gives t = 0.59 s and d = 1.7 m. d) 3.0 m − d = 1.3 m

2.86: a) The helicopter accelerates from rest for 10.0 s at a constant 5.0 m s 2 . It thus reaches an upward velocity of v y = v0 y + a y t = (5.0 m s 2 )(10.0 s) = 50.0 m s and a height of y = 1 a y t 2 = 1 (5.0 m s 2 )(10.0 s) 2 = 250 m at the moment the engine is shut 2 2 off. To find the helicopter's maximum height use 2 2 v y = v0 y + 2a y ( y − y 0 ) Taking y0 = 250 m , where the engine shut off, and since v 2 = 0 at the maximum height: y y max − y 0 =
2 − v0 y

2g

(50.0 m s) 2 y max = 250 m − = 378 m 2( −9.8 m s 2 ) or 380 m to the given precision. b) The time for the helicopter to crash from the height of 250 m where Powers stepped out and the engine shut off can be found from: 1 y = y0 + v0 yt + a y t 2 = 250 m + (50.0 m s)t + 1 (−9.8 m s 2 )t 2 = 0 2 2 where we now take the ground as y = 0 . The quadratic formula gives solutions of t = 3.67 s and 13.88 s, of which the first is physically impossible in this situation. Powers' position 7.0 seconds after the engine shutoff is given by: 1 y = 250 m + (50.0 m s)(7.0 s) + (−9.8 m s 2 )(49.0 s 2 ) = 359.9 m 2 at which time his velocity is 2 v y = v0 y + gt = 50.0 m s + (−9.80 m s )(7.0 s) = −18.6 m s Powers thus has 13.88 − 7.0 = 6.88 s more time to fall before the helicopter crashes, at his constant downward acceleration of 2.0 m s 2 . His position at crash time is thus: 1 y = y0 + v0 y t + a y t 2 2 1 = 359.9 m + (−18.6 m s)(6.88 s) + (−2.0 m s 2 )(6.88 s) 2 2 = 184.6 m or 180 m to the given precision.

2.87: Take +y to be downward. Last 1.0 s of fall: y − y 0 = v0 y t + 1 a y t 2 gives h 4 = v0 y (1.0 s) + (4.9 m s 2 )(1.0 s) 2 2 v0y is his speed at the start of this time interval. Motion from roof to y − y 0 = 3h 4 : v 0 y = 0, v y = ?
2 2 v y = v0 y + 2a y ( y − y 0 ) gives v y = 2(9.80 m s 2 )(3h 4) = 3.834 h m s This is vy for the last 1.0 s of fall. Using this in the equation for the first 1.0 s gives h 4 = 3.834 h + 4.9 Let h = u 2 and solve for u : u = 16.5. Then h = u 2 = 270 m.

2.88: a)

t fall + t sound return = 10.0 s t f + t s = 10.0 s d Rock = d Sound 1 2 gt f = vs t s 2 1 (9.8 m s 2 )t f2 = (330 m s)t s 2 (2) (1)

Combine (1) and (2): t f = 8.84 s, t s = 1.16 s m h = vs t s = (330 )(1.16 s) = 383 m s b) You would think that the rock fell for 10 s, not 8.84 s, so you would have thought it fell farther. Therefore your answer would be an overestimate of the cliff's height. 2.89: a) Let +y be upward. y − y 0 = −15.0 m, t = 3.25 s, a y = −9.80 m s 2 , v0 y = ? y − y 0 = v0 y t + 1 a y t 2 gives v0 y = 11.31 m s 2 Use this v0y in v y = v0 y + a y t to solve for v y : v y = −20.5 m s b) Find the maximum height of the can, above the point where it falls from the scaffolding: 2 v y = 0, v 0 y = +11.31 m s , a y = −9.80 m s , y − y 0 = ?
2 2 v y = v 0 y + 2a y ( y − y 0 ) gives y − y 0 = 6.53 m The can will pass the location of the other painter. Yes, he gets a chance.

2.90: a) Suppose that Superman falls for a time t, and that the student has been falling for a time t0 before Superman’s leap (in this case, t0 = 5 s). Then, the height h of the building is related to t and t0 in two different ways: 1 − h = v 0 y t − gt 2 2 1 2 = − g( t + t0 ) , 2 where v0y is Superman’s initial velocity. Solving the second t gives t = Solving the first for v0y gives v0y = −
2h g

− t0 .

h g + t , and substitution of numerical values gives t 2 t = 1.06 s and v0y = –165 m s , with the minus sign indicating a downward initial velocity. b)

c) If the skyscraper is so short that the student is already on the ground, then 1 2 h = gt 0 = 123 m. 2

2.91: a) The final speed of the first part of the fall (free fall) is the same as the initial speed of the second part of the fall (with the Rocketeer supplying the upward acceleration), and assuming the student is a rest both at the top of the tower and at the v2 v2 ground, the distances fallen during the first and second parts of the fall are 1 and 1 , 2g 10 g where v1 is the student's speed when the Rocketeer catches him. The distance fallen in free fall is then five times the distance from the ground when caught, and so the distance above the ground when caught is one-sixth of the height of the tower, or 92.2 m. b) The student falls a distance 5H 6 in time t = 5H 3 g , and the Rocketeer falls the same distance in time t–t0, where t0=5.00 s (assigning three significant figures to t0 is more or less arbitrary). Then, 5H 1 = v0 (t − t 0 ) + g (t − t 0 ) 2 , or 6 2 5H 6 1 v0 = − g (t − t 0 ). (t − t 0 ) 2 At this point, there is no great advantage in expressing t in terms of H and g algebraically; t − t 0 = 5(553 m) 29.40 m s 2 − 5.00 s = 4.698 s , from which v 0 = 75.1 m s . c)

2.92: a) The time is the initial separation divided by the initial relative speed, H/v0. More precisely, if the positions of the balls are described by y1 = v0 t − (1 2) gt 2 , y 2 = H − (1 2) gt 2 , setting y1 = y2 gives H = v0t. b) The first ball will be at the highest point of its motion if at the collision time t found in part (a) its velocity has been reduced from v0 to 0, or gt = 2 gH/v0 = v0, or H = v 0 / g .

2.93: The velocities are v A = α + 2 βt and vB = 2γt − 3δt 2 a) Since vB is zero at t = 0, car A takes the early lead. b) The cars are both at the origin at t = 0. The non-trivial solution is found by setting xA = xB, cancelling the common factor of t, and solving the quadratic for t= 1 ( β − γ ) ± ( β − γ ) 2 − 4αδ . 2δ

[

]

Substitution of numerical values gives 2.27 s, 5.73 s. The use of the term “starting point” can be taken to mean that negative times are to be neglected. c) Setting vA = vB leads to a different quadratic, the positive solution to which is t=− 1 (2 β − 2γ ) − (2 β − 2γ ) 2 − 12αδ . 6δ

[

]

Substitution of numerical results gives 1.00 s and 4.33 s. d) Taking the second derivative of xA and xB and setting them equal, yields, 2 β = 2γ − 6δt . Solving, t = 2.67 s . 2.94: a) The speed of any object falling a distance H – h in free fall is 2 g ( H − h). b) The acceleration needed to bring an object from speed v to rest over a distance h is 2 g ( H − h) v2 H  = = g  − 1. 2h 2h h 

2.95: For convenience, let the student's (constant) speed be v0 and the bus's initial position be x0. Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero, and the initial velocity of the bus is taken to be zero. The positions of the student x1 and the bus x2 as functions of time are then x1 = v0 t , x2 = x0 + (1 2)at 2 . a) Setting x1 = x 2 and solving for the times t gives t= 1 2 v0 ± v0 − 2ax0 a 1 = (5.0 m s) ± (5.0 m s) 2 − 2(0.170 m s 2 )(40.0 m) 2 (0.170 m s ) = 9.55 s, 49.3 s.

(

(

)

)

The student will be likely to hop on the bus the first time she passes it (see part (d) for a discussion of the later time). During this time, the student has run a distance v0 t = (5 m s)(9.55 s) = 47.8 m. b) The speed of the bus is (0.170 m s )(9.55 s) = 1.62 m s . c) The results can be verified by noting that the x lines for the student and the bus intersect at two points:
2

d) At the later time, the student has passed the bus, maintaining her constant speed, but the accelerating bus then catches up to her. At this later time the bus's velocity is 0.170 m s 2 ( 49.3 s ) = 8.38 m s .

(

)

2.96: The time spent above ymax/2 is

1 2

the total time spent in the air, as the time is

proportional to the square root of the change in height. Therefore the ratio is 1/ 2 1 = = 2 .4 . 1 − 1/ 2 2 −1

2.97: For the purpose of doing all four parts with the least repetition of algebra, quantities will be denoted symbolically. That is, 1 1 2 let y1 = h + v0t − gt 2 , y2 = h − g ( t − t0 ) . In this case, t0 = 1.00 s . Setting 2 2 2 y1 = y2 = 0, expanding the binomial ( t − t0 ) and eliminating the common term
1 2 2 gt 2 yields v0t = gt0t − 1 gt0 , which can be solved for t; 2

t=

2 gt0 t 1 = 0 . v gt0 − v0 2 1 − gt00 1 2

Substitution of this into the expression for y1 and setting y1 = 0 and solving for h as a function of v0 yields, after some algebra, 2 1   gt − v  1 2 2 0 0 h = gt0 . 2 ( gt0 − v0 ) 2 a) Using the given value t0 = 1.00 s and g = 9.80 m s 2 , 2  4.9 m s − v0  h = 20.0 m = ( 4.9 m )   9 .8 m s − v  .  0   This has two solutions, one of which is unphysical (the first ball is still going up when the second is released; see part (c)). The physical solution involves taking the negative square root before solving for v0 , and yields 8.2 m s.

b) The above expression gives for i), 0.411 m and for ii) 1.15 km. c) As v0 approaches 9.8 m s , the height h becomes infinite, corresponding to a relative velocity at the time the second ball is thrown that approaches zero. If v0 > 9.8 m s , the first ball can never catch the second ball. d) As v0 approaches 4.9 m/s, the height approaches zero. This corresponds to the first ball being closer and closer (on its way down) to the top of the roof when the second ball is released. If v0 > 4.9 m s , the first ball will already have passed the roof on the way down before the second ball is released, and the second ball can never catch up.

2.98: a) Let the height be h and denote the 1.30-s interval as ∆ t; the simultaneous equations h = 1 gt 2 , 2 h = 1 g (t − ∆t ) 2 can be solved for t. Eliminating h and taking the 2 3 2 square root,
t t − ∆t

=

3 2

, and t = 1− ∆t2 3 , and substitution into h = 1 gt 2 gives h = 246 m. 2

This method avoids use of the quadratic formula; the quadratic formula is a generalization of the method of “completing the square”, and in the above form, 2 h = 1 g (t − ∆t ) 2 , the square is already completed. 3 2 b) The above method assumed that t >0 when the square root was taken. The negative root (with ∆t = 0) gives an answer of 2.51 m, clearly not a “cliff”. This would correspond to an object that was initially near the bottom of this “cliff” being thrown upward and taking 1.30 s to rise to the top and fall to the bottom. Although physically possible, the conditions of the problem preclude this answer.

Chapter 3

3.1: a)

v x ,ave = v y ,ave

(5.3 m) − (1.1 m) = 1.4 m s , (3.0 s) (−0.5 m) − (3.4 m) = = −1.3 m s . (3.0 s)

b) vave = (1.4 m s) 2 + (−1.3 m s) 2 = 1.91 m s , or 1.9 m s to two significant figures, θ = arctan( −1.3 ) = −43° . 1.4 3.2: a) x = (v x ,ave )Δt = ( −3.8 m s)(12.0 s) = −45.6 m and y = (v y ,ave )Δt = (4.9 m s)(12.0 s) = 58.8 m. b) r = x 2 + y 2 = (−45.6 m) 2 + (58.8 m) 2 = 74.4 m.  ˆ 3.3: The position is given by r = [4.0 cm + (2.5 cm s 2 )t 2 ]i + (5.0 cm s)tˆ . j ˆ , and (a) r (0) = [4.0 cm]i ˆ ˆ r(2s) = [4.0 cm + (2.5 cm s 2 )(2 s) 2 ]i + (5.0 cm s)(2 s) ˆ = (14.0 cm) i + (10.0 cm) ˆ . Then j j  ˆ + (10 cm − 0 ) ˆ j ˆ using the definition of average velocity, vave = (14 cm − 4 cm )2is = (5 cm s)i + (5 cm s) ˆ. j v ave = 7.1 cm s at an angle of 45° .   ˆ ˆ b) v = dr = (2)(2.5 cm s)ti + (5 cm s) ˆ = (5 cm s)ti + (5 cm s) ˆ . Substituting for j j

t = 0,1s , and 2 s, gives:    ˆ ˆ v (0) = (5 cm s) ˆ, v (1 s) = (5 cm s)i + (5 cm s) ˆ , and v (2 s) = (10 cm s)i + (5 cm s) ˆ . j j j  The magnitude and direction of v at each time therefore are: t = 0 : 5.0 cm s at 90° ; t = 1.05 : 7.1 cm s at 45°; t = 2.05 : 11 cm s at 27° . c)

dt

 ˆ 3.4: v = 2bti + 3ct 2 ˆ . This vector will make a 45° -angle with both axes when the x- and j y-components are equal; in terms of the parameters, this time is 2b 3c . 3.5: a)

b)

a x ,ave = a y ,ave
2

( −170 m s) − (90 m s) 2 = −8.7 m s , (30.0 s) (40 m s) − (110 m s) 2 = = −2.3 m s . (30.0 s)
2 −2 3 ( −8..7 ) = 14.8° + 180° = 195°.

c)

(−8.7 m s ) 2 + (− 2.3 m s ) 2 = 9.0 m s 2 , arctan

3.6: a) a x = (0.45 m s ) cos 31.0° = 0.39 m s , a y = (0.45 m s ) sin 31.0° = 0.23 m s , so v x = 2.6 m s + (0.39 m s )(10.0 s) = 6.5 m s and
2

2

2

2

2

v y = −1.8 m s + (0.23 m s )(10.0 s) = 0.52 m s .
b) v = (6.5 m s) 2 + (0.52 m s) 2 = 6.48 m s , at an angle of arctan ( 06.52 ) = 4.6° above the .5 horizontal. c)

2

3.7: a)

b)

 2 ˆ ˆ v = αi − 2 βtˆ = (2.4 m s)i − [(2.4 m s )t ] ˆ j j  2 a = −2 βˆ = (−2.4 m s ) ˆ. j j  ˆ c) At t = 2.0 s , the velocity is v = (2.4 m s)i − ( 4.8 m s) ˆ ; the magnitude is j

. (2.4 m s) 2 + (−4.8 m s) 2 = 5.4 m s , and the direction is arctan ( −24.48 ) = −63° . The

acceleration is constant, with magnitude 2.4 m s 2 in the − y -direction. d) The velocity vector has a component parallel to the acceleration, so the bird is speeding up. The bird is turning toward the − y -direction, which would be to the bird’s right (taking the + z direction to be vertical). 3.8:

3.9: a) Solving Eq. (3.18) with y = 0 , v0 y = 0 and t = 0.350 s gives y 0 = 0.600 m . b) v x t = 0.385 m c) v x = v0 x = 1.10 m s, v y = − gt = −3.43 m s, v = 3.60 m s , 72.2° below the horizontal.

3.10: a) The time t is given by t =

2h g

= 7.82 s .

b) The bomb’s constant horizontal velocity will be that of the plane, so the bomb travels a horizontal distance x = v x t = (60 m s)(7.82 s) = 470 m . c) The bomb’s horizontal component of velocity is 60 m/s, and its vertical component is − gt = −76.7 m s . d)

e) Because the airplane and the bomb always have the same x-component of velocity and position, the plane will be 300 m above the bomb at impact.

3.11: Take + y to be upward. Use Chirpy’s motion to find the height of the cliff. 2 v 0 y = 0, a y = −9.80 m s , y − y 0 = −h, t = 3.50 s y − y 0 = v 0 y t + 1 a y t 2 gives h = 60.0 m 2 Milada: Use vertical motion to find time in the air. 2 v 0 y = v0 sin 32.0°, y − y 0 = −60.0 m, a y = −9.80 m s , t = ? y − y 0 = v 0 y t + 1 a y t 2 gives t = 3.55 s 2 Then v0 x = v0 cos 32.0°, a x = 0, t = 3.55 s gives x − x 0 = 2.86 m . 3.12: Time to fall 9.00 m from rest: 1 2 gt 2 1 9.00 m = (9.8 m s 2 )t 2 2 t = 1.36 s y= Speed to travel 1.75 m horizontally: x = v0 t 1.75 m = v 0 (1.36 s) v 0 = 1.3 m s 3.13: Take +y to be upward. Use the vertical motion to find the time in the air: v 0 y = 0, a y = −9.80 m s 2 , y − y 0 = −(21.3 m − 1.8 m) = −19.5 m, t = ? y − y 0 = v 0 y t + 1 a y t 2 gives t = 1.995 s 2 Then x − x0 = 61.0 m, a x = 0, t = 1.995 s, v0 x = ? x − x0 = v0 x t + 1 a x t 2 gives v 0 x = 30.6 m s. 2 b) v x = 30.6 m s since a x = 0 v y = v 0 y + a y t = −19.6 m s
2 2 v = v x + v y = 36.3 m s

3.14: To make this prediction, the student needs the ball’s horizontal velocity at the moment it leaves the tabletop and the time it will take for the ball to reach the floor (or rather, the rim of the cup). The latter can be determined simply by measuring the height of the tabletop above the rim of the cup and using y = 1 gt 2 to calculate the falling time. 2 The horizontal velocity can be determined (although with significant uncertainty) by timing the ball’s roll for a measured distance before it leaves the table, assuming that its speed doesn’t change much on the hard tabletop. The horizontal distance traveled while the ball is in flight will simply be horizontal velocity × falling time. The cup should be placed at this distance (or a slightly shorter distance, to allow for the slowing of the ball on the tabletop and to make sure it clears the rim of the cup) from a point vertically below the edge of the table.

3.15: a) Solving Eq. (3.17) for v y = 0 , with v 0 y = (15.0 m s) sin 45.0° , = 1.08 s. 2 9.80 m s b) Using Equations (3.20) and (3.21) gives at t1 , ( x, y ) = (6.18 m, 4.52 m) : t 2 , (11.5 m, 5.74 m) : t 3 , (16.8 m, 4.52 m) . c) Using Equations (3.22) and (3.23) gives at t1 , (v x , v y ) = (10.6 m s , 4.9 m s) : t 2 , (10.6 m s , 0) t 3 : (10.6 m s , − 4.9 m s), for velocities, respectively, of 11.7 m s @ 24.8°, 10.6 m s @ 0° and 11.7 m s @ −24.8°. Note that vx is the same for all times, and that the y-component of velocity at t3 is negative that at t1 . d) The parallel and perpendicular components of the acceleration are obtained from      a ⋅v   (a ⋅ v ) v    a|| = , a|| = , a⊥ = a − a|| . 2 v v    ˆ, so a ⋅ v = − gv , and the components of acceleration For projectile motion, a = − gj y parallel and perpendicular to the velocity are t1 : −4.1 m s 2 , 8.9 m s 2 . t 2 : 0, 9.8 m s 2 . t 3 : 4.1 m s 2 , 8.9 m s 2 . e) T= (15.0 m s) sin 45°

f) At t1, the projectile is moving upward but slowing down; at t2 the motion is instantaneously horizontal, but the vertical component of velocity is decreasing; at t3, the projectile is falling down and its speed is increasing. The horizontal component of velocity is constant.

3.16: a) Solving Eq. (3.18) with y = 0, y 0 = 0.75 m gives t = 0.391s . b) Assuming a horizontal tabletop, v 0 y = 0 , and from Eq. (3.16), v0 x = ( x − x 0 ) / t = 3.58 m s . c) On striking the floor, v y = − gt = − 2 gy 0 = −3.83 m s , and so the ball has a velocity of magnitude 5.24 m s , directed 46.9° below the horizontal. d)

Although not asked for in the problem, this y vs. x graph shows the trajectory of the tennis ball as viewed from the side.

2 3.17: The range of a projectile is given in Example 3.11, R = v0 sin 2α 0 g .

a) (120 m s) 2 sin 110° (9.80 m s ) = 1.38 km .b) (120 m s) 2 sin 110° (1.6 m s ) = 8.4 km .

2

2

3.18: a) The time t is
v2

vy 0 g

=

16.0 m s 9.80 m s 2

= 1.63 s .

b) 1 gt 2 = 1 v y 0 t = 2yg0 = 13.1 m . 2 2 c) Regardless of how the algebra is done, the time will be twice that found in part (a), or 3.27 s d) vx is constant at 20.0 m s , so (20.0 m s)(3.27 s) = 65.3 m . e)

3.19: a) v 0 y = (30.0 m s) sin 36.9° = 18.0 m s ; solving Eq. (3.18) for t with y0 = 0 and

y = 10.0 m gives
= 0.68 s, 2.99 s 2 9.80 m s b) The x-component of velocity will be (30.0 m s) cos 36.9° = 24.0 m s at all times. The y-component, obtained from Eq. (3.17), is 11.3 m s at the earlier time and − 11.3 m s at the later. c) The magnitude is the same, 30.0 m s , but the direction is now 36.9° below the horizontal. t= (18.0 m s) ± (18.0 m s) 2 − 2(9.80 m s )(10.0 m)
2

3.20: a) If air resistance is to be ignored, the components of acceleration are 0 2 horizontally and − g = −9.80 m s vertically. b) The x-component of velocity is constant at v x = (12.0 m s) cos 51.0° = 7.55 m s . The y-component is v 0 y = (12.0 m s) sin 51.0° = 9.32 m s at release and

v 0 y − gt = (10.57 m s) − (9.80 m s )(2.08 s) = −11.06 m s when the shot hits.
c) v0 x t = (7.55 m s)(2.08 s) = 15.7 m . d) The initial and final heights are not the same. e) With y = 0 and v0y as found above, solving Eq. (3.18) for y0 = 1.81 m . f)

2

3.21: a) The time the quarter is in the air is the horizontal distance divided by the horizontal component of velocity. Using this time in Eq. (3.18), x gx 2 y − y 0 = v0 y − 2 v0 x 2v 0 x = tan α 0 x − gx 2 2 v0 2 cos 2 α 0
2

(9.80 m s )(2.1 m) 2 = tan 60°(2.1 m) − = 1.53 m rounded. 2(6.4 m s) 2 cos 2 60° b) Using the same expression for the time in terms of the horizontal distance in Eq. (3.17), 2 (9.80 m s )(2.1 m) gx v y = v 0 sin α 0 − = (6.4 m s) sin 60° − = −0.89 m s . v 0 cosα 0 (6.4 m s) cos 60°

3.22: Substituting for t in terms of d in the expression for ydart gives   gd . ydart = d  tan α 0 − 2 2  2v0 cos α 0    Using the given values for d and α 0 to express this as a function of v0 ,  26.62 m 2 s 2  . y = (3.00 m) 0.90 − 2   v0   Then, a) y = 2.14 m , b) y = 1.45 m , c) y = −2.29 m . In the last case, the dart was fired with so slow a speed that it hit the ground before traveling the 3-meter horizontal distance. 3.23: a) With v y = 0 in Eq. (3.17), solving for t and substituting into Eq. (3.18) gives ( y − y0 ) =
2 v0 y

2g

=

2 v 0 sin 2 α 0 (30.0 m/s ) 2 sin 2 33.0° = = 13.6 m 2g 2(9.80 m/s 2 )

b) Rather than solving a quadratic, the above height may be used to find the time the rock takes to fall from its greatest height to the ground, and hence the vertical component of velocity, v y = 2 yg = 2(28.6 m)(9.80 m/s 2 ) = 23.7 m/s , and so the speed of the rock is (23.7 m/s) 2 + ((30.0 m/s)(cos 33.0°)) 2 = 34.6 m/s . c) The time the rock is in the air is given by the change in the vertical component of velocity divided by the acceleration –g; the distance is the constant horizontal component of velocity multiplied by this time, or (−23.7 m/s − ((30.0 m/s)sin33. 0°)) x = (30.0 m/s)cos33.0° = 103 m. (−9.80 m/s 2 ) d)

3.24: a)

v 0 cosαt = 45.0 m cosα = 45.0 m = 0.600 ( 25.0 m/s)(3.00 s)

α = 53.1°
b) v x = (25.0 m/s) cos 53.1° = 15.0 m/s vy= 0

v = 15.0 m/s a = 9.80 m/s 2 downward
c) Find y when t = 3.00s y = v0 sin αt − 1 2 gt 2
1 (9.80 m/s 2 )(3.00 s) 2 2

= (25.0 m/s)(sin53.1°)(3.00s) −
= 15.9 m v x = 15.0 m/s = constant
2 2

v y = v 0 sin α − gt = (25.0 m/s)(sin 53.1°) − (9.80 m/s 2 )(3.00 s) = −9.41

v = v x + v y = (15.0 m/s ) 2 + (−9.41 m/s 2 = 17.7 m/s

3.25: Take + y to be downward. a) Use the vertical motion of the rock to find the initial height. t = 6.00 s, v0 y = +20.0 s, a y = +9.80 m/s 2 , y − y0 = ? y − y 0 = v0 y t + 1 a y t 2 gives y − y 0 = 296 m 2 b) In 6.00 s the balloon travels downward a distance y − y0 = (20.0 s)(6.00 s) = 120 m . So, its height above ground when the rock hits is 296 m − 120 m = 176 m . c) The horizontal distance the rock travels in 6.00 s is 90.0 m. The vertical component of the distance between the rock and the basket is 176 m, so the rock is (176 m) 2 + (90 m) 2 = 198 m from the basket when it hits the ground. d) (i) The basket has no horizontal velocity, so the rock has horizontal velocity 15.0 m/s relative to the basket. Just before the rock hits the ground, its vertical component of velocity is v y = v0 y + a y t = 20.0 s + (9.80 m/s 2 )(6.00 s) = 78.8 m/s , downward, relative to the ground. The basket is moving downward at 20.0 m/s, so relative to the basket the rock has downward component of velocity 58.8 m/s. e) horizontal: 15.0 m/s; vertical: 78.8 m/s 3.26: a) horizontal motion: x − x0 = v 0 x t so t =
60.0 m (v0 cos 43° ) t

vertical motion (take + y to be upward): y − y 0 = v0 y t + 1 a y t 2 gives 25.0 m = (v 0 sin 43.0°)t + 1 (−9.80 m/s 2 )t 2 2 2 Solving these two simultaneous equations for v0 and t gives v0 = 3.26 m/s and t = 2.51 s . b) v y when shell reaches cliff: v y = v0 y + a y t = (32.6 m/s) sin 43.0° − (9.80 m/s 2 )(2.51 s) = −2.4 m/s The shell is traveling downward when it reaches the cliff, so it lands right at the edge of the cliff. 3.27: Take + y to be upward. Use the vertical motion to find the time it takes the suitcase to reach the ground: v 0 y = v0 sin 23°, a y = −9.80 m/s 2 , y − y 0 = −114 m, t = ? y − y 0 = v0 y t + 1 a y t 2 gives t = 9.60 s 2 The distance the suitcase travels horizontally is x − x0 = v0 x = (v0 cos 23.0°)t = 795 m

3.28: For any item in the washer, the centripetal acceleration will be proportional to the square of the frequency, and hence inversely proportional to the square of the rotational period; tripling the centripetal acceleration involves decreasing the period by a factor of 3 , so that the new period T ′ is given in terms of the previous period T by T ′ = T / 3 . 3.29: Using the given values in Eq. (3.30), 4π 2 (6.38 × 10 6 m) = 0.034 m/s 2 = 3.4 × 10 −3 g. 2 ((24 h)(3600 s/h)) (Using the time for the siderial day instead of the solar day will give an answer that differs in the third place.) b) Solving Eq. (3.30) for the period T with arad = g , a rad = T= 4π 2 (6.38 × 106 m) = 5070 s ~ 1.4 h. 9.80 m/s 2

3.30: 550 rev/min = 9.17 rev/s , corresponding to a period of 0.109 s. a) From Eq. (3.29), π v = 2TR = 196 m/s . b) From either Eq. (3.30) or Eq. (3.31), arad = 1.13 × 104 m/s 2 = 1.15 × 103 g . 3.31: Solving Eq. (3.30) for T in terms of R and arad , a) 4π 2 (7.0 m)/(3.0)(9.80 m/s 2 ) = 3.07 s . b) 1.68 s.

3.32: a) Using Eq. (3.31), arad = 5.91 × 10−3 m/s 2 .

2πR T

= 2.97 × 10 4 m/s . b) Either Eq. (3.30) or Eq. (3.31) gives

c) v = 4.78 × 104 m/s , and a = 3.97 × 10−2 m/s 2 .

3.33: a) From Eq. (3.31), a = (7.00 m/s) 2 /(15.0 m) = 3.50 m/s 2 . The acceleration at the bottom of the circle is toward the center, up. b) a = 3.50 m/s 2 , the same as part (a), but is directed down, and still towards the center. c) From Eq. (3.29), T = 2πR / v = 2π (15.0 m)/(7.00 m/s) = 12.6 s .

3.34: a) arad = (3 m/s) 2 /(14 m) = 0.643 m/s 2 , and atan = 0.5 m/s 2 . So, a = ((0.643 m/s 2 ) 2 + (0.5 m/s 2 ) 2 )1 / 2 = 0.814 m/s 2 , 37.9° to the right of vertical. b)

3.35: b) No. Only in a circle would arad point to the center (See planetary motion in Chapter 12). c) Where the car is farthest from the center of the ellipse. 3.36: Repeated use of Eq. (3.33) gives a) 5.0 = m/s to the right, b) 16.0 m/s to the left, and c) 13.0 = m/s to the left. 3.37: a) The speed relative to the ground is 1.5 m/s + 1.0 m/s = 2.5 m/s , and the time is 35.0 m/2.5 m/s = 14.0 s. b) The speed relative to the ground is 0.5 m/s, and the time is 70 s. 3.38: The walker moves a total distance of 3.0 km at a speed of 4.0 km/h, and takes a time of three fourths of an hour (45.0 min). The boat’s speed relative to the shore is 6.8 km/h downstream and 1.2 km/h upstream, so the total time the rower takes is 1.5 km 1.5 km + = 1.47 hr = 88 min. 6.8 km/h 1.2 km/h 3.39: The velocity components are − 0.50 m/s + (0.40 m/s)/ 2 east and (0.40 m/s)/ 2 south, for a velocity relative to the earth of 0.36 m/s, 52.5° south of west.

3.40: a) The plane’s northward component of velocity relative to the air must be 80.0 km/h, so the heading must be arcsin 80.8 = 14° north of west. b) Using the angle found in 320 part (a), (320 km/h) cos 14° = 310 km/h . Equivalently, (320 km/h) 2 − (80.0 km/h) 2 = 310 km/h .

3.41: a) (2.0 m/s) 2 + (4.2 m/s) 2 = 4.7 m/s, arctan 2.0 = 25.5° , south of east. 4.2 b) 800 m/4.2 m/s = 190 s . c) 2.0 m/s × 190 s = 381 m . 3.42: a) The speed relative to the water is still 4.2 m/s; the necessary heading of the boat is arcsin 2.0 = 28° north of east. b) (4.2 m/s) 2 − (2.0 m/s) 2 = 3.7 m/s , east. d) 4.2 800 m/3.7 m/s = 217 s , rounded to three significant figures. 3.43: a)

b) x : −(10 m/s) cos 45° = −7.1 m/s. y := −(35 m/s) − (10 m/s)sin 45° = −42.1 m/s . c) (−7.1 m/s) 2 + (−42.1 m/s) 2 = 42.7 m/s, arctan −−42..11 = 80° , south of west. 7

3.44: a) Using generalizations of Equations 2.17 and 2.18, α v x = v0 x + α t 3 , v y = v0 y + βt − γ2 t 2 , and x = v0 x t + 12 t 4 , y = v 0 y t + β t 2 − γ6 t 3 . b) Setting 3 2 v y = 0 yields a quadratic in t , 0 = v 0 y + βt − γ2 t 2 , which has as the positive solution 1 β + β 2 + 2v0γ = 13.59 s, γ keeping an extra place in the intermediate calculation. Using this time in the expression for y(t) gives a maximum height of 341 m. t=

[

]

3.45: a) The ax = 0 and a y = −2 β , so the velocity and the acceleration will be perpendicular only when v y = 0 , which occurs at t = 0 . b) The speed is v = (α 2 + 4 β 2t 2 )1 / 2 , dv / dt = 0 at t = 0 . (See part d below.) c) r and v are perpendicular when their dot product is 0: (αt )(α) + (15.0 m − βt 2 ) × (−2 βt ) = α 2t − (30.0 m) βt + 2 β 2t 3 = 0 . Solve this for t: t=±
( 30.0 m)(0.500 m/s 2 ) − (1.2 m/s) 2 2 ( 0.500 m/s 2 ) 2

= +5.208 s , and 0 s, at which times the student is at (6.25 m,

1.44 m) and (0 m, 15.0 m), respectively. d) At t = 5.208 s , the student is 6.41 m from the origin, at an angle of 13° from the xaxis. A plot of d (t ) = ( x (t ) 2 + y (t ) 2 )1 / 2 shows the minimum distance of 6.41 m at 5.208 s:

e) In the x - y plane the student’s path is:

  ˆ ˆ 3.46: a) Integrating, r = (αt − β t 3 )i + ( γ2 t 2 ) ˆ . Differentiating, a = (−2 β )i + γ ˆ . j j 3 b) The positive time at which x = 0 is given by t 2 = 3α β . At this time, the y-coordinate is γ 3αγ 3(2.4 m/s)(4.0 m/s 2 ) y = t2 = = = 9.0 m 2 2β 2(1.6 m/s 3 ) .

3.47: a) The acceleration is v 2 ((88 km/h)(1 m/s)/(3.6 km/h)) 2 a= = = 0.996 m/s 2 ≈ 1 m/s 2 2x 2(300 m) 15 m b) arctan( 460 m −300 m ) = 5.4° . c) The vertical component of the velocity is 1 15 (88 km/h) ( 3.6 m/s ) 160m = 2.3 m/s . d) The average speed for the first 300 m is 44 km/h, so km/h m the elapsed time is 300 m 160 m + = 31.1 s, (44 km/h)(1 m/s)(3.6 km/h) (88 km/h)(1 m/s)cos 5.4°/(3.6 km/h) or 31 s to two places.

3.48: a)

The equations of motions are: y = h + (v0 sin α )t − x = (v0 cos α )t v y = v0 sin α − gt vx = v0 cos α Note that the angle of 36.9 results in sin 36.9° = 3/5 and cos 36.9° = 4/5 . b) At the top of the trajectory, v y = 0 . Solve this for t and use in the equation for y to find the maximum height: t = reduces to y = h + y =h+
( 25 gh / 8 )(3 / 5 ) 2 2g v 0 sin α 2g
2 2

1 2 gt 2

v0 sin α g

. Then, y = h + (v 0 sin α )

(

v0 sin α g

) − g(
1 2

v0 sin α 2 g

)

, which

. Using v0 = 25 gh / 8 , and sin α = 3 / 5 , this becomes

9 25 = h + 16 h , or y = 16 h . Note: This answer assumes that y0 = h . Taking

9 y0 = 0 will give a result of y = 16 h (above the roof). c) The total time of flight can be found from the y equation by setting y = 0 , assuming y0 = h , solving the quadratic for t and inserting the total flight time in the x equation to

find the range. The quadratic is t=
( 3 / 5) v0 ± ( − ( 3 / 5 ) v0 ) 2 − 4 ( 1 g )( − h ) 2 2( 1 g ) 2

1 2

3 gt 2 − 5 v0 − h = 0 . Using the quadratic formula gives ( 3 / 5) 25 gh / 8 ± g
9 25 gh 16 gh • 8 + 8 25

. Substituting v0 = 25 gh / 8 gives t =
1 2

Collecting terms gives t: t = meaningful and so t = 4
h 2g

(

9h 2g

±

25 h 2g

) = (3
1 2

. Then, using x = (v 0

) . Only the positive root is cos α )t , x = ( ) ( 4 ) = 4h .
h 2g

.

±5

h 2g

25 gh 4 8 5

h 2g

3.49: The range for a projectile that lands at the same height from which it was launched v 2 sin 2 α is R = 0 g . Assuming α = 45° , and R = 50 m, v0 = gR = 22 m/s .

3.50: The bird’s tangential velocity can be found from circumfere nce 2π (8.00 m) 50.27 m vx = = = = 10.05 m/s time of rotation 5.00 s 5.00 s Thus its velocity consists of the components vx = 10.05 m/s and v y = 3.00 m/s . The speed relative to the ground is then 2 2 v = vx + v y = 10.052 + 3.002 = 10.49 m/s or 10.5 m/s (b) The bird’s speed is constant, so its acceleration is strictly centripetal–entirely in the horizontal direction, toward the center of its spiral path–and has magnitude v 2 (10.05 m/s) 2 ac = x = = 12.63 m/s 2 or 12.6 m/s 2 r 8.00 m (c) Using the vertical and horizontal velocity components: 3.00 m/s θ = tan −1 = 16.6° 10.05 m/s 3.51: Take + y to be downward. Use the vertical motion to find the time in the air: v0 y = 0, a y = 9.80 m/s 2 , y − y0 = 25 m, t = ? y − y0 = v0 y t + 1 a y t 2 gives t = 2.259 s 2 During this time the dart must travel 90 m horizontally, so the horizontal component of its velocity must be x − x0 90 m v0 x = = = 40 m/s t 2.25 s

3.52: a) Setting y = −h in Eq. (3.27) (h being the stuntwoman’s initial height above the ground) and rearranging gives 2v 2 sin α 0 cos α 0 2v 2 x2 − 0 x − 0 x h = 0, g g The easier thing to do here is to recognize that this can be put in the form 2v 0 x v 0 y 2v 2 x2 − x − 0 x h = 0, g g the solution to which is v 2 x = 0 x v0 y + v0 y + 2 gh = 55.5 m (south). g b) The graph of vx (t ) is a horizontal line.

[

]

3.53: The distance is the horizontal speed times the time of free fall, 2y 2(90 m) vx = (64.0 m/s) = 274 m. g (9.80 m/s 2 ) 3.54: In terms of the range R and the time t that the balloon is in the air, the car’s original distance is d = R + vcar t . The time t can be expressed in terms of the range and the horizontal component of velocity, t =
R v0 cos α0

, so d = R 1 + v0 vcar α0 . Using cos

(

)

2 R = v0 sin 2α0 / g and the given values yields d = 29.5 m .

2 3.55: a) With α 0 = 45° , Eq. (3.27) is solved for v0 =

gx 2 x− y

. In this case, y = −0.9 m is the

change in height. Substitution of numerical values gives v0 = 42.8 m/s . b) Using the above algebraic expression for v0 in Eq. (3.27) gives  x  y = x−  188 m  (188.9 m)    Using x = 116 m gives y = 44.1 m above the initial height, or 45.0 m above the ground, which is 42.0 m above the fence. 3.56: The equations of motions are y = (v0 sin α )t − 1 / 2 gt 2 and x = (v0 cos α )t , assuming the match starts out at x = 0 and y = 0 . When the match goes in the wastebasket for the minimum velocity, y = 2 D and x = 6 D . When the match goes in the wastebasket for the maximum velocity, y = 2 D and x = 7 D . In both cases, sin α = cos α = 2 / 2. . To reach the minimum distance: 6 D = equation for t gives t = 2D = 6D − 1 g 2
2 2 2

v0t , and 2 D =

2 2

v0t − 1 gt 2 . Solving the first 2

(

6D 2 v0

) . Solving this for v
2

6D 2 v0

. Substituting this into the second equation gives
0

gives v0 = 3 gD .
2 2

To reach the maximum distance: 7 D = equation for t gives t = gives 2 D = 7 D − 1 g 2

v0t , and 2 D =

2 2

v0 t − 1 gt 2 . Solving the first 2

(

7D 2 v0 2

7D 2 v0

) . Solving this for v

. Substituting this into the second equation
0

gives v0 = 49 gD / 5 = 3.13 gD , which,

as expected, is larger than the previous result. 3.57: The range for a projectile that lands at the same height from which it was launched v 2 sin 2 α v 2 sin 2 α is R = 0 g , and the maximum height that it reaches is H = 0 2 g . We must find R when H = D and v0 = 6 gD . Solving the height equation for sin α , D = sin α = (1 / 3)1 / 2 . Then, R =
6 gD sin(70.72 ° ) g 6 gD sin 2 α 2g

, or

, or R = 5.6569 D = 4 2 D .

3.58: Equation 3.27 relates the vertical and horizontal components of position for a given set of initial values. a) Solving for v0 gives gx 2 / 2 cos 2 α 0 . x tan α 0 − y Insertion of numerical values gives v0 = 16.6 m/s . b) Eliminating t between Equations 3.20 and 3.23 gives v y as a function of x ,
2 v0 =

gx . v0 cos α 0 Using the given values yields v x = v0 cos α 0 = 8.28 m/s, v y = −6.98 m/s, so v y = v0 sin α 0 −
.98 v = (8.28 m/s ) 2 + ( −6.98 m/s ) 2 = 10.8 m/s, at an angle of arctan ( −86.24 ) = −40.1° , with the negative sign indicating a direction below the horizontal. c) The graph of vx (t ) is a horizontal line.

3.59: a) In Eq. (3.27), the change in height is y = −h . This gives a quadratic equation in x, the solution to which is v 2 cos α 0  2 2 gh  x= 0  tan α 0 + 2  g v0 cos α 0   v0 cos α 0 2 v0 sin α 0 + v0 sin 2 α 0 + 2 gh . g If h = 0 , the square root reduces to v0 sin α 0 , and x = R . b) The expression for x = becomes x = (10.2 m) cos α 0 + [sin 2 α 0 + sin 2 α 0 + 0.98 ] The angle α 0 = 90° corresponds to the projectile being launched straight up, and there is no horizontal motion. If α 0 = 0 , the projectile moves horizontally until it has fallen the distance h.

[

]

c) The maximum range occurs for an angle less than 45° , and in this case the angle is about 36° .

3.60: a) This may be done by a direct application of the result of Problem 3.59; with α 0 = −40° , substitution into the expression for x gives 6.93 m. b)

c) Using (14.0 m − 1.9 m) instead of h in the above calculation gives x = 6.3 m , so the man will not be hit. 3.61: a) The expression for the range, as derived in Example 3.10, involves the sine of twice the launch angle, and sin (2(90° − α 0 )) = sin (180° − 2α 0 ) = sin 180° cos 2α 0 − cos 180° sin 2α 0 = sin 2α 0 , and so the range is the same. As an alternative, using sin( 90° − α 0 ) = cosα and cos(90° − α 0 ) = sin α 0 in the expression for the range that involves the product of the sine and cosine of α 0 gives the same result. b) The range equation is R =
2 v 0 sin 2α g

. In this case, v0 = 2.2 m/s and R = 0.25 m .

Hence sin 2α = (9.8 m/s 2 )(0.25 m)/( 2.2 m/s 2 ), or sin 2α = 0.5062 ; and α = 15.2° or 74.8° .

3.62: a) Using the same algebra as in Problem 3.58(a), v0 = 13.8 m/s . b) Again, the algebra is the same as that used in Problem 3.58; v = 8.4 m/s , at an angle of 9.1° , this time above the horizontal. c) The graph of vx (t ) is a horizontal line.

A graph of y (t ) vs. x(t ) shows the trajectory of Mary Belle as viewed from the side:

d) In this situation it’s convenient to use Eq. (3.27), which becomes y = (1.327) x − (0.071115 m −1 ) x 2 . Use of the quadratic formula gives x = 23.8 m . 3.63: a) The algebra is the same as that for Problem 3.58, gx 2 2 v0 = . 2 cos2 α 0 ( x tan α 0 − y ) In this case, the value for y is − 15.0 m , the change in height. Substitution of numerical values gives 17.8 m/s. b) 28.4 m from the near bank (i.e., in the water!).

3.64: Combining equations 3.25, 3.22 and 3.23 gives 2 v 2 = v0 cos 2 α 0 + (v0 sin α 0 − gt ) 2
2 = v0 (sin 2 α 0 + cos 2 α 0 ) − 2v0 sin α 0 gt + ( gt ) 2 2 = v0 − 2 g (v0 sin α 0 t − 2 = v0 − 2 gy,

1 2 gt ) 2

where Eq. (3.21) has been used to eliminate t in favor of y. This result, which will be seen in the chapter dealing with conservation of energy (Chapter 7), is valid for any y, positive, negative or zero, as long as v 2 > 0 . For the case of a rock thrown from the roof of a building of height h, the speed at the ground is found by substituting y = −h into the
2 above expression, yielding v = v0 + 2 gh , which is independent of α 0 .

3.65: Take + y to be upward. The vertical motion of the rocket is unaffected by its horizontal velocity. a) v y = 0 (at maximum height), v 0 y = +40.0 m/s, a y = −9.80 m/s 2 , y − y 0 = ?
2 2 v y = v0 y + 2a y ( y − y0 ) gives y − y0 = 81.6 m b) Both the cart and the rocket have the same constant horizontal velocity, so both travel the same horizontal distance while the rocket is in the air and the rocket lands in the cart. c) Use the vertical motion of the rocket to find the time it is in the air. v0 y = 40 m/s, a y = −9.80 m/s 2 , v y = −40 m/s, t = ?

v y = v0 y + a y t gives t = 8.164 s Then x − x0 = v0 xt = (30.0 m/s)(8.164 s) = 245 m. d) Relative to the ground the rocket has initial velocity components v0 x = 30.0 m/s and v0 y = 40.0 m/s , so it is traveling at 53.1° above the horizontal. e) (i)

Relative to the cart, the rocket travels straight up and then straight down (ii)

Relative to the ground the rocket travels in a parabola.

3.66: (a) v x (runner) = v x (ball) 6.00 m/s = (20.0 m/s ) cos θ cos θ = 0.300 θ = 72.5° Time the ball is in the air: y = v0 sin θt − 1 gt 2 2 1 − 45.0 m = (20.0 m/s)(sin 72.5°)t − (9.80 m/s 2 )t 2 2 Solve for t: t = 5.549 s . x = v0 cos θt = (20.0 m/s)(cos72.5°)(5.549 s) = 33.4 m (b)

3.67: Take + y to be downward. a) Use the vertical motion of the boulder to find the time it takes it to fall 20 m to the level of the surface of the water. v 0 y = 0, a y = 9.80 m/s 2 , y − y 0 = 20 m, t = ? y − y 0 = v 0 y t + 1 a y t 2 gives t = 2.02 s 2 The rock must travel 100 m horizontally during this time, so x − x0 100 m v0 x = = = 49 m/s. t 2.20 s b) The rock travels downward 45 m in going from the cliff to the plain. Use this vertical motion to find the time: v0 y = 0, a y = 9.80 m/s 2 , y − y0 = 45 m, t = ? y − y 0 = v 0 y t + 1 a y t 2 gives t = 3.03 s 2 During this time the rock travels horizontally x − x0 = v0 xt = (49 m/s)(3.03 s) = 150 m The rock lands 50 m past the foot of the dam. 3.68: (a) When she catches the bagels, Henrietta has been jogging for 9.00 s plus the time for the bagels to fall 43.9 m from rest. Get the time to fall: 1 y = gt 2 2 1 43.9 m = (9.80 m/s 2 )t 2 2 t = 2.99 s So she has been jogging for 9.00 s + 2.99 s = 12.0 s . During this time she has gone x = vt = (3.05 m/s)(12.0 s) = 36.6 m . Bruce must throw the bagels so they travel 36.6 m horizontally in 2.99 s x = vt 36.6 m = v(2.99 s) v = 12.2 m/s (b) 36.6 m from the building.

3.69: Take + y to be upward. a) The vertical motion of the shell is unaffected by the horizontal motion of the tank. Use the vertical motion of the shell to find the time the shell is in the air: v0 y = v0 sin α = 43.4 m/s, a y = −9.80 m/s 2 , y − y0 = 0 (returns to initial height) , t = ? y − y 0 = v 0 y t + 1 a y t 2 gives t = 8.86 s 2 Relative to tank #1 the shell has a constant horizontal velocity v0 cosα = 246.2 m/s . Relative to the ground the horizontal velocity component is 246.2 m/s + 15.0 m/s = 261.2 m/s . Relative to tank #2 the shell has horizontal velocity component 261.2 m/s − 35.0 m/s = 226.2 m/s . The distance between the tanks when the shell was fired is the (226.2 m/s) (8.86 s) = 2000 m that the shell travels relative to tank #2 during the 8.86 s that the shell is in the air. b) The tanks are initially 2000 m apart. In 8.86 s tank #1 travels 133 m and tank #2 travels 310 m, in the same direction. Therefore, their separation increases by 310 m − 183 m = 177 m . So, the separation becomes 2180 m (rounding to 3 significant figures). 3.70: The firecracker’s falling time can be found from the usual 2h t= g The firecracker’s horizontal position at any time t (taking the student’s position as x = 0 ) is x = vt − 1 at 2 = 0 when cracker hits the ground, from which we can find that t = 2v / a . 2 Combining this with the expression for the falling time: 2v 2h = a g so 2v 2 g h= 2 a

3.71: a) The height above the player’s hand will be

2 v0 y

2g

=

2 v 0 sin 2 α 0 2g

= 0.40 m , so the

maximum height above the floor is 2.23 m. b) Use of the result of Problem 3.59 gives 3.84 m. c) The algebra is the same as that for Problems 3.58 and 3.62. The distance y is 3.05 m − 1.83 m = 1.22 m , and (9.80 m/s 2 )(4.21 m) 2 = 8.65 m/s. 2 cos 2 35°((4.21 m) tan 35° − 1.22 m) d) As in part (a), but with the larger speed, 1.83 m + (8.65 m/s) 2 sin 2 35° / 2(9.80 m/s 2 ) = 3.09 m. The distance from the basket is the distance from the foul line to the basket, minus half the range, or 4.21 m − (8.655 m/s ) 2 sin 70° / 2(9.80 m/s 2 ) = 0.62 m. Note that an extra figure in the intermediate calculation was kept to avoid roundoff error. v0 = 3.72: The initial y-component of the velocity is v0 y = 2 gy , and the time the pebble is in flight is t = 2 y / g . The initial x-component is v0 x = x / t = x 2 g / 2 y . The magnitude of the initial velocity is then  x  x2 g v0 = 2 gy + = 2 gy 1 +   ,  2y  2y  
2

and the angle is arctan

( ) = arctan (2 y / x) .
v0 y v0 x

3.73: a) The acceleration is given as g at an angle of 53.1° to the horizontal. This is a 34-5 triangle, and thus, a x = (3 / 5) g and a y = (4 / 5) g during the "boost" phase of the flight. Hence this portion of the flight is a straight line at an angle of 53.1° to the horizontal. After time T, the rocket is in free flight, the acceleration is a x = 0 and a y = g , and the familiar equations of projectile motion apply. During this coasting phase of the flight, the trajectory is the familiar parabola.

b) During the boost phase, the velocities are: vx = (3 / 5) gt and v y = (4 / 5) gt , both straight lines. After t = T , the velocities are vx = (3 / 5) gT , a horizontal line, and v y = (4 / 5) gT − g (t − T ) , a negatively sloping line which crosses the axis at the time of the maximum height.

c) To find the maximum height of the rocket, set v y = 0 , and solve for t, where t = 0 when the engines are cut off, use this time in the familiar equation for y. Thus, using t = (4 / 5)T and 8 4 4 y max = y 0 + v 0 y t − 1 gt 2 , y max = 2 gT 2 + 5 gT ( 4 T ) − 1 g ( 5 T ) 2 , y max = 2 gT 2 + 16 gT 2 − 25 gT 2 . 2 5 5 2 5 25 Combining terms, y max = 18 gT 2 . 25 d) To find the total horizontal distance, break the problem into three parts: The boost phase, the rise to maximum, and the fall back to earth. The fall time back to earth can be found from the answer to part (c), (18 / 25) gT 2 = (1 / 2) gt 2 , or t = (6 / 5)T . Then, 3 3 4 multiplying these times and the velocity, x = 10 gT 2 + ( 3 gT )( 5 T ) + ( 5 gT )( 6 T ) , or 5 5
3 x = 10 gT 2 + 12 gT 2 + 18 gT 2 . Combining terms gives x = 3 gT 2 . 25 25 2

3.74: In the frame of the hero, the range of the object must be the initial separation plus the amount the enemy has pulled away in that time. Symbolically, R = x0 + vE/Ht = x0 + vE/H vRx , where vE/H is the velocity of the enemy relative to the hero, t 0 is the time of flight, v0x is the (constant) x-component of the grenade’s velocity, as measured by the hero, and R is the range of the grenade, also as measured by the hero. Using Eq. (3-29) for R, with sin 2α 0 = 1 and v0 x = v0 / 2 ,
2 v0 v 2 = x0 + v E/H 0 2 , or v 0 − ( 2v E/H )v 0 − gx0 = 0. g g This quadratic is solved for 1 2 v0 = ( 2vE/H + 2vE/H + 4 gx0 ) = 61.1 km/h, 2 where the units for g and x0 have been properly converted. Relative to the earth, the xcomponent of velocity is 90.0 km/h + (61.1 km/h) cos 45° = 133.2 km/h , the y-component, the same in both frames, is (61.1 km/h) sin 45° = 43.2 km/h , and the magnitude of the velocity is then 140 km/h.

3.75: a) x 2 + y 2 = ( R cos ωt ) 2 + ( R sin ωt ) 2 = R 2 (cos2 ωt + sin 2 ωt ) = R 2 , so the radius is R. b) and so the dot product vx = −ωR sin ωt , v y = ωR cosωt ,   r ⋅ v = xv x + yv y = ( R cos ωt )(−ωR sin ωt ) + ( R sin ωt )(ωR cos ωt ) = ωR(− cos ωt sin ωt + sin ωt cos ωt ) = 0. a x = −ω 2 R cos ωt = −ω 2 x, a y = ω 2 R sin ωt = −ω 2 y,   and so a = −ω 2 r and a = ω 2 R. c)
2 2 d) v 2 = vx + v y = ( −ωR sin ωt ) 2 + (ωR cosωt ) 2 = ω 2 R 2 (sin 2 ωt + cos2 ωt ) = ω 2 R 2 , and so

v = ωR .

e)

a = ω 2R =

(ωR ) 2
R

=

v2 . R

3.76: a) dv d 2 = vx + v 2 y dt dt 2 2 d (1 / 2) dt (vx + v y ) = 2 2 vx + v y = vx a x + v y a y
2 vx + v 2 y

.

b) Using the numbers from Example 3.1 and 3.2, dv (−1.0 m/s)( −0.50 m/s 2 ) + ( 1.3 m/s)( 0.30 m/s 2 ) = = 0.54 m/s. dt ( −1.0 m/s) 2 + ( 1.3 m/s) 2 The acceleration is due to changing both the magnitude and direction of the velocity. If the direction of the velocity is changing, the magnitude of the acceleration is larger than   2 2 the rate of change of speed. c) v ⋅ a = v x a x + v y a y , v = v x + v y , and so the above form   for dv is seen to be v ⋅ a / v. dt 3.77: a) The path is a cycloid.

b) To find the velocity components, take the derivative of x and y with respect to time: v x = Rω (1 − cos ωt ), and v y = Rω sin ωt. To find the acceleration components, take the derivative of vx and v y with respect to time: a x = Rω 2 sin ωt, and a y = Rω 2 cos ωt. c) The particle is at rest (v y = vx = 0) every period, namely at t = 0 , 2π / ω, 4π / ω,.... At that time, x = 0 , 2πR, 4πR,...; and y = 0. The acceleration is a = Rω 2 in the + y - direction. d) No, since a = Rω 2 sin ωt + Rω 2 cosωt

[(

) (
2

)]

2 1/ 2

= Rω 2 .

3.78: A direct way to find the angle is to consider the velocity relative to the air and the velocity relative to the ground as forming two sides of an isosceles triangle. The wind direction relative to north is half of the included angle, or arcsin(10 / 50) = 11.53°, east of north.

3.79: Finding the infinite series consisting of the times between meeting with the brothers is possible, and even entertaining, but hardly necessary. The relative speed of the brothers is 70 km/h, and as they are initially 42 km apart, they will reach each other in six-tenths of an hour, during which time the pigeon flies 30 km. 3.80: a) The drops are given as falling vertically, so their horizontal component of velocity with respect to the earth is zero. With respect to the train, their horizontal component of velocity is 12.0 m/s, west (as the train is moving eastward). b) The vertical component, in either frame, is ( 12.0 m/s) / ( tan 30°) = 20.8 m/s, and this is the magnitude of the velocity in the frame of the earth. The magnitude of the velocity in the frame of the train is (12.0 m/s ) / sin 30°. (12.0 m/s ) 2 + ( 20.8 m/s ) 2 = 24 m/s. This is, of course, the same as

3.81: a) With no wind, the plane would be 110 km west of the starting point; the wind has blown the plane 10 km west and 20 km south in half an hour, so the wind velocity is (20 km/h ) 2 + (40 km/h ) 2 = 44.7 km/h at a direction of arctan(40 / 20) = 63° south of west. b) arcsin( 40 / 220) = 10.5° north of west. 3.82: a) 2 D / v b) 2 Dv /(v 2 − w2 ) c) 2 D / v 2 − w2 d) 1.50 h, 1.60 h, 1.55 h. 3.83: a) The position of the bolt is 3.00 m + ( 2.50 m/s )t − 1 / 2(9.80 m/s 2 )t 2 , and the position of the floor is (2.50 m/s)t. Equating the two, 3.00 m = (4.90 m/s 2 )t 2 . Therefore t = 0.782 s. b) The velocity of the bolt is 2.50 m/s − (9.80 m/s 2 )(0.782 s) = −5.17 m/s relative to Earth, therefore, relative to an observer in the elevator v = −5.17 m/s − 2.50 m/s = −7.67 m/s. c) As calculated in part (b), the speed relative to Earth is 5.17 m/s. d) Relative to Earth, the distance the bolt travelled is (2.50 m/s )t − 1 / 2(9.80 m/s 2 )t 2 = (2.50 m/s )(0.782 s) − ( 4.90 m/s 2 )(0.782 s) 2 = −1.04 m

3.84: Air speed of plane = With wind from A to B:

5310 km 6.60 h

= 804.5 km/h tAB + t BA = 6.70 h (1) (2) (3)

Same distance both ways: 5310 km = 2655 km 2 (804.5 km/h + vw )tBA = 2655 km Solve (1), (2), and (3) to obtain wind speed vw : vw = 98.1 km/h (804.5 km/h + v w )t AB =

3.85: The three relative velocities are:  v J/G, Juan relative to the ground. This velocity is due north and has magnitude vJ/G = 8.00 m/s.  v B/G, the ball relative to the ground. This vector is 37.0° east of north and has magnitude v B/G = 12.0 m/s.  v B/J, the ball relative to Juan. We are asked to find the magnitude and direction of this vector.       The relative velocity addition equation is v B/G = v B/J + v J/G, so v B/J = v B/G − v J/G . Take + y to be north and + x to be east. vB/Jx = + vB/G sin 37.0° = 7.222 m/s vB/Jy = +vB/G cos 37.0° − vJ/G = 1.584 m/s These two components give vB/J = 7.39 m/s at 12.4° north of east. 3.86: a) v0 y = 2 gh = 2(9.80 m/s 2 )(4.90 m) = 9.80 m/s. b) v0 y / g = 1.00 s. c) The speed relative to the man is (10.8 m/s)2 − (9.80 m/s ) 2 = 4.54 m/s, and the speed relative to the hoop is 13.6 m/s (rounding to three figures), and so the man must be 13.6 m in front of the hoop at release. d) Relative to the flat car, the ball is projected at an angle 9.80 θ = tan −1 9..80 m/s = 65°. Relative to the ground the angle is θ = tan −1 4.54 m/s +m/s m/s = 35.7° 4 54 m/s 9.10

(

)

(

)

3.87: a) (150 m/s ) 2 sin 2° / 9.80 m/s 2 = 80 m.
10 b) 1000 × π (10(×80 m) 2m) = 1.6 × 10 − 3. π
−2 2

c) The slower rise will tend to reduce the time in the air and hence reduce the radius. The slower horizontal velocity will also reduce the radius. The lower speed would tend to increase the time of descent, hence increasing the radius. As the bullets fall, the friction effect is smaller than when they were rising, and the overall effect is to decrease the radius. 3.88: Write an expression for the square of the distance ( D 2 ) from the origin to the particle, expressed as a function of time. Then take the derivative of D 2 with respect to t, and solve for the value of t when this derivative is zero. If the discriminant is zero or negative, the distance D will never decrease. Following this process, sin −1 8 / 9 = 70.5°.

3.89: a) The trajectory of the projectile is given by Eq. (3.27), with α 0 = θ + φ, and the equation describing the incline is y = x tan θ. Setting these equal and factoring out the x = 0 root (where the projectile is on the incline) gives a value for x0 ; the range measured along the incline is 2  2v0   cos2 (θ + φ)  x / cos θ =  [tan(θ + φ) − tan θ] cos θ  .  g    b) Of the many ways to approach this problem, a convenient way is to use the same sort of "trick", involving double angles, as was used to derive the expression for the range along a horizontal incline. Specifically, write the above in terms of α = θ + φ, as
2  2v 0  2 R=  [sin α cosα cosθ − cos α sin θ ] . g cos 2 θ   The dependence on α and hence φ is in the second term. Using the identities

sin α cosα = (1 / 2) sin 2α and cos2 α = (1 / 2)(1 + cos 2α ), this term becomes (1 / 2)[cos θ sin 2α − sin θ cos 2α − sin θ ] = (1 / 2)[sin( 2α − θ ) − sin θ ] . This will be a maximum when sin( 2α − θ ) is a maximum, at 2α − θ = 2φ + θ = 90°, or φ = 45° − θ / 2. Note that this reduces to the expected forms when θ = 0 (a flat incline, φ = 45° and when θ = −90° (a vertical cliff), when a horizontal launch gives the greatest distance).

3.90: As in the previous problem, the horizontal distance x in terms of the angles is  gx  1 tanθ = tan(θ + φ ) −  2   2v  cos2 (θ + φ ) .  0
2 Denote the dimensionless quantity gx / 2v0 by β ; in this case

(9.80 m/s 2 )(60.0 m)cos 30.0° = 0.2486. 2(32.0 m/s ) 2 The above relation can then be written, on multiplying both sides by the product cosθ cos(θ + φ ), β cosθ sin θ cos(θ + φ ) = sin(θ + φ ) cosθ − , cos(θ + φ ) and so β cosθ sin(θ + φ ) cosθ − cos(θ + φ ) sin θ = . cos(θ + φ )

β=

The term on the left is sin((θ + φ ) − θ ) = sin φ, so the result of this combination is

sin φ cos(θ + φ ) = β cosθ .
Although this can be done numerically (by iteration, trial-and-error, or other methods), the expansion sin a cos b = 1 (sin( a + b) + sin( a − b)) allows the angle φ to be isolated; 2 specifically, then 1 (sin( 2φ + θ ) + sin( −θ )) = β cosθ , 2 with the net result that sin( 2φ + θ ) = 2 β cosθ + sin θ . a) For θ = 30°, and β as found above, φ = 19.3° and the angle above the horizontal is θ + φ = 49.3°. For level ground, using β = 0.2871, gives φ = 17.5°. b) For θ = −30°, the same β as with θ = 30° may be used (cos 30° = cos(−30°)), giving φ = 13.0° and φ + θ = −17.0°.

3.91: In a time ∆t, the velocity vector has moved through an angle (in radians) ∆φ = (see Figure 3.23). By considering the isosceles triangle formed by the two velocity  vectors, the magnitude ∆v is seen to be 2v sin(φ / 2) , so that

v∆t R

v   v∆t  10 m/s aave = 2 sin  sin(1.0 / s ⋅ ∆t) = ∆t  2 R  ∆t Using the given values gives magnitudes of 9.59 m/s 2 , 9.98 m/s 2 and 10.0 m/s 2 . The instantaneous acceleration magnitude, v 2 / R = (5.00 m/s) 2 /(2.50 m) = 10.0 m/s 2 is indeed approached in the limit at ∆t → 0. The changes in direction of the velocity vectors are given by ∆θ = v∆t and are, respectively, 1.0 rad, 0.2 rad, and 0.1 rad. Therefore, the angle R of the average acceleration vector with the original velocity vector is π + ∆θ = π / 2 + 1 / 2 rad (118.6°), π / 2 + 0.1 rad (95.7°) , and π / 2 + 0.05 rad (92.9°). 2

3.92:

The x-position of the plane is (236 m/s )t and the x-position of the rocket is (236 m/s )t + 1 / 2(3.00)(9.80 m/s 2 ) cos 30°(t − T ) 2 . The graphs of these two have the form,

If we take y = 0 to be the altitude of the airliner, then y (t ) = −1 / 2 gT 2 − gT (t − T ) + 1 / 2(3.00)(9.80 m/s 2 )(sin 30°)(t − T ) 2 for the rocket. This graph looks like

By setting y = 0 for the rocket, we can solve for t in terms of T , 0 = −(4.90 m/s 2 )T 2 − (9.80 m/s 2 )T (t − T ) + (7.35 m/s 2 )(t − T ) 2 . Using the quadratic formula for the variable x = t − T, we find x = t −T =
( 9.80 m/s 2 )T + ( 9.80 m/s 2T ) 2 + ( 4 )( 7.35 m/s 2 )( 4.9 )T 2 2 ( 7.35 m/s 2 )

or t = 2.72 T . Now, using

the condition that xrocket − xplane = 1000 m, we find (236 m/s)t + (12.7 m/s 2 ) × (t − T ) 2 − (236 m/s )t = 1000 m, or (1.72T ) 2 = 78.6 s 2 . Therefore T = 5.15 s. 3.93: a) Taking all units to be in km and h, we have three equations. We know that heading upstream vc / w − vw / G = 2 where vc / w is the speed of the curve relative to water and vw/G is the speed of the water relative to the ground. We know that heading downstream for a time t , (vc / w + vw / G )t = 5. We also know that for the bottle vw / G (t + 1) = 3. Solving these three equations for vw / G = x, vc / w = 2 + x, therefore ( 2 + x + x)t = 5 or (2 + 2 x)t = 5. Also t = 3 / x − 1, so (2 + 2 x)( 3 − 1) = 5 or x

2 x 2 + x − 6 = 0. The positive solution is x = vw / G = 1.5 km/h. b) vc / w = 2 km/h + vw/G = 3.5 km/h.

Chapter 4

4.1: a) For the magnitude of the sum to be the sum of the magnitudes, the forces must be parallel, and the angle between them is zero. b) The forces form the sides of a right isosceles triangle, and the angle between them is 90° . Alternatively, the law of cosines may be used as F 2 + F 2 = 2 F − 2 F 2 cos θ , from which cos θ = 0 , and the forces are perpendicular. c) For the sum to have 0 magnitude, the forces must be antiparallel, and the angle between them is 180° . 4.2: In the new coordinates, the 120-N force acts at an angle of 53° from the − x -axis, or 233° from the + x -axis, and the 50-N force acts at an angle of 323° from the + x axis. a) The components of the net force are R x = (120 N) cos 233° + (50 N) cos 323° = −32 N R y = ( 250 N) + (120 N) sin 233° + (50 N) sin 323° = 124 N.
2 2 b) R = Rx + Ry = 128 N, arctan( 124 ) = 104° . The results have the same magnitude, −32

(

)

2

and the angle has been changed by the amount (37°) that the coordinates have been rotated. 4.3: The horizontal component of the force is (10 N) cos 45° = 7.1 N to the right and the vertical component is (10 N) sin 45° = 7.1 N down. 4.4: a) Fx = F cos θ , where θ is the angle that the rope makes with the ramp ( θ = 30° in  F 60 N this problem), so F = F = cosxθ = cos.030° = 69.3 N. b) Fy = F sin θ = Fx tan θ = 34.6 N.

4.5: Of the many ways to do this problem, two are presented here. Geometric: From the law of cosines, the magnitude of the resultant is R = (270 N) 2 + (300 N) 2 + 2( 270 N)(300 N) cos 60° = 494 N. The angle between the resultant and dog A’s rope (the angle opposite the side corresponding to the 250-N force in a vector diagram) is then  sin 120°(300 N)   = 31.7°. arcsin   ( 494 N )    Components: Taking the + x -direction to be along dog A’s rope, the components of the resultant are Rx = (270 N) + (300 N) cos 60° = 420 N Ry = (300 N) sin 60° = 259.8 N, so R = (420 N) 2 + (259.8 N) 2 = 494 N, θ = arctan

( 259.8 ) = 31.7°. 420

4.6: a) F1x + F2 x = (9.00 N) cos 120° + (6.00 N) cos ( −126.9°) = −8.10 N F1 y + F2 y = (9.00 N) sin 120° + (6.00 N) sin (−126.9°) = +3.00 N.
2 2 b) R = Rx + Ry = (8.10 N) 2 + (3.00 N) 2 = 8.64 N.

4.7: a = F / m = (132 N) / (60 kg) = 2.2 m / s 2 (to two places). 4.8: F = ma = (135 kg)(1.40 m/s 2 ) = 189 N. 4.9: m = F / a = (48.0 N)/(3.00 m/s 2 ) = 16.00 kg.

4.10: a) The acceleration is a = m=
F a

2x t2

=

2 (11.0 m ) (5.00 s) 2

= 0.88 m / s 2 . The mass is then

=

80.0 N 0.88 m/s 2

= 90.9 kg.

b) The speed at the end of the first 5.00 seconds is at = 4.4 m/s , and the block on the frictionless surface will continue to move at this speed, so it will move another vt = 22.0 m in the next 5.00 s.

4.11: a) During the first 2.00 s, the acceleration of the puck is F / m = 1.563 m/s 2 (keeping an extra figure). At t = 2.00 s , the speed is at = 3.13 m/s and the position is at 2 / 2 = vt / 2 = 3.13 m . b) The acceleration during this period is also 1.563 m/s 2 , and the speed at 7.00 s is 3.13 m/s + (1.563 m/s 2 )(2.00 s) = 6.26 m/s . The position at t = 5.00 s is x = 3.13 m + (3.13 m/s)(5.00 s − 2.00 s) = 125 m , and at t = 7.00 s is 12.5 m + (3.13 m/s)(2.00 s) + (1/2)(1.563 m/s 2 )(2.00 s) 2 = 21.89 m, or 21.9 m to three places. 4.12: a) a x = F / m = 140 N / 32.5 kg = 4.31 m/s 2 . b) With v0 x = 0, x = 1 at 2 = 215 m . 2 c) With v0 x = 0, vx = axt = 2 x / t = 43.0 m/s .  4.13: a) ∑ F = 0 b), c), d)

4.14: a) With v0 x = 0 , ax = b) t =
vx ax 2 vx (3.00 × 106 m/s ) 2 = = 2.50 × 1014 m/s 2 . 2 x 2(1.80 × 10− 2 m)

=

3.00 ×10 6 m / s 2.50 ×1014 m / s 2

= 1.20 × 10−8 s . Note that this time is also the distance divided by

the average speed. c) F = ma = (9.11 × 10−31 kg )(2.50 × 1014 m/s 2 ) = 2.28 × 10−16 N. 4.15: F = ma = w( a / g ) = (2400 N)(12 m/s 2 )(9.80 m/s 2 ) = 2.94 × 103 N.

4.16:

a=

F F F  160  2 2 = = g = (9.80 m/s ) = 22.0 m/s . m w/ g w 71.2  

4.17: a) m = w / g = (44.0 N) /(9.80 m/s 2 ) = 4.49 kg b) The mass is the same, 4.49 kg, and the weight is (4.49 kg )(1.81 m/s 2 ) = 8.13 N.

4.18: a) From Eq. (4.9), m = w / g = (3.20 N) /(9.80 m / s 2 ) = 0.327 kg. b) w = mg = (14.0 kg )(9.80 m / s 2 ) = 137 N. 4.19: F = ma = (55 kg )(15 m / s 2 ) = 825 N. The net forward force on the sprinter is exerted by the blocks. (The sprinter exerts a backward force on the blocks.) 4.20: a) the earth (gravity) b) 4 N, the book c) no d) 4 N, the earth, the book, up e) 4 N, the hand, the book, down f) second g) third h) no i) no j) yes k) yes l) one (gravity) m) no 4.21: a) When air resistance is not neglected, the net force on the bottle is the weight of the bottle plus the force of air resistance. b) The bottle exerts an upward force on the earth, and a downward force on the air. 4.22: The reaction to the upward normal force on the passenger is the downward normal force, also of magnitude 620 N, that the passenger exerts on the floor. The reaction to the passenger’s weight is the gravitational force that the passenger exerts on the earth, F upward and also of magnitude 650 N. ∑ = 620 N−650 N = −0.452 m/s 2 . The passenger’s
m 650 N /9.80 m/s 2

acceleration is 0.452 m / s 2 , downward.

4.23: aE =

F mg (45 kg )(9.80 m / s 2 ) = = = 7.4 × 10 − 23 m / s 2 . 24 mE mE (6.0 × 10 kg)

4.24: (a) Each crate can be considered a single particle:

FAB (the force on mA due to mB ) and FBA (the force on mB due to mA ) form an action-reaction pair. (b) Since there is no horizontal force opposing F, any value of F, no matter how small, will cause the crates to accelerate to the right. The weight of the two crates acts at a right angle to the horizontal, and is in any case balanced by the upward force of the surface on them. 4.25: The ball must accelerate eastward with the same acceleration as the train. There must be an eastward component of the tension to provide this acceleration, so the ball hangs at an angle relative to the vertical. The net force on the ball is not zero.

4.26: The box can be considered a single particle.

For the truck:

The box’s friction force on the truck bed and the truck bed’s friction force on the box form an action-reaction pair. There would also be some small air-resistance force action to the left, presumably negligible at this speed.

4.27: a)

b) For the chair, a y = 0 so n − mg − F sin 37° = 0 n = 142 N 4.28: a)

∑F

y

= ma y gives

b)

T = mg sin θ = (65.0 kg )(9.80 m / s 2 ) sin 26.0° = 279 N

4.29: tricycle and Frank

T is the force exerted by the rope and f g is the force the ground exerts on the tricycle. spot and the wagon

T ′ is the force exerted by the rope. T and T ′ form a third-law action-reaction pair,   T = −T ′. = =
2 ( 0.130 m ) 350 m / s

4.30: a) The stopping time is result.)

x vave

x ( v0 / 2 )

= 7.43 × 10 −4 s.

( 350 m 2 b) F = ma = (1.80 × 10 −3 kg) (7.43×10 -/4s )s) = 848 N. (Using a = v0 / 2 x gives the same

  4.31: Take the + x -direction to be along F1 and the + y -direction to be along R . Then  F2 x = −1300 N and F2 y = 1300 N , so F2 = 1838 N , at an angle of 135° from F1 .
4.32: Get g on X: 1 2 gt 2 1 10.0 m = g (2.2 s) 2 2 g = 4.13 m / s 2 y= wX = mg X = (0.100 kg)(4.03 m / s 2 ) = 0.41 N

4.33: a) The resultant must have no y-component, and so the child must push with a force with y-component (140 N) sin 30° − (100 N) sin 60° = −16.6 N. For the child to exert the smallest possible force, that force will have no x-component, so the smallest possible force has magnitude 16.6 N and is at an angle of 270° , or 90° clockwise from the + x -direction.
F + b) m = ∑ = 100 N cos260°m1402 N cos 30° = 85.6 kg. w = mg = (85.6 kg)(9.80 m / s 2 ) = 840 N. . a .0 / s

4.34: The ship would go a distance 2 2 v0 v0 mv 2 (3.6 × 107 kg)(1.5 m / s) 2 = = 0 = = 506.25 m, 2a 2( F / m) 2 F 2(8.0 × 10 4 N) so the ship would hit the reef. The speed when the tanker hits the reef is also found from
2 v = v0 − (2 Fx / m) = (1.5 m/s) 2 −

2(8.0 × 10 4 N)(500 m) = 0.17 m/s, (3.6 × 107 kg )

so the oil should be safe.
2 4.35: a) Motion after he leaves the floor: v 2 = v0 y + 2a y ( y − y0 ). y

v y = 0 at the maximum height, y − y 0 = 1.2 m, a y = −9.80 m/s 2 , so v0 y = 4.85 m/s. b) aav = ∆v / ∆t = (4.85 m / s) /(0.300 s) = 16.2 m/s 2 . c)

Fav − w = maav Fav = w + ma av = 890 N + (890 N / 9.80 m / s 2 )(16.2 m / s 2 ) Fav = 2.36 × 10 3 N 4.36:
2 v0 (12.5 m / s) 2 F = ma = m = (850 kg) = 3.7 × 10 6 N. −2 2x 2(1.8 × 10 m)

4.37: a)

Fnet = F − mg (upward) b) When the upward force has its maximum magnitude Fmax (the breaking strength), the net upward force will be Fmax − mg and the upward acceleration will be F − mg Fmax 75.0 N a = max = −g= − 9.80 m / s 2 = 5.83 m / s 2 . m m 4.80 kg 4.38: a) w = mg = 539 N b)

 Downward velocity is decreasing so a is upward and the net force should be upward. Fair > mg , so the net force is upward.

c) Taking the upward direction as positive, the acceleration is F F − mg Fair 620 N a = = air = − 9.80 m / s 2 = − 9.80 m / s 2 = 1.47 m / s 2 . m m m 55.0 kg

4.39: a) Both crates moves together, so a = 2.50 m / s 2 b)

T = m1 a = (4.00 kg )(2.50 m / s 2 ) = 10.0 N c)

d) F − T = m2 a

 F > T and the net force is to the right, in the direction of a .

F = T + m2 a = 10.0 N + (6.00 kg )(2.50 m / s 2 ) = 25.0 N 4.40: a) The force the astronaut exerts on the rope and the force that the rope exerts on the astronaut are an action-reaction pair, so the rope exerts a force of 80.0 N on the 80.0 F astronaut. b) The cable is under tension. c) a = m = 105.0 N = 0.762 m / s 2 . d) There is no kg net force on the massless rope, so the force that the shuttle exerts on the rope must be 80.0 N (this is not an action-reaction pair). Thus, the force that the rope exerts on the 80 0 F shuttle must be 80.0 N. e) a = m = 9.05×.10Nkg = 8.84 × 10 −4 m / s 2 . 4 4.41: a) x (0.025 s) = (9.0 × 10 3 m / s 2 )(0.025 s) 2 − (8.0 × 10 4 m / s 3 )(0.025 s) 3 = 4.4 m. b) Differentiating, the velocity as a function of time is v(t ) = (1.80 × 10 4 m / s 2 )t − (2.40 × 10 5 m / s 3 )t 2 , so v(0.025 s) = (1.80 × 10 4 m / s 2 )(0.025 s) − (2.40 × 10 5 m / s 3 )(0.025 s) 2 = 3.0 × 10 2 m / s. c) The acceleration as a function of time is a(t ) = 1.80 × 104 m / s 2 − (4.80 × 105 m / s3 ) t , so (i) at t = 0, a = 1.8 × 10 4 m / s 2 , and (ii) a(0.025 s) = 6.0 × 103 m / s 2 , and the forces are (i) ma = 2.7 × 104 N and (ii) ma = 9.0 × 103 N.

4.42: a) The velocity of the spacecraft is downward. When it is slowing down, the acceleration is upward. When it is speeding up, the acceleration is downward.

b)

speeding up: w > F and the net force is downward slowing down: w < F and the net force is upward c) Denote the y-component of the acceleration when the thrust is F1 by a1 and the ycomponent of the acceleration when the thrust is F2 by a2 . The forces and accelerations are then related by F1 − w = ma1 , F2 − w = ma2 . Dividing the first of these by the second to eliminate the mass gives F1 − w a1 = , F2 − w a2 and solving for the weight w gives a F − a2 F1 w= 1 2 . a1 − a2 In this form, it does not matter which thrust and acceleration are denoted by 1 and which by 2, and the acceleration due to gravity at the surface of Mercury need not be found. Substituting the given numbers, with + y upward, gives w= (1.20 m / s 2 )(10.0 × 10 3 N) − ( −0.80 m / s 2 )(25.0 × 10 3 N ) = 16.0 × 10 3 N. 2 2 1.20 m / s − (−0.80 m / s )

In the above, note that the upward direction is taken to be positive, so that a2 is negative. Also note that although a2 is known to two places, the sums in both numerator and denominator are known to three places.

4.43:

a) The engine is pulling four cars, and so the force that the engine exerts on the first car  is 4m a . b), c), d): Similarly, the forces the cars exert on the car behind are    3m a , 2m a and − m a . e) The direction of the acceleration, and hence the direction of the forces, would change but the magnitudes would not; the answers are the same. 4.44: a) If the gymnast climbs at a constant rate, there is no net force on the gymnast, so the tension must equal the weight; T = mg . b) No motion is no acceleration, so the tension is again the gymnast’s weight.  c) T − w = T − mg = ma = m a (the acceleration is upward, the same direction as the  tension), so T = m( g + a ) .  d) T − w = T − mg = ma = −m a (the acceleration is downward, the same opposite as  the tension), so T = m( g − a ) .

4.45: a)

The maximum acceleration would occur when the tension in the cables is a maximum, a= Fnet T − mg T 28,000 N = = −g= − 9.80 m / s 2 = 2.93 m / s 2 . m m m 2200 kg 28,000 N − 1.62 m / s 2 = 11.1 m / s 2 . 2200 kg

b)

4.46: a) His speed as he touches the ground is v = 2 gh = 2(9.80 m / s 2 )(3.10 m) = 7.80 m / s. b) The acceleration while the knees are bending is v 2 (7.80 m / s) 2 a= = = 50.6 m / s 2 . 2y 2(0.60 m)

c)

The net force that the feet exert on the ground is the force that the ground exerts on the feet (an action-reaction pair). This force is related to the weight and acceleration by F − w = F − mg = ma, so F = m(a + g ) = (75.0 kg)(50.6 m / s 2 + 9.80 m / s 2 ) = 4532 N . As F a fraction of his weight, this force is mg = a + 1 = 6.16 (keeping an extra figure in the g

(

)

intermediate calculation of a). Note that this result is the same algebraically as

(

3.10 m 0.60 m

+1 .

)

4.47: a)

b) The acceleration of the hammer head will be the same as the nail, 2 a = v0 / 2 x = (3.2 m / s) 2 / 2(0.45 cm) = 1.138 × 103 m / s 2 . The mass of the hammer head is its weight divided by g , 4.9 N / 9.80 m / s 2 = 0.50 kg , and so the net force on the hammer head is (0.50 kg )(1.138 × 10 3 m / s 2 ) = 570 N. This is the sum of the forces on the hammer head; the upward force that the nail exerts, the downward weight and the downward 15-N force. The force that the nail exerts is then 590 N, and this must be the magnitude of the force that the hammer head exerts on the nail. c) The distance the nail moves is .12 m, so the acceleration will be 4267 m / s 2 , and the net force on the hammer head will be 2133 N. The magnitude of the force that the nail exerts on the hammer head, and hence the magnitude of the force that the hammer head exerts on the nail, is 2153 N, or about 2200 N.

4.48:

a) The net force on a point of the cable at the top is zero; the tension in the cable must be equal to the weight w. b) The net force on the cable must be zero; the difference between the tensions at the top and bottom must be equal to the weight w, and with the result of part (a), there is no tension at the bottom. c) The net force on the bottom half of the cable must be zero, and so the tension in the cable at the middle must be half the weight, w / 2 . Equivalently, the net force on the upper half of the cable must be zero. From part (a) the tension at the top is w, the weight of the top half is w / 2 and so the tension in the cable at the middle must be w − w/ 2 = w/ 2 . d) A graph of T vs. distance will be a negatively sloped line.

4.49: a)

b) The net force on the system is 200 N − (15.00 kg )(9.80 m / s 2 ) = 53.0 N (keeping three figures), and so the acceleration is (53.0 N) /(15.0 kg) = 3.53 m / s 2 , up. c) The net force on the 6-kg block is (6.00 kg )(3.53 m / s 2 ) = 21.2 N , so the tension is found from F − T − mg = 21.2 N , or T = (200 N) − (6.00 kg)(9.80 m / s 2 ) − 21.2 N = 120 N . Equivalently, the tension at the top of the rope causes the upward acceleration of the rope and the bottom block, so T − (9.00 kg) g = (9.00 kg) a , which also gives T = 120 N . d) The same analysis of part (c) is applicable, but using 6.00 kg + 2.00 kg instead of the mass of the top block, or 7.00 kg instead of the mass of the bottom block. Either way gives T = 93.3 N .

4.50: a)

b) The athlete’s weight is mg = (90.0 kg )(9.80 m / s 2 ) = 882 N . The acceleration of the barbell is found from vav = 0.60 m / 1.6 s = 0.375 m / s . Its final velocity is thus (2)(0.375 m/s) = 0.750 m/s , and its acceleration is v − v0 0.750 m / s a= = = 0.469 m / s 2 t 1.65 The force needed to lift the barbell is given by: Fnet = Flift − wbarbell = ma The barbell’s mass is (490 N) / (9.80 m/s 2 ) = 50.0 kg , so Flift = wbarbell + ma = 490 N + (50.0 kg )(0.469 m / s 2 ) = 490 N + 23 N = 513 N The athlete is not accelerating, so: Fnet = Ffloor − Flift − wathlete = 0 Ffloor = Flift + wathlete = 513 N + 882 N = 1395 N

4.51: a)

L is the lift force b) ∑ Fy = ma y Mg − L = M ( g / 3) L = 2 Mg / 3 c) L − mg = m(g / 2) , where m is the mass remaining. L = 2Mg / 3 , so m = 4M / 9 . Mass 5M / 9 must be dropped overboard.

4.52: a) m = mass of one link

The downward forces of magnitude 2ma and ma for the top and middle links are the reaction forces to the upward force needed to accelerate the links below. b) (i) The weight of each link is mg = (0.300 kg )(9.80 m / s 2 ) = 2.94 N . Using the freebody diagram for the whole chain: F 12 N − 3(2.94 N) 3.18 N a = net = = = 3.53 m / s 2 or 3.5 m / s 2 3m 0.900 kg 0.900 kg (ii) The second link also accelerates at 3.53 m / s 2 , so: Fnet = Ftop − ma − 2mg = ma Ftop = 2ma + 2mg = 2(0.300 kg)(3.53 m / s 2 ) + 2(2.94 N) = 2.12 N + 5.88 N = 8.0 N

4.53: Differentiating twice, the acceleration of the helicopter as a function of time is  ˆ ˆ a = (0.120 m / s 3 )ti − (0.12 m / s 2 )k , and at t = 5.0s , the acceleration is  ˆ ˆ a = (0.60 m / s 2 )i − (0.12 m / s 2 )k. The force is then  w  (2.75 × 10 5 N) ˆ ˆ F = ma = a = (0.60 m / s 2 )i − (0.12 m / s 2 )k 2 g (9.80 m / s ) 4 ˆ ˆ = (1.7 × 10 N)i − (3.4 × 10 3 N)k.

[

]

4.54: The velocity as a function of time is v(t ) = A − 3Bt 2 and the acceleration as a function of time is a (t ) = −6 Bt , and so the Force as a function of time is F (t ) = ma(t ) = −6mBt . 4.55:  1 t 1  ˆ k 2 4 ˆ v (t ) = ∫ a dt =  k1ti + t j.  m 0 m 4  

4.56: a) The equation of motion, − Cv 2 = m dv cannot be integrated with respect to time, dt as the unknown function v (t ) is part of the integrand. The equation must be separated before integration; that is, C dv − dt = 2 m v Ct 1 1 − =− + , m v v0 where v0 is the constant of integration that gives v = v0 at t = 0 . Note that this form shows that if v0 = 0 , there is no motion. This expression may be rewritten as dx  1 Ct  v= = +  , dt  v0 m    which may be integrated to obtain m  Ctv0  ln 1 + . C  m    To obtain x as a function of v, the time t must be eliminated in favor of v; from the expression obtained after the first integration, Ctv0 = vv0 − 1 , so m x − x0 = x − x0 = b) By the chain rule, m  v0  ln  . C v
−1

dv dv dv dv = = v, dt dx dt dx and using the given expression for the net force,  dv  − Cv 2 =  v  m  dx  C dv − dx = m v  v C − ( x − x0 ) = ln   v  m  0 x − x0 = m  v0  ln  . C v

4.57: In this situation, the x-component of force depends explicitly on the y-component of position. As the y-component of force is given as an explicit function of time, v y and y can be found as functions of time. Specifically, a y = (k3 / m)t , so v y = (k3 / 2m)t 2 and y = (k3 / 6m)t 3 , where the initial conditions v0 y = 0, y0 = 0 have been used. Then, the expressions for a x , vx and x are obtained as functions of time: k1 k 2 k 3 3 + t m 6m 2 k k k v x = 1 t + 2 32 t 4 m 24m k k k x = 1 t 2 + 2 32 t5. 2m 120m ax = In vector form, k k  k ˆ  k  r =  1 t 2 + 2 32 t5 i +  3 t3  ˆ j 120m  2m   6m   k k k ˆ  k  v =  1 t + 2 32 t 4  i +  3 t 2  ˆ. j 24m m   2m 

Chapter 5

5.1: a) The tension in the rope must be equal to each suspended weight, 25.0 N. b) If the mass of the light pulley may be neglected, the net force on the pulley is the vector sum of the tension in the chain and the tensions in the two parts of the rope; for the pulley to be in equilibrium, the tension in the chain is twice the tension in the rope, or 50.0 N. 5.2: In all cases, each string is supporting a weight w against gravity, and the tension in each string is w. Two forces act on each mass: w down and T (= w) up. 5.3: a) The two sides of the rope each exert a force with vertical component T sin θ , and the sum of these components is the hero’s weight. Solving for the tension T, w (90.0 kg) (9.80 m s 2 ) T= = = 2.54 × 103 N . 2 sin θ 2 sin 10.0° b) When the tension is at its maximum value, solving the above equation for the angle θ gives  (90.0 kg) (9.80 m s 2   w θ = arcsin   = arcsin   2(2.50 × 10 4 N)  = 1.01°.   2T    5.4: The vertical component of the force due to the tension in each wire must be half of the weight, and this in turn is the tension multiplied by the cosine of the angle each wire makes with the vertical, so if the weight is w, w = 34w cos θ and θ = arccos 2 = 48°. 2 3 5.5: With the positive y-direction up and the positive x-direction to the right, the freebody diagram of Fig. 5.4(b) will have the forces labeled n and T resolved into x- and ycomponents, and setting the net force equal to zero, Fx = T cos α − n sin α = 0 Fy = n cosα + T sin α − w = 0. Solving the first for n = T cot α and substituting into the second gives  cos 2 α sin 2 α  cos 2α T T + T sin α = T   sin α + sin α  = sin α = w  sin α   and so n = T cot α = w sin α cot α = w cos α, as in Example 5.4.

5.6: w sin α = mg sin α = (1390 kg) (9.80 m s 2 ) sin 17.5° = 4.10 × 103 N.

5.7: a) TB cos θ = W , or TB = W cos θ =

( 4090 kg)(9.8 m s 2 ) cos 40°

= 5.23 × 104 N.

b) TA = TB sin θ = (5.23 × 10 4 N) sin 40° = 3.36 × 10 4 N. 5.8: a) TC = w, TA sin 30° + TB sin 45° = TC = w, and TA cos 30° − TB cos 45° = 0. Since

sin 45° = cos 45°, adding the last two equations gives TA (cos 30° + sin 30°) = w, and so w cos TA = 1.366 = 0.732 w. Then, TB = TA cos 30° = 0.897 w. 45° b) Similar to part (a), TC = w, − TA cos 60° + TB sin 45° = w, and w TA sin 60° − TB cos 45° = 0. Again adding the last two, TA = (sin 60° − cos 60°) = 2.73w, and
TB = TB
sin 60° cos 45°

= 3.35w.

5.9: The resistive force is w sin α = (1600 kg)(9.80 m s 2 )(200 m 6000 m) = 523 N. . 5.10: The magnitude of the force must be equal to the component of the weight along the incline, or W sin θ = (180 kg)(9.80 m s 2 ) sin 11.0° = 337 N. 5.11: a) W = 60 N, T sin θ = W , so T = (60 N) sin45 °, or T = 85 N. b) F1 = F2 = T cos θ , F1 = F2 = 85 N cos 45° = 60 N. 5.12: If the rope makes an angle θ with the vertical, then sin θ = 01..110 = 0.073 (the 51 denominator is the sum of the length of the rope and the radius of the ball). The weight is then the tension times the cosine of this angle, or w mg (0.270 kg)(9.80 m s 2 ) T= = = = 2.65 N. cos θ cos(arcsin(.073)) 0.998 The force of the pole on the ball is the tension times sin θ , or (0.073)T = 0.193 N. 5.13: a) In the absence of friction, the force that the rope between the blocks exerts on block B will be the component of the weight along the direction of the incline, T = w sin α . b) The tension in the upper rope will be the sum of the tension in the lower rope and the component of block A’s weight along the incline, w sin α + w sin α = 2 w sin α. c) In each case, the normal force is w cos α. d) When α = 0, n = w, when α = 90°, n = 0.

5.14: a) In level flight, the thrust and drag are horizontal, and the lift and weight are vertical. At constant speed, the net force is zero, and so F = f and w = L. b) When the plane attains the new constant speed, it is again in equilibrium and so the new values of the thrust and drag, F ′ and f ′ , are related by F ′ = f ′ ; if F ′ = 2 F , f ′ = 2 f . c) In order to increase the magnitude of the drag force by a factor of 2, the speed must increase by a factor of 2 . 5.15: a)

The tension is related to the masses and accelerations by T − m1 g = m1a1 T − m2 g = m2 a2 . b) For the bricks accelerating upward, let a1 = −a2 = a (the counterweight will accelerate down). Then, subtracting the two equations to eliminate the tension gives ( m2 −m1 ) g = (m1 + m2 )a, or  28.0 kg − 15.0 kg  m2 − m1 2 = 9.80 m s 2   28.0 kg + 15.0kg  = 2.96 m s .  m2 + m1   c) The result of part (b) may be substituted into either of the above expressions to find the tension T = 191 N. As an alternative, the expressions may be manipulated to eliminate a algebraically by multiplying the first by m2 and the second by m1 and adding (with a2 = − a1 ) to give T ( m1 + m2 ) − 2m1m2 g = 0, or a=g 2m1m2 g 2(15.0 kg) (28.0 kg) (9.80 m s 2 ) = = 191 N. m1 + m2 (15.0 kg + 28.0 kg) In terms of the weights, the tension is 2m2 2m1 T = w1 = w2 . m1 + m2 m1 + m2 If, as in this case, m2 > m1 , 2m2 > m1 + m2 and 2m1 < m1 + m2 , so the tension is greater than w1 and less than w2 ; this must be the case, since the load of bricks rises and the counterweight drops. T=

5.16: Use Second Law and kinematics: a = g sin θ , 2ax = v 2 , solve for θ . g sin θ = v 2 2 x, or θ = arcsin v 2 2 gx = arcsin[(2. 5 m s) 2 [(2)(9.8 m s 2 )(1.5 m)]], θ = 12.3°. 5.17: a)

(

)

b) In the absence of friction, the net force on the 4.00-kg block is the tension, and so the acceleration will be (10.0 N) (4.00 kg) = 2.50 m s 2 . c) The net upward force on the suspended block is T − mg = ma, or m = T ( g + a ). The block is accelerating downward, so a = −2.50 m s 2 , and so m = (10.0 N) (9.80 m s 2 − 2.50 m s 2 ) = 1.37 kg . d) T = ma + mg , so T < mg , because a < 0. 5.18: The maximum net force on the glider combination is 12,000 N − 2 × 2500 N = 7000 N,
7000 N so the maximum acceleration is amax = 1400 kg = 5.0 m s 2 .

a) In terms of the runway length L and takoff speed v, a = L >
2 2

v2 2L

< amax , so

v (40 m s) = = 160 m. 2amax 2(5.0 m s 2 ) b) If the gliders are accelerating at amax , from T − Fdrag = ma, T = ma + Fdrag = (700 kg)(5.0 m s 2 ) + 2500 N = 6000 N. Note that this is exactly half of the maximum tension in the towrope between the plane and the first glider. 5.19: Denote the scale reading as F, and take positive directions to be upward. Then, w F  F − w = ma = a, or a = g  − 1. g w  2 2 a) a = (9.80 m s )((450 N) (550 N) − 1) = −1.78 m s , down. b) a = (9.80 m s 2 )((670 N) (550 N) − 1) = 2.14 m s 2 , up. c) If F = 0, a = − g and the student, scale, and elevator are in free fall. The student should worry.

2 5.20: Similar to Exercise 5.16, the angle is arcsin( gtL2 ), , but here the time is found in

terms of velocity along the table, t =

x v0

, x being the length of the table and v0 the

velocity component along the table. Then, 2  2L   2 Lv0   = arcsin  arcsin   gx 2    g( x v )2    0    2(2.50 × 10 − 2 m)(3.80 m s) 2   = 1.38°. = arcsin    9.80 m s 2 (1.75 m) 2  

(

)

5.21:

5.22:

5.23:

a) For the net force to be zero, the applied force is F = f k = μk n = μk mg = (0.20) (11.2 kg) (9.80 m s 2 ) = 22.0 N. b) The acceleration is µ k g , and 2ax = v 2 , so x = v 2 2 µ k g , or x = 3.13 m.

5.24: a) If there is no applied horizontal force, no friction force is needed to keep the box in equilibrium. b) The maximum static friction force is, from Eq. (5.6), μs n = μs w = (0.40) (40.0 N) = 16.0 N, so the box will not move and the friction force balances the applied force of 6.0 N. c) The maximum friction force found in part (b), 16.0 N. d) From Eq. (5.5), µ k n = (0.20)(40.0 N) = 8.0 N e) The applied force is enough to either start the box moving or to keep it moving. The answer to part (d), from Eq. (5.5), is independent of speed (as long as the box is moving), so the friction force is 8.0 N. The acceleration is ( F − f k ) m = 2.45 m s 2 . 5.25: a) At constant speed, the net force is zero, and the magnitude of the applied force must equal the magnitude of the kinetic friction force,  F = f k = μk n = μk mg = (0.12) (6.00 kg) (9.80 m s 2 ) = 7 N.  b) F − f k = ma, so  F = ma + f k = ma = μk mg = m(a + µ k g ) = (6.00 kg)(0.180 m s 2 + (0.12)9.80 m s 2 ) = 8 N. c) Replacing g = 9.80 m s 2 with 1.62 m s 2 gives 1.2 N and 2.2 N.

5.26: The coefficient of kinetic friction is the ratio fnk , and the normal force has magnitude 85 N + 25 N = 110 N. The friction force, from FH − f k = ma = w a is g f k = FH − w  − 0.9 m s 2  a = 20 N − 85 N   9.80 m s 2  = 28 N  g  

28 (note that the acceleration is negative), and so µ k = 110N = 0.25. N

5.27: As in Example 5.17, the friction force is µ k n = µ k w cosα and the component of the weight down the skids is w sin α . In this case, the angle α is arcsin( 2.00 20.0) = 5.7°.
cos µ 25 The ratio of the forces is µ ksin α α = tankα = 0..10 > 1, so the friction force holds the safe back, 0 and another force is needed to move the safe down the skids. b) The difference between the downward component of gravity and the kinetic friction force is w (sin α − μk cos α ) = (260 kg) (9.80 m s 2 ) (sin 5.7° − (0.25) cos 5.7°) = −381 N.

5.28: a) The stopping distance is v2 v2 (28.7 m s) 2 = = = 53 m. 2a 2µ k g 2(0.80) (9.80 m s 2 ) b) The stopping distance is inversely proportional to the coefficient of friction and proportional to the square of the speed, so to stop in the same distance the initial speed should not exceed µ 0.25 v k, wet = (28.7 m s) == 16 m s . µ k, dry 0.80

5.29: For a given initial speed, the distance traveled is inversely proportional to the coefficient of kinetic friction. From Table 5.1, the ratio of the distances is then 0..44 = 11. 0 04 5.30: (a) If the block descends at constant speed, the tension in the connecting string must be equal to the hanging block’s weight, wB . Therefore, the friction force µ k wA on block A must be equal to wB , and wB = µ k wA . (b) With the cat on board, a = g ( wB − µ k 2wA ) ( wB + 2wA ). 5.31:

a) For the blocks to have no acceleration, each is subject to zero net force. Considering the horizontal components,  T = f A , F = T + f B , or  F = f A + fB.  Using f A = µ k gm A and f B = µ k gmB gives F = µ k g (mA + mB ) . b) T = f A = µ k gmA .

5.32:
2 2 2 2 a v0 − v 2 v0 − 1 v0 3 v0 4 = = = , g 2 Lg 2 Lg 8 Lg where L is the distance covered before the wheel’s speed is reduced to half its original

µr =

( 3 speed. Low pressure, L = 18.1 m; 8 (18.13.50 m s)m s 2 ) = 0.0259. High pressure, m)(9.80 .50 m ) 3 L = 92.9 m; 8 ( 92.(93m)(9.80sm s 2 ) = 0.00505.
2

2

5.33: Without the dolly: n = mg and F − µ k n = 0 ( a x = 0 since speed is constant). F 160 N m= = = 34.74 kg μk g (0.47) (9.80 m s 2 ) With the dolly: the total mass is 34.7 kg + 5.3 kg = 40.04 kg and friction now is rolling friction, f r = µ r mg. F − µ r mg = ma a= F − µ r mg = 3.82 m s 2 m

5.34: Since the speed is constant and we are neglecting air resistance, we can ignore the 2.4 m/s, and Fnet in the horizontal direction must be zero. Therefore f r = µ r n = = Fhoriz = 200 N before the weight and pressure changes are made. After the changes, (0.81) (1.42n) = Fhoriz , because the speed is still constant and Fnet = 0 . We can simply divide the two equations: (0.81µ r )(1.42n) Fhoriz = 200 N. μr n (0.81) (1.42) (200 N) = Fhoriz = 230 N

5.35: First, determine the acceleration from the freebody diagrams.

There are two equations and two unknowns, a and T: − µ k mA g + T = mAa mB g − T = mB a Add and solve for a : a = g ( mB − µ k mA ) (mB + mA ), a = 0.79 m s 2 . (a) v = (2ax )1 2 = 0.22 m s. (b) Solving either equation for the tension gives T = 11.7 N. 5.36: a) The normal force will be w cos θ and the component of the gravitational force along the ramp is w sin θ . The box begins to slip when w sin θ > µ s w cos θ , or tan θ > µ s = 0.35, so slipping occurs at θ = arctan(0.35) = 19.3° , or 19° to two figures. b) When moving, the friction force along the ramp is µ k wcos θ , the component of the gravitational force along the ramp is wsin θ , so the acceleration is ( w sin θ ) − wµ k cos θ ) m = g (sin θ − µ k cos θ ) = 0.92 m s 2 . (c) 2ax = v 2 , so v = (2ax)1 2 , or v = [(2)(0.92 m s 2 )(5 m)]1 2 = 3 m.  5.37: a) The magnitude of the normal force is mg + F sin θ. The horizontal component   of F , F cos θ must balance the frictional force, so   F cosθ = μk (mg + F sin θ );  solving for F gives  μk mg F = cos θ − μk sin θ b) If the crate remains at rest, the above expression, with µs instead of µ k , gives the force that must be applied in order to start the crate moving. If cot θ < µ s , the needed force is infinite, and so the critical value is µ s = cot θ.

5.38: a) There is no net force in the vertical direction, so n + F sin θ − w = 0, or n = w − F sin θ = mg − F sin θ. The friction force is f k = µ k n = µ k (mg − F sin θ ). The net horizontal force is F cos θ − f k = F cos θ − µ k (mg − F sin θ ) , and so at constant speed, µ k mg F= cos θ + µ k sin θ b) Using the given values, (0.35)(90 kg)(9.80 m s 2 ) F= = 293 N, (cos 25° + (0.35) sin 25°) or 290 N to two figures. 5.39: a)

b) The blocks move with constant speed, so there is no net force on block A; the tension in the rope connecting A and B must be equal to the frictional force on block A, µ k = (0.35) (25.0 N) = 9 N. c) The weight of block C will be the tension in the rope connecting B and C; this is found by considering the forces on block B. The components of force along the ramp are the tension in the first rope (9 N, from part (a)), the component of the weight along the ramp, the friction on block B and the tension in the second rope. Thus, the weight of block C is wC = 9 N + wB (sin 36.9° + µ k cos 36.9°) = 9 N + (25.0 N)(sin 36.9° + (0.35)cos 36.9°) = 31.0 N, or 31 N to two figures. The intermediate calculation of the first tension may be avoided to obtain the answer in terms of the common weight w of blocks A and B, wC = w( μk + (sin θ + µ k cos θ )), giving the same result. (d) Applying Newton’s Second Law to the remaining masses (B and C) gives: a = g ( wc − µ k wB cos θ − wB sin θ) ( wB + wc ) = 1.54 m s 2 .

5.40: Differentiating Eq. (5.10) with respect to time gives the acceleration k a = vt  e −( k m ) t = ge −( k m ) t , m where Eq. (5.9), vt = mg k has been used. Integrating Eq. (5.10) with respect to time with y0 = 0 gives y = ∫ vt [1 − e − ( k m ) t ] dt
0 t

 m  m = vt t +  e − ( k m ) t  − vt   k  k   m  = vt t − 1 − e − ( k m ) t  .  k 

(

)

5.41: a) Solving for D in terms of vt , D= b) vt = mg (80 kg) (9.80 m s 2 ) = = 0.44 kg m . vt2 (42 m s) 2 mg (45 kg)(9.80 m s 2 ) = = 42 m s . D (0.25 kg m)

5.42: At half the terminal speed, the magnitude of the frictional force is one-fourth the weight. a) If the ball is moving up, the frictional force is down, so the magnitude of the net force is (5/4)w and the acceleration is (5/4)g, down. b) While moving down, the frictional force is up, and the magnitude of the net force is (3/4)w and the acceleration is (3/4)g, down. 5.43: Setting Fnet equal to the maximum tension in Eq. (5.17) and solving for the speed v gives Fnet R (600 N)(0.90 m) v= = = 26.0 m s , m (0.80 kg) or 26 m/s to two figures. 5.44: This is the same situation as Example 5.23. Solving for µ s yields

µs =

v2 (25.0 m s) 2 = = 0.290. Rg (220 m)(9.80 m s 2 )

5.45: a) The magnitude of the force F is given to be equal to 3.8w. “Level flight” means that the net vertical force is zero, so F cos β = (3.8) w cos β = w, , and β = arccos(1 3.8) = 75° . (b) The angle does not depend on speed. 5.46: a) The analysis of Example 5.22 may be used to obtain tan β = (v 2 gR ), but the subsequent algebra expressing R in terms of L is not valid. Denoting the length of the π horizontal arm as r and the length of the cable as l , R = r + l sin β. The relation v = 2TR is still valid, so tan β = T=
4π 2 R gT 2

=

4π 2 ( r + l sin β ) gT 2

. Solving for the period T,

4π2 (r + l sin β) 4π2 (3.00 m + (5.00 m)sin 30°) = = 6.19 s. g tan β (9.80 m s 2 ) tan 30° Note that in the analysis of Example 5.22, β is the angle that the support (string or cable) makes with the vertical (see Figure 5.30(b)). b) To the extent that the cable can be considered massless, the angle will be independent of the rider’s weight. The tension in the cable will depend on the rider’s mass. 5.47: This is the same situation as Example 5.22, with the lift force replacing the tension in the string. As in that example, the angle β is related to the speed and the turning radius by tan β =
v2 gR

. Solving for β ,

 v2   (240 km h × ((1m s) (3.6 km h))) 2   = 20.7°. β = arctan  gR  = arctan    9.80 m s 2 (1200 m )    

(

)

5.48: a) This situation is equivalent to that of Example 5.23 and Problem 5.44, so 2 v2 π µs = Rg . Expressing v in terms of the period T, v = 2TR , so µS = 4Tπ2 gR . A platform speed of 40.0 rev/min corresponds to a period of 1.50 s, so 4π 2 (0.150 m) μs = = 0.269. (1.50 s) 2 (9.80 m s 2 ) b) For the same coefficient of static friction, the maximum radius is proportional to the square of the period (longer periods mean slower speeds, so the button may be moved further out) and so is inversely proportional to the square of the speed. Thus, at the higher 0 2 speed, the maximum radius is (0.150 m) ( 40..0 ) = 0.067 m . 60

5.49: a) Setting arad = g in Eq. (5.16) and solving for the period T gives T = 2π R 400 m = 2π = 40.1 s, g 9.80 m s 2

so the number of revolutions per minute is (60 s min) (40.1 s) = 1.5 rev min . b) The lower acceleration corresponds to a longer period, and hence a lower rotation rate, by a factor of the square root of the ratio of the accelerations, T ′ = (1.5 rev min) × 3.70 9.8 = 0.92 rev min. . 5.50: a) 2π R T = 2π(50.0 m) (60.0 s) = 5.24 m s . b) The magnitude of the radial force is mv 2 R = m4π 2 R T 2 = w(4π 2 R gT 2 ) = 49 N (to the nearest Newton), so the apparent weight at the top is 882 N − 49 N = 833 N, and at the bottom is 882 N + 49 N = 931 N . c) For apparent weightlessness, the radial acceleration at the top is equal to g in magnitude. Using this in Eq. (5.16) and solving for T gives R 50.0 m T = 2π = 2π = 14 s. g 9.80 m s 2 d) At the bottom, the apparent weight is twice the weight, or 1760 N. 5.51: a) If the pilot feels weightless, he is in free fall, and a = g = v 2 R , so v = Rg = (150 m)(9.80 m s 2 ) = 38.3 m s , or 138 km h . b) The apparent weight is the sum of the net inward (upward) force and the pilot’s weight, or  a w + ma = w 1 +   g  
2   (280 km h) 1 +  = ( 700 N )  (3.6 (km h) (m s))2 (9.80 m s 2 )(150 m)    = 3581 N, or 3580 N to three places.

5.52: a) Solving Eq. (5.14) for R, R = v 2 a = v 2 4 g = (95.0 m s) 2 (4 × 9.80 m s 2 ) = 230 m. b) The apparent weight will be five times the actual weight, 5 mg = 5 (50.0 kg) (9.80 m s 2 ) = 2450 N to three figures.

5.53: For no water to spill, the magnitude of the downward (radial) acceleration must be at least that of gravity; from Eq. (5.14), v > gR = (9.80 m s 2 )(0.600 m) = 2.42 m s .

5.54: a) The inward (upward, radial) acceleration will be

v2 R

=

( 4.2 m s) 2 (3.80 m)

= 4.64 m s . At the

2

bottom of the circle, the inward direction is upward. b) The forces on the ball are tension and gravity, so T − mg = ma,  4.64 m s 2  a  T = m(a + g ) = w  + 1 = (71.2 N) g   9.80 m s 2 + 1 = 105 N.      5.55: a)

T1 is more vertical so supports more of the weight and is larger. You can also see this from ∑ Fx = max : T2 cos 40° − T1 cos 60° = 0  cos 40°  T1 =  T2 = 1.532T2  cos 60°  b) T1 is larger so set T1 = 5000 N. Then T2 = T1 1.532 = 3263.5 N , ∑ Fy = ma y T1 sin 60° + T2 sin 40° = w w = 6400 N

5.56:

The tension in the lower chain balances the weight and so is equal to w. The lower pulley must have no net force on it, so twice the tension in the rope must be equal to w, and so the tension in the rope is w 2 . Then, the downward force on the upper pulley due to the rope is also w, and so the upper chain exerts a force w on the upper pulley, and the tension in the upper chain is also w. 5.57: In the absence of friction, the only forces along the ramp are the component of the  weight along the ramp, w sin α , and the component of F along the ramp,  F cos α = F cos α . These forces must sum to zero, so F = w tan α . Considering horizontal and vertical components, the normal force must have horizontal component equal to n sin α , which must be equal to F; the vertical component must balance the weight, n cosα = w . Eliminating n gives the same result. 5.58: The hooks exert forces on the ends of the rope. At each hook, the force that the hook exerts and the force due to the tension in the rope are an action-reaction pair. The vertical forces that the hooks exert must balance the weight of the rope, so each hook exerts an upward vertical force of w 2 on the rope. Therefore, the downward force that the rope exerts at each end is Tend sin θ = w 2 , so Tend = w (2 sin θ ) = Mg (2 sin θ ). b) Each half of the rope is itself in equilibrium, so the tension in the middle must balance the horizontal force that each hook exerts, which is the same as the horizontal component of the force due to the tension at the end; Tend cos θ = Tmiddle , so Tmiddle = Mg cos θ (2 sin θ ) = Mg (2tan θ ) . (c) Mathematically speaking, θ ≠ 0 because this would cause a division by zero in the equation for Tend or Tmiddle . Physically speaking, we would need an infinite tension to keep a non-massless rope perfectly straight.

5.59: Consider a point a distance x from the top of the rope. The forces acting in this point are T up and M + m ( LL− x ) g downwards. Newton’s Second Law becomes
M T − M + m ( LL− x ) g = M + m ( LL− x ) a. Since a = F − (M ++mm ) g , T = M + MF x = 0, T = F , and at x = L, T = M + m = M (a + g ) as expected.

(

)

(

(

)

)

(

m( L − x) L

)(

F M +m

). . At

5.60: a) The tension in the cord must be m2 g in order that the hanging block move at constant speed. This tension must overcome friction and the component of the gravitational force along the incline, so m2 g = ( m1 g sin α + μk m1 g cos α ) and m2 = m1 (sin α + μk cosα ) . b) In this case, the friction force acts in the same direction as the tension on the block of mass m1 , so m2 g = (m1 g sin α − μ k m1 g cos α ) , or m2 = m1 (sin α − μ k cos α ) . c) Similar to the analysis of parts (a) and (b), the largest m2 could be is m1 (sin α + µs cos α ) and the smallest m2 could be is m1 (sin α − µs cos α ) . 5.61: For an angle of 45.0° , the tensions in the horizontal and vertical wires will be the same. a) The tension in the vertical wire will be equal to the weight w = 12.0 N ; this must be the tension in the horizontal wire, and hence the friction force on block A is also 12.0 N . b) The maximum frictional force is μs wA = (0.25)(60.0 N) = 15 N ; this will be the tension in both the horizontal and vertical parts of the wire, so the maximum weight is 15 N. 5.62: a) The most direct way to do part (a) is to consider the blocks as a unit, with total weight 4.80 N. Then the normal force between block B and the lower surface is 4.80 N, and the friction force that must be overcome by the force F is μ k n = (0.30)(4.80 N) = 1.440 N, or 1.44 N, to three figures. b) The normal force between block B and the lower surface is still 4.80 N, but since block A is moving relative to block B, there is a friction force between the blocks, of magnitude (0.30)(1.20 N) = 0.360 N, so the total friction force that the force F must overcome is 1.440 N + 0.360 N = 1.80 N . (An extra figure was kept in these calculations for clarity.)

 5.63: (Denote F by F.) a) The force normal to the surface is n = F cos θ ; the vertical component of the applied force must be equal to the weight of the brush plus the friction force, so that F sin θ = w + μk F cos θ , and w 12.00 N F= = = 16.9 N, sin θ − μk cosθ sin 53.1° − (0.51) cos 53.1° keeping an extra figure. b) F cosθ = (16.91 N)cos 53.1° = 10.2 N . 5.64: a)

∑ F = ma = m(62.5 g ) = 62.5mg = (62.5)(210 × 10 − 6 g )(980 cm s 2 )

= 13 dynes = 1.3 × 10− 4 N This force is 62.5 times the flea’s weight. b) Fmax = mamax = m(140 g ) = 140mg = 29 dynes = 2.9 × 10 − 4 N Occurs at approximately 1.2 ms. c) ∆v = v − v0 = v − 0 = v = area under a-t graph. Approximate area as shown:

A = A(1) + A(2) + A(3) 1 = (1.2 ms)(77.5 g) + (1.2 ms)(62.5 g) 2 1 + (0.05 ms)(140 g) 2 = 120 cm s = 1.2 m s

5.65: a) The instrument has mass m = w g = 1.531 kg . Forces on the instrument:

∑ Fy = ma y T − mg = ma T − mg a= = 13.07 m s 2 m v0 y = 0, v y = 330 m s , a y = 13.07 m s 2 , t = ? v y = v0 y + a y t gives t = 25.3 s Consider forces on the rocket; rocket has the same a y . Let F be the thrust of the rocket engines. F − mg = ma F = m( g + a ) = (25,000 kg) (9.80 m s 2 + 13.07 m s 2 ) = 5.72 × 105 N b) y − y0 = v0 y t + 1 a y t 2 gives y − y0 = 4170 m. 2 5.66: The elevator’s acceleration is: 3 dv(t ) a= = 3.0 m s 2 + 2(0.20 m s )t = 3.0 m s 2 + (0.40 m s3 )t dt At t = 4.0 s, a = 3.0 m s 2 + (0.40 m s 3 )(4.0 s) = 4.6 m s 2 . From Newton’s Second Law, the net force on you is Fnet = Fscale − w = ma Fscale = apparent weight = w + ma = (72 kg )(9.8 m s 2 ) + (72 kg)(4.6 m s 2 ) = 1036.8 N or 1040 N

5.67: Consider the forces on the person:

∑ Fy = ma y n − mg = ma n = 1.6mg so a = 0.60 g = 5.88 m s 2 y − y 0 = 3.0 m, a y = 5.88 m s , v0 y = 0, v y = ?
2 2 v y = v 0 y + 2a y ( y − y 0 ) gives v y = 5.0 m s 2

5.68: (a) Choosing upslope as the positive direction: Fnet = −mg sin 37° − f k = −mg sin 37° − µ k mg cos37° = ma and a = −(9.8 m s 2 ) (0.602 + (0.30)(0.799)) = −8.25 m s 2 2 Since we know the length of the slope, we can use v 2 = v0 + 2a ( x − x0 ) with x0 = 0 and v = 0 at the top. 2 v0 = −2ax = −2(−8.25 m s 2 )(8.0 m) = 132 m 2 s 2 v0 = 132 m 2 s 2 = 11.5 m s or 11m s (b) For the trip back down the slope, gravity and the friction force operate in opposite directions: Fnet = − mg sin 37° + μk mg cos 37° = ma a = g (− sin 37° + 0.30 cos37°) = (9.8 m s 2 )((−0.602) + (0.30)(0.799)) = −3.55 m s 2 Now v0 = 0, x0 = −8.0 m, x = 0, and
2 v 2 = v0 + 2a ( x − x0 ) = 0 + 2(−3.55 m s 2 )(−8.0 m)

= 56.8 m 2 s 2 v = 56.8 m 2 s 2 = 7.54 m s or 7.5 m s

5.69: Forces on the hammer:

∑ Fy = ma y gives T sin 74° − mg = 0 so T sin 74° = mg ∑ Fx = ma x gives T cos 74° − ma Divide the second equation by the first: a 1 = and a = 2.8 m s 2 g tan74°

5.70:

It’s interesting to look at the string’s angle measured from the perpendicular to the top of the crate. This angle is of course 90° —angle measured from the top of the crate. The free-body diagram for the washer then leads to the following equations, using Newton’s Second Law and taking the upslope direction as positive: − m w g sin θ slope + T sin θ string = m w a − mw g cosθ slope + T cos θ string = 0 T sin θ string = m w (a + g sin θ slope ) T cosθ string = m w g cos θ slope Dividing the two equations: tanθstring = a + g sin θslope g cos 0slope

For the crate, the component of the weight along the slope is −mc g sin θ slope and the normal force is mc g cos θslope . Using Newton’s Second Law again: − mc g sin θ slope + µ k mc g cos θ slope = mc a a + g sin θ slope g cos θ slope

µk =

which leads to the interesting observation that the string will hang at an angle whose tangent is equal to the coefficient of kinetic friction: µ k = tan θstring = tan(90° − 68°) = tan 22° = 0.40

5.71: a) Forces on you:

∑ Fy = ma y gives n = mg cosα ∑ Fx = max mg sin α − f k = ma a = g (sin α − μk cos α ) = −3.094 m s 2 Find your stopping distance v x = 0, a x = −3.094 m s 2 , v 0 x = 20 m s, x − x 0 = ?
2 2 v x = v0 x + 2a x ( x − x0 ) gives x − x0 = 64.6 m, which is greater than 40 m. You don’t stop before you reach the hole, so you fall into it. b) a x = −3.094 m s 2 , x − x 0 = 40 m, v x = 0, v0 x = ? 2 2 v x = v0 x + 2a x ( x − x0 ) gives v0 x = 16 m s .

5.72: The key idea in solving this problem is to recognize that if the system is accelerating, the tension that block A exerts on the rope is different from the tension that block B exerts on the rope. (Otherwise the net force on the rope would be zero, and the rope couldn’t accelerate.) Also, treat the rope as if it is just another object. Taking the “clockwise” direction to be positive, the Second Law equations for the three different parts of the system are: Block A (The only horizontal forces on A are tension to the right, and friction to the left): − µ k m A g + T A = m A a. Block B (The only vertical forces on B are gravity down, and tension up): m B g − TB = m B a. Rope (The forces on the rope along the direction of its motion are the tensions at either end and the weight of the portion of the rope that hangs vertically): m R ( d ) g + TB − T A = m R a. L To solve for a and eliminate the tensions, add the left hand sides and right hand sides of the three equations: − µ k m A g + m B g + m R ( d ) g = (m A + m B + m R )a, or a = g (Bm A +RmLB + mkR )A . L mB + mR ( d ) L (a) When µ k = 0, a = g ( m A + m B + m R ) . As the system moves, d will increase, approaching
m +m
d

−µ m

L as a limit, and thus the acceleration will approach a maximum value of a = g ( m Am+Bm+Bm+Rm R ) .

(b) For the blocks to just begin moving, a > 0, so solve 0 = [mB + mR ( d ) − µ s mA ] for d. L Note that we must use static friction to find d for when the block will begin to move. 1.0 Solving for d, d = mLR ( µ s mA − mB ), or d = .160 m (.25( 2 kg) − .4 kg) = .63 m. . kg (c) When m R = .04 kg, d =
1.0 m .04 kg

(.25( 2 kg) − .4 kg) = 2.50 m. This is not a physically

possible situation since d > L. The blocks won’t move, no matter what portion of the rope hangs over the edge. 5.73: For a rope of length L, and weight w, assume that a length rL is on the table, so that a length (1 − r) L is hanging. The tension in the rope at the edge of the table is then (1 − r ) w, and the friction force on the part of the rope on the table is f s = µs rw. This must be the same as the tension in the rope at the edge of the table, so µ s rw = (1 − r ) w and r = 1 (1 + µ s ). Note that this result is independent of L and w for a uniform rope. The fraction that hangs over the edge is 1 − r = µ s (1 + µ s ) ; note that if µs = 0, r = 1 and 1 − r = 0.

5.74: a) The normal force will be mg cos α + F sin α , and the net force along (up) the ramp is F cosα − mg sin α − µ s (mg cos α + F sin α ) = F (cosα − µ s sin α ) − mg (sin α + µ s cosα ). In order to move the box, this net force must be greater than zero. Solving for F, sin α + µs cos α F > mg . cos α − µs sin α Since F is the magnitude of a force, F must be positive, and so the denominator of this expression must be positive, or cosα > µs sin α , and µs < cot α . b) Replacing µs with µ k with in the above expression, and making the inequality an equality, sin α + µ k cosα F = mg . cos α − µ k sin α 5.75: a) The product µs g = 2.94 m s 2 is greater than the magnitude of the acceleration of the truck, so static friction can supply sufficient force to keep the case stationary relative to the truck; the crate accelerates north at 2.20 m s 2 , due to the friction force of ma = 66.0 N. b) In this situation, the static friction force is insufficient to maintain the case at rest relative to the truck, and so the friction force is the kinetic friction force, µ k n = µ k mg = 59 N. 5.76: To answer the question, v0 must be found and compared with 20 m s (72 km hr).
2 The kinematics relationship 2ax = −v0 is useful, but we also need a. The acceleration must be large enough to cause the box to begin sliding, and so we must use the force of static friction in Newton’s Second Law: − µs mg ≤ ma, or a = − µs g . Then, 2 2(− μs g ) x = −v0 , or v0 = 2 µs gx = 2(.30)(9.8 m s 2 )(47 m) . Hence,

v0 = 16.6 m s = 60 km h , which is less than 72 km/h, so do you not go to jail.

5.77: See Exercise 5.40. a) The maximum tension and the weight are related by Tmax cos β = µ k ( w − Tmax sin β ), and solving for the weight w gives  cos β  w = Tmax  + sin β  .  μ   k  This will be a maximum when the quantity in parentheses is a maximum. Differentiating with respect to β ,  d  cos β sin β  + sin β  = − + cos β = 0,  µ  dβ  k µk  or tanθ = µ k , where θ is the value of β that maximizes the weight. Substituting for µ k in terms of θ ,  cos θ  w = Tmax   sin θ cos θ + sin θ      cos2 θ + sin 2 θ   = Tmax    sin θ   T = max . sin θ b) In the absence of friction, any non-zero horizontal component of force will be enough to accelerate the crate, but slowly. 5.78: a) Taking components along the direction of the plane’s descent, f = w sin α and L = w cosα . b) Dividing one of these relations by the other cancels the weight, so tan α = f L . c) The distance will be the initial altitude divided by the tangent 1300 of α . f = L tan α and L = w cos α , therefore sin α = f w = 12 ,900N g and so α = 5.78°. N This makes the horizontal distance (2500 m) tan(5.78°) = 24.7 km. d) If the drag is reduced, the angle α is reduced, and the plane goes further. 5.79: If the plane is flying at a constant speed of 36.1 m s, then ∑ F = 0, or T − w sin α − f = 0. The rate of climb and the speed give the angle α, α = arcsin( 5 36.1) = 7.96° . Then, T = w sin α + f . T = (12,900 N) sin 7.96° + 1300 N = 3087 N. Note that in level flight (α = 0), the thrust only needs to overcome the drag force to maintain the constant speed of 36.1 m s .

5.80: If the block were to remain at rest relative to the truck, the friction force would need to cause an acceleration of 2.20 m s 2 ; however, the maximum acceleration possible due to static friction is (0.19)(9.80 m s 2 ) = 1.86 m s 2 , and so the block will move relative to the truck; the acceleration of the box would be µ k g = (0.15)(9.80 m s 2 ) = 1.47 m s 2 . The difference between the distance the truck moves and the distance the box moves (i.e., the distance the box moves relative to the truck) will be 1.80 m after a time t= 2∆x = a truck − a box 2(1.80 m) (2.20 m s 2 − 1.47 m s )
2

= 2.22 s.

In this time, the truck moves 1 a truck t 2 = 1 (2.20 m s 2 ) (2.221 s) 2 = 5.43 m. Note that an 2 2 extra figure was kept in the intermediate calculation to avoid roundoff error. 5.81: The friction force on block A is µ k wA = (0.30)(1.40 N) = 0.420 N, as in Problem 568. This is the magnitude of the friction force that block A exerts on block B, as well as the tension in the string. The force F must then have magnitude F = µ k ( wB + wA ) + µ k wA + T = µ k ( wB + 3wA ) = (0.30)(4.20 N + 3(1.40 N)) = 2.52 N. Note that the normal force exerted on block B by the table is the sum of the weights of the blocks.

5.82: We take the upward direction as positive. The explorer’s vertical acceleration is − 3.7 m s 2 for the first 20 s. Thus at the end of that time her vertical velocity will be v y = at = (−3.7 m s 2 )(20 s) = −74 m s. She will have fallen a distance  − 74 m s  d = vav t =  (20 s) = −740 m 2   and will thus be 1200 − 740 = 460 m above the surface. Her vertical velocity must reach zero as she touches the ground; therefore, taking the ignition point of the PAPS as y0 = 0,
2 2 v y = v0 + 2a ( y − y0 ) 2 v 2 − v0 y

a=

2( y − y0 )

=

0 − (−74 m s) 2 = 5.95 m s 2 or 6.0 m s 2 − 460

which is the vertical acceleration that must be provided by the PAPS. The time it takes to reach the ground is given by t= v − v0 0 − (−74 m s) = = 12.4 s a 5.95 m s 2

Using Newton’s Second Law for the vertical direction FPAPSv + mg = ma FPAPSv = ma − mg = m(a + g ) = (150kg)(5.95 − ( −3.7)) m s 2 = 1447.5 N or 1400 N which is the vertical component of the PAPS force. The vehicle must also be brought to a stop horizontally in 12.4 seconds; the acceleration needed to do this is v − v 0 0 − 33 m s 2 a= = = 2.66 m s 2 t 12.4 s and the force needed is FPAPSh = ma = (150 kg)(2.66 m s 2 ) = 399 N or 400 N, since there are no other horizontal forces.

5.83: Let the tension in the cord attached to block A be TA and the tension in the cord attached to block C be TC . The equations of motion are then TA − mA g = mAa TC − µ k mB g − TA = mB a mC g − TC = mC a. a) Adding these three equations to eliminate the tensions gives a (mA + mB + mC ) = g (mC − mA − µ k mB ), solving for mC gives m A ( a + g ) + mB ( a + µ k g ) , g−a and substitution of numerical values gives mC = 12.9 kg. b) TA = mA ( g + a ) = 47.2 N, TC = mC ( g − a) = 101 N. mC = 5.84: Considering positive accelerations to be to the right (up and to the right for the lefthand block, down and to the right for the right-hand block), the forces along the inclines and the accelerations are related by T − (100 kg) g sin 30° = (100 kg)a, (50 kg) g sin 53° − T = (50 kg)a, where T is the tension in the cord and a the mutual magnitude of acceleration. Adding these relations, (50 kg sin 53° − 100 kg sin 30°) g = (50 kg + 100 kg) a, or a = −0.067 g . a) Since a comes out negative, the blocks will slide to the left; the 100-kg block will slide down. Of course, if coordinates had been chosen so that positive accelerations were to the left, a would be

+ 0.067 g . b) 0.067(9.80 m s 2 ) = 0.658 m s . c) Substituting the value of a (including the proper sign, depending on choice of coordinates) into either of the above relations involving T yields 424 N.

2

5.85: Denote the magnitude of the acceleration of the block with mass m1 as a; the block of mass m2 will descend with acceleration a 2 . If the tension in the rope is T, the equations of motion are then T = m1a m2 g − 2T = m2 a 2 . Multiplying the first of these by 2 and adding to eliminate T, and then solving for a gives a= m2 g 2m2 =g . 2m1 + m2 2 4m1 + m2

The acceleration of the block of mass m2 is half of this, or g m2 (4m1 + m2 ). 5.86: Denote the common magnitude of the maximum acceleration as a. For block A to remain at rest with respect to block B, a < µs g . The tension in the cord is then T = ( mA + mB )a + µ k g (mA + mB ) = (mA + mB )(a + µ k g ). This tension is related to the mass mC by T = mC ( g − a ). Solving for a yields a=g mC − µ k (mA + mB ) < µs g . mA + mB + mC

Solving the inequality for mC yields mC < (mA + mB )( µs + µ k ) . 1 − µs

5.87: See Exercise 5.15 (Atwood’s machine). The 2.00-kg block will accelerate upward kg − at g 5.00 kg + 2.00 kg = 3g 7 , and the 5.00-kg block will accelerate downward at 3g 7. Let the 5.00 2.00 kg initial height above the ground be h0 ; when the large block hits the ground, the small block will be at a height 2h0 , and moving upward with a speed given by
2 2 v0 = 2ah0 = 6 gh0 7 . The small block will continue to rise a distance v0 2 g = 3h0 7 , and so the maximum height reached will be 2h0 + 3h0 7 = 17 h0 7 = 1.46 m , which is 0.860 m above its initial height.

5.88: The floor exerts an upward force n on the box, obtained from n − mg = ma, or n = m(a + g ). The friction force that needs to be balanced is

µ k n = µ k m(a + g ) (0.32) (28.0 kg)(1.90 m s 2 + 9.80 m s ) = 105 N.
5.89: The upward friction force must be f s = µs n = mA g , and the normal force, which is the only horizontal force on block A, must be n = mA a, and so a = g µs . An observer on the cart would “feel” a backwards force, and would say that a similar force acts on the block, thereby creating the need for a normal force. 5.90: Since the larger block (the trailing block) has the larger coefficient of friction, it will need to be pulled down the plane; i.e., the larger block will not move faster than the smaller block, and the blocks will have the same acceleration. For the smaller block, (4.00 kg) g (sin30 ° − (0.25) cos 30°) − T = (4.00 kg)a, or 11.11 N − T = (4.00 kg)a, and similarly for the larger, 15.44 N + T = (8.00 kg)a, a) Adding these two relations, 26.55 N = (12.00 kg)a, a = 2.21 m s 2 (note that an extra figure was kept in the intermediate calculation to avoid roundoff error). b) Substitution into either of the above relations gives T = 2.27 N. Equivalently, dividing the second relation by 2 and subtracting from the first gives 3 T = 11.11 N − 15.44 N , giving the same result. c) The 2 2 string will be slack. The 4.00-kg block will have a = 2.78 m s 2 and the 8.00-kg block will have a = 1.93 m s 2 , until the 4.00-kg block overtakes the 8.00-kg block and collides with it. 5.91: a) Let nB be the normal force between the plank and the block and nA be the normal force between the block and the incline. Then, nB = w cos θ and n A = nB + 3w cos θ = 4 w cos θ. The net frictional force on the block is µ k (n A + n B ) = µ k 5w cosθ . To move at constant speed, this must balance the component of the block’s weight along the incline, so 3w sin θ = µ k 5w cos θ , and 3 µ k = 3 tan θ = 5 tan 37° = 0.452. 5

2

5.92: (a) There is a contact force n between the man (mass M) and the platform (mass m). The equation of motion for the man is T + n − Mg = Ma, where T is the tension in the rope, and for the platform, T − n − mg = ma . Adding to eliminate n, and rearranging, T = 1 ( M + m)(a + g ). This result could be found directly by considering the man2 platform combination as a unit, with mass m + M , being pulled upward with a force 2T due to the two ropes on the combination. The tension T in the rope is the same as the force that the man applies to the rope. Numerically, 1 2 T = (70.0 kg + 25.0 kg)(1.80 m s 2 + 9.80 m s ) = 551 N. 2 (b) The end of the rope moves downward 2 m when the platform moves up 1 m, so arope = −2aplatform . Relative to the man, the acceleration of the rope is 3a = 5.40 m s 2 , downward. 5.93: a) The only horizontal force on the two-block combination is the horizontal  component of F , F cos α. The blocks will accelerate with a = F cosα (m1 + m2 ). b) The normal force between the blocks is m1 g + Fsin α , for the blocks to move together, the product of this force and µs must be greater than the horizontal force that the lower block exerts on the upper block. That horizontal force is one of an action-reaction pair; the reaction to this force accelerates the lower block. Thus, for the blocks to stay together, m2 a ≤ µs (m1 g + F sin α ). Using the result of part (a), Fcosα m2 ≤ µs (m1 g + F sin α ). m1 + m2 Solving the inequality for F gives the desired result.

5.94: The banked angle of the track has the same form as that found in Example 5.24, 2 v0 tan β = gR , where v0 is the ideal speed, 20 m s in this case. For speeds larger than v0 , a frictional force is needed to keep the car from skidding. In this case, the inward force will consist of a part due to the normal force n and the friction force f ; n sin β + f cos β = marad . The normal and friction forces both have vertical components; since there is no vertical acceleration, n cos β − f sin β = mg . Using f = µs n and arad =
v2 R

=

(1.5v 0 ) 2 R

= 2.25 g tan β , these two relations become n sin β + µ s n cos β = 2.25 mg tan β ,
n cos β − µs n sin β = mg .

Dividing to cancel n gives

sin β + µs cos β = 2.25 tan β . cos β − µs sin β Solving for µs and simplifying yields 1.25 sin β cos β µs = . 1 + 1.25 sin 2 β Using β = arctan

(

( 20 m s) 2 ( 9.80 m s 2 )(120 m)

) = 18.79° gives µ

s

= 0.34.

5.95: a) The same analysis as in Problem 5.90 applies, but with the speed v an unknown. The equations of motion become n sin β + µs n cos β = mv 2 R , n cos β − µs n sin β = mg. Dividing to cancel n gives sin β + µs cos β v 2 = . cos β − µs sin β Rg Solving for v and substituting numerical values gives v = 20.9 m s (note that the value for the coefficient of static friction must be used). b) The same analysis applies, but the friction force must be directed up the bank; this has the same algebraic effect as replacing f with − f , or replacing µs with − µs (although coefficients of friction may certainly never be negative). The result is sin β − µ s cos β v 2 = ( gR ) , cos β + µ s sin β and substitution of numerical values gives v = 8.5 m s.

5.96: (a) 80 mi h is 35.7 m s in SI units. The centripetal force needed to keep the car on the road is provided by friction; thus 2 µ s mg = mv r (35.7 m s) 2 v2 r= = = 171 m or 170 m µ s g (0.76)(9.8 m s 2 ) (b) If µs = 0.20 : v 2 = rμs g = (171 m) (0.20) (9.8 m s 2 ) = 335.2 m 2 s 2 v = 18.3 m s or about 41 mi h (c) If µs = 0.37 : v 2 = (171 m) (0.37) (9.8 m s 2 ) = 620 m 2 s 2 v = 24.9 m s or about 56 mi h The speed limit is evidently designed for these conditions. 5.97: a) The static friction force between the tires and the road must provide the centripetal acceleration for motion in the circle. v2 µ s mg = m r v1 v m, g, and r are constant so = 2 , where 1 refers to dry road and 2 to wet µ s1 µ s2 road. μs 2 = 1 μs1 , so v2 = (27 m s) 2 = 19 m s 2 b) Calculate the time it takes you to reach the curve v0 x = 27 m s , vx = 19 m s, x − x0 = 800 m, t = ?  v + vx  x − x0 =  0 x  t gives t = 34.7 s  2  During this time the other car will travel x − x0 = v0 xt = (36 m s) (34.7s) = 1250 m. The other car will be 50 m behind you as you enter the curve, and will be traveling at nearly twice your speed, so it is likely it will skid into you. 5.98: The analysis of this problem is the same as that of Example 5.22; solving for v in terms of β and R, v = gR tan β = (9.80 m s 2 ) (50.0) tan 30.0° = 16.8 m s, about 60.6 km h.

5.99: The point to this problem is that the monkey and the bananas have the same weight, and the tension in the string is the same at the point where the bananas are suspended and where the monkey is pulling; in all cases, the monkey and bananas will have the same net force and hence the same acceleration, direction and magnitude. a) The bananas move up. b) The monkey and bananas always move at the same velocity, so the distance between them stays the same. c) Both the monkey and bananas are in free fall, and as they have the same initial velocity, the distance bewteen them doesn’t change. d) The bananas will slow down at the same rate as the monkey; if the monkey comes to a stop, so will the bananas. 5.100: The separated equation of motion has a lower limit of 3vt instead of 0; specifically, v  v vt − v dv 1 k ∫ v − vt = ln − 2vt = ln  2v t − 2  = − m t , or     3v t 1  v = 2v t  + e −( k m ) t  . 2  Note that the speed is always greater than vt .

5.101: a) The rock is released from rest, and so there is initially no resistive force and a0 = (18.0 N) (3.00 kg) = 6.00 m s 2 . b) (18.0 N − (2.20 N ⋅ s m) (3.00 m s)) (3.00 kg) = 3.80 m s 2 . c) The net force must be 1.80 N, so kv = 16.2 N and v = (16.2 N) (2.20 N ⋅ s m) = 7.36 m s . d) When the net force is equal to zero, and hence the acceleration is zero, kvt = 18.0 N and vt = (18.0 N) (2.20 N ⋅ s m) = 8.18 m s. e) From Eq. (5.12),   3.00 kg y = (8.18 m s) (2.00 s) − (1 − e − ( ( 2.20 N ⋅ s m) ( 3.00 kg))(2.00 s) ) 2.20 N ⋅ s m   = +7.78 m. From Eq. (5.10), v = (8.18 m s)[1 − e − ( ( 2.2 N⋅ s m ) ( 3.00 kg))(2.00 s) ] = 6.29 m s. From Eq. (5.11), but with a0 instead of g, a = (6.00 m s 2 )e − ( ( 2.20 N ⋅s m) ( 3.00 kg))(2.00 s) = 1.38 m s 2 . . f) 1− v = 0.1 = e − ( k m )t , so vt m ln (10) = 3.14 s. k

t=

5.102: (a) The retarding force of the surface is the only horizontal force acting. Thus F F − kv1 2 dv a = net = R = = m m m dt dv k = − dt 12 v m v t dv k ∫ v1 2 = m ∫ dt v0 0 2v1 2 which gives
v v0

=−

kt m

v1 2 kt k 2t 2 v = v0 − 0 + m 4m 2 For the rock’s position: dx v1 2 kt k 2t 2 = v0 − 0 + dt m 4m 2 v1 2 ktdt k 2t 2 dt dx = v0 dt − 0 + m 4m 2 and integrating gives x = v0t − (b)
1 v0 2 kt k 2t 2 + m 2m 2 This is a quadratic equation in t; from the quadratic formula we can find the single solution: 2mv1 2 0 t= k (c) Substituting the expression for t into the equation for x: 1 3 2mv1 2 v0 2 k 4m 2v0 k 2 8m3v0 2 0 x = v0 ⋅ − ⋅ + ⋅ k 2m k2 12m 2 k3 3 2mv0 2 = 3k 1 v0 2 kt 2 k 2t 3 + 2m 12m 2

v = 0 = v0 −

5.103: Without buoyancy, kv t = mg , so k =

mg mg = . vt 0.36 s With buoyancy included there is the additional upward buoyancy force B, so B − kvt = mg  0.24 m B = mg − kvt = mg 1 −  0.36 m  s  = mg 3 s 

5.104: Recognizing the geometry of a 3-4-5 right triangle simplifies the calculation. For instance, the radius of the circle of the mass’ motion is 0.75 m. 4 4 a) Balancing the vertical force, TU 5 − TL 5 = w, so 5 5 TL = TU − w = 80.0 N − (4.00 kg) (9.80 m s 2 ) = 31.0 N. 4 4 2 3 3 b) The net inward force is F = 5 TU + 5 TL = 66.6 N. Solving F = marad = m 4π 2 R for T the period T, mR (4.00 kg) (0.75 m) = 2π = 1.334 s, F (66.6 N) or 0.02223 min, so the system makes 45.0 rev/min. c) When the lower string becomes slack, the system is the same as the conical pendulum considered in Example 5.22. With T = 2π

cos β = 0.800, the period is T = 2π (1.25 m) (0.800) (9.80 m s 2 ) = 2.007 s, which is the same as 29.9 rev min. d) The system will still be the same as a conical pendulum, but the block will drop to a smaller angle.

5.105: a) Newton’s 2nd law gives dv mg = vt m y = mg − kv y , where dt k vy t dv y k ∫ v − vt = − m ∫ dt v0 y 0 This is the same expression used in the derivation of Eq. (5.10), except the lower limit in the velocity integral is the initial speed v0 instead of zero. Evaluating the integrals and rearranging gives v = v0e − kt m + vt (1 − e − kt m ) Note that at t = 0 this expression says v y = v0 and at t → α it says v y → vt . b) The downward gravity force is larger than the upward fluid resistance force so the acceleration is downward, until the fluid resistance force equals gravity when the terminal speed is reached. The object speeds up until v y = vt . Take + y to be downward.

c) The upward resistance force is larger than the downward gravity force so the acceleration is upward and the object slows down, until the fluid resistance force equals gravity when the terminal speed is reached. Take + y to be downward.

5.106: (a) To find find the maximum height and time to the top without fluid resistance: 2 2 v y = v0 y + 2a ( y − y0 ) 0 − (6.0 m s) 2 y − y0 = = = 1.84 m or 1.8 m 2a 2(−9.8 m s 2 ) v − v0 0 − 6.0 m s t= = = 0.61 s a − 9.8 m s 2 (b) Starting from Newton’s Second Law for this situation dv m = mg − kv dt we rearrange and integrate, taking downward as positive as in the text and noting that the velocity at the top of the rock’s “flight” is zero: 0 dv k ∫ v − vt = − m t v ln( v − vt ) 0 = ln v − vt − 2.0 m s = ln = ln( 0.25) = −1.386 v − vt − 6.0 m s − 2.0 m s
2 2 v y − v0 y

From Eq. 5.9, m k = vt g = (2.0 m s 2 ) (9.8 m s 2 ) = 0.204 s , and t = − m (−1.386) = (0.204 s) (1.386) = 0.283 s to the top. Equation 5.10 in the text gives us k dx = vt (1 − e − ( k m ) t ) = vt − vt e − ( k m ) t dt

∫ dx = ∫ v dt − ∫ v e
t t 0 0 0

x

t

t

− ( k m) t

dt

vt m − ( k m) t (e − 1) k = (2.0 m s) (0.283 s) + (2.0 m s) (0.204 s)(e−1.387 − 1) = vt t + = 0.26 m

5.107: a) The forces on the car are the air drag force f D = Dv 2 and the rolling friction force µ r mg. Take the velocity to be in the + x -direction. The forces are opposite in direction to the velocity. ∑ Fx = max gives − Dv 2 − µ r mg = ma We can write this equation twice, once with v = 32 m s and a = − 0.42 m s 2 and once with v = 24 m s and a = −0.30 m/s 2 . Solving these two simultaneous equations in the unknowns D and µ r gives µ r = 0.015 and D = 0.36 N ⋅ s 2 m 2 . b) n = mg cos β and the component of gravity parallel to the incline is mg sin β , where β = 2.2°. For constant speed, mg sin 2.2° − µ r mg cos 2.2° − Dv 2 = 0. Solving for v gives v = 29 m s . c) For angle β , mg sin β − µ r mg cos β − Dv 2 = 0 and v = mg (sin β − µ r cos β ) D The terminal speed for a falling object is derived from Dv t2 − mg = 0, so vt = mg D. v vt = sin β − µ r cos β And since μr = 0.015, v vt = sin β − (0.015) cos β 5.108: (a) One way of looking at this is that the apparent weight, which is the same as the upward force on the person, is the actual weight of the person minus the centripetal force needed to keep him moving in its circular path:  mv 2 (12 m s) 2  wapp = mg − = (70 kg) (9.8 m s 2 ) − R 40 m    = 434 N (b) The cart will lose contact with the surface when its apparent weight is zero; i.e., when the road no longer has to exert any upward force on it: mv 2 mg − =0 R v 2 = Rg = (40 m) (9.8 m s 2 ) = 392 m 2 s 2 v = 19.8 m s or 20 m s The answer doesn’t depend on the cart’s mass, because the centripetal force needed to hold it on the road is proportional to its mass and so is its weight, which provides the centripetal force in this situation.

5.109: a) For the same rotation rate, the magnitude of the radial acceleration is proportional to the radius, and for twins of the same mass, the needed force is proportional to the radius; Jackie is twice as far away from the center, and so must hold on with twice as much force as Jena, or 120 N. b) ∑ FJackie = mv 2 r . v=
(120 N) (3.6 m) 30 kg

= 3.8 m s.

5.110: The passenger’s velocity is v = 2π R t = 8.80 m s. The vertical component of the seat’s force must balance the passenger’s weight and the horizontal component must provide the centripetal force. Therefore: Fseat sin θ = mg = 833 N mv 2 = 188 N R Therefore tan θ = 833 N 188 N = 4.43; θ = 77.3° above the horizontal. The magnitude of the net force exerted by the seat (note that this is not the net force on the passenger) is Fseat cos θ =  mv 2  2 2 Fseat = (mg ) +   R  = (833 N) + (188 N)    = 854 N (b) The magnitude of the force is the same, but the horizontal component is reversed.
2 2

5.111: a)

b) The upward friction force must be equal to the weight, so µs n = µs m(4π 2 R T 2 ) ≥ mg and gT 2 (9.80 m s 2 ) (1s 0.60 rev)2 = = 0.28. 4π 2 R 4π 2 (2.5 m) c) No; both the weight and the required normal force are proportional to the rider’s mass.

µs >

5.112: a) For the tires not to lose contact, there must be a downward force on the tires. Thus, the (downward) acceleration at the top of the sphere must exceed mg, so m vR > mg , and v > gR = (9.80 m s 2 ) (13.0 m) = 11.3 m s . b) The (upward) acceleration will then be 4g, so the upward normal force must be 5 mg = 5(110 kg) (9.80 m s 2 ) = 5390 N.
2

5.113: a) What really happens (according to a nosy observer on the ground) is that you slide closer to the passenger by turning to the right. b) The analysis is the same as that of Example 5.23. In this case, the friction force should be insufficient to provide the inward radial acceleration, and so μs mg < mv 2 R, or v2 (20 m s) 2 = = 120 m μs g (0.35) (9.80 m s 2 ) to two places. Why the passenger is not wearing a seat belt is another question. R< 5.114: The tension F in the string must be the same as the weight of the hanging block, and must also provide the resultant force necessary to keep the block on the table in 2 uniform circular motion; Mg = F = m vr , so v = gr M m. 5.115: a) The analysis is the same as that for the conical pendulum of Example 5.22, and so

b) For the bead to be at the same elevation as the center of the hoop, β = 90° and cos β = 0, which would mean T = 0, the speed of the bead would be infinite, and this is not possible. c) The expression for cos β gives cos β = 2.48, which is not possible. In deriving the expression for cos β , a factor of sin β was canceled, precluding the possibility that β = 0. For this situation, β = 0 is the only physical possibility.

 gT 2   (9.80 m s 2 )(1 4.00 s) 2   2  = arccos  = 81.0°. β = arccos    4π L  4π 2 (0.100 m)    

5.116: a) Differentiating twice, ax = −6 βt and a y = −2δ, so Fx = max = (2.20 kg) (−0.72 N s)t = −(1.58 N/s)t Fy = ma y = (2.20 kg) (−2.00 m s 2 ) = −4.40 N. b)

c) At t = 3.00 s, Fx = −4.75 N and Fy = −4.40 N, so
− at an angle of arctan ( −4..40 ) = 223°. 4 75

F = (−4.75 N) 2 + (−4.40 N) 2 = 6.48 N,

5.117:

5.118: See Example 5.25. 2 m 2 a) FA = m g + vR = (1.60 kg) 9.80 m s 2 + (125..000 ms ) = 61.8 N.

ms 2 v − (125..000 m ) b) FB R indicates that the track pushes down on the car. The magnitude of this force is 30.4 N.
2

( ) ( = m( g − ) = (1.60 kg)(9.80 m s

2

) ) = −30.4 N. , where the minus sign

5.119: The analysis is the same as for Problem 5.95; in the case of the cone, the speed is related to the period by v = 2π R T = 2πh tan β T , or T = 2πh tan β v. The maximum and minimum speeds are the same as those found in Problem 5.95, cos β + μs sin β vmax = gh tan β sin β − μs cos β cos β − μs sin β . sin β + μs cos β The minimum and maximum values of the period T are then h tan β sin β − μs cos β Tmin = 2π g cos β + μs sin β vmin = gh tan β Tmax = 2π h tan β sin β + μs cos β . g cos β − μs sin β

5.120: a) There are many ways to do these sorts of problems; the method presented is fairly straightforward in terms of application of Newton’s laws, but involves a good deal of algebra. For both parts, take the x-direction to be horizontal and positive to the right, and the y-direction to be vertical and positive upward. The normal force between the block and the wedge is n; the normal force between the wedge and the horizontal surface will not enter, as the wedge is presumed to have zero vertical acceleration. The horizontal acceleration of the wedge is A, and the components of acceleration of the block are a x and a y . The equations of motion are then MA = −n sin α max = n sin α ma y = n cos α − mg. Note that the normal force gives the wedge a negative acceleration; the wedge is expected to move to the left. These are three equations in four unknowns, A, a x , a y and n. Solution is possible with the imposition of the relation between A, ax and a y . An observer on the wedge is not in an inertial frame, and should not apply Newton’s laws, but the kinematic relation between the components of acceleration are not so restricted. To such an observer, the vertical acceleration of the block is a y , but the horizontal acceleration of the block is ax − A. To this observer, the block descends at an angle α , so the relation needed is ay = − tan α. ax − A At this point, algebra is unavoidable. Symbolic-manipulation programs may save some solution time. A possible approach is to eliminate ax by noting that a x = − M A (a result m that anticipates conservation of momentum), using this in the kinematic constraint to eliminate a y and then eliminating n. The results are: A= ax = − gm ( M + m) tan α + ( M tan α )

gM ( M + m) tan α + ( M tan α ) − g ( M + m) tan α ay = ( M + m) tan α + ( M tan α ) (b) When M >> m, A → 0, as expected (the large block won’t move). Also,

ax →

g tan α + (1 tan α )

tan = g tan 2αα+1 = g sin α cos α, which is the acceleration of the block

( g sin α in this case), with the factor of cos α giving the horizontal component. Similarly, a y → − g sin 2 α. (c) The trajectory is a spiral.

5.121: If the block is not to move vertically, the acceleration must be horizontal. The common acceleration is a = g tan θ , so the applied force must be ( M + m)a = ( M + m) g tan θ. 5.122: The normal force that the ramp exerts on the box will be n = w cos α − T sin θ. The rope provides a force of T cos θ up the ramp, and the component of the weight down the ramp is w sin α. Thus, the net force up the ramp is F = T cos θ − w sin α − μk ( w cos α − T sin θ ) = T (cos θ + μk sin θ ) − w(sin α + μk cos α ). The acceleration will be the greatest when the first term in parantheses is greatest; as in Problems 5.77 and 5.123, this occurs when tan θ = µ k . 5.123: a) See Exercise 5.38; F = µ k w (cos θ + µ k sin θ ). b)

c) The expression for F is a minimum when the denominator is a maximum; the calculus is identical to that of Problem 5.77 (maximizing w for a given F gives the same result as minimizing F for a given w), and so F is minimized at tan θ = µ k . For µ k = 0.25, θ = 14.0°, keeping an extra figure.

5.124: For convenience, take the positive direction to be down, so that for the baseball released from rest, the acceleration and velocity will be positive, and the speed of the baseball is the same as its positive component of velocity. Then the resisting force, directed against the velocity, is upward and hence negative. a)

b) Newton’s Second Law is then ma = mg − Dv 2 . Initially, when v = 0, the acceleration is g, and the speed increases. As the speed increases, the resistive force increases and hence the acceleration decreases. This continues as the speed approaches the terminal speed. Eq. (5.13). c) At terminal velocity, a = 0, so vt =
mg D dv dt

, in agreement with
g v t2

d) The equation of motion may be rewritten as

=

(vt2 − v 2 ). This is a

separable equation and may be expressed as dv g ∫ vt2 − v 2 = v t2 ∫ dt , or v 1 arctanh v vt  t  gt = 2,  v  t

so v = vt tanh ( gt vt ). Note: If inverse hyperbolic functions are unknown or undesirable, the integral can be done by partial fractions, in that 1 1  1 1  =  v − v + v + v , 2 2 vt − v 2vt  t t  and the resulting logarithms in the integrals can be solved for v (t ) in terms of exponentials.

5.125: Take all accelerations to be positive downward. The equations of motion are straightforward, but the kinematic relations between the accelerations, and the resultant algebra, are not immediately obvious. If the acceleration of pulley B is aB , then aB = −a3 , and aB is the average of the accelerations of masses 1 and 2, or a1 + a2 = 2aB = −2a3 . There can be no net force on the massless pulley B, so TC = 2TA . The five equations to be solved are then m1 g − TA = m1a1 m2 g − TA = m2 a2 m3 g − TC = m3a3 a1 + a2 + 2a3 = 0 2TA − TC = 0. These are five equations in five unknowns, and may be solved by standard means. A symbolic-manipulation program is of great use here. a) The accelerations a1 and a2 may be eliminated by using 2a3 = −( a1 + a2 ) = −(2 g − TA ((1 m1 ) + (1 m2 ))). The tension TA may be eliminated by using TA = (1 2)TC = (1 2)m3 ( g − a3 ). Combining and solving for a3 gives − 4m1m2 + m2 m3 + m1m3 . 4m1m2 + m2 m3 + m1m3 b) The acceleration of the pulley B has the same magnitude as a3 and is in the opposite direction. T T m c) a1 = g − A = g − C = g − 3 ( g − a3 ). m1 2m1 2m1 Substituting the above expression for a3 gives 4m m − 3m2 m3 + m1m3 a1 = g 1 2 . 4m1m2 + m2 m3 + m1m3 d) A similar analysis (or, interchanging the labels 1 and 2) gives 4m m − 3m1m3 + m2 m3 a2 = g 1 2 . 4m1m2 + m2 m3 + m1m3 e) & f) Once the accelerations are known, the tensions may be found by substitution into the appropriate equation of motion, giving 4m1 m2 m3 8m1 m2 m3 TA = g , TC = g . 4m1 m2 + m2 m3 + m1 m3 4m1 m2 + m2 m3 + m1 m3 g) If m1 = m2 = m and m3 = 2m, all of the accelerations are zero, TC = 2mg and TA = mg. All masses and pulleys are in equilibrium, and the tensions are equal to the weights they support, which is what is expected. a3 = g

5.126: In all cases, the tension in the string will be half of F. a) F 2 = 62 N, which is insufficient to raise either block; a1 = a2 = 0. b) F 2 = 62 N. The larger block (of weight 196 N) will not move, so a1 = 0, but the smaller block, of weight 98 N, has a net upward force of 49 N applied to it, and so will 49 accelerate upwards with a2 = 10.0 N = 4.9 m s 2 . kg c) F 2 = 212 N, so the net upward force on block A is 16 N and that on block B is 16 114 N 114 N, so a1 = 20.0N = 0.8 m s 2 and a2 = 10.0 kg = 11.4 m s 2 . kg

5.127: Before the horizontal string is cut, the ball is in equilibrium, and the vertical component of the tension force must balance the weight, so TA cos β = w, or TA = w cos β. At point B, the ball is not in equilibrium; its speed is instantaneously 0, so there is no radial acceleration, and the tension force must balance the radial component of the weight, so TB = w cos β , and the ratio (TB TA ) = cos2 β.

Chapter 6

6.1: a) (2.40 N) (1.5 m) = 3.60 J b) ( −0.600 N)(1.50 m) = −0.900 J c) 3.60 J − 0.720 J = 2.70 J . 6.2: a) “Pulling slowly” can be taken to mean that the bucket rises at constant speed, so the tension in the rope may be taken to be the bucket’s weight. In pulling a given length of rope, from Eq. (6.1), W = Fs = mgs = (6.75 kg ) (9.80 m / s 2 )(4.00 m) = 264.6 J. b) Gravity is directed opposite to the direction of the bucket’s motion, so Eq. (6.2) gives the negative of the result of part (a), or − 265 J . c) The net work done on the bucket is zero. 6.3: (25.0 N)(12.0 m) = 300 J . 6.4: a) The friction force to be overcome is f = µ k n = µ k mg = (0.25)(30.0 kg )(9.80 m / s 2 ) = 73.5 N, or 74 N to two figures. b) From Eq. (6.1), Fs = (73.5 N)(4.5 m) = 331 J . The work is positive, since the worker is pushing in the same direction as the crate’s motion. c) Since f and s are oppositely directed, Eq. (6.2) gives

− fs = −(73.5 N)(4.5 m) = −331 J.
d) Both the normal force and gravity act perpendicular to the direction of motion, so neither force does work. e) The net work done is zero.

6.5: a) See Exercise 5.37. The needed force is

µ k mg (0.25)(30 kg)(9.80 m / s 2 ) F= = = 99.2 N, cos φ − µ k sin φ cos 30° − (0.25) sin 30°
keeping extra figures. b) Fs cos φ = (99.2 N)(4.50 m) cos 30° = 386.5 J , again keeping an extra figure. c) The normal force is mg + F sin φ , and so the work done by friction is − (4.50 m)(0.25)((30 kg)(9.80 m / s 2 ) + (99.2 N) sin 30°) = −386.5 J . d) Both the normal force and gravity act perpendicular to the direction of motion, so neither force does work. e) The net work done is zero.

6.6: From Eq. (6.2), Fs cos φ = (180 N)(300 m) cos15.0° = 5.22 × 104 J. 6.7: 2 Fs cos φ = 2(1.80 × 10 6 N)(0.75 × 10 3 m) cos14° = 2.62 × 10 9 J, two places. 6.8: The work you do is:   ˆ ˆ F ⋅ s = ((30 N)i − (40 N) ˆ) ⋅ ((−9.0m)i − (3.0m) ˆ) j j = (30 N)(−9.0 m) + (−40 N)(−3.0 m) = −270 N ⋅ m + 120 N ⋅ m = −150 J 6.9: a) (i) Tension force is always perpendicular to the displacement and does no work. (ii) Work done by gravity is − mg ( y 2 − y1 ). When y1 = y 2 , Wmg = 0 . b) (i) Tension does no work. (ii) Let l be the length of the string. Wmg = − mg ( y 2 − y1 ) = −mg (2l ) = −25.1 J The displacement is upward and the gravity force is downward, so it does negative work. or 2.6 × 10 9 J to

6.10: a) From Eq. (6.6),  1  1 m / s  5 K = (1600 kg) (50.0 km/h )   = 1.54 × 10 J.  2 3.6 km / h      b) Equation (6.5) gives the explicit dependence of kinetic energy on speed; doubling the speed of any object increases the kinetic energy by a factor of four. 6.11: For the T-Rex, K = 1 (7000 kg)((4 km / hr ) 3.1 m / s ) 2 = 4.32 × 103 J . The person’s 2 6 km/hr velocity would be v = 2(4.32 × 10 3 J)/70 kg = 11.1 m/s , or about 40 km/h. 6.12: (a) Estimate: v ≈ 1m/s (walking) v ≈ 2 m/s (running) m ≈ 70 kg Walking: KE = 1 mv 2 = 1 (70 kg )(1 m / s) 2 = 35 J 2 2 Running: KE = 1 (70 kg )(2 m / s) 2 = 140 J 2 (b) Estimate: v ≈ 60 mph = 88 ft / s ≈ 30 m / s m ≈ 2000 kg 1 KE = 2 (2000 kg )(30 m / s) 2 = 9 × 10 5 J (c) KE = Wgravity = mgh Estimate h ≈ 2 m KE = (1 kg )(9.8 m / s 2 )(2 m) ≈ 20 J 6.13: Let point 1 be at the bottom of the incline and let point 2 be at the skier. Wtot = K 2 − K 1 1 2 K 1 = mv 0 , K 2 = 0 2 Work is done by gravity and friction, so Wtot = Wmg + W f . Wmg = −mg ( y 2 − y1 ) = −mgh W f = − fs = −( µ k mg cosα )(h / sin α ) = − µ k mgh / tan α Substituting these expressions into the work-energy theorem and solving for v0 gives v 0 = 2 gh(1 + µ k / tan α )
2

6.14:

(a)

W = ∆KE 1 1 2 − mgh = mvf2 − mv0 2 2 v 0 = vf2 + 2 gh = (25.0 m / s) 2 + 2 (9.80 m / s 2 )(15.0 m) = 30.3 m / s

(b)

W = ∆KE 1 1 2 − mgh = mvf2 − mv0 2 2 2 2 v − vf (30.3 m / s) 2 − 02 h= 0 = 2g 2(9.80 ms / s 2 ) = 46.8 m

6.15: a) parallel to incline: force component = mg sin α , down incline; displacement = h/ sin α , down incline W|| = (mg sin α )(h / sin α ) = mgh perpendicular to incline: no displacement in this direction, so W⊥ = 0 . Wmg = W|| + W⊥ = mgh , same as falling height h. b) Wtot = K 2 − K1 gives mgh = 1 mv 2 and v = 2 gh , same as if had been dropped 2 from height h. The work done by gravity depends only on the vertical displacement of the object. When the slope angle is small, there is a small force component in the direction of the displacement but a large displacement in this direction. When the slope angle is large, the force component in the direction of the displacement along the incline is larger but the displacement in this direction is smaller. c) h = 15.0 m , so v = 2 gh = 17.1 s 6.16: Doubling the speed increases the kinetic energy, and hence the magnitude of the work done by friction, by a factor of four. With the stopping force given as being independent of speed, the distance must also increase by a factor of four. 6.17: Barring a balk, the initial kinetic energy of the ball is zero, and so W = (1 / 2)mv 2 = (1 / 2)(0.145 kg)(32.0 m/s) 2 = 74.2 J.

6.18: As the example explains, the boats have the same kinetic energy K at the finish line, 2 2 2 2 so (1 / 2)m A v A = (1 / 2)m B v B , or, with m B = 2m A , v A = 2v B . a) Solving for the ratio of the speeds, v A / v B = 2 . b) The boats are said to start from rest, so the elapsed time is the distance divided by the average speed. The ratio of the average speeds is the same as the ratio of the final speeds, so the ratio of the elapsed times is t B / t A = v A / v B = 2 . 6.19: a) From Eq. (6.5), K 2 = K 1 / 16 , and from Eq. (6.6), W = −(15 / 16) K1 . b) No; kinetic energies depend on the magnitudes of velocities only. 6.20: From Equations (6.1), (6.5) and (6.6), and solving for F, F=
2 2 2 2 ∆K 1 m(v 2 − v1 ) 1 (8.00 kg)((6.00 m / s) − (4.00 m / s) ) = 2 = 2 = 32.0 N. s s (2.50 m)

6.21:

s=

∆K 1 (0.420 kg)((6.00 m / s) 2 − (2.00 m / s) 2 ) = 2 = 16.8 cm F (40.0 N)

6.22: a) If there is no work done by friction, the final kinetic energy is the work done by the applied force, and solving for the speed, 2W 2 Fs 2(36.0 N)(1.20 m) v= = = = 4.48 m / s. m m (4.30 kg ) b) The net work is Fs − f k s = ( F − µ k mg ) s , so v= = 2( F − µ k mg ) s m

2(36.0 N − (0.30)(4.30 kg)(9.80 m / s 2 ))(1.20 m) (4.30 kg) = 3.61 m / s. (Note that even though the coefficient of friction is known to only two places, the difference of the forces is still known to three places.)

6.23: a) On the way up, gravity is opposed to the direction of motion, and so W = −mgs = −(0.145 kg )(9.80 m / s 2 )(20.0 m) = −28.4 J . W 2(−28.4 J) = (25.0 m / s) 2 + = 15.26 m / s . m (0.145 kg ) c) No; in the absence of air resistance, the ball will have the same speed on the way down as on the way up. On the way down, gravity will have done both negative and positive work on the ball, but the net work will be the same. b) v2 = v12 + 2 6.24: a) Gravity acts in the same direction as the watermelon’s motion, so Eq. (6.1) gives W = Fs = mgs = (4.80 kg)(9.80 m / s 2 )(25.0 m) = 1176 J. b) Since the melon is released from rest, K 1 = 0 , and Eq. (6.6) gives K = K 2 = W = 1176 J. 6.25: a) Combining Equations (6.5) and (6.6) and solving for v2 algebraically, v 2 = v12 + 2 Wtot 2(10.0 N)(3.0 m) = ( 4.00 m / s) 2 + = 4.96 m / s. m (7.00 kg )

Keeping extra figures in the intermediate calculations, the acceleration is a = (10.0 kg ⋅ m / s 2 ) /(7.00 kg) = 1.429 m / s 2 . From Eq. (2.13), with appropriate change in notation, 2 v 2 = v12 + 2as = ( 4.00 m / s) 2 + 2(1.429 m / s 2 )(3.0 m), giving the same result. 6.26: The normal force does no work. The work-energy theorem, along with Eq. (6.5), gives 2K 2W v= = = 2 gh = 2 gL sin θ , m m where h = L sin θ is the vertical distance the block has dropped, and θ is the angle the plane makes with the horizontal. Using the given numbers, v = 2(9.80 m / s 2 )(0.75 m) sin 36.9° = 2.97 m / s.

6.27: a) The friction force is µ k mg , which is directed against the car’s motion, so the net 2 work done is − µ k mgs . The change in kinetic energy is ∆K = − K 1 = −(1 / 2) mv 0 , and so
2 s = v0 / 2µ k g . b) From the result of part (a), the stopping distance is proportional to the square of the initial speed, and so for an initial speed of 60 km/h, s = (91.2 m)(60.0 / 80.0) 2 = 51.3 m . (This method avoids the intermediate calculation of µ k , which in this case is about 0.279.)

6.28: The intermediate calculation of the spring constant may be avoided by using Eq. (6.9) to see that the work is proportional to the square of the extension; the work needed to compress the spring 4.00 cm is (12.0 J)

(

4.00 cm 2 3.00 cm

)

= 21.3 J .

6.29: a) The magnitude of the force is proportional to the magnitude of the extension or compression; (160 N)(0.015 m / 0.050 m) = 48 N, (160 N)(0.020 m / 0.050 m) = 64 N. b) There are many equivalent ways to do the necessary algebra. One way is to note  169 N   = (0.050 m) 2 = 4 J , that to stretch the spring the original 0.050 m requires 1  2 0.050 m    so that stretching 0.015 m requires (4 J )(0.015 / 0.050) 2 = 0.360 J and compressing 0.020 m requires (4 J )(0.020 / 0.050) 2 = 0.64 J . Another is to find the spring constant k = (160 N) ÷ (0.050 m) = 3.20 × 10 3 N / m , from which (1 / 2)(3.20 × 10 3 N / m)(0.015 m) 2 = 0.360 J and (1 / 2)(3.20 × 10 3 N / m)(0.020 m) 2 = 0.64 J . 6.30: The work can be found by finding the area under the graph, being careful of the sign of the force. The area under each triangle is 1/2 base × height . a) 1 / 2 (8 m)(10 N) = 40 J . b) 1 / 2 (4 m)(10 N) = +20 J . c) 1 / 2 (12 m)(10 N) = 60 J . 6.31: Use the Work-Energy Theorem and the results of Problem 6.30. (2)(40 J) a) v = = 2.83 m / s 10 kg b) At x = 12 m , the 40 Joules of kinetic energy will have been increased by 20 J, so v= (2)(60 J ) = 3.46 m / s . 10 kg

6.32: The work you do with your changing force is

∫

6.9

0

F ( x ) dx = ∫

6.9

0

(−20.0 N) dx − ∫

6.9

0

3.0

N xdx m

N 2 )( x / 2) | 6.9 0 m = −138 N ⋅ m − 71.4 N ⋅ m = −209.4 J or − 209 J = (−20.0 N) x | 6.9 −(3.0 0 The work is negative because the cow continues to advance as you vainly attempt to push her backward. 6.33: Wtot = K 2 − K 1 1 2 K 1 = mv 0 , K 2 = 0 2 Work is done by the spring force. Wtot = − 1 kx 2 , where x is the amount the spring is 2 compressed. 1 1 2 − kx 2 = − mv 0 and x = v 0 m / k = 8.5 cm 2 2 6.34: a) The average force is (80.0 J) /(0.200 m) = 400 N , and the force needed to hold the platform in place is twice this, or 800 N. b) From Eq. (6.9), doubling the distance quadruples the work so an extra 240 J of work must be done. The maximum force is quadrupled, 1600 N. Both parts may of course be done by solving for the spring constant k = 2(80.0 J) ÷ (0.200 m) 2 = 4.00 × 10 3 N/m , giving the same results.

6.35: a) The static friction force would need to be equal in magnitude to the spring force, m 0 µ s mg = kd or μs = ((020.0 N / )(9)(80.086/ m2)) = 1.76 , which is quite large. (Keeping extra figures in .100 kg . m s the intermediate calculation for d gives a different answer.) b) In Example 6.6, the relation 1 1 µ k mgd + kd 2 = mv12 2 2 was obtained, and d was found in terms of the known initial speed v1 . In this case, the condition on d is that the static friction force at maximum extension just balances the spring force, or kd = µ s mg . Solving for v12 and substituting, k v12 = d 2 + 2 gdµ k d m k  µ mg   µ mg  =  s  + 2µ k g  s  m k   k  mg 2 2 (µ s + 2µ s µ k ) k  (0.10 kg)(9.80 m / s 2 ) 2  ((0.60) 2 + 2(0.60)(0.47)), =   (20.0 N / m)   =
2

from which v1 = 0.67 m / s .

6.36: a) The spring is pushing on the block in its direction of motion, so the work is positive, and equal to the work done in compressing the spring. From either Eq. (6.9) or Eq. (6.10), W = 1 kx 2 = 1 (200 N / m)(0.025 m) 2 = 0.06 J . 2 2 b) The work-energy theorem gives 2(0.06 J ) 2W v= = = 0.18 m / s. m (4.0 kg) 6.37: The work done in any interval is the area under the curve, easily calculated when the areas are unions of triangles and rectangles. a) The area under the trapezoid is 4.0 N ⋅ m = 4.0 J . b) No force is applied in this interval, so the work done is zero. c) The area of the triangle is 1.0 N ⋅ m = 1.0 J , and since the curve is below the axis ( Fx < 0) , the work is negative, or − 1.0 J . d) The net work is the sum of the results of parts (a), (b) and (c), 3.0 J. (e) + 1.0 J − 2.0 J = −1.0 J .

6.38: a) K = 4.0 J , so v = 2 K m = 2 (4.0 J) (2.0 kg ) = 2.00 m / s . b) No work is done between x = 3.0 m and x = 4.0 m , so the speed is the same, 2.00 m/s. c)

K = 3.0 J , so v = 2 K / m = 2(3.0 J ) /(2.0 kg) = 1.73 m / s .

6.39: a) The spring does positive work on the sled and rider; (1 / 2)kx 2 = (1 / 2)mv 2 , or v = x k / m = (0.375 m) (4000 N / m) /(70 kg) = 2.83 m / s . b) The net work done by
2 the spring is (1 / 2)k ( x12 − x 2 ) , so the final speed is

v=

k 2 2 ( x1 − x 2 ) = m

(4000 N / m ((0.375 m) 2 − (0.200 m) 2 ) = 2.40 m / s. (70 kg )

6.40: a) From Eq. (6.14), with dl = Rdφ , F cosφ dl = 2 wR ∫ cosφ dφ = 2 wR sin θ 0 . P1 0    In an equivalent geometric treatment, when F is horizontal, F ⋅ dl = Fdx , and the total work is F = 2 w times the horizontal distance, in this case (see Fig. 6.20(a)) R sin θ0 , 2w giving the same result. b) The ratio of the forces is w tan θ 0 = 2 cot θ0 . c) 2wR sin θ0 sin θ0 θ =2 = 2 cot 0 . wR (1 − cos θ0 ) (1 − cos θ0 ) 2 W =∫
P2

θ0

6.41: a) The initial and final (at the maximum distance) kinetic energy is zero, so the positive work done by the spring, (1 / 2)kx 2 , must be the opposite of the negative work done by gravity, − mgL sin θ , or 2mgL sin θ 2(0.0900 kg )(9.80 m / s 2 )(1.80 m) sin 40.0° = = 5.7 cm. k (640 N / m) At this point the glider is no longer in contact with the spring. b) The intermediate calculation of the initial compression can be avoided by considering that between the point 0.80 m from the launch to the maximum distance, gravity does a negative amount of work given by − (0.0900 kg)(9.80 m / s 2 )(1.80 m − 0.80 m) sin 40.0° = −0.567 J , and so the kinetic energy of the glider at this point is 0.567 J. At this point the glider is no longer in contact with the spring. x=

6.42: The initial and final kinetic energies of the brick are both zero, so the net work done on the brick by the spring and gravity is zero, so (1 2)kd 2 − mgh = 0 , or d = 2mgh / k = 2(1.80 kg )(9.80 m / s 2 )(3.6 m) /(450 N / m) = 0.53 m. The spring will provide an upward force while the spring and the brick are in contact. When this force goes to zero, the spring is at its uncompressed length. 6.43: Energy = ( power)(time ) = (100 W )(3600 s) = 3.6 × 10 5 J 1 K = mv 2 so v = 2 K/m = 100 s for m = 70 kg. 2 6.44: Set time to stop: ΣF = ma : µ k mg = ma a = μk g = (0.200)(9.80 m / s 2 ) = 1.96 m / s 2 v = v0 + at 0 = 8.00 m / s − (1.96 m / s 2 )t t = 4.08 s 2 KE 1 mv 2 P= = t t 1 (20.0 kg )(8.00 m / s 2 ) = 2 = 157 W 4.08 s

6.45: The total power is (165 N)(9.00 m / s) = 1.485 × 10 3 W , so the power per rider is 742.5 W, or about 1.0 hp (which is a very large output, and cannot be sustained for long periods). (1.0 × 1019 J / yr ) = 3.2 × 1011 W. 7 (3.16 × 10 s / yr) 3.2 × 1011 W = 1.2 kW/person . 2.6 × 10 8 folks 3.2 × 1011 W = 8.0 × 10 8 m 2 = 800 km 2 . 3 2 (0.40)1.0 × 10 W / m

6.46: a) b) c)

6.47: The power is P = F ⋅ v . F is the weight, mg, so P = (700 kg) (9.8 m s 2 ) (2.5 m s) = 17.15 kW. So, 17.15 kW 75 kW. = 0.23, or about 23% of the engine power is used in climbing. 6.48: a) The number per minute would be the average power divided by the work (mgh) required to lift one box, (0.50 hp) (746 W hp) = 1.41 s, (30 kg) (9.80 m s 2 ) (0.90 m) or 84.6 min. b) Similarly, (100 W) = 0.378 s, (30 kg) (9.80 m s 2 ) (0.90 m) or 22.7 min.

6.49: The total mass that can be raised is (40.0 hp) (746 W hp)(16.0 s) = 2436 kg, (9.80 m s 2 ) (20.0 m) so the maximum number of passengers is
1836 kg 65.0 kg

= 28.

6.50: From any of Equations (6.15), (6.16), (6.18) or (6.19), Wh (3800 N) (2.80 m) P= = = 2.66 × 103 W = 3.57 hp. t (4.00 s)

6.51: F =

(0.70) Pave (0.70) (280,000 hp)(746 W hp) = = 8.1 × 106 N. v (65 km h) ((1 km h) (3.6 m s))

6.52: Here, Eq. (6.19) is the most direct. Gravity is doing negative work, so the rope  must do positive work to lift the skiers. The force F is gravity, and F = Nmg , where N is the number of skiers on the rope. The power is then P = ( Nmg ) (v) cos φ  1m s  = (50) (70 kg) (9.80 m s 2 ) (12.0 km h)   3.6 km h  cos (90.0° − 15.0°)    4 = 2.96 × 10 W. Note that Eq. (1.18) uses φ as the angle between the force and velocity vectors; in this case, the force is vertical, but the angle 15.0° is measured from the horizontal, so φ = 90.0° − 15.0° is used.

6.53: a) In terms of the acceleration a and the time t since the force was applied, the speed is v = at and the force is ma, so the power is P = Fv = (ma ) (at ) = ma 2t. b) The power at a given time is proportional to the square of the acceleration, tripling the acceleration would mean increasing the power by a factor of nine. c) If the magnitude of the net force is the same, the acceleration will be the same, and the needed power is proportional to the time. At t = 15.0 s , the needed power is three times that at 5.0 s, or 108 W. 6.54: dK d  1 2  =  mv  dt dt  2  dv = mv dt = mva = mav = Fv = P. 6.55: Work done in each stroke is W = Fs and Pav = W t = 100 Fs t t = 1.00 s, F = 2mg and s = 0.010 m. Pav = 0.20 W.

6.56:

Let M = total mass and T = time for one revolution 1 KE = ∫ (dm)v 2 2 M dm = dx L 2πx v= T 1  M   2πx  KE = ∫  dx    2 L  T  0 1M =  2 L   4π  2   T
2 L 2

L 2  ∫ x dx  0 1  M   4π 2   L3  2 =    2    = π 2 ML2 T 2 2 L  T  3  3    5 revolutions in 3 seconds → T = 3 5 s 2 KE = π 2 (12.0 kg) (2.00 m) 2 (3 5 s) 2 = 877 J. 3 6.57: a) (140 N) (3.80 m) = 532 J b) ( 20.0 kg) (9.80 m s 2 ) (3.80 m) (− sin 25°) = −315 J c) The normal force does no work. d) W f = − f k s = − μk ns = − μk mgs cos θ = −(0.30) (20.0 kg) (9.80 m s 2 ) (3.80 m) cos 25° = −203 J e) 532 J − 315 J − 203 J = 15 J (14.7 J to three figures). f) The result of part (e) is the kinetic energy at the top of the ramp, so the speed is v = 2 K m = 2(14.7 J) (20.0 kg) = 1.21 m s .

6.58: The work per unit mass is (W m) = gh. a) The man does work, (9.8 N kg) (0.4 m) = 3.92 J kg. b) (3.92 J kg) (70 J kg) × 100 = 5.6%. c) The child does work, (9.8 N kg) (0.2 m) = 1.96 J kg. (1.96 J kg) (70 J kg) × 100 = 2.8%. d) If both the man and the child can do work at the rate of 70 J kg, and if the child only needs to use 1.96 J kg instead of 3.92 J kg, the child should be able to do more pull ups.
1 6.59: a) Moving a distance L along the ramp, sin = L, sout = L sin α, so IMA = sin α . b) If AMA = IMA, ( Fout Fin ) = ( sin sout ) and so ( Fout ) ( sout ) = ( Fin ) ( sin ) , or Wout = Win . c)

d) E= Wout ( Fout )(sout ) Fout Fin AMA = = = . Win ( Fin )(sin ) sin sout IMA

6.60: a) m = b) n =

w − Wg s (7.35 × 103 J) = = = 41.7 kg. g g (9.80 m s 2 ) (18.0 m)

Wn 8.25 × 103 J = = 458 N. s 18.0 m c) The weight is mg = Wsg = 408 N, so the acceleration is the net force divided by the 458 N − 408 N = 1.2 m s 2 . mass, 41.7 kg 6.61: a)  2π (6.66 × 106 m) 1 2 1  2πR  1 mv = m  = (86,400 kg)   (90.1 min) (60 s min ) 2 2  T  2    = 2.59 × 1012 J.   2 2 b) (1 2) mv = (1 2) (86,400 kg) ((1.00 m) (3.00 s)) = 4.80 × 103 J.
2 2

6.62: a)

W f = − f k s = − μk mg cos θs = −(0.31) (5.00 kg) (9.80 m s 2 ) cos 12.0°(1.50 m) = −22.3 J

(keeping an extra figure) b) (5.00 kg) (9.80 m s 2 ) sin 12.0° (1.50 m) = 15.3 J. c) The normal force does no work. d) 15.3 J − 22.3 J = −7.0 J. e) K 2 = K1 + W = (1 2) (5.00 kg) (2.2 m s) 2 − 7.0 J = 5.1 J, and so v2 = 2(5.1 J ) /(5.00 kg) = 1.4 m / s . 6.63: See Problem 6.62: The work done is negative, and is proportional to the distance s that the package slides along the ramp, W = mg (sin θ − μk cos θ ) s . Setting this equal to the (negative) change in kinetic energy and solving for s gives (1 / 2) mv12 v12 s=− = mg (sin θ − μk cos θ ) 2 g (sin θ − μk cos θ ) = (2.2 m/s) 2 = 2.6 m. 2(9.80 m / s 2 )(sin 12° − (0.31) cos12°)

As a check of the result of Problem 6.62, (2.2 m / s) 1 − (1.5 m) /(2.6 m) = 1.4 m / s . 6.64: a) From Eq. (6.7),
2 1 1 dx  1 = −k −  = k  − . 2 x  x1 x1 x  x  x1  2 x1  The force is given to be attractive, so Fx < 0 , and k must be positive. If x2 > x1 ,

W = ∫ Fx dx = −k ∫

x2

x2

x

1 x2

<

1 x1

,

and W < 0 . b) Taking “slowly” to be constant speed, the net force on the object is zero, so the force applied by the hand is opposite Fx , and the work done is negative of that found in part (a), or k x11 − x12 , which is positive if x2 > x1 . c) The answers have the same

(

)

magnitude but opposite signs; this is to be expected, in that the net work done is zero. 6.65: F = mg ( RE / r ) 2
2 2 RE  mgR  R 2 W = − ∫ Fds = − ∫  2 E  dr = −mgRE (−(1 / r ) |∞ E ) = mgRE  1 ∞   r  Wtot = K 2 − K1 , K1 = 0

This gives K 2 = mgRE = 1.25 × 1012 J
2 K 2 = 1 mv2 so v2 = 2 K 2 / m = 11,000 m / s 2

6.66: Let x be the distance past P. when x = 12.5 m , µ k = 0.600

µ k = 0.100 + Ax

A = 0.500 / 12.5 m = 0.0400 /m

(a)

W = ∆KE : Wf = KEf − KEi 1 − ∫ μk mgdx = 0 − mvi2 2 xf 1 g ∫ (0.100 + Ax)dx = vi2 0 2 2  x  1 g (0.100) xf + A f  = vi2 2 2 

 x2  1 (9.80 m / s 2 ) (0.100) xf + (0.0400 / m) f  = (4.50 m / s) 2 2 2  Solve for xf : xf = 5.11m (b) µ k = 0.100 + (0.0400 / m)(5.11 m) = 0.304 (c) Wf = KEf − KEi 1 − μk mgx = 0 − mv12 2 (4.50 m / s) 2 x = vi2 / 2 μk g = = 10.3 m 2(0.100)(9.80 m / s 2 )

3 6.67: a) αxa = (4.00 N m3 )(1.00 m)3 = 4.00 N. 3 b) αxb = (4.00 N m3 )(2.00 m)3 = 32.0 N. c) Equation 6.7 gives the work needed to move an object against the force; the work done by the force is the negative of this, x2 α 4 − ∫ αx 3dx = − ( x2 − x14 ). x1 4 With x1 = xa = 1.00 m and x2 = xb = 2.00 m , W = −15.0 J , this work is negative.

6.68: From Eq. (6.7), with x1 = 0 ,
W = ∫ Fdx = ∫ (kx − bx 2 + cx 3 )dx = k 2 b 3 c 4 x2 − x2 + x2 0 0 2 3 4 2 2 3 3 4 = (50.0 N / m) x2 − (233 N / m ) x2 + (3000 N / m ) x2
x2 x2

a) When x2 = 0.050 m , W = 0.115 J , or 0.12 J to two figures. b) When x2 = −0.050 m, W = 0.173 J , or 0.17 J to two figures. c) It’s easier to stretch the spring; the quadratic − bx 2 term is always in the − x -direction, and so the needed force, and hence the needed work, will be less when x2 > 0 . 6.69: a) T = marad = m vR = (0.120kg) (0.7040m/s) = 0.147 N , or 0.15 N to two figures. b) At ( 0 . m)
2 2

the later radius and speed, the tension is (0.120kg) (2.8010m/s) = 9.41 N , or 9.4 N to two ( 0 . m) figures. c) The surface is frictionless and horizontal, so the net work is the work done by the cord. For a massless and frictionless cord, this is the same as the work done by the person, and is equal to the change in the block’s kinetic energy, 2 K 2 − K1 = (1 / 2)m(v2 − v12 ) = (1 / 2)(0.120 kg )((2.80 m / s) 2 − (0.70 m / s) 2 ) = 0.441 J . Note that in this case, the tension cannot be perpendicular to the block’s velocity at all times; the cord is in the radial direction, and for the radius to change, the block must have some non-zero component of velocity in the radial direction. 6.70: a) This is similar to Problem 6.64, but here α > 0 (the force is repulsive), and x2 < x1 , so the work done is again negative; 1 1 W = α  −  = (2.12 × 10− 26 N ⋅ m 2 ((0.200 m −1 ) − (1.25 × 109 m −1 )) x x  2   1 = −2.65 × 10−17 J. Note that x1 is so large compared to x 2 that the term (6.13)) and solving for v2 , v2 = v12 + 2W 2(−2.65 × 10 −17 J) = (3.00 × 105 m / s) 2 + = 2.41 × 105 m / s. − 27 m (1.67 × 10 kg)
1 x1

2

is negligible. Then, using Eq.

α b) With K 2 = 0, W = − K1 . Using W = − x 2 ,

α 2α 2(2.12 × 10−26 N ⋅ m 2 ) = = = 2.82 × 10−10 m. K1 mv12 (1.67 × 10 − 27 kg )(3.00 × 105 m / s) 2 c) The repulsive force has done no net work, so the kinetic energy and hence the speed of the proton have their original values, and the speed is 3.00 × 105 m / s . x2 =

6.71: The velocity and acceleration as functions of time are dx v(t ) = = 2αt + 3βt 2 , a(t ) = 2α + 6 βt dt 2 a) v(t = 4.00 s) = 2(0.20 m / s )(4.00s) + 3(0.02 m / s3 )(4.00s) 2 = 2.56 m / s. b) ma = (6.00 kg )(2(0.20 m / s 2 ) + 6(0.02 m / s3 )(4.00 s) = 5.28 N. c) W = K 2 − K1 = K 2 = (1 / 2)(6.00 kg)(256 m / s) 2 = 19.7 J. 6.72: In Eq. (6.14), dl = dx and φ = 31.0° is constant, and so W = ∫ F cosφ dl = ∫ F cosφ dx
P1 x1 P2 x2

= (5.00 N / m 2 ) cos 31.0°∫

1.50 m

1.00 m

x 2 dx = 3.39 J.

The final speed of the object is then 2W 2(3.39 J) v2 = v12 + = (4.00 m / s) 2 + = 6.57 m / s. m (0.250 kg )
2 6.73: a) K 2 − K1 = (1 / 2)m(v2 − v12 ) = (1 / 2)(80.0 kg )((1.50 m / s) 2 − (5.00 m / s) 2 ) = −910 J. b) The work done by gravity is − mgh = −(80.0 kg)(9.80 m/s 2 )(5.20 m) = −4.08 ×103 J , so the work done by the rider is −910 J − (−4.08 × 103 J) = 3.17 ×103 J . ∞

6.74: a) W = ∫

∞

x0

b b dx = n x (− n − 1)) x n −1
1−n

=
x0

b . n (n − 1) x0 −1

Note that for this part, for n > 1, x integral must be used,

→ 0 as x → ∞ . b) When 0 < n < 1 , the improper

 b  n n W = lim  ( x 2 −1 − x0 −1 ) , x2 →∞ ( n − 1)   n −1 and because the exponent on the x 2 is positive, the limit does not exist, and the integral diverges. This is interpreted as the force F doing an infinite amount of work, even though F → 0 as x2 → ∞. 6.75: Setting the (negative) work done by the spring to the needed (negative) change in 2 kinetic energy, 1 kx 2 = 1 mv0 , and solving for the spring constant, 2 2 k=
2 mv0 (1200 kg )(0.65 m / s) 2 = = 1.03 × 105 N / m. x2 (0.070 m) 2

6.76: a) Equating the work done by the spring to the gain in kinetic energy, 2 2 1 1 2 kx 0 = 2 mv , so v= k 400 N / m x0 = (0.060 m) = 6.93 m / s. m 0.0300 kg

2 b) Wtot must now include friction, so 1 mv 2 = Wtot = 1 kx0 − fx0 , where f is the magnitude 2 2 of the friction force. Then, k 2 2f v= x0 − x0 m m

400 N / m 2(6.00 N) (0.06 m) 2 − (0.06 m) = 4.90 m / s. 0.0300 kg (0.0300 kg) c) The greatest speed occurs when the acceleration (and the net force) are zero, or 6. f kx = f , x = k = 40000 Nm = 0.0150 m . To find the speed, the net work is N/ =
2 Wtot = 1 k ( x0 − x 2 ) − f ( x0 − x) , so the maximum speed is 2

vmax = =

k 2 2f ( x0 − x 2 ) − ( x0 − x) m m 400 N / m 2(6.00 N) ((0.060 m) 2 − (0.0150 m) 2 ) − (0.060 m − 0.0150 m) (0.0300 kg ) (0.0300 kg)

= 5.20 m/s, which is larger than the result of part (b) but smaller than the result of part (a). 6.77: Denote the initial compression of the spring by x and the distance from the initial position by L. Then, the work done by the spring is 1 kx 2 and the work done by friction is 2 − µ k mg ( x + L) ; this form takes into account the fact that while the spring is compressed, the frictional force is still present (see Problem 6.76). The initial and final kinetic energies are both zero, so the net work done is zero, and 1 kx 2 = µ k mg ( x + L) . Solving for 2 L, (1 / 2)kx 2 (1 / 2)(250 N / m)(0.250 m) 2 L= −x= − (0.250 m) = 0.813 m, µ k mg (0.30)(2.50 kg )(9.80 m / s 2 ) or 0.81 m to two figures. Thus the book moves .81 m + .25 m = 1.06 m , or about 1.1 m. 6.78: The work done by gravity is Wg = −mgL sin θ (negative since the cat is moving up), and the work done by the applied force is FL, where F is the magnitude of the applied force. The total work is Wtot = (100 N)(2.00 m) − (7.00 kg )(9.80 m / s 2 )(2.00 m) sin 30° = 131.4 J. The cat’s initial kinetic energy is v2 =
1 2

mv12 = 1 (7.00 kg)(2.40 m / s) 2 = 20.2 J , and 2 2(20.2 J + 131.4 J ) = 6.58 m / s. (7.00 kg)

2( K1 + W ) = m

6.79: In terms of the bumper compression x and the initial speed v0 , the necessary relations are 1 2 1 2 kx = mv0 , kx < 5 mg . 2 2 Combining to eliminate k and then x, the two inequalties are v2 mg 2 x> and k < 25 2 . 5g v a) Using the given numbers, (20.0 m / s) 2 x> = 8.16 m, 5(9.80 m / s 2 ) (1700 kg )(9.80 m / s 2 ) 2 = 1.02 × 10 4 N / m. 2 (20.0 m / s) b) A distance of 8 m is not commonly available as space in which to stop a car. k < 25 6.80: The students do positive work, and the force that they exert makes an angle of 30.0° with the direction of motion. Gravity does negative work, and is at an angle of 60.0° with the chair’s motion, so the total work done is Wtot = ((600 N) cos 30.0° − (85.0 kg )(9.80 m / s 2 ) cos 60.0°)(2.50 m) = 257.8 J , and so the speed at the top of the ramp is 2Wtot 2(257.8 J) v2 = v12 + = (2.00 m / s) 2 + = 3.17 m / s. m (85.0 kg) Note that extra figures were kept in the intermediate calculation to avoid roundoff error. 6.81: a) At maximum compression, the spring (and hence the block) is not moving, so the block has no kinetic energy. Therefore, the work done by the block is equal to its initial kinetic energy, and the maximum compression is found from 1 kX 2 = 1 mv 2 , or 2 2 m v= k b) Solving for v in terms of a known X, k v= X = m X= 5.00 kg (6.00 m s) = 0.600 m. 500 N m 500 N m (0.150 m) = 1.50 m s. 5.00 kg

6.82: The total work done is the sum of that done by gravity (on the hanging block) and that done by friction (on the block on the table). The work done by gravity is (6.00 kg) gh and the work done by friction is − μk (8.00 kg) gh, so Wtot = (6.00 kg − (0.25)(8.00 kg) (9.80 m s 2 ) (1.50 m) = 58.8 J. This work increases the kinetic energy of both blocks; 1 Wtot = (m1 + m2 )v 2 , 2 so 2(58.8 J) v= = 2.90 m s. (14.00 kg) 6.83: See Problem 6.82. Gravity does positive work, while friction does negative work. Setting the net (negative) work equal to the (negative) change in kinetic energy, 1 (m1 − μk m2 ) gh = − ( m1 + m2 )v 2 , 2 and solving for µ k gives μk = = m1 + (1 2) (m1 + m2 )v 2 gh m2

(6.00 kg) + (1 2) (14.00 kg) (0.900 m s) 2 ((9.80 m s 2 ) (2.00 m)) (8.00 kg) = 0.79. 6.84: The arrow will acquire the energy that was used in drawing the bow (i.e., the work done by the archer), which will be the area under the curve that represents the force as a function of distance. One possible way of estimating this work is to approximate the F vs. x curve as a parabola which goes to zero at x = 0 and x = x0 , and has a maximum of F0 at x =
x0 2

, so that F ( x) =

4 F0
2 x0

x( x0 − x). This may seem like a crude approximation to the

figure, but it has the ultimate advantage of being easy to integrate; x0 x 4F 0 4 F  x 2 x3  2 Fdx = 20 ∫ ( x0 x − x 2 ) dx = 20  x0 0 − 0  = F0 x0 . ∫ x0 0 x0  2 3 3   0 With F0 = 200 N and x0 = 0.75 m, W = 100 J. The speed of the arrow is then
2W m

=

2 (100 J) (0.025 kg)

= 89 m s . Other ways of finding the area under the curve in Fig. (6.28)

should give similar results.

6.85: f k = 0.25 mg so W f = Wtot = −(0.25 mg ) s, where s is the length of the rough patch. Wtot = K 2 − K1
2 2 2 2 K1 = 1 mv0 , K 2 = 1 mv2 = 1 m(0.45v0 ) = 0.2025 1 mv0 2 2 2 2

(

)

2 The work-energy relation gives − (0.25mg ) s = (0.2025 − 1) 1 mv0 2 The mass divides out and solving gives s = 1.5 m.

6.86: Your friend’s average acceleration is v − v0 6.00 m/s a= = = 2.00 m/s 2 t 3.00 s Since there are no other horizontal forces acting, the force you exert on her is given by Fnet = ma = (65.0 kg)(2.00 m/s 2 ) = 130 N Her average velocity during your pull is 3.00 m/s, and the distance she travels is thus 9.00 m. The work you do is Fx = (130 N)(9.00 m) = 1170 J , and the average power is therefore 1170 J/3.00 s = 390 W. The work can also be calculated as the change in the kinetic energy. 6.87: a) (800 kg)(9.80 m/s 2 )(14.0 m) = 1.098 × 105 J, or 1.10 × 105 J to three figures. b) (1 / 2)(800 kg)(18.0 m/s 2 ) = 1.30 × 105 J. c) 1.10 × 10 5 J + 1.30 × 10 5 J = 3.99 kW. 60 s

6.88:

P = Fv = mav = m(2α + 6 βt )(2αt + 3 βt 2 ) = m(4α 2t + 18αβt 2 + 18 β 2t 3 ) = (0.96 N/s)t + (0.43 N/s 2 )t 2 + (0.043 N/s3 )t 3 .

At t = 400 s, the power output is 13.5 W. 6.89: Let t equal the number of seconds she walks every day. Then, (280 J/s)t + (100 J/s)(86400 s − t ) = 1.1 × 10 7 J. Solving for t, t = 13,111 s = 3.6 hours.

6.90: a) The hummingbird produces energy at a rate of 0.7 J/s to 1.75 J/s. At 10 beats/s, the bird must expend between 0.07 J/beat and 0.175 J/beat. b) The steady output of the athlete is 500 W/70 kg = 7 W/kg, which is below the 10 W/kg necessary to stay aloft. Though the athlete can expend 1400 W/70 kg = 20 W/kg for short periods of time, no human-powered aircraft could stay aloft for very long. Movies of early attempts at human-powered flight bear out this observation.
d d 6.91: From the chain rule, P = dt W = dt (mgh) = dm gh, for ideal efficiency. Expressing dt the mass rate in terms of the volume rate and solving gives

(2000 × 106 W) m3 = 1.30 × 103 . (0.92)(9.80 m/s 2 )(170 m)(1000 kg/m3 ) s

6.92: a) The power P is related to the speed by Pt = K = 1 mv 2 , so v = 2 b) a) a= dv d = dt dt 2 Pt = m 2P d t= m dt 2P 1 P = .. m 2 t 2mt

2 Pt m

.

x − x0 = ∫ v dt =

2P 1 t 2 dt = m ∫

2P 2 3 8P 2 3 t2 = t2. m 3 9m 3

6.93: a) (7500 × 10−3 kg 3 )(1.05 × 103 kg/m3 )(9.80 m/s 2 )(1.63 m) = 1.26 × 105 J. b) (1.26 × 105 J)/(86,400 s) = 1.46 W.

6.94: a) The number of cars is the total power available divided by the power needed per car, 13.4 × 106 W = 177, (2.8 × 103 N)(27 m/s) rounding down to the nearest integer. b) To accelerate a total mass M at an acceleration a and speed v, the extra power needed is Mav. To climb a hill of angle α , the extra power needed is Mg sin αv. These will be nearly the same if a ~ g sin α ; if g sin α ~ g tan α ~ 0.10 m/s 2 , the power is about the same as that needed to accelerate at 0.10 m/s 2 . c) (1.10 × 106 kg)(9.80 m/s 2 )(0.010)(27 m/s) = 2.9 MW. d) The power per car needed is that used in part (a), plus that found in part (c) with M being the mass of a single car. The total number of cars is then 13.4 × 106 W − 2.9 × 106 W = 36, (2.8 × 103 N + (8.2 × 104 kg)(9.80 m/s 2 )(0.010))(27 m/s) rounding to the nearest integer. 6.95: a) P0 = Fv = (53 × 103 N)(45 m/s) = 2.4 MW. b) P = mav = (9.1 × 105 kg)(1.5 m/s 2 )(45 m/s) = 61 MW. 1 c) Approximating sin α , by tan α , and using the component of gravity down the incline as mg sin α , P2 = (mg sin α )v = (9.1 × 105 kg)(9.80 m/s 2 )(0.015)(45 m/s) = 6.0 MW. 6.96: a) Along this path, y is constant, and the displacement is parallel to the force, so 2 W = αy ∫ xdx = (2.50 N/m 2 )(3.00 m) (2.002m) = 15.0 J. b) Since the force has no y-component, no work is done moving in the y-direction. c) Along this path, y varies with position along the path, given by y = 1.5 x, so Fx = α (1.5 x) x = 1.5αx 2 , and W = ∫ Fx dx = 1.5α ∫ x 2 dx = 1.5(2.50 N/m 2 ) (200 m) 3 = 10.0 J. 3

6.97: a)

P = Fv = ( Froll + Fair )v = ((0.0045)(62.0 kg)(9.80 m/s 2 ) + (1 / 2)(1.00)(0.463 m 2 )(1.2 kg/m 3 )(12.0 m/s) 2 )(12.0 m/s) = 513 W.

6.98: a)

P 28.0 × 103 W F= = = 1.68 × 103 N. v (60.0 km/h)((1 m/s)/(3.6 km/h))

b) The speed is lowered by a factor of one-half, and the resisting force is lowered by a factor of (0.65 + 0.35 / 4), and so the power at the lower speed is

(28.0 kW)(0.50)(0.65 + 0.35/4) = 10.3 kW = 13.8 hp.
c) Similarly, at the higher speed,

(28.0 kW)(2.0)(0.65 + 0.35 × 4) = 114.8 kW = 154 hp.

6.99: a)

(8.00 hp)(746 W/hp) = 358 N. (60.0 km/h)((1 m/s)/(3.6 km/h))

b) The extra power needed is mgv|| = (1800 kg)(9.80 m/s 2 ) 60.6 km/h sin(arctan(1 / 10)) = 29.3 kW = 39.2 hp, 3.6 km/h m/s

so the total power is 47.2 hp. (Note: If the sine of the angle is approximated by the tangent, the third place will be different.) c) Similarly, mgv|| = (1800 kg)(9.80 m/s 2 ) 60.0 km/h sin(arctan(0.010)) = 2.94 kW = 3.94 hp, 3.6 km/h m/s

This is the rate at which work is done on the car by gravity. The engine must do work on the car at a rate of 4.06 hp. d) In this case, approximating the sine of the slope by the tangent is appropriate, and the grade is (8.00 hp)(746 W/hp) = 0.0203, (1800 kg)(9.80 m/s 2 )(60.0 km/h)((1 m/s)/(3.6 km/h)) very close to a 2% grade.

6.100: Use the Work–Energy Theorem, W = ∆KE , and integrate to find the work. 1 2 ∆KE = 0 − mv0 and W = ∫ (−mg sin α − μmg cos α )dx. 2 0 Then,
x   Ax 2 W = −mg ∫ (sin α + Ax cos α )dx, W = −mg sin αx + cos α . 2   0 x

Set W = ∆KE .   1 2 Ax 2 − mv0 = −mg sin αx + cos α . 2 2   To eliminate x, note that the box comes to a rest when the force of static friction balances the component of the weight directed down the plane. So, mg sin α = Ax mg cos α; solve this for x and substitute into the previous equation. sin α x= . A cos α Then,
2    sin α   A     1 2 sin α sin α +  A cos α  cos α , v0 = + g   2 A cos α 2      

2 and upon canceling factors and collecting terms, v0 =

3 g sin 2 α . Or the box will remain A cos α

2 stationary whenever v0 ≥

3 g sin 2 α . A cos α

6.101: a) Denote the position of a piece of the spring by l; l = 0 is the fixed point and l = L is the moving end of the spring. Then the velocity of the point corresponding to l, 1 denoted u, is u (l ) = v L (when the spring is moving, l will be a function of time, and so u is an implicit function of time). The mass of a piece of length dl is dm = M dl , and so L dK = and Mv 2 2 Mv 2 K = ∫ dK = l dl = . 2 L3 ∫ 6 0 b) 1 kx 2 = 1 mv 2 , so v = (k m) x = (3200 N m) (0.053 kg) (2.50 × 10−2 m) = 6.1 m s. . 2 2 c) With the mass of the spring included, the work that the spring does goes into the kinetic energies of both the ball and the spring, so 1 kx 2 = 1 mv 2 + 1 Mv 2 . Solving for v, 2 2 6 v= d) Algebraically, 1 2 (1 2) kx 2 mv = = 0.40 J and 2 (1 + M 3m) 1 (1 2)kx 2 Mv 2 = = 0.60 J. 6 (1 + 3m M ) k (3200 N m) x= (2.50 × 10− 2 m) = 3.9 m s. m+M 3 (0.053 kg) + (0.243 kg) 3
L

1 1 Mv 2 2 dmu 2 = l dl , 2 2 L3

6.102: In both cases, a given amount of fuel represents a given amount of work W0 that the engine does in moving the plane forward against the resisting force. In terms of the range R and the (presumed) constant speed v, β  W0 = RF = Rαv 2 + 2 . v   In terms of the time of flight T , R = vt , so β  W0 = vTF = T αv 3 + . v  a) Rather than solve for R as a function of v, differentiate the first of these relations with respect to v, setting dW0 = 0 to obtain dR F + R dF = 0. For the maximum range, dR = 0, so dv dv dv dv
dF dv

= 0. Performing the differentiation,
14

dF dv

= 2αv − 2 β v 3 = 0, which is solved for
14

 3.5 × 105 N ⋅ m 2 s 2  β v=  =  0.30 N ⋅ s 2 m 2  = 32.9 m s = 118 km h.  α    d b) Similarly, the maximum time is found by setting dv ( Fv ) = 0; performing the differentiation, 3αv 2 − β v 2 = 0, which is solved for  β  v=   3α 
14

 3.5 × 105 N ⋅ m 2 s 2  =  3(0.30 N ⋅ s 2 m 2    

14

= 25 m s = 90 km h.

6.103: a) The walk will take one-fifth of an hour, 12 min. From the graph, the oxygen consumption rate appears to be about 12 cm3 kg ⋅ min, and so the total energy is (12 cm 3 kg ⋅ min) (70 kg) (12 min) (20 J cm 3 ) = 2.0 × 105 J. b) The run will take 6 min. Using an estimation of the rate from the graph of about 33 cm3 kg ⋅ min gives an energy consumption of about 2.8 × 105 J. c) The run takes 4 min, and with an estimated rate of about 50 cm 3 kg ⋅ min, the energy used is about 2.8 × 105 J. d) Walking is the most efficient way to go. In general, the point where the slope of the line from the origin to the point on the graph is the smallest is the most efficient speed; about 5 km h.

  6.104: From F = ma , Fx = max , Fy = ma y and Fz = maz . The generalization of Eq. (6.11) is then dv dv dv a x = vx x , a y = v y y , a z = v z z . dx dy dz The total work is then
Wtot = ∫
( x2 , y 2 , z 2 ) ( x1 , y1 , z1 )

Fx dx + Fy dy + Fz dz

y2 z2 dv  x 2 dv  dv = m ∫ vx x dx + ∫ v y y dy + ∫ vz z dz   x1 y1 z1 dx dy dz    vx 2 vy2 vz 2 = m ∫ vx dvx + ∫ v y dv y + ∫ vz dvz   v  v y1 vz 2  x1  1 2 2 = m(vx 2 − vx1 + v 2 2 − v 21 + vz22 − vz21 y y 2 1 2 1 = mv2 − mv12 . 2 2

Chapter 7

7.1: From Eq. (7.2), mgy = (800 kg) (9.80 m s 2 ) (440 m) = 3.45 × 106 J = 3.45 MJ . 7.2: a) For constant speed, the net force is zero, so the required force is the sack’s weight, (5.00 kg)(9.80 m s 2 ) = 49 N. b) The lifting force acts in the same direction as the sack’s motion, so the work is equal to the weight times the distance, (49.00 N) (15.0 m) = 735 J; this work becomes potential energy. Note that the result is independent of the speed, and that an extra figure was kept in part (b) to avoid roundoff error. 7.3: In Eq. (7.7), taking K1 = 0 (as in Example 6.4) and U 2 = 0, K 2 = U 1 + Wother . Friction does negative work − fy, so K 2 = mgy − fy; solving for the speed v2 , v2 = 2(mg − f ) y = m 2((200 kg) (9.80 m s 2 ) − 60 N) (3.00 m) = 7.55 m s . (200 kg)

7.4: a) The rope makes an angle of arcsin ( 3.0 m ) = 30° with the vertical. The needed 6.0 m horizontal force is then w tan θ = (120 kg) (9.80 m s 2 ) tan 30° = 679 N, or 6.8 × 102 N to two figures. b) In moving the bag, the rope does no work, so the worker does an amount of work equal to the change in potential energy, (120 kg) (9.80 m s 2 ) (6.0 m) (1 − cos 30°) = 0.95 × 103 J. Note that this is not the product of the result of part (a) and the horizontal displacement; the force needed to keep the bag in equilibrium varies as the angle is changed. 7.5: a) In the absence of air resistance, Eq. (7.5) is applicable. With y1 − y2 = 22.0 m, solving for v2 gives v2 = v12 + 2 g ( y2 − y1 ) = (12.0 m s) 2 + 2(9.80 m s 2 )(22.0 m) = 24.0 m s. b) The result of part (a), and any application of Eq. (7.5), depends only on the magnitude of the velocities, not the directions, so the speed is again 24.0 m s. c) The ball thrown upward would be in the air for a longer time and would be slowed more by air resistance.

7.6: a) (Denote the top of the ramp as point 2.) In Eq. (7.7), K 2 = 0, Wother = −(35 N) × (2.5 m) = −87.5 J, and taking U1 = 0 and U 2 = mgy2 = (12 kg) (9.80 m s 2 ) (2.5 m sin 30°) = 147 J, v1 =
2 (147 J + 87.5 J) 12 kg

= 6.25 m s, or

6.3 m s to two figures. Or, the work done by friction and the change in potential energy are both proportional to the distance the crate moves up the ramp, and so the initial speed 2.5 m is proportional to the square root of the distance up the ramp; (5.0 m s) 1.6 m = 6.25 m s . b) In part a), we calculated Wother and U 2 . Using Eq. (7.7), K 2 = 1 (12 kg) (11.0 m s) 2 − 87.5 J − 147 J = 491.5 J 2 v2 =
2 K2 m

=

2 ( 491 .5 J) (12 kg)

= 9.05 m s .

7.7:

As in Example 7.7, K 2 = 0, U 2 = 94 J, and U 3 = 0. The work done by friction is

− (35 N) (1.6 m) = −56 J, and so K 3 = 38 J, and v3 =

2 ( 38 J) 12 Kg

= 2.5 m s .

7.8: The speed is v and the kinetic energy is 4K. The work done by friction is proportional to the normal force, and hence the mass, and so each term in Eq. (7.7) is proportional to the total mass of the crate, and the speed at the bottom is the same for any mass. The kinetic energy is proportional to the mass, and for the same speed but four times the mass, the kinetic energy is quadrupled. 7.9: In Eq. (7.7), K1 = 0, Wother is given as − 0.22 J, and taking U 2 = 0, K 2 = mgR − 0.22 J, so  0.22 J   = 2.8 m/s. v2 = 2 (9.80 m s 2 ) (0.50 m) −  0.20 kg   

7.10: (a) The flea leaves the ground with an upward velocity of 1.3 m/s and then is in free-fall with acceleration 9.8 m s 2 downward. The maximum height it reaches is 2 2 therefore (v y − v0 y ) 2( − g ) = 9.0 cm. The distance it travels in the first 1.25 ms can be ignored. (b) 1 W = KE = mv 2 2 1 = (210 × 10 −6 g) (130 cm s) 2 2 = 1.8 ergs = 1.8 × 10 −7 J 7.11: Take y = 0 at point A. Let point 1 be A and point 2 be B. K1 + U1 + Wother = K 2 + U 2 U1 = 0, U 2 = mg (2 R) = 28,224 J, Wother = W f 1 2 2 mv1 = 37,500 J, K 2 = 1 mv2 = 3840 J 2 2 The work - energy relation then gives W f = K 2 + U 2 − K1 = −5400 J. K1 = 7.12: Tarzan is lower than his original height by a distance l (cos 30 − cos 45), so his speed is v = 2 gl (cos 30° − cos 45°) = 7.9 m s , a bit quick for conversation.

7.13: a) The force is applied parallel to the ramp, and hence parallel to the oven’s  motion, and so W = Fs = (110 N) (8.0 m) = 880 J. b) Because the applied force F is parallel to the ramp, the normal force is just that needed to balance the component of the weight perpendicular to the ramp, n = w cos α, and so the friction force is f k = µ k mg cos α and the work done by friction is Wf = − µ k mg cos α s = −(0.25) (10.0 kg) (9.80 m s 2 ) cos 37°(8.0 m) = −157 J, keeping an extra figure. c) mgs sin α = (10.0 kg)(9.80 m s 2 )(8.0 m) sin 37° = 472 J , again keeping an extra figure. d) 880 J − 472 J − 157 J = 251 J. e) In the direction up the ramp, the net force is F − mg sin α − µ k mg cos α = 110 N − (10.0 kg)(9.80 m s 2 )(sin 37° + (0.25) cos 37°) = 31.46 N, so the acceleration is (31.46 N) 10.0 kg) = 3.15 m s 2 . The speed after moving up the ramp is v = 2as = 2(3.15 m s 2 ) (8.0 m) = 7.09 m s , and the kinetic energy is (1 2)mv 2 = 252 J. (In the above, numerical results of specific parts may differ in the third place if extra figures are not kept in the intermediate calculations.) 7.14: a) At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the bottom of the circular arc) is mgl (1 − cos θ ), where l is the length of the string and θ is the angle the string makes with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so mgl (1 − cos θ ) = 1 mv 2 , or 2 v = 2 gl (1 − cos θ ) = 2(9.80 m s 2 ) (0.80 m) (1 − cos 45°) = 2.1 m s . b) At 45° from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the radial component of the weight, or mg cos θ = (0.12 kg) (9.80 m s 2 ) cos 45° = 0.83 N. c) At the bottom of the circle, the tension is the sum of the weight and the radial acceleration, 2 mg + mv2 l = mg (1 + 2(1 − cos 45°)) = 1.86 N, or 1.9 N to two figures. Note that this method does not use the intermediate calculation of v. 7.15: Of the many ways to find energy in a spring in terms of the force and the distance, one way (which avoids the intermediate calculation of the spring constant) is to note that the energy is the product of the average force and the distance compressed or extended. a) (1 2)(800 N)(0.200 m) = 80.0 J. b) The potential energy is proportional to the square of the compression or extension; (80.0 J) (0.050 m 0.200 m) 2 = 5.0 J.

7.16:

U = 1 ky 2 , where y is the vertical distance the spring is stretched when the weight 2

w = mg is suspended. y = mg , and k = F , where x and F are the quantities that k x “calibrate” the spring. Combining, 1 ( mg ) 2 1 ((60.0 kg) (9.80 m s 2 )) 2 U= = = 36.0 J 2 F x 2 (720 N 0.150 m)

7.17:

a) Solving Eq. (7.9) for x, x =

2U k

=

2 ( 3.20 J) (1600 N m)

= 0.063 m.

b) Denote the initial height of the book as h and the maximum compression of the spring by x. The final and initial kinetic energies are zero, and the book is initially a height x + h above the point where the spring is maximally compressed. Equating initial and final potential energies, 1 kx 2 = mg ( x + h). This is a quadratic in x, the solution to 2 which is mg  2kh  x= 1 ± 1 +  k  mg  (1.20 kg)(9.80 m s 2 )  2(1600 N m) (0.80 m)  = 1 ± 1 +  (1600 N m) (1.20 kg) (9.80 m s 2 )   = 0.116 m, − 0.101 m. The second (negative) root is not unphysical, but represents an extension rather than a compression of the spring. To two figures, the compression is 0.12 m. 7.18: a) In going from rest in the slingshot’s pocket to rest at the maximum height, the potential energy stored in the rubber band is converted to gravitational potential energy; U = mgy = (10 × 10−3 kg)(9.80 m s 2 ) (22.0 m) = 2.16 J. b) Because gravitational potential energy is proportional to mass, the larger pebble rises only 8.8 m. c) The lack of air resistance and no deformation of the rubber band are two possible assumptions. 7.19: The initial kinetic energy and the kinetic energy of the brick at its greatest height are both zero. Equating initial and final potential energies, 1 kx 2 = mgh, where h is the 2 greatest height. Solving for h, kx 2 (1800 N m) (0.15 m) 2 h= = = 1.7 m. 2mg 2(1.20 kg) (9.80 m s 2 )

7.20:

As in Example 7.8, K1 = 0 and U1 = 0.0250 J. For v2 = 0.20 m s,
2 ( 0.0210 J) 5.00 N m

K 2 = 0.0040 J, so U 2 = 0.0210 J = 1 kx 2 , so x = ± 2

= ±0.092 m. In the absence

of friction, the glider will go through the equilibrium position and pass through x = −0.092 m with the same speed, on the opposite side of the equilibrium position. 7.21: v2 = a) In this situation, U 2 = 0 when x = 0, so K 2 = 0.0250 J and
2 ( 0.0250 J) 0.200 kg

= 0.500 m s. b) If v2 = 2.50 m s,
2 ( 0.625 J) 5.00 N m

K 2 = (1 2) (0.200 kg)(2.50 m s) 2 = 0.625 J = U 1 , so x1 =

= 0.500 m. Or,

because the speed is 5 times that of part (a), the kinetic energy is 25 times that of part (a), and the initial extension is 5 × 0.100 m = 0.500 m. 7.22: a) The work done by friction is Wother = − μk mg∆x = −(0.05) (0.200 kg) (9.80 m s 2 ) (0.020 m) = −0.00196 J, so K 2 = 0.00704 J and v2 =
2 ( 0.00704 J) 0.200 kg

= 0.27 m s . b) In this case Wother = −0.0098 J, so
2 ( 0.0152 J) 0.200 kg

K 2 = 0.0250 J − 0.0098 J = 0.0152 J, and v2 =

= 0.39 m s.

c) In this case, K 2 = 0, U 2 = 0, so U1 + Wother = 0 = 0.0250 J − μk (0.200 kg) (9.80 m s 2 ) × (0.100 m), or μ k = 0.13. 7.23: a) In this case, K1 = 625,000 J as before, Wother = −17,000 J and
2 U 2 = (1 2)ky2 + mgy 2

= (1 2)(1.41 × 105 N m) (−1.00 m) 2 + (2000 kg) (9.80 m s 2 ) (−1.00) = 50,900 J. The kinetic energy is then K 2 = 625,000 J − 50,900 J − 17,000 J = 557,100 J , corresponding to a speed v2 = 23.6 m s. b) The elevator is moving down, so the friction force is up (tending to stop the elevator, which is the idea). The net upward force is then 2 − mg + f − kx = −(2000 kg)(9.80 m s ) + 17,000 N − (1.41 × 10 5 N m)(−1.00 m) = 138,400 N, for an upward acceleration of 69.2 m s .
2

7.24:

From 1 kx 2 = 1 mv 2 , the relations between m, v, k and x are 2 2 kx 2 = mv 2 , kx = 5mg .
v2 5g

Dividing the first by the second gives x = k = 25 mg2 , so a) & b), v x=
2

, and substituting this into the second gives

(2.50 m s) 2 = 0.128 m, 2 5(9.80 m s )
2

(1160 kg )(9.80 m s ) 2 k = 25 = 4.46 × 10 5 N m . 2 (2.50 m s)

7.25: a) Gravity does negative work, − (0.75 kg )(9.80 m s )(16 m) = −118 J. b) Gravity does 118 J of positive work. c) Zero d) Conservative; gravity does no net work on any complete round trip. a) & b) − (0.050 kg)(9.80 m s )(5.0 m) = −2.5 J.
2

2

7.26:

c) Gravity is conservative, as the work done to go from one point to another is pathindependent.

 7.27: a) The displacement is in the y-direction, and since F has no y-component, the work is zero. b)  P2  x2 12 N/m 2 3 F ⋅ dl = −12 ∫ x 2 dx = − ( x2 − x13 ) = −0.104 J. ∫P1 x1 3 c) The negative of the answer to part (b), 0.104 m3 d) The work is independent of path, and the force is conservative. The corresponding potential energy is 2 3 2 U = (12 N 3m ) x = (4 N m ) x 3 .

 7.28: a) From (0, 0) to (0, L), x = 0 and so F = 0 , , and the work is zero. From (0, L) to     (L, L), F and dl are perpendicular, so F ⋅ dl = 0. and the net work along this path is   zero. b) From (0, 0) to (L, 0), F ⋅ dl = 0. From (L, 0) to (L, L), the work is that found in the example, W2 = CL2 , so the total work along the path is CL2 . c) Along the diagonal   2 path, x = y, and so F ⋅ dl = Cy dy ; integrating from 0 to L gives CL . (It is not a 2 coincidence that this is the average to the answers to parts (a) and (b).) d) The work depends on path, and the field is not conservative. 7.29: a) When the book moves to the left, the friction force is to the right, and the work is − (1.2 N)(3.0 m) = −3.6 J. b) The friction force is now to the left, and the work is again − 3.6 J. c) − 7.2 J. d) The net work done by friction for the round trip is not zero, and friction is not a conservative force. 7.30: The friction force has magnitude µ k mg = (0.20)(30.0 kg )(9.80 m/s 2 ) = 58.8 N. a) For each part of the move, friction does − (58.8 N)(10.6 m) = −623 J, so the total work done by friction is − 1.2 kN. b) − (58.8 N)(15.0 m) = −882 N. 7.31: The magnitude of the friction force on the book is 2 µ k mg = (0.25)(1.5 kg )(9.80 m s ) = 3.68 N. a) The work done during each part of the motion is the same, and the total work done is − 2(3.68 N)(8.0 m) = −59 J (rounding to two places). b) The magnitude of the displacement is 2 (8.0 m) , so the work done by friction is − 2 (8.0 m)(3.68 N) = −42 N. c) The work is the same both coming and going, and the total work done is the same as in part (a), − 59 J. d) The work required to go from one point to another is not path independent, and the work required for a round trip is not zero, so friction is not a conservative force.
2 2 a) 1 k ( x12 − x2 ) b) − 1 k ( x12 − x2 ). The total work is zero; the spring force is 2 2

7.32:

2 2 2 conservative c) From x1 to x3 , W = − 1 k ( x3 − x12 ). From x3 to x2 , W = 1 k ( x2 − x3 ). 2 2 2 The net work is − 1 k ( x2 − x12 ). This is the same as the result of part (a). 2

7.33:

From Eq. (7.17), the force is Fx = −

6C dU d  1 = C6  6  = − 76 . dx dx  x  x The minus sign means that the force is attractive.

7.34:

From Eq. (7.15), Fx = − dU = −4αx 3 = −(4.8 J m ) x 3 , and so dx
4

Fx ( −0.800 m) = −(4.8 J m )(−0.80 m) 3 = 2.46 N.
∂U ∂x

4

7.35:

= 2kx + k ′y,

∂U ∂y

= 2ky + k ′x and ∂∂U = 0 , so from Eq. (7.19), z  ˆ F = −(2kx + k ′y )i − (2ky + k ′x) ˆ. j

7.36:
∂U ∂x

 ˆ From Eq. (7.19), F = − ∂∂U i − ∂∂U ˆ, since U has no z-dependence. x y j
−2α x3

=

and

∂U ∂y

=

−2α y3

, so
 −2 ˆ −2 F = −α  3 i + 3 x y  ˆ . j  

7.37:

a) Fr = − ∂∂U = 12 ra − 6 rb7 . 13 r

b) Setting Fr = 0 and solving for r gives rmin = (2a b)1 / 6 . This is the minimum of potential energy, so the equilibrium is stable. c) a b U (rmin ) = 12 − 6 rmin rmin = = a b − 1 / 6 12 ((2a / b) ) ((2a / b)1 / 6 )6

ab 2 b 2 b2 − =− . 4a 2 2a 4a To separate the particles means to remove them to zero potential energy, and requires the negative of this, or E0 = b 2 4a . d) The expressions for E0 and rmin in terms of a and b are b2 6 2a E0 = rmin = . 4a b 6 Multiplying the first by the second and solving for b gives b = 2 E0 rmin , and substituting
12 this into the first and solving for a gives a = E0 rmin . Using the given numbers,

a = (1.54 × 10 −18 J)(1.13 × 10 −10 m)12 = 6.68 × 10 −138 J ⋅ m12 b = 2(1.54 × 10 −18 J)(1.13 × 10 −10 m) 6 = 6.41 × 10 −78 J ⋅ m 6 . (Note: the numerical value for a might not be within the range of standard calculators, and the powers of ten may have to be handled seperately.) 7.38: a) Considering only forces in the x-direction, Fx = − dU , and so the force is zero dx when the slope of the U vs x graph is zero, at points b and d. b) Point b is at a potential minimum; to move it away from b would require an input of energy, so this point is stable. c) Moving away from point d involves a decrease of potential energy, hence an increase in kinetic energy, and the marble tends to move further away, and so d is an unstable point.

7.39: a) At constant speed, the upward force of the three ropes must balance the force, so the tension in each is one-third of the man’s weight. The tension in the rope is the 2 force he exerts, or (70.0 kg )(9.80 m s ) 3 = 229 N. b) The man has risen 1.20 m, and so the increase in his potential energy is (70.0 kg )(9.80 m s )(1.20 m) = 823 J. In moving up a given distance, the total length of the rope between the pulleys and the platform changes by three times this distance, so the length of rope that passes through the man’s hands is 3 × 1.20 m = 3.60 m, and (229 N)(3.6 m) = 824 J. First find the acceleration: 2 v 2 − v0 (3.00 m s) 2 2 a= = = 3.75 m s 2( x − x0 ) 2(1.20 m) Then, choosing motion in the direction of the more massive block as positive: Fnet = Mg − mg = ( M + m)a = Ma + ma M ( g − a ) = m( g + a ) M g + a (9.80 + 3.75) m s = = = 2.24 m g − a (9.80 − 3.75) m s 2 M = 2.24 m Since M + m = 15.0 kg : 2.24m + m = 15.0 kg m = 4.63 kg M = 15.0 kg − 4.63 kg = 10.4 kg a) K1 + U1 + Wother = K 2 + U 2
2 2

7.40:

7.41:

U1 = U 2 = K 2 = 0 Wother = W f = − µ k mgs, with s = 280 ft = 85.3 m The work-energy expression gives 1 mv12 − µ k mgs = 0 2 v1 = 2 µ k gs = 22.4 m s = 50 mph ; the driver was speeding. a) 15 mph over speed limit so \$150 ticket.

7.42: a) Equating the potential energy stored in the spring to the block's kinetic energy, 1 kx 2 = 1 mv 2 , or 2 2 k 400 N m x= (0.220 m) = 3.11 m s . m 2.00 kg b) Using energy methods directly, the initial potential energy of the spring is the final gravitational potential energy, 1 kx 2 = mgL sin θ , or 2 v=
2 1 kx 2 2 ( 400 N m )(0.220 m) L= = = 0.821 m. mg sin θ (2.00 kg )(9.80 m s 2 ) sin 37.0° 1 2

7.43: The initial and final kinetic energies are both zero, so the work done by the spring is the negative of the work done by friction, or 1 kx 2 = μk mgl, where l is the distance the 2 block moves. Solving for μk ,

µk =

(1 / 2)kx 2 (1 / 2)(100 N m)(0.20 m) 2 = = 0.41. 2 mgl (0.50 kg)(9.80 m s )(1.00 m)

7.44:

Work done by friction against the crate brings it to a halt: f k x = potential energy of compressed spring fk =

360 J = 64.29 N 5.60 m The friction force working over a 2.00-m distance does work f k x = (−64.29 N)(2.00 m) = −128.6 J. The kinetic energy of the crate at this point is thus 360 J − 128.6 J = 231.4 J, and its speed is found from mv 2 = 231.4 J 2 2 2(231.4 J) v2 = = 9.256 m 2 s 50.0 kg v = 3.04 m s

7.45: a) mgh = (0.650 kg )(9.80 m s )(2.50 m) = 15.9 J b) The second height is 0.75(2.50 m) = 1.875 m, so second mgh = 11.9 J ; loses 15.9 J − 11.9 J = 4.0 J on first bounce. This energy is converted to thermal energy. a) The third height is 0.75(1.875 m) = 1.40 m, , so third mgh = 8.9 J ; loses 11.9 J − 8.9 J = 3.0 J on second bounce.

2

7.46:

2 a) U A − U B = mg ( h − 2 R ) = 1 mv A . From previous considerations, the speed at the 2

top must be at least

gR . Thus, mg (h − 2 R ) > 1 5 mgR, or h > R. 2 2
2

b) U A − U C = (2.50) Rmg = K C , so
2

vC = (5.00) gR = (5.00)(9.80 m s )(20.0 m) = 31.3 m s .
C The radial acceleration is arad = vR = 49.0 m s . The tangential direction is down, the normal force at point C is horizontal, there is no friction, so the only downward force is 2 gravity, and a tan = g = 9.80 m s .

2

7.47: a) Use work-energy relation to find the kinetic energy of the wood as it enters the rough bottom: U1 = K 2 gives K 2 = mgy1 = 78.4 J. Now apply work-energy relation to the motion along the rough bottom: K1 + U1 + Wother = K 2 + U 2 Wother = W f = − µ k mgs, K 2 = U 1 = U 2 = 0 ; K1 = 78.4 J 78.4 J − µ k mgs = 0 ; solving for s gives s = 20.0 m. The wood stops after traveling 20.0 m along the rough bottom. b) Friction does − 78.4 J of work.

7.48:

(a)

KE Bottom + Wf = PETop

1 2 mv0 − µ k mg cos θ d = mgh 2 d = h sin θ 1 2 h v0 − µ k g cos θ = gh 2 sin θ 1 2 cos 40° 2 (15 m s) 2 − (0.20)(9.8 m s ) h = (9.8 m s ) h 2 sin 40° h = 9.3 m (b) Compare maximum static friction force to the weight component down the plane. 2 f s = µ s mg cos θ = (0.75)(28 kg)(9.8 m s ) cos 40° = 158 N mg sin θ = (28 kg)(9.8 m s )(sin 40°) = 176 N > f s so the rock will slide down. (c) Use same procedure as (a), with h = 9.3 m PETop + Wf = KE Bottom
2

mgh − µ k mg cos θ

h 1 2 = mvB sin θ 2 vB = 2 gh − 2µ k gh cos θ sin θ = 11.8 m s

7.49: a) K1 + U1 + Wother = K 2 + U 2 Let point 1 be point A and point 2 be point B. Take y = 0 at point B.
2 mgy1 + 1 mv12 = 1 mv2 , with h = 20.0 m and v1 = 10.0 m s 2 2

v2 = v12 + 2 gh = 22.2 m s b) Use K1 + U1 + Wother = K 2 + U 2 , with point 1 at B and point 2 where the spring has its maximum compression x. U 1 = U 2 = K 2 = 0 ; K1 = 1 mv12 with v1 = 22.2 m s 2 Wother = W f + Wel = − µ k mgs − 1 kx 2 , with s = 100 m + x 2 The work-energy relation gives K1 + Wother = 0.
1 2

mv12 − µ k mgs − 1 kx 2 = 0 2

Putting in the numerical values gives x 2 + 29.4 x − 750 = 0. The positive root to this equation is x = 16.4 m. b) When the spring is compressed x = 16.4 m the force it exerts on the stone is Fel = kx = 32.8 N. The maximum possible static friction force is max f s = µ s mg = (0.80)(15.0 kg )(9.80 m s ) = 118 N. The spring force is less than the maximum possible static friction force so the stone remains at rest.
2

First get speed at the top of the hill for the block to clear the pit. 1 y = gt 2 2 1 2 20 m = (9.8 m s )t 2 2 t = 2.0 s 40 m vTop t = 40 m → vTop = = 20 m s 20 s Energy conservation: KE Bottom = PETop + KETop 1 2 1 2 mvB = mgh + mvT 2 2
2 vB = vT + 2 gh

7.50:

= (20 m s) 2 + 2(9.8 m s )(70 m) = 42 m s

2

7.51: K1 + U1 + Wother = K 2 + U 2 Point 1 is where he steps off the platform and point 2 is where he is stopped by the cord. Let y = 0 at point 2. y1 = 41.0 m. Wother = − 1 kx 2 , where x = 11.0 m is the amount 2 the cord is stretched at point 2.The cord does negative work. K1 = K 2 = U 2 = 0 , so mgy1 − 1 kx 2 = 0 and k = 631 N/m. 2 Now apply F = kx to the test pulls: F = kx so x = F k = 0.602 m.

7.52: For the skier to be moving at no more than 30.0 m s ; his kinetic energy at the bottom of the ramp can be no bigger than mv 2 (85.0 kg )(30.0 m s) 2 = = 38,250 J 2 2 Friction does − 4000 J of work on him during his run, which means his combined PE and KE at the top of the ramp must be no more than 38,250 J + 4000 J = 42,250 J. His KE at the top is mv 2 (85.0 kg)(2.0 m s) 2 = = 170 J 2 2 His PE at the top should thus be no more than 42,250 J − 170 J = 42,080 J, which gives a height above the bottom of the ramp of 42,080 J 42,080 J h= = = 50.5 m. 2 mg (85.0 kg)(9.80 m s ) 7.53: The net work done during the trip down the barrel is the sum of the energy stored in the spring, the (negative) work done by friction and the (negative) work done by gravity. Using 1 kx 2 = 1 ( F 2 k ) , the performer’s kinetic energy at the top of the barrel is 2 2 K= 1 (4400 N) 2 2 − (40 N)(4.0 m) − (60 kg)(9.80 m s )(2.5 m) = 7.17 × 10 3 J, 2 1100 N m
2 ( 7.17×103 J ) 60 kg

and his speed is

= 15.5 m s .

7.54: To be at equilibrium at the bottom, with the spring compressed a distance x0 , the spring force must balance the component of the weight down the ramp plus the largest value of the static friction, or kx0 = w sin θ + f . The work-energy theorem requires that the energy stored in the spring is equal to the sum of the work done by friction, the work done by gravity and the initial kinetic energy, or 1 2 1 kx0 = ( w sin θ − f ) L + mv 2 , 2 2 where L is the total length traveled down the ramp and v is the speed at the top of the 2 ramp. With the given parameters, 1 kx0 = 248 J and kx0 = 1.10 × 103 N. Solving for k 2 gives k = 2440 N m . 7.55: The potential energy has decreased by 2 2 (12.0 kg)(9.80 m s )(2.00 m) − (4.0 kg) × (9.80 m s )(2.00 m) = 156.8 J. The kinetic 1 energy of the masses is then ( m1 + m2 )v 2 = (8.0 kg)v 2 = 156.8 J, so the common speed is 2 v=
(156 .8 J) 8.0 kg

= 4.43 m s , or 4.4 m s to two figures.

7.56: a) The energy stored may be found directly from 1 2 ky2 = K1 + Wother − mgy2 = 625,000 J − 51,000 J − (−58,000 J ) = 6.33 × 105 J. 2 b) Denote the upward distance from point 2 by h. The kinetic energy at point 2 and at the height h are both zero, so the energy found in part (a) is equal to the negative of the work done by gravity and friction, 2 − (mg + f )h = −((2000 kg)(9.80 m s ) + 17,000 N)h = (36,600 N)h, so h=
6.33×10 5 J 3.66 ×10 4 J

= 17.3 m. c) The net work done on the elevator between the highest point of

the rebound and the point where it next reaches the spring is (mg − f )(h − 3.00 m ) = 3.72 × 104 J. Note that on the way down, friction does negative work. The speed of the elevator is then
2 ( 3.72 × 10 4 J ) 2000 kg

= 6.10 m s . d) When the elevator

next comes to rest, the total work done by the spring, friction, and gravity must be the negative of the kinetic energy K 3 found in part (c), or 1 2 2 K 3 = 3.72 × 10 4 J = −(mg − f ) x3 + kx3 = −(2,600 N) x3 + (7.03 × 10 4 N m) x3 . 2 (In this calculation, the value of k was recalculated to obtain better precision.) This is a quadratic in x3 , the positive solution to which is x3 = 1 2(7.03 × 10 4 N m) × 2.60 × 10 3 N + (2.60 × 10 3 N) 2 + 4(7.03 × 10 4 N m)(3.72 × 10 4 J ) = 0.746 m,

[

]

corresponding to a force of 1.05 × 105 N and a stored energy of 3.91 × 104 J . It should be noted that different ways of rounding the numbers in the intermediate calculations may give different answers.

7.57: The two design conditions are expressed algebraically as ky = f + mg = 3.66 × 104 N (the condition that the elevator remains at rest when the spring is compressed a distance y; y will be taken as positive) and 2 2 1 1 (the condition that the change in energy is the work 2 mv + mgy − fy = 2 kx Wother = − fy ). Eliminating y in favor of k by y =
3.66 ×10 4 N k

1 (3.66 × 104 N ) 2 (1.70 × 104 N )(3.66 × 104 N ) + 2 k k (1.96 × 104 N)(3.66 × 104 N) = 62.5 × 104 J + . k This is actually not hard to solve for k = 919 N m , and the corresponding x is 39.8 m. This is a very weak spring constant, and would require a space below the operating range of the elevator about four floors deep, which is not reasonable. b) At the lowest point, the spring exerts an upward force of magnitude f + mg . Just before the elevator stops, however, the friction force is also directed upward, so the net force is ( f + mg ) + f − mg = 2 f , and the upward acceleration is 2mf = 17.0 m s2 . 7.58: One mass rises while the other falls, so the net loss of potential energy is (0.5000 kg − 0.2000 kg )(9.80 m s 2 )(0.400 m) = 1.176 J. This is the sum of the kinetic energies of the animals. If the animals are equidistant from the center, they have the same speed, so the kinetic energy of the combination is 1 mtot v 2 , 2 and 2(1.176 J) v= = 1.83 m s . (0.7000 kg ) 7.59: a) The kinetic energy of the potato is the work done by gravity (or the potential
2

energy lost), 1 mv 2 = mgl , or v = 2 gl = 2(9.80 m s )(2.50 m ) = 7.00 m s . 2 b) v2 T − mg = m = 2mg , l so T = 3mg = 3(0.100 kg)(9.80 m s ) = 2.94 N.
2

7.60:

a) The change in total energy is the work done by the air, 1 2  ( K 2 + U 2 ) − ( K1 + U1 ) = m (v2 − v12 ) + gy 2  2   (1 2) ((18.6 m s) 2 − (30.0 m s) 2   = (0.145 kg)  − (40.0 m s) 2 ) + (9.80 m s 2 )(53.6 m)    = −80.0 J.

b) Similarly,  (1 2)((11.9 m s) 2 + ( −28.7 m s) 2   ( K 3 + U 3 ) − ( K 2 + U 2 ) = (0.145 kg)  − (18.6 m s) 2 ) − (9.80 m s 2 )(53.6 m )    = −31.3 J. c) The ball is moving slower on the way down, and does not go as far (in the x-direction), and so the work done by the air is smaller in magnitude. 7.61: a) For a friction force f, the total work done sliding down the pole is mgd − fd . This is given as being equal to mgh, and solving for f gives (d − h) h  f = mg = mg 1 −  . d  d When h = d , f = 0 , as expected, and when h = 0, f = mg ; there is no net force on the fireman. b) (75 kg)(9.80 m s2 )(1 − 1..0 m ) = 441 N . c) The net work done is 25 m

( mg − f )( d − y ) , and this must be equal to 1 mv 2 . Using the above expression for f, 2
1 2 mv = ( mg − f )(d − y ) 2 h = mg  (d − y ) d  y  = mgh 1 − ,  d from which v = 2 gh (1 − y d ) . When y = 0 v = 2 gh , which is the original condition. When y = d , v = 0 ; the fireman is at the top of the pole.

7.62: a) The skier’s kinetic energy at the bottom can be found from the potential energy at the top minus the work done by friction, K1 = mgh − WF = (60.0 kg )(9.8 N kg)(65.0 m) − 10.500 J, or K1 = 38,200 J − 10,500 J = 27,720 J . Then v1 =
2K m

=

2 ( 27 , 7 20 J ) 60 kg

= 30.4 m s .

b) K 2 + K1 − (WF + WA ) = 27,720 J − (µk mgd + f air d ), K 2 = 27,720 J − [(.2)(588 N) ×

(82 m) + (160 N)(82 m)], , or K 2 = 27,720 J − 22,763 J = 4957 J . Then,
v2 = 2K = m 2(4957 J ) = 12.85 m s ≈ 12.9 m s . 60 kg

c) Use the Work-Energy Theorem to find the force. W = ∆KE , F = KE d = (4957 J ) (2.5 m) = 1983 N ≈ 2000 N. 7.63: The skier is subject to both gravity and a normal force; it is the normal force that causes her to go in a circle, and when she leaves the hill, the normal force vanishes. The vanishing of the normal force is the condition that determines when she will leave the hill. As the normal force approaches zero, the necessary (inward) radial force is the radial component of gravity, or mv 2 R = mg cos α , where R is the radius of the snowball. The speed is found from conservation of energy; at an angle α , she has descended a vertical distance R(1 − cos α ) , so 1 mv 2 = mgR(1 − cos α) , or v 2 = 2 gR (1 − cos α ) . Using this in 2 2 the previous relation gives 2(1 − cos α) = cos α , or α = arccos  = 48.2° . This result 3 does not depend on the skier’s mass, the radius of the snowball, or g. 7.64: If the speed of the rock at the top is vt, then conservation of energy gives the 2 speed vb from 1 mvb = 1 mvt2 + mg (2 R ) , R being the radius of the circle, and so 2 2
2 vb = vt2 + 4 gR . The tension at the top and bottom are found from Tt + mg =

mvt2 R

and

Tb − mg =

2 mvb

R

, so Tb − Tt =

m R

2 (vb − vt2 ) + 2mg = 6mg = 6 w .

7.65: a) The magnitude of the work done by friction is the kinetic energy of the 2 package at point B, or µ k mgL = 1 mv B , or 2

µk =
b)

2 (1 2)v B (1 2)(4.80 m s) 2 = = 0.392. gL (9.80 m s 2 )(3.00 m)

Wother = K B − U A = 1 (0.200 kg )(4.80 m s) 2 − (0.200 kg )(9.80 m s )(1.60 m) 2 = −0.832 J.
2

Equivalently, since K A = K B = 0, U A + W AB + WBC = 0 , or WAB = −U A − WBC = mg (−(1.60 m) − (0.300)(−3.00 m)) = −0.832 J.
2 7.66: Denote the distance the truck moves up the ramp by x. K1 = 1 mv0 , 2 U 1 = mgL sin α , K 2 = 0 , U 2 = mgx sin β and Wother = − µ r mgx cos β . From Wother = ( K 2 + U 2 ) − ( K1 + U 1 ) , and solving for x,

x=

(v 2 2 g ) + L sin α K1 + mgL sin α = 0 . mg (sin β + µr cos β) sin β + µr cos β

7.67: a) Taking U (0) = 0 , U ( x) = ∫ Fx dx =
0 x

α 2 β 3 2 x + x = (30.0 N m) x 2 + (6.00 N m ) x 3 . 2 3

b)

K 2 = U1 − U 2 = ((30.0 N m)(1.00 m) 2 + (6.00 N m )(1.00 m) 3 ) − ((30.0 N m)(0.50 m) 2 + (6.00 N m )(0.50 m) 3 ) = 27.75 J,
2 2

and so v2 =

2 ( 27.75 J ) 0.900 kg

= 7.85 m s .

7.68: The force increases both the gravitational potential energy of the block and the potential energy of the spring. If the block is moved slowly, the kinetic energy can be taken as constant, so the work done by the force is the increase in potential energy, ∆U = mga sin θ + 1 k (aθ ) 2 . 2

7.69: v2 =

2 With U 2 = 0, K1 = 0, K 2 = 1 mv2 = U 1 = 1 kx 2 + mgh , and solving for v2 , 2 2

kx 2 + 2 gh = m

(1900 N m)(0.045 m) 2 2 + 2(9.80 m s )(1.20 m) = 7.01 m s (0.150 kg)

7.70: a) In this problem, use of algebra avoids the intermediate calculation of the spring constant k. If the original height is h and the maximum compression of the spring is d, then mg (h + d ) = 1 kd 2 . The speed needed is when the spring is compressed d , and from 2 2 conservation of energy, mg (h + d 2) − 1 k (d 2) 2 = 1 mv 2 . Substituting for k in terms of 2 2 h+d , d  mg (h + d ) 1 2  mg  h +  − = mv , 2 4 2  which simplifies to 1  3 v 2 = 2 g  h + d . 4  4 Insertion of numerical values gives v = 6.14 m s . b) If the spring is compressed a distance x, 1 kx 2 = mgx , or x = 2 mg . Using the expression from part (a) that gives k in 2 k terms of h and d, d2 d2 x = (2mg ) = = 0.0210 m. 2mg (h + d ) h + d 7.71: The first condition, that the maximum height above the release point is h, is expressed as 1 kx 2 = mgh . The magnitude of the acceleration is largest when the spring is 2 compressed to a distance x; at this point the net upward force is kx − mg = ma , so the second condition is expressed as x = (m k )( g + a) . a) Substituting the second expression into the first gives 1 m k   ( g + a) 2 = mgh, 2 k
2

or

k=

m( g + a ) 2 . 2 gh
2 gh g +a

b) Substituting this into the expression for x gives x =

.

7.72: Following the hint, the force constant k is found from w = mg = kd , or k = mg . d When the fish falls from rest, its gravitational potential energy decreases by mgy; this becomes the potential energy of the spring, which is 1 ky 2 = 1 mg y 2 . Equating these, 2 2 d 1 mg 2 y = mgy, or y = 2d . 2 d

9.73: a)

2 2 ∆arad = ω2 r − ω0 r = (ω2 − ω0 )r

= [ ω − ω0 ] [ ω + ω0 ] r  ω − ω0  [ (ω + ω0 )t ] r =  t   = [ α] [2 (θ − θ0 )r. b) From the above, αr = ∆arad (85.0 m s 2 − 25.0 m s 2 ) = = 2.00 m s 2 . 2∆θ 2(15.0 rad)

c) Similar to the derivation of part (a), 1 1 2 1 ∆K = ω2 I − ω0 I = [α ][2∆θ ]I = Iα∆θ. 2 2 2 d) Using the result of part (c), I= ∆K (45.0 J − 20.0 J) = = 0.208 kg ⋅ m 2 . α∆θ ((2.00 m s 2 ) /(0.250 m))(15.0 rad)

7.74: a) From either energy or force considerations, the speed before the block hits the spring is v = 2 gL(sin θ − µ k cosθ ) = 2(9.80 m s 2 )(4.00 m)(sin 53.1° − (0.20) cos 53.1°) = 7.30 m s . b) This does require energy considerations; the combined work done by gravity and friction is mg ( L + d )(sin θ − µk cos θ ) , and the potential energy of the spring is 1 kd 2 , 2 where d is the maximum compression of the spring. This is a quadratic in d, which can be written as k d2 − d − L = 0. 2mg (sin θ − µ k cos θ ) The factor multiplying d 2 is 4.504 m −1 , and use of the quadratic formula gives d = 1.06 m . c) The easy thing to do here is to recognize that the presence of the spring determines d, but at the end of the motion the spring has no potential energy, and the distance below the starting point is determined solely by how much energy has been lost to friction. If the block ends up a distance y below the starting point, then the block has moved a distance L + d down the incline and L + d − y up the incline. The magnitude of the friction force is the same in both directions, µk mg cos θ , and so the work done by friction is − µk (2 L + 2d − y ) mg cos θ . This must be equal to the change in gravitational potential energy, which is − mgy sin θ . Equating these and solving for y gives 2 µ k cosθ 2µ k y = (L + d ) = (L + d ) . sin θ + µ k cosθ tan θ + µ k Using the value of d found in part (b) and the given values for µ k and θ gives y = 1.32 m . a) K B = Wother − U B = (20.0 N)(0.25 m) − (1 2)(40.0 N m)(.25 m) 2 = 3.75 J,
2 ( 3.75 J ) 0.500 kg

7.75: so v B =

= 3.87 m s , or 3.9 m s to two figures. b) At this point (point C),
2 ( 5.00 J ) 40.0 N m

K C = 0 , and so U C = Wother and xc = −

= −0.50 m (the minus sign denotes a

displacement to the left in Fig. (7.65)), which is 0.10 m from the wall.

7.76: The kinetic energy K ′ after moving up the ramp the distance s will be the energy initially stored in the spring, plus the (negative) work done by gravity and friction, or 1 K ′ = kx 2 − mg (sin α + µ k cos α ) s. 2 Minimizing the speed is equivalent to minimizing K ′ , and differentiating the above expression with respect to α and setting dK ′ = 0 gives dα 0 = −mgs (cos α − µ k sin α ),  1  , α = arctan  . Pushing the box straight up (α = 90°) maximizes the µ   k vertical displacement h, but not s = h sin α . or tan α =
1 µk

7.77: Let x1 = 0.18 m , x2 = 0.71 m . The spring constants (assumed identical) are then known in terms of the unknown weight w, 4kx1 = w . The speed of the brother at a given height h above the point of maximum compression is then found from 1 1  w 2 (4k ) x2 =   v 2 + mgh, 2 2 g   or v2 = ( 4k ) g 2 x2 − 2 gh = w  x2  g  2 − 2h , x   1 

so v = (9.80 m s 2 ) ((0.71 m) 2 (0.18 m) − 2(0.90 m)) = 3.13 m s , or 3.1 m s to two figures. b) Setting v = 0 and solving for h, 2 2kx2 x2 h= = 2 = 1.40 m, mg 2 x1 or 1.4 m to two figures. c) No; the distance x1 will be different, and the ratio
2 x2 x1

=

( x1 + 0.53 m ) 2 x1

= x1 1 +

(

0.53 m 2 x1

)

will be different. Note that on a small planet, with lower g,

x1 will be smaller and h will be larger.

7.78:

2 a) a x = d 2 x dt 2 = − ω0 x,

2 Fx = ma x = −m ω0 x 2 Fy = ma y = −m ω0 y

2 2 a y = d 2 y dt 2 = − ω0 y = − ω0 y,

2 b) U = − ∫ Fx dx + ∫ Fy dy = mω0 ∫ xdx + ∫ ydy =

[

]

[

]

1 2 mω0 ( x 2 + y 2 ) 2

c)

v x = dx dt = − x0 ω0 sin ω0t = − x0 ω0 ( y y0 ) v y = dy dt = + y0 ω0 cos ω0t = + y0 ω0 ( x x0 )

(i) When x = x0 and y = 0, v x = 0 and v y = y0 ω0 1 1 2 2 1 2 2 1 2 2 2 2 2 m(v x + v y ) = my0 ω0 , U = ω0 mx0 and E = K + U = mω0 ( x0 + y0 ) 2 2 2 2 (ii) When x = 0 and y = y0 , v x = − x0 ω0 and v y = 0 K= 1 2 2 1 1 2 2 2 2 2 ω0 mx0 , U = mω0 y0 and E = K + U = mω0 ( x0 + y0 ) 2 2 2 Note that the total energy is the same. K= 7.79: a) The mechanical energy increase of the car is K 2 − K1 = 1 (1500 kg )(37 m s) 2 = 1.027 × 10 6 J. 2 Let α be the number of gallons of gasoline consumed. α (1.3 × 108 J )(0.15) = 1.027 × 10 6 J α = 0.053 gallons b) (1.00 gallons ) α = 19 accelerati ons

7.80:

(a) Stored energy = mgh = (ρV ) gh = ρ A(1 m) gh = (1000 kg m 3 )(3.0 × 10 6 m 2 )(1 m)(9.8 sm )(150 m) 2 = 4.4 × 1012 J.

(b) 90% of the stored energy is converted to electrical energy, so (0.90) (mgh) = 1000 kW h (0.90) ρ V gh = 1000 kW h V = (1000 kW h ) ( 3600 s ) 1h (0.90)(1000 kg m 3 ) (150 m) (9.8 m s 2 )

= 2.7 × 103 m 3 Change in level of the lake: A∆h = Vwater V 2.7 × 10 3 m 3 ∆h = = = 9.0 × 10 −4 m 6 2 A 3.0 × 10 m 7.81: The potential energy of a horizontal layer of thickness dy, area A, and height y is dU = (dm) gy . Let ρ be the density of water. dm = ρdV = ρA dy, so dU = ρAgy dy. The total potential energy U is U = ∫ dU = ρAg ∫ ydy = 1 ρAh 2 . 2
0 0 h h

A = 3.0 × 10 m and h = 150 m, so U = 3.3 × 1014 J = 9.2 × 10 7 kWh.
6 2

7.82:

a) Yes; rather than considering arbitrary paths, consider that   ∂  Cy 3  ˆ   j. F = −  − ∂y  3    

b) No; consider the same path as in Example 7.13 (the field is not the same). For this      force, F = 0 along Leg 1, F ⋅ dl = 0 along legs 2 and 4, but F ⋅ dl ≠ 0 along Leg 3.

7.83:

  a) Along this line, x = y , so F ⋅ dl = −αy 3 dy , and y2 α 4 4 ∫y1 Fy dy = − 4 ( y2 − y1 ) = −50.6 J.   b) Along the first leg, dy = 0 and so F ⋅ dl = 0 . Along the second leg, x = 3.00 m ,

so Fy = −(7.50 N m 2 ) y 2 , and

∫
7.84: a)

y2

y1

3 Fy dy = −(7.5 3 N m 2 )( y 2 − y13 ) = −67.5 J.

c) The work done depends on the path, and the force is not conservative.

 b) (1): x = 0 along this leg, so F = 0 and W = 0 . (2): Along this leg, y = 1.50 m , so     F ⋅ dl = (3.00 N m) xdx , and W = (1.50 N m)((1.50 m) 2 − 0) = 3.38 J (3) F ⋅ dl = 0 , so  W = 0 (4) y = 0 , so F = 0 and W = 0 . The work done in moving around the closed path is 3.38 J. c) The work done in moving around a closed path is not zero, and the force is not conservative.

7.85:

a) For the given proposed potential U ( x), −
2

dU dx

= −kx + F , so this is a possible

potential function. For this potential, U (0) = − F 2k , not zero. Setting the zero of potential is equivalent to adding a constant to the potential; any additive constant will not change the derivative, and will correspond to the same force. b) At equilibrium, the force is zero; solving − kx + F = 0 for x gives x0 = F k . U ( x0 ) = − F 2 k , and this is a minimum of U, and hence a stable point. c)

d) No; Ftot = 0 at only one point, and this is a stable point. e) The extreme values of x correspond to zero velocity, hence zero kinetic energy, so U ( x± ) = E , where x± are the extreme points of the motion. Rather than solve a quadratic, note that 2 2 1 k , so U ( x± ) = E becomes 2 k(x − F k) − F 1  F F2 k  x± −  − F k = 2  k k F F x± − = ±2 , k k F F x+ = 3 x− = − . k k f) The maximum kinetic energy occurs when U (x) is a minimum, the point x0 = F k found in part (b). At this point K = E − U = ( F 2 k ) − (− F 2 k ) = 2 F 2 k , so v = 2F mk .
2

7.86: a) The slope of the U vs. x curve is negative at point A, so Fx is positive (Eq. (7.17)). b) The slope of the curve at point B is positive, so the force is negative. c) The kinetic energy is a maximum when the potential energy is a minimum, and that figures to be at around 0.75 m. d) The curve at point C looks pretty close to flat, so the force is zero. e) The object had zero kinetic energy at point A, and in order to reach a point with more potential energy than U (A) , the kinetic energy would need to be negative. Kinetic energy is never negative, so the object can never be at any point where the potential energy is larger than U (A) . On the graph, that looks to be at about 2.2 m. f) The point of minimum potential (found in part (c)) is a stable point, as is the relative minimum near 1.9 m. g) The only potential maximum, and hence the only point of unstable equilibrium, is at point C.

7.87:

a) Eliminating β in favor of α and x0 ( β = α x0 ) ,

2 2 α β α x0 α α  x0   x0  U ( x) = 2 − = 2 2 − = 2   −   . x x x0 x x0 x x0  x   x    α U ( x0 ) = x 2 (1 − 1) = 0 . U (x) is positive for x < x0 and negative for x > x0 ( α and β
0

must be taken as positive).

b)
2  2α    x   x   2 U =  2   0  −  0   .  mx   x m  0     x    The proton moves in the positive x-direction, speeding up until it reaches a maximum speed (see part (c)), and then slows down, although it never stops. The minus sign in the square root in the expression for v(x) indicates that the particle will be found only in the region where U < 0 , that is, x > x0 . c) The maximum speed corresponds to the maximum kinetic energy, and hence the minimum potential energy. This minimum occurs when dU = 0 , or dx

v( x) = −

dU α = 3 dx x0 occurs at a point where
dU dx

  x0  3  x0  2  − 2  +    = 0,   x  x   
0

which has the solution x = 2x0 . U (2 x0 ) = − 4α 2 , so v = x
2 x1 = 3x0 , and U (3x0 ) = − 9 2α 2 mx0

α 2 2mx0

. d) The maximum speed

= 0 , and from Eq. (7.15), the force at this point is zero. e)
α 2 x0

((

x0 x

)−( )

x0 2 x

− 2 9 . The particle is confined to the region where U ( x) < U ( x1 ) . The

)

; v ( x) =

2 m

(U ( x1 ) − U ( x)) =

2 m

[( ) − ((
−2 α 2 9 x0

α 2 x0

x0 2 x

)

−

x0 x

)] =

Chapter 8

8.1: a) (10,000 kg)(12.0 m s) = 1.20 × 10 5 kg ⋅ m s. b) (i) Five times the speed, 60.0 m s. (ii)

5 (12.0 m s ) = 26.8 m s.

8.2: See Exercise 8.3 (a); the iceboats have the same kinetic energy, so the boat with the larger mass has the larger magnitude of momentum by a factor of (2m) (m) = 2.

8.3: a)

K=

1 1 m2v 2 1 p 2 mv 2 = = . 2 2 m 2 m p12 p2 = 2 , so the larger mass m1 m2

b) From the result of part (a), for the same kinetic energy,

baseball has the greater momentum; ( pbird pball ) = 0.040 0.145 = 0.525 . From the result of part (b), for the same momentum K 1 m1 = K 2 m2 , so K 1 w1 = K 2 w2 ; the woman, with the smaller weight, has the larger kinetic energy. ( K man K woman ) = 450 700 = 0.643. 8.4: From Eq. (8.2), p x = mvx = ( 0.420 kg ) ( 4.50 m s ) cos 20.0° = 1.78 kg m s p y = mv y = ( 0.420 kg ) ( 4.50 m s ) sin 20.0° = 0.646 kg m s.

8.5: The y-component of the total momentum is

( 0.145 kg ) (1.30 m s ) + ( 0.0570 kg ) ( − 7.80 m s ) = −0.256 kg ⋅ m s.
This quantity is negative, so the total momentum of the system is in the − y -direction. 8.6: From Eq. (8.2), p y = −( 0.145 kg ) ( 7.00 m s ) = −1.015 kg ⋅ m s, and

px = ( 0.045 kg ) ( 9.00 m s ) = 0.405 kg ⋅ m s, so the total momentum has magnitude p=
2 px + p 2 = y

and is at an angle arctan ( ) = −68° , using the value of the arctangent function in the fourth quadrant ( p x > 0, p y < 0).
−1.015 + .405

( − 0.405 kg ⋅ m s ) 2 + ( − 1.015 kg ⋅ m s ) 2

= 1.09 kg ⋅ m s,

8.7:

∆p ∆t

=

( 0.0450 kg )( 25.0 m s )
2.00×10 − 3 s

= 563 N. The weight of the ball is less than half a newton, so the

weight is not significant while the ball and club are in contact. 8.8: a) The magnitude of the velocity has changed by

( 45.0 m s ) − ( − 55.0 m s ) = 100.0 m s,
and so the magnitude of the change of momentum is (0.145 kg) (100.0 m s) = 14.500 kg m s, to three figures. This is also the magnitude of the impulse. b) From Eq. (8.8), the magnitude of the average applied force is 14.500 kg.m/s =7.25 × 10 3 N. 2.00×10 − 3 s 8.9: a) Considering the +x-components, p2 = p1 + J = (0.16 kg)( 3.00 m s) + (25.0 N) × (0.05 s) = 1.73 kg ⋅ m s , and the velocity is 10.8 m s in the +x-direction. b) p 2 = 0.48 kg ⋅ m s + (–12.0 N)(0.05 s) = –0.12 kg ⋅ m s , and the velocity is +0.75 m s in the –x-direction.

  8.10: a) F t=(1.04 × 10 5 kg ⋅ m s) j . b) (1.04 × 10 5 kg ⋅ m s) ˆ . j 5 (1.04×10 kg. m s ) ˆ j = (1.10 m s) ˆ. d) The initial velocity of the shuttle is not known; the j c)
( 95 , 000 kg)

change in the square of the speed is not the square of the change of the speed.

8.11: a) With t1 = 0,
2 3 J x = ∫ Fx dt = (0.80 × 10 7 N s)t 2 − (2.00 × 10 9 N s 2 )t 2 , 0 t2

which is 18.8 kg ⋅ m s , and so the impulse delivered between t=0 and ˆ t = 2.50 × 10 −3 s is (18.8 kg ⋅ m s)i . b)
2

J y = −(0.145 kg) (9.80 m s ) (2.50 × 10 −3 s), and the impulse is
(−3.55 × 10−3 kg ⋅ m s) ˆ j ˆ (7.52 × 103 N)i .    d) p2 = p1 + j ˆ ˆ = −(0.145kg)(40.0i + 5.0 ˆ)m/s + (18.8i − 3.55 × 10 −3 ˆ) j j ˆ = (13.0 kg.m/s) i − (0.73 kg.m/s) ˆ. j The velocity is the momentum divided by the mass, or (89.7 m/s) ˆ i − (5.0m/s) ˆ. j 8.12: The change in the ball’s momentum in the x-direction (taken to be positive to the right) is (0.145 kg) (−(65.0 m s) cos 30 − 50.0m/s) = −15.41 kg ⋅ m/s, so the xcomponent of the average force is − 15.41 kg ⋅ m/s = −8.81 × 103 N, −3 1.75 × 10 s and the y-component of the force is (0.145kg)(65.0 m/s) sin 30° = 2.7 × 103 N. (1.75 × 10 − 3 s) c) Jx =7.52 × 10 3 N, so the average force is t2

2

8.13: a)

J = ∫ Fdt = A(t 2 − t1 ) +
t1

t2

B 3 3 (t 2 − t1 ), 3

3 or J = At 2 + ( B / 3)t 2 if t 1 = 0.

b) v =

p J A B 3 = = t2 + t2 . m m m 3m

8.14: The impluse imparted to the player is opposite in direction but of the same magnitude as that imparted to the puck, so the player’s speed is ( 0.16 kg) (20.0 m s ) = 4.27 cm/s, in the direction opposite to the puck’s. ( 75.0 kg)

8.15: a) You and the snowball now share the momentum of the snowball 0.400 kg) when thrown so your speed is ((70.0 kg + (10.0 m s) = 5.68 cm s. b) The change in 0.400 kg) the snowball’s momentum is (0.400 kg) (18.0 m s) = 7.20 kg ⋅ m s), so your speed is
7.20 kg ⋅ m s 70.0 kg

= 10.3 cm/s.

8.16: a) The final momentum is (0.250 kg )(−0.120 m s) + (0.350)(0.650 m s) = 0.1975 kg ⋅ m s, taking positive directions to the right. a) Before the collision, puck B was at rest, so all of the momentum is due to puck A’s motion, and p 0.1975 kg ⋅ m/s = = 0.790 m/s. mA 0.250 kg 1 1 1 2 2 2 ∆K = K 2 − K 1 = m A v A 2 + m B v B 2 − m A v A1 2 2 2 v A1 = = 1 1 (0.250 kg ) (−0.120 m s) 2 + (0.350 kg )(0.650 m s) 2 2 2 1 − (0.250 kg )(−0.7900 m s) 2 2

b)

= −0.0023 J

8.17: The change in velocity is the negative of the change in Gretzky’s momentum, divided by the defender’s mass, or vB 2 = vB1 − mA (v A2 − v A1 ) mB 756 N (1.50 m s − 13.0 m s) 900 N

= −5.00 m s − = 4.66 m s.

Positive velocities are in Gretzky’s original direction of motion, so the defender has changed direction. b) K 2 − K 1 = 1 1 2 2 2 2 m A (v A2 − v A1 ) + m B (v B 2 − v B1 ) 2 2 (756 N)((1.50 m/s) 2 − (13.0 m/s) 2 )  1 =  2  2(9.80 m/s )  + (900 N)((4.66 m/s) 2 − (−5.00 m/s) 2 )   = −6.58 kJ.

8.18: Take the direction of the bullet’s motion to be the positive direction. The total momentum of the bullet, rifle, and gas must be zero, so (0.00720 kg)(601 m/s − 1.85 m/s) + (2.80 kg)(−1.85 m/s) + pgas = 0, and p gas = 0.866 kg ⋅ m s . Note that the speed of the bullet is found by subtracting the speed of the rifle from the speed of the bullet relative to the rifle. 8.19: a) See Exercise 8.21; v A =

(

3.00 kg 1.00 kg

)(0.800 m s) = 3.60 m s .

b) (1 2) (1.00 kg) (3.60 m/s) 2 + (1 / 2)(3.00 kg)(1.200 m/s) 2 = 8.64 J.

8.20: In the absence of friction, the horizontal component of the hat-plus-adversary system is conserved, and the recoil speed is (4.50 kg)(22.0 m s) cos 36.9° = 0.66 m s . (120 kg)

8.21: a) Taking v A and v B to be magnitudes, conservation of momentum is m expressed as m Av A = mB vB , so v B = A v A . mB b)
2 2 K A (1 / 2)m A v A mAv A m = = = B. 2 2 K B (1 / 2)m B v B m B ((m A / m B )v A ) mA

(This result may be obtained using the result of Exercise 8.3.) 8.22:
214

Po decay :

214

Po →4 α + 210 X

Setvα : KEα = vα = =

1 2 mα vα 2

2 KEα mα 2(1.23 × 10 −12 J) = 1.92 × 107 m/s − 27 6.65 × 10 kg

Momentum conservation: 0 = mα vα − mx vx vx = = mα vα mv = α α mx 210mp (6.65 × 10− 27 kg)(1.92 × 107 m/s) (210)(1.67 × 10− 27 kg)

= 3.65 × 105 m/s

8.23: Let the +x-direction be horizontal, along the direction the rock is thrown. There is no net horizontal force, so Px is constant. Let object A be you and object B be the rock. 0 = −m A v A + m B v B cos 35.0° vA = m B v B cos 35.0° = 2.11 m/s mA

8.24: Let Rebecca’s original direction of motion be the x-direction. a) From conservation of the x-component of momentum, (45.0 kg)(13.0 m/s) = (45.0 kg)(8.0 m/s)cos 53.1° + (65.0 kg)vx , So v x = 5.67 m s. If Rebecca’s final motion is taken to have a positive y - component, then vy = − Daniel’s final speed is
2 2 vx + v y =

(45.0 kg)(8.0 m s) sin 53.1° = −4.43 m s. (65.0 kg)

(5.67 m s) 2 + ( −4.43 m s) 2 = 7.20 m s,

.42 and his direction is arctan ( −54.67 ) = −38° from the x - axis, which is 91.1° from the direction of Rebecca’s final motion.

1 1 1 b) ∆K = (45.0 kg) (8.0 m s) 2 + (65.0 kg) (7.195 m s) 2 − (45.0) (13.0 m s) 2 2 2 2 = −680 J. Note that an extra figure was kept in the intermediate calculation.

8.25: (mKim + mKen )(3.00 m s) = mKim ( 4.00 m s) + mKen (2.25 m s), so mKim (3.00 m s) − (2.25 m s) = = 0.750, mKen (4.00 m s) − (3.00 m s) and Kim weighs (0.750)(700 N) = 525 N.

8.26: The original momentum is (24,000 kg )(4.00 m s) = 9.60 × 104 kg ⋅ m s, the final mass is 24,000 kg + 3000 kg = 27,000 kg, and so the final speed is 9.60 × 10 4 kg ⋅ m s = 3.56 m s. 2.70 × 10 4 kg

8.27: Denote the final speeds as v A and v B and the initial speed of puck A as v 0 , and omit the common mass. Then, the condition for conservation of momentum is v0 = v A cos 30.0° + vB cos 45.0 0 = v A sin 30.0° − vB sin 45.0. The 45.0° angle simplifies the algebra, in that sin 45.0° = cos 45.0°, and so the v B terms cancel when the equations are added, giving vA = v0 = 29.3 m s cos 30.0° + sin 30.0°
vA 2

From the second equation, vB =

= 20.7 m s. b) Again neglecting the common mass,

2 2 K 2 (1 2)(v A + vB ) (29.3 m s) 2 + (20.7 m s) 2 = = = 0.804, 2 K1 (1 2)v0 (40.0 m s) 2

so 19.6% of the original energy is dissipated.

8.28: a) From m1v1 + m2v2 = m1v + m2v = (m1 + m2 )v, v = velocities to the right, v1 = −3.00. m s and

m1v1 + m 2 v 2 m1 + m 2

. Taking positive

v2 = 1.20 m s , so v = −1.60 m s .

1 b) ∆K = (0.500 kg + 0.250 kg )(−1.60 m s) 2 2 1 1 − (0.500 kg )(−3.00 m s) 2 − (0.250 kg )(1.20 m s) 2 2 2 = −1.47 J. 8.29: For the truck, M = 6320 kg, and V = 10 m s, for the car, m = 1050 kg and v = −15 m s (the negative sign indicates a westbound direction). a) Conservation of momentum requires ( M + m)v′ = MV + mv , or v′ = (6320 kg )(10 m s) + (1050 kg)(−15 m s) = 6.4 m s eastbound. (6320 kg + 1050 kg)

b) V =

− mv − (1050 kg)(−15 m s) = = 2.5 m s. M 6320 kg

c) ∆KE = −281 kJ for part (a) and ∆KE = −138 kJ for part (b). 8.30: Take north to be the x-direction and east to be the y-direction (these choices are arbitrary). Then, the final momentum is the same as the intial momentum (for a sufficiently muddy field), and the velocity components are (110 kg )(8.8 m s) = 5.0 m s (195 kg ) (85 kg )(7.2 m s) vy = = 3.1 m s. (195 kg ) The magnitude of the velocity is then (5.0 m s) 2 + (3.1 m s) 2 = 5.9 m s , at an angle or 31 arctan ( 5..0 ) = 32° east of north. vx =

8.31: Use conservation of the horizontal component of momentum to find the velocity of the combined object after the collision. Let +x be south. P x is constant gives (0.250 kg )(0.200 m s) − (0.150 kg )(0.600 s) = (0.400 kg)v2 x v2 x = −10.0 cm s (v2 = 10.0 cm s , north) K1 = 1 (0.250 kg )(0.200 s) 2 + 1 (0.150 kg )(0.600 s) 2 = 0.0320 J 2 2 K 2 = 1 (0.400 kg)(0.100 s) 2 = 0.0020 J 2 ∆K = K 2 − K1 = −0.0300 J Kinetic energy is converted to thermal energy due to work done by nonconservative forces during the collision. 8.32: (a) Momentum conservation tells us that both cars have the same change in momentum, but the smaller car has a greater velocity change because it has a smaller mass. M∆V = m∆v M ∆v (small car) = ∆V (large car) m 3000 kg = ∆V = 2.5∆V (large car) 1200 kg (b) The occupants of the small car experience 2.5 times the velocity change of those in the large car, so they also experience 2.5 times the acceleration. Therefore they feel 2.5 times the force, which causes whiplash and other serious injuries. 8.33: Take east to be the x-direction and north to be the y-direction (again, these choices are arbitrary). The components of the common velocity after the collision are (1400 kg) (−35.0 km h ) = −11.67 km h (4200 kg) (2800 kg) (−50.0 km h ) vy = = −33.33 km h. (4200 kg) vx = The velocity has magnitude (−11.67 km h) 2 + (−33.33 km h) 2 = 35.3 km h and is at a −33 33 direction arctan ( −11..67 ) = 70.7° south of west.

8.34: The initial momentum of the car must be the x-component of the final momentum as the truck had no intial x-component of momentum, so vcar = px (m + mtruck )v cos θ = car mcar mcar 2850 kg = (16.0 m s ) cos (90° − 24°) 950 kg = 19.5 m s.

Similarly, v truck =

2850 (16.0 m s) sin 66° = 21.9 m s. 1900

8.35: The speed of the block immediately after being struck by the bullet may be found from either force or energy considerations. Either way, the distance s is related to the speed v block by v 2 = 2 μk gs. The speed of the bullet is then m + mbullet v bullet = block 2µ k gs mbullet 1.205 kg 2(0.20)(9.80 m s 2 )(0.230 m) −3 5.00 × 10 kg = 229 m s, = or 2.3 × 10 2 m s to two places. 8.36: a) The final speed of the bullet-block combination is 12.0 × 10 −3 kg (380 m s) = 0.758 m s. 6.012 kg Energy is conserved after the collision, so (m + M ) gy = 1 (m + M )V 2 , and 2 V= y= b) 1 V 2 1 (0.758 m s) 2 = = 0.0293 m = 2.93 cm. 2 g 2 (9.80 m s 2 )

K 1 = 1 mv 2 = 1 (12.0 × 10 −3 kg)(380 m s) 2 = 866 J. 2 2

c) From part a), K 2 = 1 (6.012 kg)(0.758 m s) 2 = 1.73 J. 2

8.37: Let +y be north and +x be south. Let vS 1 and v A1 be the speeds of Sam and of Abigail before the collision. mS = 80.0 kg, m A = 50.0 kg, v S 2 = 6.00 m s, v A 2 = 9.00 m s. Px is constant gives mS vS 1 = mS vS 2 cos 37.0° + mAv A 2 cos 23.0° vS 1 = 9.67 m s (Sam) Py is constant gives mAv A1 = mS vS 2 sin 37.0° − mAv A 2 sin 23.0° v A1 = 2.26 m s (Abigail)
2 2 b) K1 = 1 mS vS 1 + 1 mAv A1 = 4101 J 2 2 2 2 K 2 = 1 mS vS 2 + 1 mAv A2 = 3465 J 2 2

∆K = K 2 − K1 = −640 J 8.38: (a) At maximum compression of the spring, v 2 = v10 = V . Momentum conservation gives (2.00 kg)(2.00 m s) = (12.0 kg)V V = 0.333 m s 1 1 2 Energy conservation : m2 v 0 = (m2 + m10 )V 2 + U spr 2 2 1 1 (2.00 kg )(2.00 m s) 2 = (12.0 kg)(0.333 m s ) 2 + U spr 2 2 U spr = 3.33 J (b) The collision is elastic and Eqs. (8.24) and (8.25) may be used: v 2 = −1.33 m s, v10 = +0.67 m s 8.39: In the notation of Example 8.10, with the smaller glider denoted as A, conservation of momentum gives (1.50)v A 2 + (3.00)v B 2 = −5.40 m s. The relative velocity has switched direction, so v A 2 − v B 2 = −3.00 m s. Multiplying the second of these relations by (3.00) and adding to the first gives (4.50)v A2 = −14.4 m s, or v A 2 = −3.20 m s, with the minus sign indicating a velocity to the left. This may be substituted into either relation to obtain v B 2 = −0.20 m s; or, multiplying the second relation by (1.50) and subtracting from the first gives (4.50)v B 2 = −0.90 m s, which is the same result.

8.40: a) In the notation of Example 8.10, with the large marble (originally moving to the right) denoted as A, (3.00)v A2 + (1.00)v B 2 = 0.200 m s. The relative velocity has switched direction, so v A2 − v B 2 = −0.600 m s. Adding these eliminates v B 2 to give (4.00)v A 2 = −0.400 m s, or v A 2 = −0.100 m s, with the minus sign indicating a final velocity to the left. This may be substituted into either of the two relations to obtain v B 2 = 0.500 m s; or, the second of the above relations may be multiplied by 3.00 and subtracted from the first to give (4.00)v B 2 = 2.00 m s, the same result. b) ∆PA = −0.009 kg ⋅ m s, ∆PB = 0.009 kg ⋅ m s c) ∆K A = −4.5 × 10 − 4 , ∆K B = 4.5 × 10− 4. Because the collision is elastic, the numbers have the same magnitude. 8.41: Algebraically, v B 2 = 20 m s. This substitution and the cancellation of common factors and units allow the equations in α and β to be reduced to 2 = cos α + 1.8 cos β 0 = sin α − 1.8 sin β. Solving for cosα and sin α , squaring and adding gives

(2 −

1.8 cos β
1.2 1.8

) +(
2

1.8 .sin β

)

2

= 1.

Minor algebra leads to cos β =

, or β = 26.57°. Substitution of this result into the first

of the above relations gives cos α = 4 , and α = 36.87°. 5
u 8.42: a) Using Eq. (8.24), v A = 1 u − 2 u = 1 . b) The kinetic energy is proportional to the V 1 +2 u 3

square of the speed, so
1 3

KA K

=

1 9

c) The magnitude of the speed is reduced by a factor of

after each collision, so after N collisions, the speed is N, consider N 1 1 or   = 59,000  3 3N = 59,000

( 1 ) N of its original value. To find 3

N ln( 3) = ln( 59,000) ln( 59,000) N= = 10. ln( 3) to the nearest integer. Of course, using the logarithm in any base gives the same result.

8.43: a) In Eq. (8.24), let mA = m and mB = M . Solving for M gives v − vA M =m v + vA In this case, v = 1.50 × 10 7 m s, and v A = −1.20 × 10 7 m s, with the minus sign indicating 1. 1.20 a rebound. Then, M = m 1.5050 )+−1.20 ) = 9m. Either Eq. (8.25) may be used to find +
v vB = 5 = 3.00 × 106 m s, or Eq. (8.23), which gives

vB = (1.50 × 107 m s) + (−1.20 × 107 m s), the same result. 8.44: From Eq. (8.28), xcm = ycm (0.30 kg)( 0.20 m) + (0.40 kg)(0.10 m) + (0.20 kg )(−0.30 m) = +0.044 m, (0.90 kg) (0.30 kg)(0.30 m) + (0.40 kg )(−0.40 m) + (0.20 kg)(0.60 m) = = 0.056 m. (0.90 kg )

8.45: Measured from the center of the sun, (1.99 × 10 30 kg)(0) + (1.90 × 10 27 kg)(7.78 × 1011 m) = 7.42 × 10 8 m. 1.99 × 10 30 kg + 1.90 × 10 27 kg The center of mass of the system lies outside the sun. 8.46: a) Measured from the rear car, the position of the center of mass is, from Eq. (8.28), (1800 kg )(40.0 m) = 24.0 m, which is 16.0 m behind the leading car. (1200 kg + 1800 kg) b) (1200 kg )(12.0 m s) + (1800 kg)(20.0 m s) = 5.04 × 10 4 kg ⋅ m s. c) From Eq. (8.30), (1200 kg )(12.0 m s) + (1800 kg )(20.0 m s) = 16.8 m s. (1200 kg + 1800 kg ) d) (1200 kg + 1800 kg)(16.8 m s) = 5.04 × 10 4 kg ⋅ m s. vcm =

8.47: a) With x1 = 0 in Eq. (8.28), m1 = m2 (( x 2 / x cm ) − 1) = (0.10 kg )((8.0 m)/(2.0 m) − 1) = 0.30 kg. ˆ ˆ b) P = M v cm = (0.40 kg )(5.0 m s) i = (2.0 kg ⋅ m s) i . c) In Eq. (8.32),    ˆ v 2 = 0, so v 1 = P /(0.30 kg ) = (6.7 m s)i . 8.48: As in Example 8.15, the center of mass remains at rest, so there is zero net momentum, and the magnitudes of the speeds are related by m1v1 = m2 v 2 , or v 2 = (m1 / m2 )v1 = (60.0 kg / 90.0 kg )(0.70 m s) = 0.47 m s. 8.49: See Exercise 8.47(a); with y1 = 0, Eq. (8.28) gives m1 = m2 (( y 2 / y cm ) − 1) = (0.50 kg )((6.0 m) /(2.4 m) − 1) = 0.75 kg, so the total mass of the system is 1.25 kg.   d ˆ a cm = dt v cm = (1.50 m s 3 ) ti .   ˆ ˆ c) F = macm = (1.25 kg ) (1.50 m s3 ) (3.0 s) i = (5.63 N)i . b) 8.50: pz = 0, so Fz = 0. The x -component of force is Fx = dpx = (−1.50 N s)t. dt dp y dt = 0.25 N

Fy =

8.51: a) From Eq. (8.38), F = (1600 m s)(0.0500 kg s) = 80.0 N. b) The absence of atmosphere would not prevent the rocket from operating. The rocket could be steered by ejecting the fuel in a direction with a component perpendicular to the rocket’s velocity, and braked by ejecting in a direction parallel (as opposed to antiparallel) to the rocket’s velocity.

8.52: It turns out to be more convenient to do part (b) first; the thrust is the force that accelerates the astronaut and MMU, F = ma = (70 kg + 110 kg)(0.029 m s 2 ) = 5.22 N. a) Solving Eq. (8.38) for dm , dm = F dt (5.22 N)(5.0 s) = = 53 gm. v ex 490 m s

8.53: Solving for the magnitude of dm in Eq. (8.39), (6000 kg)(25.0 m s 2 ) ma dm = dt = (1 s) = 75.0 kg. v ex (2000 m s) 8.54: Solving Eq. (8.34) for vex and taking the magnitude to find the exhaust speed,
m vex = a dm dt = 15.0 m s 2

(

) (160 s) = 2.4 km s. In this form, the quantity
m ∆m

m dm dt

is

approximated by

m ∆m ∆t

=

∆t = 160 s.

8.55: a) The average thrust is the impulse divided by the time, so the ratio of the average (10.0 thrust to the maximum thrust is (13.3 N) N ⋅s) s) = 0.442. b) Using the average force in Eq. (1.70 (8.38), vex =
F dt dm

=

10.0 N ⋅s 0.0125 kg

= 800 m s. c) Using the result of part (b) in Eq. (8.40),

v = (800 m s) ln (0.0258 0.0133) = 530 m s .

8.56: Solving Eq. (8.4) for the ratio
 v m0 = exp  v m  ex

m0 m

, with v0 = 0 ,

  8.00 km s   = exp   2.10 km s  = 45 .1.     

8.57: Solving Eq. (8.40) for

m mo

, the fraction of the original rocket mass that is not fuel,

 .   −3 5 a) For v = 1.00 × 10 c = 3.00 × 10 m s , exp( − (3.00 × 105 m s (2000 m s)) = 7.2 × 10−66 . b) For v = 3000 m s , exp( −(3000 m s) (2000 m s)) = 0.22.

 v m = exp  −  v m0  ex

8.58: a) The speed of the ball before and after the collision with the plate are found from the heights. The impulse is the mass times the sum of the speeds,

J = m(v1 + v2 ) = m( 2 gy1 + 2 gy2 ) = (0.040 kg) 2(9.80 m s 2
J b) ∆t = (0.47 N ⋅ s/2.00 × 10 s) = 237 N. −3

(

2.00 m + 1.60 m = 0.47 N ⋅ s.

)

  ˆ j j j 8.59: p = ∫ F dt = ( αt 3 3) ˆ + ( βt + γt 2 2) ˆ = (8.33 N s 2 t 3 )i + (30.0 Nt + 2.5 N st 2 ) ˆ  ˆ After 0.500 s, p = (1.04 kg ⋅ m s) i + (15.63 kg ⋅ m s) ˆ, and the velocity is j   ˆ v = p m = (0.52 m s) i + (7.82 m s) ˆ. j 8.60: a) J x = Fx t = (−380 N) (3.00 × 10 −3 s) = −1.14 N ⋅ s. J y = Fy t = (110 N) (3.00 × 10 −3 s) = 0.33 N ⋅ s. b) v2 x = v1x + J x m = ( 20.0 m s ) +

( − 1.14 N.s) = 0.05 m s ( 0.560 N ) (9.80 m s 2 )

v2 y = v1 y + J y m = (−4.0 m s) +

(0.33 N.s) = 1.78 m s. ((0.560 N) (9.80 m s 2 ))

8.61: The total momentum of the final combination is the same as the initial momentum; for the speed to be one-fifth of the original speed, the mass must be five times the original mass, or 15 cars. 8.62: The momentum of the convertible must be the south component of the total momentum, so (800 kg ⋅ m s)cos 60.0° = 2.67 m s. (1500 kg) Similarly, the speed of the station wagon is vcon = vsw = (800 kg ⋅ m s) sin 60.0° = 3.46 m s. (2000 kg)

8.63: The total momentum must be zero, and the velocity vectors must be three vectors of the same magnitude that sum to zero, and hence must form the sides of an equilateral triangle. One puck will move 60° north of east and the other will move 60° south of east. 8.64: a) mAv Ax + mB vBx + mC vCx = mtot vx , therefore (0.100 kg)(0.50 m s) − (0.020 kg)(−1.50 m s) − (0.030 kg)(−0.50 m s)cos60 ° 0.050kg vCx = 1.75 m s vCx = Similarly, (0.100kg)(0 m s) − (0.020kg)(0 m s) − (0.030 kg)(−0.50 m s)sin 60° 0.050 kg vCy = 0.26 m s vCy = b) ∆K = 1 (0.100 kg)(0.5 m s) 2 − 1 (0.020 kg)(1.50 m s) 2 − 1 (0.030 kg)(−0.50 m s) 2 2 2 2 − 1 (0.050 kg) × [(1.75 m s) 2 + (0.26 m s) 2 ] = −0.092 J 2 8.65: a) To throw the mass sideways, a sideways force must be exerted on the mass, and hence a sideways force is exerted on the car. The car is given to remain on track, so some other force (the tracks on the car) act to give a net horizontal force of zero on the car, which continues at 5.00 m s east. b) If the mass is thrown with backward with a speed of 5.00 m s relative to the initial motion of the car, the mass is at rest relative to the ground, and has zero momentum. The 200 kg speed of the car is then (5.00 m s) (( 175 kg )) = 5.71 m s, and the car is still moving east. c) The combined momentum of the mass and car must be same before and after the )+ mass hits the car, so the speed is ( 200 kg )( 5.00 m (s225( 25.)0 kg )( −6.00 m s ) = 3.78 m s, with the car still kg moving east.

8.66: The total mass of the car is changing, but the speed of the sand as it leaves the car is the same as the speed of the car, so there is no change in the velocity of either the car or the sand (the sand acquires a downward velocity after it leaves the car, and is stopped on the tracks after it leaves the car). Another way of regarding the situation is that vex in Equations (8.37), (8.38) and (8.39) is zero, and the car does not accelerate. In any event, the speed of the car remains constant at 15.0 m/s. In Exercise 8.24, the rain is given as falling vertically, so its velocity relative to the car as it hits the car is not zero. 8.67: a) The ratio of the kinetic energy of the Nash to that of the Packard is
2 mN vN 2 mP vP

=

( 840 kg )( 9 m s ) 2 ( 1620 kg )( 5 m s ) 2
mN vN mP vP

= 1.68. b) The ratio of the momentum of the Nash to that of the

Packard is

= ((840 kg)(9 m/s )) = 0.933, therefore the Packard has the greater magnitude 1620 kg)(5 m/s

of momentum. c) The force necessary to stop an object with momentum P in time t is F = − P / t. Since the Packard has the greater momentum, it will require the greater force to stop it. The ratio is the same since the time is the same, therefore FN / FP = 0.933. d) By the work-kinetic energy theorem, F = ∆k . Therefore, since the Nash has the greater d kinetic energy, it will require the greater force to stop it in a given distance. Since the distance is the same, the ratio of the forces is the same as that of the kinetic energies, FN / FP = 1.68. 8.68: The recoil force is the momentum delivered to each bullet times the rate at which the bullets are fired,  1000 bullets/mi n  Fave = (7.45 × 10 −3 kg) (293 m/s)   = 36.4 N. 60 s/ min  

8.69: (This problem involves solving a quadratic. The method presented here formulates the answer in terms of the parameters, and avoids intermediate calculations, including that of the spring constant.) Let the mass of the frame be M and the mass putty be m. Denote the distance that the frame streteches the spring by x0 , the height above the frame from which the putty is dropped as h , and the maximum distance the frame moves from its initial position (with the frame attached) as d. The collision between the putty and the frame is completely inelastic, and the m common speed after the collision is v0 = 2 gh m + M . After the collision, energy is conserved, so that 1 1 2 2 ( m + M )v 0 + (m + M ) gd = k ((d + x0 ) 2 − x 0 , or 2 2 2 1 m 1 mg 2 ( 2 gh) + ( m + M ) gd = ((d + x0 ) 2 − x0 , 2 m+M 2 x0 where the above expression for v0 , and k = mg x0 have been used. In this form, it is seen that a factor of g cancels from all terms. After performing the algebra, the quadratic for d becomes m m2  d 2 − d  2 x0 = 0,  − 2hx0 M m+M  which has as its positive root
2  m   h  m2 m  . d = x0   +   + 2  x0  M ( m + M )    M  M     For this situation, m = 4/3 M and h/x0 = 6, so

d = 0.232 m.

8.70: a) After impact, the block-bullet combination has a total mass of 1.00 kg, and the k speed V of the block is found from 1 M total V 2 = 1 kX 2 , or V = m X . The spring constant k 2 2 is determined from the calibration; k =
0.75 N 2 .50 ×10 −3 m

= 300 N m. Combining,

300 N/m 15.0 × 10 − 2 m = 2.60 m s. 1.00 kg b) Although this is not a pendulum, the analysis of the inelastic collision is the same; V = v= M total 1.00 Kg V = ( 2.60 m s ) = 325 m s . m 8.0 × 10 −3 Kg

(

)

8.71: a) Take the original direction of the bullet’s motion to be the x-direction, and the direction of recoil to be the y-direction. The components of the stone’s velocity after impact are then  6.00 × 10 −3 Kg  vx =   0.100 Kg ( 350 m s ) = 21.0 m s ,     6.00 × 10 −3 Kg  v y = −  0.100 Kg ( 250 m s ) = 15.0 m s ,    and the stone’s speed is

( 21.0 m s ) 2 + (15.0 m s ) 2 = 25.8 m s , at an angle of arctan ( 15..00 ) = 35.5°. b) K1 = 12 ( 6.00 × 10 −3 kg ) ( 350 m/s ) 2 = 368 J 21 2 K 2 = 1 ( 6.00 × 10 −3 kg )( 250 m/s ) + 1 ( 0.100 kg ) ( 25.8 m 2 / s 2 ) = 221 J, so the collision is 2 2

not perfectly elastic.

8.72: a) The stuntman’s speed before the collision is v 0s = 2 gy = 9.9 m / s. The speed after the collision is v= ms 80.0 kg ( 9.9 m/s ) = 5.3 m/s. v0 s = ms + m v 0.100 kg

b) Momentum is not conserved during the slide. From the work-energy theorem, the distance x is found from 1 mtotal v 2 = μk mtotal gx, or 2

( 5.28 m / s ) v2 x= = = 5.7 m. 2 µ kg 2( 0.25) 9.80 m / s 2
2

(

)

Note that an extra figure was needed for V in part (b) to avoid roundoff error. 8.73: Let v be the speed of the mass released at the rim just before it strikes the second mass. Let each object have mass m. Conservation of energy says 1 mv 2 = mgR; v = 2 gR 2 This is speed v1 for the collision. Let v2 be the speed of the combined object just after the collision. Conservation of momentum applied to the collision gives mv1 = 2mv 2 so v 2 = v1 2 = gR 2 Apply conservation of energy to the motion of the combined object after the collision. Let y3 be the final height above the bottom of the bowl.
1 2

( 2m ) v22 = ( 2m ) gy3

2 v2 1  gR  y3 = =   = R/4 2g 2g  2  Mechanical energy is lost in the collision, so the final gravitational potential energy is less than the initial gravitational potential energy.

8.74: Collision: Momentum conservation gives mv0 = mv1 + (3m)v3 v 0 = v1 + 3v3 Energy Conservation: 1 2 1 2 1 2 mv 0 = mv1 + ( 3m ) v3 2 2 2 2 2 2 v 0 = v1 + 3v3 Solve (1) and (2) for v3 : v3 = 2.50 m s Energy conservation after collision: 1 ( 3m) v32 = ( 3m ) gh = ( 3m ) gl (1 − cosθ ) 2 Solve for θ : θ = 68.8° 8.75: First consider the motion after the collision. The combined object has mass   mtot = 25.0 kg. Apply ΣF = ma to the object at the top of the circular loop, where the object has speed v3 .
2 v3 T + mg = m R The minimum speed v3 for the object not to fall out of the circle is given by setting

(1)

(2)

T = 0 . This gives v3 = Rg , where R = 3.50 m.

Next, use conservation of energy with point 2 at the bottom of the loop and point 3 at the top of the loop. Take y = 0 at the point 2. Only gravity does work, so K2 + U 2 = K3 + U3
1 2

2 2 mtot v2 = 1 mtot v3 + mtot g ( 2 R ) 2

Use v3 = Rg and solve for v 2 : v 2 = 5 gR = 13.1 m s Now apply conservation of momentum to the collision between the dart and the sphere. Let v1 be the speed of the dart before the collision. ( 5.00 kg ) v1 = ( 25.0 kg )(13.1 m s ) v 1 = 65.5 m s

8.76: Just after the collision:

∑ F = ma

2 v8 T − m8 g = m8 R

1600 N − ( 8.00 kg ) 9.80 m s 2 = ( 8.00 kg ) v8 = 16.0 m s

(

)

v82 1.35 m

Energy and momentum are conserved during the elastic collision.

( 2.00 kg ) v0 = ( 2.00 kg ) v 2 + ( 8.00 kg )(16.0 m s )
v0 = v 2 + 64.0 m s 1 1 1 2 2 2 m2 v0 = m2 v 2 + m8 v8 2 2 2 2 ( 2.00 kg ) v0 = ( 2.00 kg ) v 22 + ( 8.00 kg )(16.0 m s ) 2
2 2 v0 = v 2 + 1024 m 2 s 2

m 2 v0 = m2 v 2 + m8 v8

(1)

(2)

Solve (1) and (2) for v0 : v0 = 40.0 m s 8.77: a) The coefficient of friction, from either force or energy consideration, is μk = v 2 2 gs, where v is the speed of the block after the bullet passes through. The speed of the block is determined from the momentum lost by the bullet, 4.00 × 10−3 kg ( 280 m s ) = 1.12 kg ⋅ m s, and so the coefficient of kinetic friction is

(

)

μk = b)
1 2

( (1.12 kg ⋅ m s )

(0.80 kg ) ) = 0.22. 2 2 9.80 m s ( 0.45m )
2

(

)

( 4.00 × 10

−3

kg ( 400 m s ) − (120 m s ) = 291 J. c) From the calculation of the
2 2 p2 2m

)(

)

momentum in part (a), the block’s initial kinetic energy was

=

(1.12 kg ⋅ m s ) 2 2 ( 0.80 kg)

= 0.784 J.

8.78: The speed of the block after the bullet has passed through (but before the block has begun to rise; this assumes a large force applied over a short time, a situation characteristic of bullets) is V = 2 gy = 2(9.80 m/s 2 )(0.45 × 10 −2 m) = 0.297 m/s. The final speed v of the bullet is then p mv 0 − MV M = = v0 − V m m m 1.00 kg = 450 m/s − (0.297 m/s) = 390.6 m/s, 5.00 × 10 −3 kg or 390 m/s to two figures. v= 8.79: a ) Using the notation of Eq. ( 8.24 ), K0 − K2 = 1 2 1 2 mv − mv A 2 2   m − M 2  1  = mv 2 1 −   m+M   2     (m + M ) 2 − (m − M ) 2   = K0    (m + M ) 2    4mM  = K0   (m + M ) 2  .    b) Of the many ways to do this calculation, the most direct way is to differentiate the expression of part (a) with respect to M and set equal to zero;  d  M  0 = (4mK 0 )  (m + M ) 2 , or  dM   1 2M − 2 (m + M ) ( m + M )3 0 = (m + M ) − 2 M m = M. 0= c) From Eq.(8.24), with mA = mB = m, v A = 0; the neutron has lost all of its kinetic energy.

8.80: a) From the derivation in Sec. 8.4 of the text we have VA = MA − MB 2M A V0 and VB = MA + MB MA + MB

The ratio of the kinetic energies of the two particles after the collision is
2 M AVA M A  VA  M  M − MB  (M A − M B )2   = A A  = = M BVB2 M B  VB  M B  2M A  4M A M B     (M A − M B )2 or KEA = KEB 4M A M B 1 2 1 2 2 2

b) i) For M A = M B , KE A = 0; i.e., the two objects simply exchange kinetic energies. ii) For M A = 5M B , KE A (4 M B ) 2 4 = = KE B 4(5M B )( M B ) 5 i.e., M A gets 4/9 or 44% of the total. c) We want
2 2 KE A ( M A − M B ) 2 M A − 2M A M B + M B =1= = KE B 4M A M B 4M A M B

which reduces to
2 2 M A − 6 M A M B + M B = 0,

from which, using the quadratic formula, we get the two possibilities M A = 5.83 M B and M A = 0.172 M B.

8.81: a) Apply conservation of energy to the motion of the package from point 1 as it leaves the chute to point 2 just before it lands in the cart. Take y = 0 at point 2, so y1 = 4.00 m. Only gravity does work, so K1 + U1 = K 2 + U 2
1 2 2 mv12 + mgy1 = 1 mv2 2

v2 = v12 + 2 gy1 = 9.35 m/s b) In the collision between the package and the cart momentum is conserved in the horizontal direction. (But not in the vertical direction, due to the vertical force the floor exerts on the cart.) Take +x to be to the right. Let A be the package and B be the cart. Px is constant gives ma v A1x + m B v B1x = (m A + M B )v 2 x v B1x = −5.00 m/s v A1x = (3.00 m/s) cos 37.0ο ( The horizontal velocity of the package is constant during its free-fall.) Solving for v 2 x gives v 2 x = −3.29 m/s. The cart is moving to the left at 3.29 m/s after the package lands in it. 8.82: Even though one of the masses is not known, the analysis of Section (8.4) leading to Eq. (8.26) is still valid, and vred = 0.200 m/s + 0.050 m/s = 0.250 m/s. b) The mass mred may be found from either energy or momentum considerations. From momentum conservation, (0.040 kg)(0.200 m/s − 0.050 m/s) mred = = 0.024 kg. (0.250 m/s) As a check, note that K1 = 1 (0.040 kg)(0.200 m/s) 2 = 8.0 × 10−4 J, and 2 K 2 = 1 (0.040 kg)(0.050 m/s) 2 + 1 (0.024 kg)(0.250 m/s) 2 = 8.0 × 10− 4 J, 2 2 so K1 = K 2 , as it must for a perfectly elastic collision.

8.83: a) In terms of the primed coordinates,     2 ′ v A = (v ′ + v cm ) ⋅ (v A + v cm ) A       = v ′ ⋅ v ′ + v cm ⋅ v cm + 2v ′ ⋅ v cm A A A   2 2 = v ′ + vcm + 2v ′ ⋅ v cm , A A
2 with a similar expression for v B . The total kinetic energy is then 1 1 2 2 K = m A v A + mB v B 2 2     1 1 2 2 2 2 ′ = mA v′ + vcm + 2v ′ ⋅ vcm + mB v′ + vcm + 2v B ⋅ vcm A A B 2 2  2  2 1 1 2 ′ = (mA + mB )vcm + mAv ′ + mB v B A 2 2     ′ + 2[ mAv ′ .vcm + mB v B .vcm ] . A

(

)

(

)

(

)

The last term in brackets can be expressed as    ′ 2(mAv ′ + mB v B ) ⋅ vcm , A and the term      ′ ′ m A v ′ + m B v B = m A v ′ + m B v B − ( m A + m B ) v cm A A = 0, and so the term in square brackets in the expression for the kinetic energy vanishes, showing the desired result. b) In any collision for which other forces may be neglected 2 the velocity of the center of mass does not change, and the 1 Mvcm in the kinetic energy 2 will not change. The other terms can be zero (for a perfectly inelastic collision, which is 2 not likely), but never negative, so the minimum possible kinetic energy is 1 Mvcm . 2 8.84: a) The relative speed of approach before the collision is the relative speed at which the balls separate after the collision. Before the collision, they are approaching with relative speed 2 v , and so after the collision they are receding with speed 2 v . In the limit that the larger ball has the much larger mass, its speed after the collision will be unchanged (the limit as mA >> mB in Eq. (8.24)), and so the small ball will move upward with speed 3 v . b) With three times the speed, the ball will rebound to a height time times greater than the initial height.

8.85: a) If the crate had final speed v , J&J have speed 4.00 m s − v relative to the ice, and so (15.0 kg) v = (120.0 kg)(4.00 m/s − v ). Solving for v, v = b) After Jack jumps, the speed of the crate is
( 75.0 kg) (135.0 kg) (120.0 kg) (4.00 m s) (135 .0 kg)

= 3.56 m s .

(4.00 m s) = 2.222 m s , and the

momentum of Jill and the crate is 133.3 kg ⋅ m s . After Jill jumps, the crate has a speed v and Jill has speed 4.00 m s − v, and so 133.3 kg ⋅ m s = (15.0 kg )v − (45.0 kg)(4.00 m s − v), and solving for v gives v = 5.22 m s. c) Repeating the calculation for part (b) with Jill jumping first gives a final speed of 4.67 m s.

8.86: (a) For momentum to be conserved, the two fragments must depart in opposite directions. We can thus write M AVA = − M BVB Since M A = M − M B , we have

( M − M B )VA = − M BVB VA − MB = VB M − M B

Then for the ratio of the kinetic energies
2 2 KE A 1 M AVA M A MB M = 2 = = B 2 2 1 KE B M B (M − M B ) MA 2 M BVB

The ratio of the KE’s is simply the inverse ratio of the masses. From the two equations M KEA = B KEB and KEA + KEB = Q MA We can solve for KE B to find

 MA  KEA = Q − KEB = Q1 −  M +M    A B  Q QM A KEB = = MB 1+ MA MA + MB (b) If M B = 4M A , then M A will get 4 times as much KE as M B , or 80% of Q for M A and 20% for M B .

8.87: Let the proton be moving in the + x -direction with speed v p after the decay. The initial momentum of the neutron is zero, so to conserve momentum the electron must be moving in the − x -direction after the collision; let its speed be ve . Px is constant gives 0 = − me ve + m p v p . ve = ( m p me ) v – p The total kinetic energy after decay is K tot = 1 me ve2 + 1 m p v 2 . Using the p 2 2 momentum equation to replace ve gives K tot = 1 m p v 2 ( 1+ m p me ). p 2 Thus Kp K tot = 1 1 = = 5.44 × 10−4 = 0.0544 % 1 + mp me 1836
0.0176 1.0176 1 and 1.0176 , so the kinetic energies are

8.88: The ratios that appear in Eq. ( 8.42 ) are a)
0.0176 1.0176

1 (6.54 × 10−13 J) = 1.13 × 10-14 J and b) 1.0176 (6.54 × 10−13 J) = 6.43 × 10−13 J.

Note that the energies do not add to 6.54 × 10 −13 J exactly, due to roundoff. 8.89: The “missing momentum” is 5.60 × 10−22 kg ⋅ m s − (3.50 × 10 −25 kg)(1.14 × 10 3 m/s ) = 1.61 × 10−22 kg ⋅ m s . Since the electron has momentum to the right, the neutrino’s momentum must be to the left.

8.90: a ) For the x - and y - directions, respectively, and m as the common mass of a proton, mv A1 = mv A 2 cosα + mv2 cos β 0 = mv A2 sin α − mvB 2 sin β or v A1 = v A 2 cosα + v B 2 cos β 0 = v A2 sin α − vB 2 sin β . b ) After minor algebra,
2 2 2 v A1 = v A2 + vB 2 + 2v A2vB 2 ( cos α cos β − sin α sin β ) 2 2 = v A 2 + vB 2 + 2v A2vB 2 cos(α + β ).

c ) For a perfectly elastic collision, 1 2 1 2 1 2 2 2 2 mv A1 = mv A 2 + mv B 2 or v A1 = v A2 + v B 2 . 2 2 2 Substitution into the above result gives cos(α + β) = 0. d) The only positive angle with zero cosine is π (90°). 2 8.91: See Problem 8.90. Puck B moves at an angle 65.0 °(i.e. 90° − 25° = 65°) from the original direction of puck A’s motion, and from conservation of momentum in the y - direction, vB 2 = 0.466v A 2 . Substituting this into the expression for conservation of momentum in the x − direction, v A2 = v A1 (cos 25.0ο + 0.466 cos 65ο ) = 13.6 m s , and so v B 2 = 6.34 m/s. As an alternative, a coordinate system may be used with axes along the final directions of motion (from Problem 8.90, these directions are known to be perpendicular). The initial direction of the puck’s motion is 25.0° from the final direction, so v A 2 = v A1 cos 25.0°and vB 2 = v A1 cos 65.0, giving the same results.

8.92: Since mass is proportional to weight, the given weights may be used in determining velocities from conservation of momentum. Taking the positive direction to the left, (800 N)(5.00 m/s) cos 30.0ο − (600 N)(7.00 m/s) cos 36.9ο v= = 0.105 m s 1000 N 8.93: a ) From symmetry, the center of mass is on the vertical axis, a distance ( L 2) cos(α 2) from the apex. b) The center of mass is on the (vertical) axis of symmetry, a distance 2( L / 2) / 3 = L 3 from the center of the bottom of the . c) Using the wire frame as a coordinate system, the coordinates of the center of mass are equal, and each is equal to ( L 2) / 2 = L 4 . The distance of this point from the corner is

(a), with α = 45° d ) By symmetry, the center of mass is in the center of the equilateral
triangle, a distance ( L 2)(tan 60°) = L 12 = (0.289) L above the center of the base.

(1 8 )L = (0.353) L. This may also be found from consideration of the situation of part

8.94: The trick here is to notice that the final configuration is the same as if the canoe ( assumed symmetrical ) has been rotated about its center of mass. Intially, the center of mass is a distance
( 45.0 kg) (1.5 m) (105 kg)

= 0.643 m from the center of the canoe, so in rotating about

this point the center of the canoe would move 2 × 0.643 m = 1.29 m. 8.95: Neglecting friction, the total momentum is zero, and your speed will be one-fifth of the slab’s speed, or 0.40 m / s.

8.96: The trick here is to realize that the center of mass will continue to move in the original parabolic trajectory, “landing” at the position of the original range of the projectile. Since the explosion takes place at the highest point of the trajectory, and one fragment is given to have zero speed after the explosion, neither fragment has a vertical component of velocity immediately after the explosion, and the second fragment has twice the velocity the projectile had before the explosion. a) The fragments land at positions symmetric about the original target point. Since one lands at 1 R, the other 2 lands at 2 3 3 v0 3 (80 m / s) 2 R= sin 2α 0 = sin 120ο = 848 m. 2 2 2 g 2 (9.80 m / s ) b) In terms of the mass m of the original fragment and the speed v before the explosion, K1 = 1 mv 2and K 2 = 1 m (2v) 2 = mv 2 , so ∆K = mv 2 − 1 mv 2 = 1 mv 2 . The speed v is related 2 2 2 2 2 to v0 by v = v0 cos α 0 , so ∆K = 1 2 1 mv0 cos 2 α 0 = (20.0 kg )(80 m/s ) cos 60.0°) 2 = 1.60 × 10 4 J. 2 2

8.97: Apply conservation of energy to the explosion. Just before the explosion the sheel is at its maximum height and has zero kinetic energy. Let A be the piece with mass 1.40 kg and B be the piece with mass 0.28 kg. Let v A and v B be the speeds of the two pieces immediately after the collision.
1 2 2 2 mAv A + 1 mB vB = 860 J 2

Since the two fragments reach the ground at the same time, their velocitues just after the explosion must be horizontal. The initial momentum of the shell before the explosion is zero, so after the explosion the pieces must be moving in opposite horizontal directions and have equal magnitude of momentum: m A v A = m B v B . Use this to eliminate v A in the first equation and solve for v B :
1 2 2 mB vB (1 + mB / m A ) = 860 J and vB = 71.6 m/s.

Then v A = (mB m A )vB = 14.3 m/s. b) Use the vertical motion from the maximum height to the ground to find the time it takes the pieces to fall to the ground after the explosion. Take + y downward. v0 y = 0, a y = +9.80 m/s 2 , y − y 0 = 80.0 m, t = ? y − y 0 = v0 y t + 1 a y t 2 gives t = 4.04 s. 2 During this time the horizontal distance each piece moves is x A = v A t = 57.8 m and x B = v B t = 289.1 m. They move in opposite directions, so they are x A + x B = 347 m apart when they land.

8.98: The two fragments are 3.00 kg and 9.00 kg. Time to reach maximum height = time to fall back to the ground.

v9 = v0 sin θ − gt 0 = (150 m s) sin 55.0ο − 9.8 m s t
2

t = 12.5 s.

The heavier fragment travels back to its starting point, so it reversed its velocity. vx = v0 cos θ = (150 m s) cos 55ο = 86.0 m s to the left after the explosion; this is v9 . Now get v3 using momentum conversation. Mv0 = m3v3 + m9v9

(12 kg )( 86.0 m s ) = ( 3.00 kg ) v3 + ( 9.00kg )( − 86.0 m s )

v3 = 602 m s x3 = x +x = (86.0 m s)(12.5 s) + (602 m s)(12.5) Before explosion After explosion x3 = 8600m from where it was launched Energy released = Energy after explosion – Energy before explosion 1 1 1 2 2 2 m 3 v 3 + m 9 v 9 − ( m3 + m 9 ) v 9 2 2 2 1 1 = (3.00 kg)(602 m s) 2 + (9.00 kg )(86.0 m s) 2 2 2 1 − (12.0 kg)(86.0 m s) 2 2 = 5.33 × 10 5 J =

8.99: The information is not sufficient to use conservation of energy. Denote the emitted neutron that moves in the + y - direction by the subscript 1 and the emitted neutron that moves in the –y-direction by the subscript 2. Using conservation of momentum in the xand y-directions, neglecting the common factor of the mass of neutron, v0 = (2v0 3) cos 10° + v1 cos 45° + v 2 cos 30° 0 = (2v0 3)sin 10° + v1 sin 45° − v 2 sin 30°. With sin 45° = cos 45° , these two relations may be subtracted to eliminate v1, and rearrangement gives v0 (1 − (2 3) cos 10° + (2 3)sin 10°) = v2 (cos 30° + sin 30°), from which v 2 = 1.01 × 10 3 m s or 1.0 × 10 3 m s to two figures. Substitution of this into either of the momentum relations gives v1 = 221 m s. All that is known is that there is no z - component of momentum, and so only the ratio of the speeds can be determined. The ratio is the inverse of the ratio of the masses, so v Kr = (1.5)v Ba .

8.100: a) With block B initially at rest, vcm =

mA m A + mB

v A1 . b) Since there is no net external

force, the center of mass moves with constant velocity, and so a frame that moves with the center of mass is an inertial reference frame. c) The velocities have only x components, and the x -components are u A1 = v A1 − vcm = m Am+Bm B v A1 , u B1 = −vcm = − m Am+Am B u A1 . Then, Pcm = mAu A1 + mBu B1 = 0 . d) Since there is zero momentum in the center-of-mass frame before the collision, there can be no momentum after the collision; the momentum of each block after the collision must be reversed in direction. The only way to conserve kinetic energy is if the momentum of each has the same magnitude so in the center-of-mass frame, the blocks change direction but have the same speeds. Symbolically, u A2 = −u A1 , u B 2 = −u B1 . e) The velocities all have only x –components; these components are u A1 = 0..200 6.00 m s = 2.00 m s, u B1 = − 0..400 6.00 m s = −4.00 m s, u A 2 = −2.00 m s, u B 2 = 4.00 m s, 0 600 0 600 and v A 2 = +2.00 m s, v B 2 = 8.00 m s. and Equation (8.24) predicts v A 2 = + 1 v A1 and Eq. 3 (8.25) predicts v B 2 = 4 u A1 , which are in agreement with the above. 3

8.101: a) If the objects stick together, their relative speed is zero and ∈= 0 . b) From Eq. (8.27), the relative speeds are the same, and ∈= 1 . c) Neglecting air resistance, the speeds before and after the collision are
2

2 gh and

2gH 1 , and ∈ =
k +1

2 gH1 2 gh

=

H1 h . d)

From part (c), H1 = ∈2 h = ( 0.85) (1.2 m ) = 0.87 m. e) H H n = ∈2 n h. f) (1.2 m ) ( 0.85) = 8.9 cm.
16

= H k ∈2 , and by induction

8.102: a) The decrease in potential energy ( − ∆ < 0 ) means that the kinetic energy increases. In the center of mass frame of two hydrogen atoms, the net momentum is necessarily zero and after the atoms combine and have a common velocity, that velocity must have zero magnitude, a situation precluded by the necessarily positive kinetic energy. b) The initial momentum is zero before the collision, and must be zero after the collision. Denote the common initial speed as v0 , the final speed of the hydrogen atom as v , the final speed of the hydrogen molecule as V , the common mass of the hydrogen atoms as m and the mass of the hydrogen molecules as 2 m . After the collision, the two particles must be moving in opposite directions, and so to conserve momentum, v = 2V . From conservation of energy, 1 ( 2m)V 2 − ∆ + 1 mv 2 = 3 1 mv02 2 2 2 2 mV 2 − ∆ + 2mV 2 = 3 mv0 2 V2 =
2 v0 ∆ + , 2 3m

from which V = 1.203 × 10 4 m/s, or 1.20 × 10 4 m/s to two figures and the hydrogen atom speed is v = 2.41 × 10 4 m/s. 8.103: a) The wagon, after coming down the hill, will have speed After the “collision”, the speed is 300 kg (10 m/s ) = 6.9 m/s, and in the 5.0 s, the wagon 435 kg will not reach the edge. b) The “collision” is completely inelastic, and kinetic energy is not conserved. The change in kinetic energy is 2 2 1 1 2 ( 435 kg )( 6.9 m/s ) − 2 ( 300 kg )(10 m/s ) = − 4769 J, so about 4800 J is lost.

(

)

2gL sin α = 10 m/s.

8.104: a) Including the extra force, Eq. ( 8.37 ) becomes dv dm = − vex − mg , dt dt where the positive direction is taken upwards (usually a sign of good planning). b) Diving by a factor of the mass m, m a= v dm dv = − ex − g. dt m dt

c) 20 m s 2 − 9.80 m s 2 = 10.2 m s 2 . d) 3327 m s − (9.80 m/s 2 ) ( 90 ) = 2.45 km s , which is about three-fourths the speed found in Example 8.17. 8.105: a) From Eq. ( 8.40 ) , v = vex ln

b) vex ln (13,000 / 4,000 ) = (1.18) vex . c) (1.18) vex + vex In(1000 / 300 ) = ( 2.38) vex . d) Setting the result of part (c) equal to

(

13,000 kg 3,300 kg

) = (1.37) v

ex

.

7.00 km/s and solving for vex gives vex = 2.94 km/s. 8.106: a) There are two contribution to Fnet , Fnet = vex | dm dt | −v | dm dt |, or Fnet = ( vex − v ) dm dt . vex − v.

b) Fnet / | dm / dt |= (1300 N ) (150 kg s) = 8.66 m s = 31 km/h. This equal to

8.107: a) For t < 0 the rocket is at rest. For 0 ≤ t ≤ 90 s, Eq. ( 8.40 ) is valid, and v( t ) = ( 2400 m s ) ln (1 (1 − ( t / 120 s ) ) ). At t = 90 s, this speed is 3.33 km/s, and this is also the speed for t > 90 s.

b) The acceleration is zero for t < 0 and t > 90 s. For 0 ≤ t ≤ 90 s, Eq. (8.39) gives, 2 with dm = −m0 120 s, a = ( 1−20 /m/s s ) ) . dt ( t 120

c) The maximum acceleration occurs at the latest time of firing, t = 90 s, at which time 2 20 the acceleration is, from the result of part (a), ( 1−90m/s ) = 80 m/s 2 , and so the astronaut is / 120 subject to a force of 6.0 kN, about eight times her weight on earth.

8.108: The impulse applied to the cake is J = μk1mgt = mv , where m is the mass of the cake and v is its speed after the impulse is applied. The distance d that the cake moves during this time is then d = 1 µ k1 gt 2 . While sliding on the table, the cake must lose its 2 gives r − d = 1 g 2
2 μk 1 μk 2

kinetic energy to friction, or µ k 2 mg ( r − d ) = 1 mv 2 . Simplification and substitution for v 2 t 2 , substituting for d in terms of t 2 gives r= which gives t = 0.59 s. 8.109: a) Noting than dm =
M L

1 2 µ2  1 µ gt  µ k1 + k1  = gt 2 k1 ( µ k1 + µ k 2 ) ,   2 2  µk 2  µk 2

dx avoids the intermediate variable ρ . Then, x cm = 1 M

∫

L

0

x

M L dx = . L 2

b) In this case, the mass M may be found in terms of ρ and L, specifically by using

dm = ρAdx = αAdx to find that M = αA ∫ xdx = αA L2 2 . Then,
xcm = 2 αAL2

∫

L

0

αAx 2 dx =

2 L3 2 L = . αAL2 3 3

8.110: By symmetry, x cm = 0 . Using plane polar coordinates leads to an easier integration, and using the Theorem of Pappus 2 πycm πa2 = 4 πa 3 is easiest of all, but the 3 method of Problem 8.109 involves Cartesian coordinates. For the x-coordinate, dm = ρt a 2 − x 2 dx , which is an even function of x, so from 0 to a, so

(

( )
2

)

∫ x dx = 0. For the y-coordinate, dm = ρt 2
ycm =

a 2 − y 2 dy , and the range of integration is

2 ρt a y a 2 − y 2 , dy. M ∫0

Making the substitutions M = 1 ρπa 2t , u = a 2 − y 2 , du = −2 y, and 2 y cm −2 = 2 πa −4 ∫a 2 u du = 3πa 2
o 1 2

 3 4a 2 . u  = 3π 2  a

0

8.111: a) The tension in the rope at the point where it is suspended from the table is T = ( λx ) g , where x is the length of rope over the edge, hanging vertically. In raising the rope a distance − dx , the work done is ( λg ) x ( − dx ) ( dx is negative ). The total work done is then x2 − ∫ ( λg ) x dx = ( λg ) l/4 2
0 l/4 0

λgl 2 = . 32

b) The center of mass of the hanging piece is initially a distance l/8 below the top of the table, and the hanging weight is ( λg ) ( l / 4 ) , so the work required to raise the rope is ( λg ) ( l / 4) ( l / 8) = λgl 2 / 32, as before.

8.112: a) For constant acceleration a, the downward velocity is v = at and the distance x that the drop has fallen is x = 1 at 2 . Substitution into the differential equation gives 2 1 2 1 3 2 at g = at 2 a + ( at ) = a 2 t 2 , 2 2 2 the non-zero solution of which is a = g . 3 1 2 1  9.80 m/s 2  ( 3.00 s ) 2 = 14.7 m. at =   2 2 3   c) kx = ( 2.00 g/m ) (14.7 m ) = 29.4 g. b)

Chapter 9

9.1: a)

1.50 m = 0.60 rad = 34.4°. 2.50 m (14.0 cm) = 6.27 cm. (128°)(π rad 180°)

b) c)

(1.50 m)(0.70 rad) = 1.05 m.

9.2: a) b)

rev   2π rad   1 min   1900 ×   = 199 rad s. min   rev   60 s   (35° × π rad 180°) (199 rad s) = 3.07 × 10 −3 s.

dωz = (12.0 rad s3 ) t , so at t = 3.5 s, α = 42 rad s 2 . The angular acceleration dt is proportional to the time, so the average angular acceleration between any two times is the arithmetic average of the angular accelerations. b) ω z = (6.0 rad s 3 )t 2 , so at t = 3.5 s, ωz = 73.5 rad s. The angular velocity is not linear function of time, so the average angular velocity is not the arithmetic average or the angular velocity at the midpoint of the interval. 9.3: a) α z = αz ( t) =
dω z dt

9.4: b)

a)

= −2 βt = (−1.60 rad s3 )t.

α z (3.0 s) = (−1.60 rad s3 )(3.0 s) = −4.80 rad s 2 .

ω(3.0 s) − ω(0) − 2.20 rad s − 5.00 rad s = = −2.40 rad s 2. , 3.0 s 3.0 s which is half as large (in magnitude) as the acceleration at t = 3.0 s. αav − z = 9.5: a) ωz = γ + 3 βt 2 = (0.400 rad s) + (0.036 rad s3 )t 2 b) At t = 0, ωz = γ = .50 0.400 rad s. c) At t = 5.00 s, ωz = 1.3 rad s, θ = 3.50 rad, so ωav − z = 35.00rad = 0.70 rad s. s The acceleration is not constant, but increasing, so the angular velocity is larger than the average angular velocity.

dθ dt

= 2bt − 3ct 2 and α z =

dw z dt

= 2b − 6ct. b) Setting α z = 0, t =

b 3c

.

9.8: (a) The angular acceleration is positive, since the angular velocity increases steadily from a negative value to a positive value. (b) The angular acceleration is α= ω − ω0 8.00 rad s − (−6.00 rad s) = = 2.00 rad s 2 t 7.00 s

9.10: a) ωz = ω0 z + α z t = 1.50 rad s + (0.300 rad s 2 )(2.50 s) = 2.25 rad s. b) θ = ω0 z t + 1 2 α z t 2 = (1.50 rad s)(2.50 s) + 1 (0.300 rad s 2 )(2.50 s) 2 = 4.69 rad. 2
min (200 rev min − 500 rev min ) × ( 160 s ) rev = −1.25 2 . (4.00 s) s The number of revolutions is the average angular velocity, 350 rev min, times the time interval of 0.067 min, or 23.33 rev. b) The angular velocity will decrease by another min 1 200 rev min in a time 200 revmin ⋅ 1.25 rev s 2 = 2.67 s. 60 s

9.11: a)

9.12: a) Solving Eq. (9.7) for t gives t =

ω z − ω0 z αz

.

Rewriting Eq. (9.11) as θ − θ0 = t (ω0 z + 1 α z t ) and substituting for t gives 2  ω − ω0 z  1   ω0 z + (ωz − ω0 z )  θ − θ0 =  z  α  2   z  1  ω + ω0 z  = (ωz − ω0 z ) z  α 2   1 = (ωz2 − ω2 0 z ), 2α which when rearranged gives Eq. (9.12).
2 2 b) α z = (1 2 )(1 ∆θ ) ωz − ω0 z = (1 2 ) (1 (7.00 rad)) (16.0 rad s ) − (12.0 rad s ) = 2 2

(

)

(

)

8 rad s 2 . 9.13: a) From Eq. ( 9.7 ) , with ω0 z = 0, t =

ωz αz

rad = 136.0rad ss2 = 24.0 s. .50 ( 36.0 rad s) 2 2 (1.50 rad) s 2

b ) From Eq. ( 9.12 ) , with ω0 z = 0, θ − θ0 =

= 432 rad = 68.8 rev.

9.14: a) The average angular velocity is 162 rad = 40.5 rad s, and so the initial angular 4.00 s velocity is 2ωav − z − ω2 z = ω0 z , ω0z = − 27 rad s. ∆ωz 108 rad s − (−27 rad s) b) αz = = = 33.8 rad s 2 . ∆t 4.00 s

9.15: From Eq. (9.11), 2 θ − θ0 α z t 60.0 rad (2.25 rad s )(4.00 s) ω0 z = − = − = 10.5 rad s. t 2 4.00 s 2

9.16: From Eq. (9.7), with ω0 z = 0, α z =

ωz t

rad = 140.00 s s = 23.33 rad s . The angle is most 6

2

easily found from θ = ωav − zt = (70 rad s)(6.00 s) = 420 rad. 9.17: From Eq. (9.12), with ω z = 0, the number of revolutions is proportional to the square of the initial angular velocity, so tripling the initial angular velocity increases the number of revolutions by 9, to 9.0 rev.

9.18: The following table gives the revolutions and the angle θ through which the wheel has rotated for each instant in time and each of the three situations:

t 0.05 0.10 0.15

(a ) rev' s θ 0.50 180 1.00 360 1.50 540

( b) rev' s 0.03 0.13

θ 11.3 45

( c) rev' s θ 0.44 158 0.75 270 0.94 338

0.28 101

0.20 2.00 720 0.50 180 1.00 360 –––––––––––––––––––––––––––––––––––––– The θ and ωz graphs are as follows: a)

b)

c)

(

)

time when the wheel stops is 12.3 s . c) Of the many ways to find the angular acceleration, the most direct is to use the intermediate calculation of part (b) to find that while slowing down ∆ωz = − 84 rad s so α z = −84 .rad s = − 8.17 rad s 2 . 10 3 s

9.20: a) Equation (9.7) is solved for ω0 z = ωz − α z t , which gives ωz − ave = ωz − α2z t , or θ − θ0 = ωz t − 1 α z t 2 . b) 2 2

(

ωz t

− ∆2θ = − 0.125 rad s 2 . c) ωz − α z t = 5.5 rad s. t

)

9.21: The horizontal component of velocity is rω , so the magnitude of the velocity is a) 47.1 m/s b)   π rad/s   2  (5.0 m)(90 rev / min )    + (4.0 m/s ) = 47.3 m/s.    30 rev/min   
2

9.22: a)

1.25 m s 25.0 ×10 −3 m

= 50.0 rad s ,

1.25 m 58.0 ×10 −3

= 21.55 rad s , or 21.6 rad s to three figures.

b) (1.25 m s ) (74.0 min) (60 s min) = 5.55 km. c) α z =
50.0 rad s − 21.55 rad s ( 74.0 min) (60 s min)

= 6.41 × 10 −3 rad s 2 .

9.23: a) ω2 r = (6.00 rad s) 2 (0.500 m) = 18 m s 2 . b) v = ωr = (6.00 rad s) (0.500 m) = 3.00 m s , and
v2 r

=

( 3.00 m s ) 2 ( 0.500 m )

= 18 m s 2 .

9.24: From arad = ω2 r , 400,000 × 9.80 m s 2 a = = 1.25 × 10 4 rad s, r 2.50 × 10 −2 m which is (1.25 × 10 4 rad s) 11rev 2π rad = 1.20 × 10 5 rev min. min 60 s ω=

(

)

9.25: a) arad = 0, atan = αr = ( 0.600 rad s 2 ) ( 0.300 m ) = 0.180 m s 2 and so a = 0.180 m s 2 . b)θ = π rad, so arad = ω2 r = 2 0.600 rad s 2 ( π 3 rad ) ( 0.300 m ) = 0.377 m s 2 . 3

(

)

The tangential acceleration is still 0.180 m s 2 , and so on a= still 0.180 m/s 2 9.26: a) ωz = ω0z + α z t = 0.250 rev s + 0.900 rev s 2 ( 0.200 s ) = 0.430 rev s (note that since ω0 z and α z are given in terms of revolutions, it’s not necessary to convert to radians). b) ωav − z ∆t = (0.340 rev s) (0.2s) = 0.068 rev . c) Here, the conversion to radians must be made to use Eq. (9.13), and  0.750 m  v = rω =   ( 0.430 rev/s × 2π rad rev ) = 1.01 m s. 2   d) Combining equations (9.14) and (9.15), a = a 2 rad + a 2 tan = (ω2 r ) 2 + (αr ) 2 = [((0.430 rev s × 2π rad rev) 2 (0.375 m))2 + ((0.900 rev s 2 × 2π rad rev)(0.375 m)) 2 = 3.46 m s 2 . 9.27: r= arad (3000)(9.80 m s 2 ) = π rad s ω2 (5000 rev min) 30 rev min

( 0.180 m s ) + ( 0.377 m s )
2 2

2 2

= 0.418 m s 2 .

c) For an angle of 120°, a rad = 0.754 m s 2 , and a = 0.775 m s 2 , since a tan is

(

)

]

1 2

(

(

))

2

= 10.7 cm,

so the diameter is more than 12.7 cm, contrary to the claim.

9.28: a) Combining Equations (9.13) and (9.15), v arad = ω2 r = ω2   = ωv. ω a rad 0.500 m s b) From the result of part (a), ω = v = 2.00 m s = 0.250 rad s .

9.29: a) ωr = (1250 rev min ) b)
v2 r

(

rad s π 30 rev min

)(

12.7 ×10 −3 m 2

) = 0.831 m s .

=

( 0.831 m s )

2

(12.7 ×10 −3 m ) 2

= 109 m s 2 .

9.30: a) α =

ω= v = = 250 rad s and at t = 0, v = 50.0 m s + (−10.0 m s 2 ) r (0 − 3.00 s) = 80.0 m s , so ω = 400 rad s . c) ωavet = (325 rad s)(3.00 s)
= 975 rad = 155 rev. d) v = arad r = (9.80 m s 2 )(0.200 m) = 1.40 m s . This speed will

a tan r 50.0 m s 0.200

.0 m = −10.200 ms = − 50.0 rad s 2 b) At t = 3.00 s, v = 50.0 m s and 0

2

be reached at time

50.0 m s −1.40 m s 10.0 m s

= 4.86 s after t = 3.00 s, or at t = 7.86 s . (There are many

equivalent ways to do this calculation.) 9.31: (a) For a given radius and mass, the force is proportional to the square of the angular velocity;

(

640 rev min 2 423 rev min

)

= 2.29 (note that conversion to rad s is not necessary for this

part). b) For a given radius, the tangential speed is proportional to the angular velocity; 640 = 1.51 (again conversion of the units of angular speed is not necessary). 423 c) (640 rev min ) arad =
v2 r

(

π rad s 30 rev min

)(

0.470 m 2

) = 15.75 m s, or 15.7 m s to three figures, and

=

(15.75 m s) ( 0.470 m 2 )

2

= 1.06 × 103 m s 2 = 108 g.

9.32: (a)

vT = Rω  7.5 rev   1 min   2π rad  2.00 cm s = R      min   60 s   1 rev  R = 2.55 cm D = 2 R = 5.09 cm aT = Rα α= aT 0.400 m s 2 = = 15.7 rad s 2 R 0.0255 m

b)

vr 5.00 m s = = 15.15 rad s. r 0.330 m The angular velocity of the front wheel is ωf = 0.600 rev s = 3.77 rad s Points on the chain all move at the same speed, so rr ωr = rf ωf rr = rr ( ωf ωr ) = 2.99 cm 9.33: The angular velocity of the rear wheel is ωr = 9.34: The distances of the masses from the axis are the moment of inertia is
2 2 L 4

, L and 34L , and so from Eq. ( 9.16 ) , 4
2

L L  3L  11 I = m  + m  + m  = mL2 . 4 4  4  16
L2 12 2
2

9.35: The moment of inertia of the cylinder is M moment of inertia of the combination is

M ( 12 + m ) L . 2

and that of each cap is m L4 , so the

9.36: Since the rod is 500 times as long as it is wide, it can be considered slender. a) From Table ( 9.2( a ) ) , I= 1 1 2 ML2 = ( 0.042 kg ) (1.50 m ) = 7.88 × 10− 3 kg ⋅ m 2 . 12 12

b) From Table ( 9.2( b ) ) , 1 1 2 I = ML2 = ( 0.042 kg ) (1.50 m ) = 3.15 × 10− 2 kg ⋅ m 2 . 3 3 c) For this slender rod, the moment of inertia about the axis is obtained by considering it as a solid cylinder, and from Table ( 9.2( f ) ) , I= 1 1 MR 2 = (0.042 kg) (1.5 × 10− 3 m) 2 = 4.73 × 10−8 kg ⋅ m 2 . 2 2

9.37: a) For each mass, the square of the distance from the axis is 2(0.200 m) 2 = 8.00 × 10−2 m 2 , and the moment of inertia is 4(0.200 kg) (0.800 × 10−2 m 2 ) = 6.40 × 10−2 kg ⋅ m 2 . b) Each sphere is 0.200 m from the axis, so the moment of inertia is 4( 0.200 kg ) ( 0.200 m ) = 3.20 × 10−2 kg ⋅ m 2 . a) The two masses through which the axis passes do not contribute to the moment of
2

inertia.

I = 2(0.2 kg) 0.2 2 m = 0.032 kg ⋅ m 2 .

(

)

2

9.38: (a) I = I bar + I balls = =

1  L M bar L2 + 2mballs   12 2

2

1 ( 4.00 kg ) ( 2.00 m ) 2 + 2( 0.500 kg ) (1.00 m ) 2 = 2.33 kg ⋅ m 2 12

1 (b) I = mbar L2 + mball L2 3 = 1 ( 4.00 kg ) ( 2.00 m ) 2 + ( 0.500 kg ) ( 2.00 m ) 2 = 7.33 kg ⋅ m 2 3

c) I = 0 because all masses are on the axis (d) I = mbar d 2 + 2mball d 2 = M Total d 2 = (5.00 kg )(0.500 m) 2 = 1.25 kg ⋅ m 2

9.39:

I = I d + I r (d = disk, r = ring )
3

disk : md = (3.00 g cm )πrd2 = 23.56 kg 1 md rd2 = 2.945 kg ⋅ m 2 2 ring : mr = (2.00 g cm3 ) π (r22 − r12 ) = 15.08 kg Id = Ir = 1 mr (r12 + r22 ) = 5.580 kg ⋅ m 2 2 I = I d + I r = 8.52 kg ⋅ m 2

( r1 = 50.0 cm, r2 = 70.0 cm )

9.40: a) In the expression of Eq. (9.16), each term will have the mass multiplied by f 3 and the distance multiplied by f , and so the moment of inertia is multiplied by f 3 ( f ) 2 = f 5 . b) (2.5)(48)5 = 6.37 × 108. 9.41: Each of the eight spokes may be treated as a slender rod about an axis through an end, so the moment of inertia of the combination is m  I = mrim R 2 + 8  spoke  R 2  3    8   = (1.40 kg ) + (0.20 kg) (0.300 m) 2 3   2 = 0.193 kg ⋅ m 9.42: a) From Eq. (9.17), with I from Table (9.2(a)), 1 1 1 rev 2π rad rev 2 K= mL2ω2 = (117 kg )(2.08 m) 2 (2400 × ) = 1.3 × 106 J. 2 12 24 min 60 s min b) From mgy = K , K 1.3 × 10 6 J y= = = 1.16 × 10 3 m = 1.16 km. 2 mg (117 kg )(9.80 m s )

(

)

9.43: a) The units of moment of inertia are [kg] [m2 ] and the units of ω are equivalent to [s −1 ] and so the product 1 Iω 2 has units equivalent to [kg ⋅ m ⋅ s −2 ] = [kg ⋅ (m s) 2 ] , 2 which are the units of Joules. A radian is a ratio of distances and is therefore unitless. b) K = π 2 Iω2 1800 , when is in rev min. 9.44: Solving Eq. (9.17) for I, 2K 2(0.025J) I= 2 = = 2.25 × 10 −3 kg ⋅ m 2 . 2 π rad s 2 ω (45 rev min × 60 rev min )

2 9.45: From Eq. (9.17), K 2 − K = 1 I (ω2 − ω12 ), and solving for I, 2

I =2 =2

( K 2 − K1 )
2 (ω2 − ω12 )

(−500 J) ((520 rev min ) 2 − (650 rev min ) 2 )

(

2 π rad s 30 rev min

)

= 0.600 kg ⋅ m 2 .

9.46: The work done on the cylinder is PL, where L is the length of the rope. Combining Equations (9.17), (9.13) and the expression for I from Table (9.2(g)), PL = 1ω 2 1 ω v2 (40.0 N)(6.00 m s) 2 v , or P = = = 14.7 N. 2g 2 g L 2(9.80 m s 2 )(5.00 m) 1 MR 2 , Eq. (9.17) 2

9.47: Expressing ω in terms of αrad , ω2 =

. Combining with I =

11 (70.0 kg )(1.20 m)(3500 m s) 2 becomes K = MRarad = = 7.35 × 10 4 J. 22 4 9.48: a) With I = MR 2 , with expression for v is v= 2 gh . 1+ M m

b) This expression is smaller than that for the solid cylinder; more of the cylinder’s mass is concentrated at its edge, so for a given speed, the kinetic energy of the cylinder is larger. A larger fraction of the potential energy is converted to the kinetic energy of the cylinder, and so less is available for the falling mass.

9.49: a) ω =

2π T

, so Eq. (9.17) becomes K = 2π2 I T 2 .

b) Differentiating the expression found in part (a) with respect to T, dK = (−4π 2 I T 3 ) dT . dt dt c) 2π 2 (8.0 kg ⋅ m 2 ) (1.5 s) 2 = 70.2 J, or 70 to two figures. d) (−4π 2 (8.0 kg ⋅ m 2 ) (1.5 s) 3 )(0.0060) = −0.56 W. 9.50: The center of mass has fallen half of the length of the rope, so the change in gravitational potential energy is 1 1 − mgL = − (3.00 kg )(9.80 m s 2 )(10.0 m) = −147 J. 2 2 9.51: (120 kg )(9.80 m s 2 )(0.700 m) = 823 J. 9.52: In Eq; (9.19), I cm = MR 2 and d = R 2 , so I P = 2 MR 2 . 2 2 4 MR 2 = MR 2 + Md 2 , so d 2 = R 2 , and the axis comes nearest to the center of 3 5 15 the sphere at a distance d = (2 15 ) R = (0.516) R. 9.53: 9.54: Using the parallel-axis theorem to find the moment of inertia of a thin rod about an axis through its end and perpendicular to the rod, I p = I cm M 2 M 2  L + Md = L + M  = L. 12 3 2
2 2

1 9.55: Ιp = Ι cm + md 2 ,so Ι = 12 Μ a 2 + b 2 + Μ ( a ) 2 + ( b ) 2 , which gives 2 2 1 1 1 Ι = Μ a 2 + b 2 + Μ a 2 + b 2 , or Ι = Μ a 2 + b 2 . 12 4 3

( (

) )

( (

)

)

(

)

1 9.56: a) Ι = 12 Μa 2

1 b) Ι = 12 Μb2

9.57:

In Eq. ( 9.19) , Ι cm =

M 12

L2 and d = ( L 2 − h ) , so
2 1 2 L   Ιp = Μ  L +  − h   2   12  

1 1  = Μ  L2 + L2 − Lh + h 2  4 12  1  = Μ  L2 − Lh + h 2  , 3  which is the same as found in Example 9.12. 9.58: The analysis is identical to that of Example 9.13, with the lower limit in the integral being zero and the upper limit being R, and the mass Μ = πLρR 2 . The result is Ι = 1 ΜR 2 , as given in Table ( 9.2(f) ). 2

9.59: With dm =

M L

dx M M x3 M 2 Ι = ∫x dx = = L. L L 3 0 3 0
2 L L

9.60: For this case, dm = γ dx. a) b) x2 Μ = ∫ dm = ∫ γx dx = γ 2 0 Ι = ∫ x 2 (γx)dx = γ
L 4 L L L

0

yL2 = 2

4 x = γL = Μ L2 . 4 2 4 0 0 This is larger than the moment of inertia of a uniform rod of the same mass and length, since the mass density is greater further away from the axis than nearer the axis.

c)

Ι = ∫ ( L − x) 2 γxdx
0

L

= γ ∫ ( L2 x − 2 Lx 2 + x 3 )dx
0

L

 x2 x3 x 4  = γ  L2 − 2 L +   2 3 4 0   L4 =γ 12 M 2 = L. 6

L

This is a third of the result of part (b), reflecting the fact that more of the mass is concentrated at the right end.
 9.61: a) For a clockwise rotation, ω will be out of the page. b ) The upward direction   crossed into the radial direction is, by the right-hand rule, counterclockwise. ω and r are    perpendicular, so the magnitude of ω × r is ωr = v. c) Geometrically, ω is    perpendicular to v , and so ω × v has magnitude ωv = arad, and from the right-hand rule, the upward direction crossed into the counterclockwise direction is inward, the  direction of arad . Algebraically,       a rad = ω × v = ω × ( ω × r )       = ω( ω ⋅ r ) − r ( ω ⋅ ω )  = −ω2 r ,   where the fact that ω and r are perpendicular has been used to eliminate their dot product.

9.62:

For planetary alignment, earth must go through 60° more than Mars: θE = θM + 60° wEt = ωMt + 60° 60° t= ωE − ωM wE = t= 9.63: a) v = 60 mph = 26.82 m s r = 12 in . = 0.3048 m ω= v = 88.0 rad s = 14.0 rev s = 840 rpm r b) same ω as in part (a) since speedometer reads same 360° 360° and wM = 1yr 1.9 yr
360 ° 1yr

 365d  60° = 0.352 yr   1yr  = 128 d  360 ° − 1.9 yr  

r = 15 in. = 0.381 m v = rω = (0.381m)(88.0 rad s) = 33.5 m s = 75 mph c) v = 50 mph = 22.35 m s r = 10 in. = 0.254 m v ω = = 88.0 rad s; . this is the same as for 60 mph with correct tires, so r speedometer read 60 mph. 9.64: a) For constant angular acceleration θ = ωα , and so arad = ω2 r = 2αθr. 2 b) Denoting the angle that the acceleration vector makes with the radial direction as β , and using Equations (9.14) and (9.15), a αr αr 1 tan β = tan = 2 = = , arad ω r 2 αθ r 2θ 1 so θ = 2 tan β = 2 tan136.9° = 0.666 rad.
2

9.65: a) ωz = b)

dθ = 2γt − 3 βt 2 = (6.40 rad s 2 )t − (1.50 rad s3 )t 2 . dt dwz αz = = 2γ − 6 β t = (6.40 rad/s2 ) − (3.00 rad/s3 )t. dt

c) An extreme of angular velocity occurs when α z = 0, which occurs at t=
γ 3β

=

= 2.13 s, and at this time ωz = (2γ)(γ / 3 β ) − (3 β )(γ / 3 β ) 2 = γ 2 / 3 β = (3.20 rad/s 2 ) 2 = 6.83rad/s. 3(0.500 rad/s3 )

9.66: a) By successively integrating Equations (9.5) and (9.3), β ωz = γt − t 2 = (1.80 rad/s 2 )t − (0.125 rad/s3 )t 2 , 2 γ β θ = t 2 − t 3 = (0.90 rad/s 2 )t 2 − (0.042 rad/s3 )t 3 . 2 6 γ b) The maximum positive angular velocity occurs when α z = 0, t = , the angular β velocity at this time is  γ β γ 1 γ 2 1 (1.80 rad/s 2 ) 2 = 6.48 rad/s.  −   = ωz = γ  = β 2 β 2 β 2 (0.25 rad/s3 )    2γ The maximum angular displacement occurs when ωz = 0, at time t = (t = 0 is an β inflection point, and θ (0) is not a maximum) and the angular displacement at this time is γ  2γ  β  2 γ  2 γ 3 2 (1.80 rad/s 2 )3 = 62.2 rad. = θ=   −   = 2 β  6  β  3 β 2 3 (0.25 rad/s3 ) 2     9.67: a) The scale factor is 20.0, so the actual speed of the car would be 35 km h = 9.72 m s b) (1 2)mv 2 = 8.51 J. c) ω =
2K I 2 3 2

= 652 rad s.

9.68: a) α =

a tan r

=

3.00 m s 2 60.0 m

= 0.050 rad s 2 . b) αt = (0.05 rad s 2 )(6.00s ) = 0.300 rad s.

c) arad = ω2 r = (0.300 rad s) 2 (60.0 m) = 5.40 m s 2 . d)

e) a = a 2 rad + a 2 tan = (5.40 m s ) 2 + (3.00 m s 2 ) 2 = 6.18 m s 2 , and the magnitude of the force is F = ma = (1240 kg )(6.18 m s 2 ) = 7.66 kN. f) arctan

2

( ) = arctan( ) = 60.9°.
a rad a tan 5.40 3.00

9.69: a) Expressing angular frequencies in units of revolutions per minute may be accomodated by changing the units of the dynamic quantities; specifically, 2W ω2 = ω12 + I  2(−4000 J )  = (300 rev min) +   16.0 kg ⋅ m 2     = 211rev min.
2

 π rad s    30 rev min    

2

b) At the initial speed, the 4000 J will be recovered; if this is to be done is 5.00 s, the power must be 4000 sJ = 800 W. 5.00

9.70: a) The angular acceleration will be zero when the speed is a maximum, which is at the bottom of the circle. The speed, from energy considerations, is v = 2 gh = 2 gR (1 − cosβ ), where β is the angle from the vertical at release, and v 2g 2(9.80 m s 2 ) = (1 − cosβ ) = (1 − cos 36.9°) = 1.25 rad s. R R (2.50 m) b) α will again be 0 when the meatball again passes through the lowest point. c) arad is directed toward the center, and arad = ω2 R, arad = (1.25 rad s 2 ) (2.50 m) = 3.93 m s 2 .

ω=

d) arad = ω2 R = (2 g R )(1 − cos β ) R = (2 g )(1 − cos β ), independent of R. 9.71: a) (60.0 rev s)(2π rad rev)(0.45 × 10−2 m) = 1.696 m s. b) ω = v = r
1.696 m s 2.00 × 10 − 2 m

= 84.8 rad s.

9.72: The second pulley, with half the diameter of the first, must have twice the angular velocity, and this is the angular velocity of the saw blade. a)  π rad s   0.208 m  (2(3450 rev min))   = 75.1 m s.  30 rev min    2   
2 2

  π rad s    0.208 m  4 2 b) arad = ω r =  2(3450 rev min)   = 5.43 × 10 m s ,  30 rev min      2      so the force holding sawdust on the blade would have to be about 5500 times as strong as gravity.

9.73: a)

2 2 ∆arad = ω2 r − ω0 r = (ω2 − ω0 )r

= [ ω − ω0 ] [ ω + ω0 ] r  ω − ω0  [ (ω + ω0 )t ] r =  t   = [ α] [2 (θ − θ0 )r. b) From the above, αr = ∆arad (85.0 m s 2 − 25.0 m s 2 ) = = 2.00 m s 2 . 2∆θ 2(15.0 rad)

c) Similar to the derivation of part (a), 1 1 2 1 ∆K = ω2 I − ω0 I = [α ][2∆θ ]I = Iα∆θ. 2 2 2 d) Using the result of part (c), I= ∆K (45.0 J − 20.0 J) = = 0.208 kg ⋅ m 2 . 2 α∆θ ((2.00 m s ) /(0.250 m))(15.0 rad)

9.74: I = I wood + I lead 2 2 mw R 2 + mL R 2 5 3 4 mw = ρwVw = ρw πR 3 3 = mL = σ L AL = σ L 4π R 2 2 4 2  I =  ρ w π R 3  R 2 + (σ L 4 π R 2 ) R 2 5 3 3  8 ρ R  = π R4  w + σL  3  5   (800 kg m 3 )(0.20 m)  8π (0.20 m) 4  + 20 kg m 2  3 5   = 0.70 kgm 2 =

9.75: I approximate my body as a vertical cylinder with mass 80 kg, length 1.7 m, and diameter 0.30 m (radius 0.15 m) I= 1 1 mR 2 = (80 kg) (0.15 m) 2 = 0.9 kg ⋅ m 2 2 2

9.76: Treat the V like two thin 0.160 kg bars, each 25 cm long. 1  1 Ι = 2 mL2  = 2  (0.160 kg)(0.250 m) 2 3   3 −3 = 6.67 × 10 kg ⋅ m 2

9.77: a) ω = 90.0 rpm = 9.425 rad s K= 1 2 2 K 2( 10.0 × 106 J) Ιω so Ι = 2 = = 2.252 × 105 kg ⋅ m 2 2 2 ω ( 9.425 rad s)

m = ρV = ρπR 2t ( ρ = 7800 kg m 3 is the density of iron and t=0.100 m is the thickness of the flywheel) 1 1 Ι = mR 2 = ρπtR 4 2 2 R = (2 I ρπt )1 4 = 3.68 m; diameter = 7.36 m b) ac = Rω2 = 327 m s 2 9.78: Quantitatively, from Table (9.2), I A = 1 mR 2 , Ι B = mR 2 and Ι C = 2 mR 2 ⋅ a) Object A 2 3 has the smallest moment of inertia because, of the three objects, its mass is the most concentrated near its axis. b) Conversely, object B’s mass is concentrated and farthest from its axis. c) Because Ι sphere = 2 5 mR 2 , the sphere would replace the disk as having the smallest moment of inertia.

9.79: a) See Exercise 9.50. 2π 2 Ι 2π 2 (0.3308)(5.97 × 10 24 kg)(6.38 × 106 m) 2 K= 2 = = 2.14 × 10 29 J. 2 T (86, 164 s) b) 1  2πR  2π 2 (5.97 × 1024 kg)(1.50 × 1011 m) 2 M = 2.66 × 1033 J.  = 2  T  (3.156 × 107 s) 2
2

c) Since the Earth’s moment on inertia is less than that of a uniform sphere, more of the Earth’s mass must be concentrated near its center. 9.80: Using energy considerations, the system gains as kinetic energy the lost potential energy, mgR. The kinetic energy is 1 1 1 1 1 K = Ιω2 + mv 2 = Ιω2 + m(ωR ) 2 = ( Ι + mR 2 ) ω 2 ⋅ 2 2 2 2 2 2 1 Using Ι = 2 mR and solving for ω, ω2 = 4g , and ω = 3R 4g . 3R

9.81: a)

Consider a small strip of width dy and a distance y below the top of the triangle. The length of the strip is x = ( y h ) b. 1 The strip has area x dy and the area of the sign is bh, so the mass of the strip is 2  xdy   yb  2 dy   2M  dm = M  1  = M    =  2  y dy  bh   h  bh   h  2  dI =
1 3 h

( dm ) x 2 = 2Mb 4
3h

2

y 3 dy

2 Mb 2 h 3 2Mb 2  1 4 h  1 y dy = y | 0  = Mb 2 4 ∫ 4  0 0 3h 3h  4  6 2 2 1 b) I = 6 Mb = 2.304 kg ⋅ m ω = 2.00 rev s = 4.00π rad s K = 1 Iω2 = 182 J 2 I = ∫ dI =

9.82: (a) The kinetic energy of the falling mass after 2.00 m is 2 KE = 1 mv 2 = 1 ( 8.00 kg ) ( 5.00 m/s ) = 100 J. The change in its potential energy while 2 2

falling is mgh = ( 8.00 kg ) 9.8 m/s 2 ( 2.00 m ) = 156.8 J The wheel must have the “missing” 56.8 J in the form of rotational KE. Since its outer rim is moving at the same speed as the falling mass, 5.00 m s : v = rω

(

)

ω=
KE =

v 5.00m/s = = 13.51 rad/s r 0.370m

1 2 Iω ; therefore 2 2 KE 2( 56.8 J ) I= 2 = = 0.6224 kg ⋅ m 2 or 0.622 kg ⋅ m 2 2 ω (13.51 rad s )

(b) The wheel’s mass is 280 N 9.8 m s 2 = 28.6 kg. The wheel with the largest possible moment of inertia would have all this mass concentrated in its rim. Its moment of inertia would be I = MR 2 = ( 28.6kg )( 0.370m ) = 3.92 kg ⋅ m 2 The boss’s wheel is physically impossible.
2

9.83: a) ( 0.160 kg ) ( − 0.500 m ) 9.80 m s 2 = −0.784 J. b) The kinetic energy of the stick is 0.784 J, and so the angular velocity is

(

)

ω=

2k 2k = = I ML2 3

2(0.784 J) = 5.42 rad s . ( 0.160 kg ) (1.00 m ) 2 3

This result may also be found by using the algebraic form for the kinetic energy, K = MgL 2, from which ω = 3g L , giving the same result. Note that ω is independent of the mass. c) v = ωL = ( 5.42 rad s ) (1.00 m ) = 5.42 m s d) 2gL = 4.43 m s ; This is 2 3 of the result of part (c).

9.84: Taking the zero of gravitational potential energy to be at the axle, the initial potential energy is zero (the rope is wrapped in a circle with center on the axle).When the rope has unwound, its center of mass is a distance πR below the axle, since the length of the rope is 2πR and half this distance is the position of the center of the mass. Initially, every part of the rope is moving with speed ω 0 R, and when the rope has unwound, and the cylinder has angular speed ω , the speed of the rope is ωR (the upper end of the rope has the same tangential speed at the edge of the cylinder). From conservation of energy, using I = (1 2) MR 2 for a uniform cylinder,  M m 2 2  M m 2 2  +  R ω0 =  +  R ω − mgπ R.  4 2  4 2 Solving for ω gives ( 4πmg R ) , 2 ω = ω0 + ( M + 2m ) and the speed of any part of the rope is v = ωR. 9.85: In descending a distance d, gravity has done work mB gd and friction has done work − µ K m A gd , and so the total kinetic energy of the system is gd ( mB − μK mA ). In terms of the speed v of the blocks, the kinetic energy is 1 1 1 K = ( m A + m B ) v 2 + Iω 2 = m A + m B + I R 2 v 2 , 2 2 2 where ω = v R, and condition that the rope not slip, have been used. Setting the kinetic energy equal to the work done and solving for the speed v, 2 gd ( mB − µ k m A ) v= . m A + mB + I R 2

(

)

(

)

9.86: The gravitational potential energy which has become kinetic energy is K = ( 4.00 kg − 2.00 kg ) 9.80 m s 2 ( 5.00 m ) = 98.0 J. In terms of the common speed v of the blocks, the kinetic energy of the system is

(

)

K=

1 1 v (m1 + m2 )v 2 + I   2 2  R

2

= v2

(0.480 kg ⋅ m 2 )  1  4.00 kg + 2.00 kg +  = v 2 (12.4 kg). 2  2 (0.160 m)  
98.0 J 12.4 kg

Solving for v gives v =

= 2.81m s.

9.87: The moment of inertia of the hoop about the nail is 2MR 2 (see Exercise 9.52), and the initial potential energy with respect to the center of the loop when its center is directly below the nail is gR ( 1 − cos β ). From the work-energy theorem, K= 1 2 Iω = Mω2 R 2 = MgR (1 − cos β), 2

from which ω = ( g R)(1 − cos β). 1 2 Iω 2
2

9.88: a) K =

11 2π rad s    =  (1000 kg)(0.90 m) 2  3000 rev min ×  22 60 rev min    = 2.00 × 107 J. b) K 2.00 × 107 J = = 1075 s, Pave 1.86 × 104 W

which is about 18 min. 1 1 1 2 M 1R12 + M 2 R2 = ((0.80 kg)(2.50 × 10 − 2 m) 2 + (1.60 kg)(5.00 × 10 − 2 m) 2 ) 2 2 2 = 2.25 × 10 −3 kg ⋅ m 2 . b) See Example 9.9. In this case, ω = v R1 , and so the expression for v becomes v= = 2 gh 1 + ( I mR 2 ) 2(9.80 m s 2 )(2.00 m) = 3.40 m s. (1 + ((2.25 × 10− 3 kg ⋅ m 2 ) (1.50 kg)(0.025 m) 2 ))

9.89:

a)

c) The same calculation, with R2 instead of R1 gives v = 4.95 m s. This does make sense, because for a given total energy, the disk combination will have a larger fraction of the kinetic energy with the string of the larger radius, and with this larger fraction, the disk combination must be moving faster.

9.90: a) In the case that no energy is lost, the rebound height h′ is related to the speed υ2 h v by h′ = 2 g , and with the form for h given in Example 9.9, h′ = 1+ M 2 m . b) Considering the system as a whole, some of the initial potential energy of the mass went into the kinetic energy of the cylinder. Considering the mass alone, the tension in the string did work on the mass, so its total energy is not conserved. 9.91: We can use Κ (cylinder) = 250 J to find ω for the cylinder and v for the mass. Ι = 1 MR 2 = 1 (10.0 kg)(0.150 m) 2 = 0.1125 kg ⋅ m 2 2 2 K = 1 Iω2 so ω = 2 K I = 66.67 rad s 2 v = Rω = 10.0 m s Use conservation of energy K 1 + U 1 = K 2 + U 2 . Take y = 0 at lowest point of the mass, so y 2 = 0 and y1 = h, the distance the mass descends. K1 = U 2 = 0 so U1 = K 2 . mgh = 1 mv 2 + 1 Iω 2 , where m = 12.0 kg 2 2 For the cylinder, I = 1 MR 2 and ω = v R , so 2 mgh = 1 mv 2 + 1 Mv 2 2 4 h= v2  M  1 +  = 7.23 m 2 g  2m 
1 2

Iω 2 = 1 Mv 2 . 4

9.92: Energy conservation: Loss of PE of box equals gain in KE of system. 1 1 2 2 mbox gh = mbox vbox + I pulley ωpulley 2 2 1 2 + I cylinder ωcylinder 2 v v ωpulley = Box and ωcylinder = Box rp rcylinder 1 1 1 v  2 mB gh = mBvB +  mP rp2   B  2 22   rp   
2

1 1 v  +  mC rC2   B  22   rC    mB gh = 1 1 1 2 2 2 mBvB + mP vB + mCvB 2 4 4 mB gh vB = 1 mB + 1 m p + 1 mC 2 4 4 =

2

(3.00 kg)(9.80 m s 2 )(1.50 m) = 3.68 m s 1.50 kg + 1 (7.00 kg) 4

9.93: a) The initial moment of inertia is I 0 = 1 MR 2 . The piece punched has a mass of 2 M 16 and a moment of inertia with respect to the axis of the original disk of
2 2 M 1  R   R   9 MR 2 .    +   = 16  2  4   2   512  

The moment of inertia of the remaining piece is then I= 1 9 247 MR 2 − MR 2 = MR 2 . 2 512 512
383 512

b) I = 1 MR 2 + M ( R / 2) 2 − 1 ( M / 16)( R / 4) 2 = 2 2

MR 2 .

9.94: a) From the parallel-axis theorem, the moment of inertia is I P = (2 5) MR 2 + ML2 , and   2  R  2  IP = 1 +    . ML2   5  L     If R = (0.05)L, the difference is (2 5)(0.05) 2 = 0.001. b) ( I rod ML2 ) = (mrod 3M ), which is 0.33% when mrod = (0.01) M .

9.95: a) With respect to O, each element ri in Eq. (9.17) is xi + y i , and so I O = ∑ mi ri = ∑ mi ( xi + y i ) = ∑ mi xi + ∑ mi y i = I x + I y .
2 2 2 2 2 i i i i

2

2

2

b) Two perpendicular axes, both perpendicular to the washer’s axis, will have the same moment of inertia about those axes, and the perpendicular-axis theorem predicts 2 2 that they will sum to the moment of inertia about the washer axis, which is M ( R1 + R2 ), 2 and so I x = I y =
M 4

( R1 + R2 ).

2

2

1 c) From Table (9.2), I = 12 m( L 2 + L2 ) = 1 mL2 . 6 1 Since I 0 = I x + I y , and I x = I y , both I x and I y must be 12 mL2 .

9.96: Each side has length a and mass

M 4

, and the moment of inertia of each side about
2

1 an axis perpendicular to the side and through its center is 12 M a 2 = Ma . The moment of 4 48 inertia of each side about the axis through the center of the square is, from the 2 2 2 perpendicular axis theorem, Ma + M ( a ) = Ma . The total moment of inertia is the sum of 48 4 2 12

the contributions from the four sides, or 4 × Ma = 12
2

Ma 2 3

.

9.97: Introduce the auxiliary variable L, the length of the cylinder, and consider thin cylindrical shells of thickness dr and radius r; the cross-sectional area of such a shell is 2π r dr , and the mass of shell is dm = 2π rLρ dr = 2π α Lr 2 dr. The total mass of the cylinder is then R R3 2 M = ∫ dm = 2π Lα ∫ r dr = 2π Lα 0 3 and the moment of inertia is R R5 3 I = ∫ r 2 dm = 2π Lα ∫ r 4 dr = 2π Lα = MR 2 . o 5 5 b) This is less than the moment of inertia if all the mass were concentrated at the edge, as with a thin shell with I = MR 2 , and is greater than that for a uniform cylinder with I = 1 MR 2 , as expected. 2 9.98: a) From Exercise 9.49, the rate of energy loss is inertia I in terms of the power P, Ι= ΡΤ 3 1 (5 × 1031 W )(0.0331 s) 3 1s = = 1.09 × 1038 kg ⋅ m 2 . 2 −13 4π dΤ dt 4π 4.22 × 10 s 5Ι = 2Μ 5(1.08 × 1038 kg ⋅ m 2 ) = 9.9 × 103 m, about 10 km. 30 2(1.4)(1.99 × 10 kg) 2πR 2π(9.9 × 103 m) = = 1.9 × 106 m s = 6.3 × 10− 3 c. Τ (0.0331 s) Μ Μ = = 6.9 × 1017 kg m 3 , 3 V (4π 3) R
4 π 2 Ι dΤ Τ 3 dt

; solving for the moment of

b) R =

c)

d)

which is much higher than the density of ordinary rock by 14 orders of magnitude, and is comparable to nuclear mass densities.

9.99: a) Following the hint, the moment of inertia of a uniform sphere in terms of the 2 mass density is I = 5 ΜR 2 = 8 π ρR 5 , and so the difference in the moments of inertia of 15 two spheres with the same density ρ but different radii
5 R2 and R1 is Ι = ρ(8π 15)( R2 − R1 ). 5

b) A rather tedious calculation, summing the product of the densities times the difference in the cubes of the radii that bound the regions and multiplying by 4π 3 , gives M = 5.97 × 10 24 kg. c) A similar calculation, summing the product of the densities times the difference in the fifth powers of the radii that bound the regions and multiplying by 8π 15, gives I = 8.02 × 1022 kg ⋅ m 2 = 0.334 MR 2 . 9.100: Following the procedure used in Example 9.14 (and using z as the coordinate 2 4 along the vertical axis) r(z) = z R , dm = πρ R2 z 2 dz and dΙ = πρ R4 z 4 dz. Then, h 2 h h Ι = ∫ dΙ = πρ R 4 2 h

∫

h

0

z 4 dz =

πρ R 4 5 h 1 z 0 = πρR 4 h . 10 h 4 10

[ ]

The volume of a right circular cone is V = 1 πR 2 h, the mass is 1 πR 2 hand so 3 3 Ι= 3  πρR 2 h  2 3   R = ΜR 2 . 10  3  10  

9.101: a) ds = r dθ = r0 dθ + βθ dθ , so s (θ ) = r0θ + β θ 2 . b) Setting s = vt = r0θ + β θ 2 2 2 gives a quadratic in θ . The positive solution is 1 2 θ(t ) = =  r0 + 2 βvt − r0  .    β  (The negative solution would be going backwards, to values of r smaller than r0 .) c) Differentiating, ωz (t ) = dθ = dt v r02 + 2 βvt ,

αz =

dω βv 2 =− dt r02 = 2 βvt

(

)

32

.

The angular acceleration α z is not constant. d) r0 = 25.0 mm; It is crucial that θ is measured in radians, so β = (1.55 μm rev ) (1 rev 2π rad ) = 0.247 μm rad. The total angle turned in 74.0 min = 4440 s is θ=  2 2.47 × 10 − 7 m/rad (1.25 m/s ) ( 4440 s )  1   2.47 × 10− 7 m/rad  + 25.0 × 10− 3 m 2 − 25.0 × 10− 3 m    5 = 1.337 × 10 rad

(

(

)

)

which is 2.13 × 10 4 rev. e)

Chapter 10

10.1: Equation (10.2) or Eq. (10.3) is used for all parts. a) (4.00 m)(10.0 N) sin 90° = 40.00 N ⋅ m, out of the page. b) (4.00 m)(10.0 N) sin 120° = 34.6 N ⋅ m, out of the page. c) (4.00 m)(10.0 N) sin 30° = 20.0 N ⋅ m, out of the page. d) (2.00 m)(10.00 N) sin 60° = 17.3 N ⋅ m, into the page. e) The force is applied at the origin, so τ = 0. f) (4.00 m)(10.0 N) sin 180° = 0. τ1 = −(8.00 N)(5.00 m) = −40.0 N ⋅ m, τ 2 = (12.0 N)(2.00 m) sin 30° = 12.0 N ⋅ m,

10.2:

where positive torques are taken counterclockwise, so the net torque is − 28.0 N ⋅ m, with the minus sign indicating a clockwise torque, or a torque into the page. 10.3: Taking positive torques to be counterclockwise (out of the page), τ1 = −(0.090 m) × (180.0 N) = −1.62 N ⋅ m, τ 2 = (0.09 m)(26.0 N) = 2.34 N ⋅ m,

τ 3 = 2 (0.090 m) (14.0 N) = 1.78 N ⋅ m, so the net torque is 2.50 N ⋅ m, with the direction counterclockwise (out of the page). Note that for τ 3 the applied force is perpendicular to the lever arm.
10.4: τ1 + τ 2 = − F1R + F2 R = ( F2 − F1 ) R = (5.30 N − 7.50 N)(0.330 m) = −0.726 N ⋅ m. a)

( )

10.5:

b) Into the plane of the page.   ˆ ˆ ˆ c) r × F = [(−0.450 m) i + (0.150 m) ˆ] × [(−5.00 n) i + (4.00 N)i ] j ˆ = (−0.450 m) (4.00 N) − (0.150 m)( −5.00 N)k ˆ = (−1.05 N ⋅ m) k

10.6: (a)

τ A = (50 N)(sin 60°)(0.2 m) = 8.7 N ⋅ m, CCW τB = 0 τ C = (50 N)(sin 30°)(0.2 m) = 5 N ⋅ m, CW τ D = (50 N)(0.2 m) = 10 N ⋅ m, CW

(b) ∑ τ = 8.7 N ⋅ m − 5 N ⋅ m − 10 N ⋅ m = −6.3 N ⋅ m, CW 10.7: I = 2 MR 2 + 2mR 2 , where M = 8.40 kg, m = 2.00 kg 3 I = 0.600 kg ⋅ m 2 ω0 = 75.0 rpm = 7.854 rad s ; ω = 50.0 rpm = 5.236 rad s ; t = 30.0 s, α = ? ω = ω0 + αt gives α = −0.08726 rad s ; Στ = Iα, τ f = Iα = −0.0524 N ⋅ m
rad s 400 rev min × 2 π rev min ∆ω 60 2 = 2.50 kg ⋅ m = 13.1 N ⋅ m. 10.8: a) τ = Iα = I ∆t ( 8.00 s ) 2

(

)(

)

 1 1 2π rad s   = 2.19 × 103 J. b) Iω2 = (2.50 kg ⋅ m 2 )  400 rev min ×  2 2 60 rev min    10.9: v = 2as = 2 0.36 m s 2 ( 2.0 m ) = 1.2 m s , the same as that found in Example 9-8. τ FR ( 40.0 N )( 0.250 m ) = = = 2.00 rad s 2 . I I 5.0 kg ⋅ m 2

2

(

)

10.10:

α=

(

)

10.11: a)

   M + 3m  m n = Mg + T = g  M +  = g 1 + 2m M  1 + 2m M    

b) This is less than the total weight; the suspended mass is accelerating down, so the tension is less than mg. c) As long as the cable remains taut, the velocity of the mass does not affect the acceleration, and the tension and normal force are unchanged.

10.12: a) The cylinder does not move, so the net force must be zero. The cable exerts a horizontal force to the right, and gravity exerts a downward force, so the normal force must exert a force up and to the left, as shown in Fig. (10.9). b) n = ( 9.0 N ) + (50 kg) 9.80 m s 2 490 N, at an angle of arctan vertical (the weight is much larger than the applied force F ).
2 2

(

(

))

9.0 ( 490 ) = 1.1° from the

10.13:

µk =

f τ R Iα MR( ω 0 t ) = = = n n Rn 2n

π rad s ( 50.0 kg ) ( 0.260 m ) ( 850 rev min ) ( 30 rev min ) = = 0.482. 2( 7.50 s ) (160 N )

10.14:

(a) Falling stone: g = 1 at 2 2

12.6 m = 1 a( 3.00 s ) 2

2

a Pulley : Σ τ = Iα : TR = 1 MR 2α = 1 MR 2 ( R ) 2 2

a = 2.80 m s 2 Stone : Σ F = ma : mg − T = ma(1) T = 1 Ma(2) 2

Solve (1) and (2):   a   10.0 kg   2.80 m/s 2  =  g −a  9.80 m/s 2 − 2.80 m/s 2      2   M = 2.00 kg M = M 2 (b)From (2): 1 1 Ma = (10.0 kg ) 2.80 m/s 2 2 2 T = 14.0 N T=

(

)

10.15:

2 ω2 = ω0 + 2α ( θ − θ0 ) gives α = −8.046 rad/s2

I = 1 mR 2 = 1 ( 8.25 kg )( 0.0750 m ) = 0.02320 kg ⋅ m 2 2 2 ω0 = 220 rpm = 23.04 rad/s; ω = 0; θ − θ0 = 5.25 rev = 33.0 rad, α = ?
2

∑ τ = Iα ∑ τ = τ f = − f k R = − μk nR − μk nR = Iα so n = Iα = 7.47 N μk R

10.16: This is the same situtation as in Example 10.3. a) T = mg (1 + 2m M ) = 42.0 N. b) v = 2 gh (1 + M 2m = 11.8 m s. c) There are many ways to find the time of fall. Rather than make the intermediate calculation of the acceleration, the time is the distance divided by the average speed, or h ( v 2 ) = 1.69 s. d) The normal force in Fig. (10.10(b)) is the sum of the tension found in part (a) and the weight of the windlass, a total 159.6 N (keeping extra figures in part ( a)). 10.17: See Example 10.4. In this case, the moment of inertia I is unknown, so 2 a1 = ( m2 g ) m1 + m2 + I R 2 . a) a1 = 2 (1.20 m ) ( 0.80 s ) = 3.75 m/s 2 , so T1 m1a1 = 7.50 N and T2 = m2 ( g − a1 ) = 18.2 N. b) The torque on the pulley is ( T2 − T1 ) R = 0.803 N ⋅ m, and the angular acceleration is α = a1 R = 50 rad/s2 , so I = τ α = 0.016 kg ⋅ m 2 .

(

(

))

10.18:

α=

τ Fl 3F = 1 2 = . I 3 Ml Ml

10.19: The acceleration of the mass is related to the tension by Macm = Mg − T , and the angular acceleration is related to the torque by Iα = τ = TR, or acm = T / M , where α = acm / RSt and I = MR 2 have been used. a) Solving these for T gives T = Mg / 2 = 0.882 N. b) Substituting the expression for T into either of the above relations gives acm = g / 2, from which

t = 2h acm = 4h g = 0.553 s. c) ω = vcm R = acm t R = 33.9 rad/s.
10.20: See Example 10.6 and Exercise 10.21. In this case, 2 K 2 = Mvcm and vcm = gh , ω = vcm R = 33.9 rad s .

10.21: From Eq. (10.11), the fraction of the total kinetic energy that is rotational is

(1 2) I cmω 2 2 (1 2) Mvcm + (1 2) I cmω 2

=

1 + ( M I cm ) v / ω
2 cm

1

2

=

1 2 , 1 + MR I cm

where vcm = Rω for an object that is rolling without slipping has been used. a) I cm = (1 2) MR 2 , so the above ratio is 1 3. b) I = (2 5)MR 2 , so the above ratio is 2 7. c) I = 2 3 MR 2 , so the ratio is 2 5. d) I = 5 8 MR 2 , so the ratio is 5 13.

f 10.22: a) The acceleration down the slope is a = g sin θ − M , the torque about the center of the shell is a 2 a 2 τ = Rf = Iα = I = MR 2 = MRa, R 3 R 3 f 2 so M = 3 a. Solving these relations a for f and simultaneously gives 5 a = g sin θ , or 3

3 3 g sin θ = (9.80 m s 2 ) sin 38.0° = 3.62 m s 2 , 5 5 2 2 f = Ma = (2.00 kg)(3.62 m s 2 ) = 4.83 N. 3 3 The normal force is Mg cos θ , and since f ≤ μ s n, 2 Ma f 2 a 2 3 g sin θ 2 μs ≥ = 3 = = 5 = tan θ = 0.313. n Mg cos θ 3 g cos θ 3 g cos θ 5 a= b) a = 3.62 m s 2 since it does not depend on the mass. The frictional force, however, is twice as large, 9.65 N, since it does depend on the mass. The minimum value of μs also does not change.

10.23:

n = mg cos α mg sin θ − μs mg cos θ = ma g (sin θ − μs cos θ ) = a (eq.1) n and mg act at the center of the ball and provide no torque. ∑ τ = τ f = μs mg cos θR; I = 2 mR 2 5
2 ∑ τ = Iα gives μs mg cos θ = 5 mR 2α 2 No slipping means α = a R , so μs g cosθ = 5 a (eq.2) We have two equations in the two unknowns a and µ s . Solving gives 5 2 2 a = 7 g sin θ and μs = 7 tan θ = 7 tan65.0° = 0.613

b) Repeat the calculation of part (a), but now I = 2 mR 2 . 3 3 2 2 a = 5 g sin θ and μs = 5 tan θ = 5 tan65.0° = 0.858 The value of µ s calculated in part (a) is not large enough to prevent slipping for the hollow ball. c) There is no slipping at the point of contact.

10.24:

vcm = Rω for no slipping a) Get v at bottom: mgh = 1 2 1 mv + Iω 2 2 2
2

1 12  v  mgh = mv 2 +  mR 2    2 25  R v= 10 gh 7

Now use energy conservation. Rotational KE does not change 1 2 mv + KERot = mgh′ + KERot 2 v 2 10 gh 5 h′ = = 7 = h 2g 2g 7

(b) mgh = mgh′ → h′ = h With friction on both halves, all the PE gets converted back to PE. With one smooth side, some of the PE remains as rotational KE.
10.25: wh − W f = K 1 = (1 / 2) I cm w 2 0 + 1 mv 2 cm 2 Solving for h with vcm = Rw h= −
w 1 2 9.80 m s 2

(

) [(0.800)(0.600 m) (25.0 rad s)
2

2

+ (0.600 m) 2 (25.0 rad s) 2 ]

w 3500 J = 11.7 m. 392 N

10.26:

a)

The angular speed of the ball must decrease, and so the torque is provided by a friction force that acts up the hill. b) The friction force results in an angular acceleration, related by Iα = fR. The equation of motion is mg sin β − f = macm, and the acceleration and angular acceleration are related by acm = Rα (note that positive acceleration is taken to be down the incline, and relation between acm and α is correct for a friction force directed uphill). Combining, I   mg sin β = ma1 + = ma ( 7 5) , 2   mR  from which acm = ( 5 7 ) g sin β. c) From either of the above relations between if f and acm , 2 2 f = macm = mg sin β ≤ μs n = μs mg cos β , 5 7 from which μs ≥ ( 2 7 ) tan β . 10.27: a) ω = α∆t = ( FR I ) ∆t = (18.0 N )( 2.40 m ) 2100 kg ⋅ m 2 (15.0 s ) = 0.3086 rad/s, or 0.309 rad/s to three figures. 2 b) W = K 2 = (1 2 ) Iω2 = (1 2 ) × 2.00 kg ⋅ m 2 ( 0.3086 rad s ) = 100 J. c) From either P = τωave or P = W ∆t , P = 6.67 W.

(

(

))

(

)

10.28:

a)

τ=

P = ω

(175 hp )( 746 W / hp ) ( 2400 rev/min )  π rad/s 

   30 rev/min 

= 519 N ⋅ m.

b) W = τ∆θ = ( 519 N ⋅ m )( 2π ) = 3261 J.

10.29: a)

τ = Iα = I

=

π rad s   ((1 2)(1.50 kg )( 0.100 m ) )(1200 rev min ) 30 rev min   
2

∆ω ∆t





2.5 s

= 0.377 N ⋅ m. b) c) d)

ω ave ∆t =

( 600 rev/min )( 2.5 s ) 25.0 rev = 157 rad.
60 s/min

τ∆θ = 59.2 J. 1 K = Iω2 2

1 = (1 / 2)(1.5 kg)(0.100 m) 2 2

(

)

  π rad/s    (1200 rev/min)      30 rev/min    = 59.2 J,

2

the same as in part (c). 10.30: From Eq. (10.26), the power output is 2π rad/s   P = τω = (4.30 N ⋅ m) 4800 rev/min ×  = 2161 W, 60 rev/min   which is 2.9 hp. 10.31: a) With no load, the only torque to be overcome is friction in the bearings (neglecting air friction), and the bearing radius is small compared to the blade radius, so any frictional torque could be neglected. b) F= τ P/ω = = R R (1.9 hp)(746 W/hp) = 65.6 N.  π rad/s  (2400 rev/min)  (0.086 m)  30 rev/min 

10.32: I = 1 mL2 = 1 (117 kg)(2.08 m) 2 = 42.2 kg ⋅ m 2 2 2 a) b) α= τ 1950 N ⋅ m = = 46.2 rad/s 2 . 2 I 42.2 kg ⋅ m

ω = 2αθ = 2(46.2 rad/s 2 )(5.0 rev × 2π rev) = 53.9 rad/s. 1 c) From either W = K = ω 2 or Eq. (10.24), 2 W = τθ = (1950 N.m)(5.00 rev × 2π rad/rev) = 6.13 × 104 J.
d), e) The time may be found from the angular acceleration and the total angle, but the instantaneous power is also found from P = τω = 105 kW(141 hp). The average power is half of this, or 52.6 kW.   π rad/s   a) τ = P / ω = (150 × 103 W)  (400 rev/min)    = 358 N ⋅ m.    30 rev/min    b) If the tension in the rope is F , F = w and so w = τ/R = 1.79 × 103 N.

10.33:

c) Assuming ideal efficiency, the rate at which the weight gains potential energy is the power output of the motor, or wv = P, so v = P w = 83.8 m/s. Equivalently, v = ωR.

10.34: As a point, the woman’s moment of inertia with respect to the disk axis is mR 2 , and so the total angular momentum is 1  L = Ldisk + Lwoman = ( I disk + I woman )ω =  M + m  R 2ω 2  1  =  110 kg + 50.0 kg (4.00 m) 2 (0.500 rev/s × 2π rad/rev) 2  3 2 = 5.28 × 10 kg ⋅ m / s. a) mvr sin φ = 115 kg ⋅ m 2 / s, with a direction from the right hand rule of into the

10.35: page.

b) dL dt = τ = ( 2 kg ) ( 9.8 N kg ) ⋅ ( 8 m ) ⋅ sin ( 90° − 36.9°) = 125 N ⋅ m = 125 kg ⋅ m 2 s , out of the page.
2

10.36: For both parts, L = Iω. Also, ω = v r , so L = I (v r ). a) L = (mr 2 )(v r ) = mvr L = (5.97 × 1024 kg)(2.98 × 104 m s) (1.50 × 1011 m) = 2.67 × 1040 kg ⋅ m 2 s b) L = (2 5mr 2 )(ω) L = (2 5)(5.97 × 10 24 kg )(6.38 × 106 m) 2 (2π rad (24.0 hr × 3600 s hr)) = 7.07 × 1033 kg ⋅ m 2 s 10.37: The period of a second hand is one minute, so the angular momentum is L = Iω = M 2 2π l 3 T −3  6.0 × 10 kg  2π (15.0 × 10− 2 m) 2 = = 4.71 × 10− 6 kg ⋅ m 2 s.   3 60 s  

10.38: The moment of inertia is proportional to the square of the radius, and so the angular velocity will be proportional to the inverse of the square of the radius, and the final angular velocity is R ω 2 = ω1  1 R  2    7.0 × 10 5 km  2π rad 3  =  (30 d)(86,400 s d  16 km  = 4.6 × 10 rad s.       
2 2

10.39: a) The net force is due to the tension in the rope, which always acts in the radial direction, so the angular momentum with respect to the hole is constant. 2 b) L1 = mω 1 r 21 , L2 = mω 2 r2 , and with L1 = L2 , ω2 = ω1 ( r1 r2 ) 2 = 7.00 rad s. c) ∆K = (1 2)m((ω2 r2 ) 2 − (ω1r1 ) 2 ) = 1.03 × 10−2 J. d) No other force does work, so 1.03 × 10 −2 J of work were done in pulling the cord.

10.40: The skater’s initial moment of inertia is 1 I1 = (0.400 kg ⋅ m 2 ) + (8.00 kg )(1.80 m) 2 = 2.56 kg ⋅ m 2 , 2 and her final moment of inertia is I 2 = (0.400 kg ⋅ m 2 ) + (8.00 kg)(25 × 10−2 m) = 0.9 kg ⋅ m 2 . Then from Eq. (10.33), ω2 = ω1 I1 2.56 kg ⋅ m 2 = (0.40 rev s) = 1.14 rev s. I2 0.9 kg ⋅ m 2

Note that conversion from rev/s to rad s is not necessary.

10.41: If she had tucked, she would have made (2) (3.6 kg ⋅ m 2 ) 18 kg ⋅ m 2 ) = 0.40 rev in the last 1.0 s, so she would have made (0.40 rev)(1.5 1.0) = 0.60 rev in the total 1.5 s. 10.42: Let I1 = I 0 = 1200kg ⋅ m 2 , I 2 = I 0 + mR 2 = 1200kg ⋅ m 2 + (40.0kg )(2.00 m) 2 = 1360kg ⋅ m 2 . Then, from Eq. (10.33), I1  2π rad  1200 kg.m 2 ω2 = ω1 =  = 0.924 rad s.  I 2  6.00 s  1360 kg.m 2

10.43:

a) From conservation of angular momentum, I1 (1 2) MR 2 = ω 1 ω2 = ω1 = ω1 1 2 2 2 (1 2) MR + mR I 0 + mR 1 + 2m M = 3.0 rad s = 1.385 rad s 1 + 2( 70 ) 120

or 1.39 rad s to three figures
2 K 2 = (1 2 ) I 0 + ( 70 kg )( 2.00 m ) ω2 = 499 J. In changing the parachutist’s horizontal component of velocity and slowing down the turntable, friction does negative work. 2

b) K1 = (1 2)(1 2)(120 kg )( 2.00 m ) ( 3.00 rad s ) = 1.80 kJ, and
2 2

(

)

10.44: Let the width of the door be l; ω= L mv( l 2 ) = I (1 3) Ml 2 + m( l 2 ) 2 ( 0.500 kg )(12.0 m s )( 0.500 m ) = = 0.223 rad s. (1 3)( 40.0 kg )(1.00 m ) 2 + ( 0.500 kg )( 0.500 m ) 2

Ignoring the mass of the mud in the denominator of the above expression gives ω = 0.225 rad s , so the mass of the mud in the moment of inertia does affect the third significant figure.  10.45: Apply conservation of angular momentum L, with the axis at the nail. Let object A be the bug and object B be the bar. Initially, all objects are at rest and L1 = 0. Just after the bug jumps, it has angular momentum in one direction of rotation and the bar is rotating with angular velocity ωB in the opposite direction. L2 = mAv Ar − I B ωB where r = 1.00 m and I B = 1 mB r 2 3 L1 = L2 gives m Av Ar = 1 mB r 2ωB 3 ωB = 3mAv A = 0.120 rad s mB r

10.46:

(a) Conservation of angular momentum: m1v0 d = − m1vd + 1 m2 L2ω 3 1  90.0 N   (2.00 m) 2 ω (3.00 kg)(10.0 m s) (1.50 m) = −(3.00 kg )(6.00 m s)(1.50 m) +  3  9.80 m s 2    ω = 5.88 rad s (b) There are no unbalanced torques about the pivot, so angular momentum is conserved. But the pivot exerts an unbalanced horizontal external force on the system, so the linear momentum is not conserved. 10.47:

10.48: a) Since the gyroscope is precessing in a horizontal plane, there can be no net vertical force on the gyroscope, so the force that the pivot exerts must be equal in magnitude to the weight of the gyroscope, F = ω = mg = ( 0.165 kg ) 9.80 m s 2 = 1.617 N, 1.62 N to three figures. b) Solving Eq. (10.36) for ω,

(

)

(1.617 N ) 4.00 × 10−2 m ωR ω= = = 188.7 rad s , IΩ 1.20 × 10− 4 kg ⋅ m 2 ( 2π ) rad 2.20 s

(

(

)

)

which is 1.80 × 10 rev min . Note that in this and similar situations, since Ω appears in the denominator of the expression for ω , the conversion from rev s and back to rev min must be made.
3

c)

10.49: a)

K (1 / 2)((1 / 2) MR 2 )ω 2 = P P 2 π rad/s (1 / 2) (1 / 2)(60,000 kg)(2.00 m) 2 ( (500 rev/min) ( 30 rev/min ) ) = 7.46 × 104 W = 2.21 × 103 s,

(

)

or 36.8 min. b) τ = IΩω  π rad/s   2π rad  = (1 / 2)(60,000 kg)(2.00 m) 2 (500 rev/min)  (1.00°/s)   30 rev/min   360°  = 1.10 × 105 N ⋅ m. 10.50: Using Eq. (10.36) for all parts, a) halved b) doubled (assuming that the added weight is distributed in such a way that r and I are not changed) c) halved (assuming that w and r are not changed) d) doubled e) unchanged.

10.51:

2 π rad a) Solving Eq. (10.36) for τ , τ = Iω Ω = (2 / 5) MR 2ω Ω. Using ω = 86,400 s and

Ω=

2π ( 26 , 000 y)(3.175 ×10 7 s/y)

and the mass and radius of the earth from Appendix F,

τ ~ 5.4 × 1022 N ⋅ m.
10.52: a) The net torque must be

2π rad/s   120 rev/min ×  ∆ω 60 rev/min  2  τ = Iα = I = (1.86 kg ⋅ m ) = 2.60 N ⋅ m. ∆t (9.00 s) This torque must be the sum of the applied force FR and the opposing frictional torques τ f at the axle and fr = μk nr due to the knife. Combining, F= 1 ( τ + τ f + μk nr ) R 1 = ( (2.60 N ⋅ m) + (6.50 N ⋅ m) + (0.60)(160 N)(0.260 m) ) 0.500 m = 68.1 N.

b) To maintain a constant angular velocity, the net torque τ is zero, and the force F ′ is 1 F ′ = 0.500 m (6.50 N ⋅ m + 24.96 N ⋅ m) = 62.9 N. c) The time t needed to come to a stop is found by taking the magnitudes in Eq. (10.27), with τ = τ f constant;
rad/s L ωI (120 rev/min × 2 π rev/min ) 1.86 kg ⋅ m 2 60 t= = = = 3.6 s. ( 6.50 N ⋅ m ) τf τf m Note that this time can also be found as t = ( 9.00 s ) 2.60 N⋅⋅m . 6.50N

(

)

10.53:

   30 rev/min  b) Rather than use the result of part (a), the magnitude of the torque is proportional to α and hence inversely proportional to | ∆t | ; equivalently, the magnitude of the change in angular momentum is the same and so the magnitude of the torque is again proportional 2s to 1 / | ∆t | . Either way, τ f = ( 5.0 N ⋅ m ) = 0.080 N ⋅ m. 125 s c) ωave ∆t = ( 50.0 rev/min ) (125 s ) (1 min/60 s ) = 104.2 rev.

a) I =

τ τ∆t = = α ∆ω

( 5.0 N ⋅ m ) ( 2.0 s ) (100 rev/min )  π rad/s 

= 0.955 kg ⋅ m 2 .

10.54: a) The moment of inertia is not given, so the angular acceleration must be found from kinematics; α= 2θ 2 s 2( 5.00 m ) = 2 = = 8.33 rad / s 2 . 2 2 t rt ( 0.30 m ) ( 2.00 s )

b) αt = 8.33 rad/s 2 ( 2.00 s ) = 16.67 rad/s. c) The work done by the rope on the flywheel will be the final kinetic energy; K = W = Fs = ( 40.0 N ) ( 5.0 m ) = 200 J. d) I= 2K 2( 200 J ) = = 1.44 kg ⋅ m 2 . 2 2 ω (16.67 rad/s)

(

)

10.55:

b) From the result of part (a), the power is ( 500 W ) ( 60..0 ) = 4.50 kW. 20 0
2 6 d) From the result of part (c), the power is ( 500 W ) ( 20.00 ) = 2.6 kW. e) No; the .00 power is proportional to the time t or proportional to the square root of the angle. 3/ 2

τ t a) P = τω = ταt = τ  t = τ 2  . I I

c) P = τω = τ 2αθ = τ 2( τ / I ) θ = τ 3 / 2 2θ / I .

ˆ j ˆ 10.56: a) From the right-hand rule, the direction of the torque is i × ˆ = k , the + z direction. b), c)

d) The magnitude of the torque is F0 ( x − x 2 l ), which has it maximum at l 2. The torque at x = l 2 is F0 l 4.

10.57:

t2 =

2θ 2θ 2θ I = = . α (τ I ) τ

The angle in radiants is π 2, the moment of inertia is (1 3) ((750 N) (9.80 m s 2 )(1.25 m))3 = 39.9 kg ⋅ m 2 and the torque is (220 N)(1.25 m) = 275 N ⋅ m. Using these in the above expression gives t 2 = 0.455 s 2 , so t = 0.675 s. 10.58: a) From geometric consideration, the lever arm and the sine of the angle  between F and r are both maximum if the string is attached at the end of the rod. b) In terms of the distance x where the string is attached, the magnitude of the torque is

Fxh x 2 + h 2 . This function attains its maximum at the boundary, where x = h, so the string should be attached at the right end of the rod. c) As a function of x, l and h, the torque has magnitude xh τ =F . ( x − l 2) 2 + h 2
This form shows that there are two aspects to increasing the torque; maximizing the lever arm l and maximizing sin φ . Differentiating τ with respect to x and setting equal to zero gives x max = (l 2)(1 + (2 h l ) 2 ). This will be the point at which to attach the string unless 2h > l, in which case the string should be attached at the furthest point to the right, x = l. 10.59: a) A distance L 4 from the end with the clay. b) In this case I = (4 3) ML2 and the gravitational torque is (3L 4)(2 Mg ) sin θ = (3Mg L 2) sin θ , so α = (9 g 8 L) sin θ . c) In this case I = (1 3) ML2 and the gravitational torque is ( L 4)(2 Mg ) sin θ = ( Mg L 2) sin θ , so α = (3 g 2 L) sin θ . This is greater than in part (b). d) The greater the angular acceleration of the upper end of the cue, the faster you would have to react to overcome deviations from the vertical.

10.60: In Fig. (10.22) and Eq. (10.22), with the angle θ measured from the vertical, sin θ = cos θ in Eq. (10.2). The torque is then τ = FR cosθ. a) W =∫
π 2

0

FR cos θ d θ = FR

b) In Eq. (6.14), dl is the horizontal distance the point moves, and so W = F ∫ dl = FR, the same as part (a). c) From K 2 = W = (MR 2 4)ω2 , ω = 4 F MR . d) The torque, and hence the angular acceleration, is greatest when θ = 0, at which point α = ( τ I ) = 2 F MR , and so the maximum tangential acceleration is 2 F M . e) Using the value for ω found in part (c), arad = ω2 R = 4 F M . 10.61: The tension in the rope must be m( g + a ) = 530 N. The angular acceleration of the cylinder is a R = 3.2 rad/s2 , and so the net torque on the cylinder must be 9.28 N ⋅ m. Thus, the torque supplied by the crank is (530 N)(0.25 m) + (9.28 N ⋅ m) = 141.8 N ⋅ m, and the force applied to the crank handle is 141 .8 N ⋅m 0.12 m = 1.2 kN to two figures. 10.62: At the point of contact, the wall exerts a friction force f directed downward and a normal force n directed to the right. This is a situation where the net force on the roll is zero, but the net torque is not zero, so balancing torques would not be correct. Balancing vertical forces, Frod cosθ = f + w + F , and balacing horizontal forces Frod sin θ = n. With f = μk n, these equations become Frod cos θ = μk n + F + w, Frod sin θ = n. (a) Eliminating n and solving for Frod gives

Frod

ω+ F (16.0 kg) (9.80 m/s2 ) + (40.0 N) = = = 266 N. cos θ − μk sin θ cos 30° − (0.25) sin 30°

b) With respect to the center of the roll, the rod and the normal force exert zero torque. The magnitude of the net torque is ( F − f ) R, and f = μk n may be found insertion of the value found for Frod into either of the above relations; i.e., f = μk Frod sin θ = 33.2 N. Then, α= τ (40.0 N − 31.54 N)(18.0 × 10−2 m) = = 4.71 rad/s 2 . I (0.260 kg ⋅ m 2 )

10.63: The net torque on the pulley is TR, where T is the tension in the string, and α = TR I . The net force on the block down the ramp is mg (sin β − μk cos β ) − T = ma. The acceleration of the block and the angular acceleration of the pulley are related by α = αR. a) Multiplying the first of these relations by I R and eliminating α in terms of a, and then adding to the second to eliminate T gives a = mg

( sin β − μk cos β ) = g ( sin β − μk cos β ) ,
m + I / R2

(1 + I / mR )
2

and substitution of numerical values given 1.12 m/s 2 . b) Substitution of this result into either of the above expressions involving the tension gives T = 14.0 N.
a 10.64: For a tension T in the string, mg − T = ma and TR = Iα = I R . Eliminating T and solving for a gives m g a=g = , 2 m+I /R 1 + I / mR 2

where m is the mass of the hanging weight, I is the moment of inertia of the disk combination I = 2.25 × 10 −3 kg ⋅ m 2 from Problem 9.89 and R is the radius of the disk to which the string is attached. a) With m = 1.50 kg, R = 2.50 × 10−2 m, a = 2.88 m/s 2 . b) With m = 1.50 kg, R = 5.00 × 10 −2 m, a = 6.13 m/s 2 . The acceleration is larger in case (b); with the string attached to the larger disk, the tension in the string is capable of applying a larger torque.

(

)

10.65: Taking the torque about the center of the roller, the net torque is fR = αI ,

I = MR 2 for a hollow cylinder, and with α = a / R, f = Ma (note that this is a relation
between magnitudes; the vectors f and a are in opposite directions). The net force is F − f = Ma, from which F = 2 Ma and so a = F 2 M and f = F 2 .
→ →

10.66: The accelerations of blocks A and B will have the same magnitude a. Since the a cord does not slip, the angular acceleration of the pulley will be α = R . Denoting the tensions in the cord as T A and TB , the equations of motion are m A g − TA = m A a TB − m B g = m B a T A − TB = I a, R2

where the last equation is obtained by dividing τ = Iα by R and substituting for α in terms of a. Adding the three equations eliminates both tensions, with the result that m A − mB a=g m A + mB + I / R 2 Then, m A − mB a α = =g . R m A R + mB R + I / R The tensions are then found from 2m A m B + m A I R 2 T A = m A ( g − a) = g m A + mB + I R 2 TB = m B ( g + a ) = g 2m B m A + m B I R 2 . m A + mB + I R 2

As a check, it can be shown that (T A − TB ) R = Iα .

10.67: For the disk, K = (3 4) Mv 2 ( see Example 10.6) . From the work-energy theorem, K 1 = MgL sin β , from which L= 3(2.50 m s) 2 3v 2 = = 0.957 m. 4 g sin β 4(9.80 m s 2 ) sin 30.0°

This same result may be obtained by an extension of the result of Exercise 10.26; for the disk, the acceleration is (2 3) g sin β , leading to the same result. b) Both the translational and rotational kinetic energy depend on the mass which cancels the mass dependence of the gravitational potential energy. Also, the moment of inertia is proportional to the square of the radius, which cancels the inverse dependence of the angular speed on the radius. 10.68: The tension is related to the acceleration of the yo-yo by (2m) g − T = ( 2m) a, and to the angular acceleration by Tb = Iα = I a . Dividing the second equation by b and b adding to the first to eliminate T yields a=g 2m 2 2 =g , α=g , 2 2 ( 2m + I b ) 2 + ( R b) 2b + R 2 b

where I = 2 1 mR 2 = mR 2 has been used for the moment of inertia of the yo-yo. The 2 tension is found by substitution into either of the two equations; e.g.,  2 T = (2m)( g − a ) = (2mg ) 1 −  2 + ( R b) 2   ( R b) 2 2mg  = 2mg = . 2  2 + ( R b) (2(b R ) 2 + 1) 

10.69: a) The distance the marble has fallen is y = h − (2 R − r ) = h + r − 2 R. The radius of the path of the center of mass of the marble is R − r , so the condition that the ball stay on the track is v 2 = g ( R − r ). The speed is determined from the work-energy theorem, mgy = (1 2)mv 2 + (1 2) Iω 2 . At this point, it is crucial to know that even for the curved track, ω = v r ; this may be seen by considering the time T to move around the circle of radius R − r at constant speed V is obtained from 2π ( R − r ) = Vt , during which time the marble rotates by an angle 2π ( R − 1) = ω T , from which ω = V r. The workr energy theorem then states mgy = (7 10)mv 2 , and combining, canceling the factors of m and g leads to (7 10)( R − r ) = h + r − 2 R, and solving for h gives h = (27 10) R − (17 10)r. b) In the absence of friction, mgy = (1 2)mv 2 , and substitution of the expressions for y and v 2 in terms of the other parameters gives (1 2)( R − r ) = h − r − 2 R, which is solved for h = (5 2) R − (3 2)r. 10.70: In the first case, F and the friction force act in opposite directions, and the friction force causes a larger torque to tend to rotate the yo-yo to the right. The net force to the right is the difference F − f , so the net force is to the right while the net torque causes a clockwise rotation. For the second case, both the torque and the friction force tend to turn the yo-yo clockwise, and the yo-yo moves to the right. In the third case, friction tends to move the yo-yo to the right, and since the applied force is vertical, the yo-yo moves to the right.
→

10.71: a) Because there is no vertical motion, the tension is just the weight of the hoop: T = Mg = ( 0.180 kg ) ( 9.8 N kg ) = 1.76 N b) Use τ = Iα to find α . The torque is RT , so α = RT / I = RT MR 2 = T / MR = Mg MR , so α = g R = 9.8 m s 2 ( 0.08 m ) = 122.5 rad/s 2 c) a = Rα = 9.8 m s 2 d) T would be unchanged because the mass M is the same, α and a would be twice as great because I is now 1 MR 2 . 2

(

)

10.72: (a) Στ = Iα and a T = Rα PR = 1 1 a  MR 2α = MR 2  T  2 2  R 2P 200 N aT = = = 50 m/s 2 M 4.00 kg

Distance the cable moves: x = 1 at 2 2 1 50 m = 50 m/s 2 t 2 → t = 1.41 s. 2 v = v0 + at = 0 + 50 m/s 2 (1.41 s ) = 70.5 m s

(

)

(

)

(b) For a hoop, I = MR 2 , which is twice as large as before, so α and a T would be 2 half as large. Therefore the time would be longer. For the speed, v 2 = v0 + 2ax, in which x is the same, so v would be smaller since a is smaller 10.73: Find the speed v the marble needs at the edge of the pit to make it to the level ground on the other side. The marble must travel 36 m horizontally while falling vertically 20 m. Use the vertical motion to find the time. Take + y to be downward. v 0 y = 0, a y = 9.80 m/s 2 , y − y0 = 20 m, t = ? y − y0 = v0 yt + 1 a y t 2 gives t = 2.02 s 2 Then x − x0 = v 0 xt gives v 0 x = 17.82 m/s. Use conservation of energy, where point 1 is at the starting point and point 2 is at the edge of the pit, where v = 17.82 m/s. Take y = 0 at point 2, so y2 = 0 and y1 = h. K1 + U 1 = K 2 + U 2 mgh = 1 mv 2 + 1 Iω 2 2 2
2 Rolling without slipping means ω = v r . I = 5 mr 2 , so 1 Iω2 = 1 mv 2 2 5 7 mgh = 10 mv 2

h= b)

7v 2 7(17.82 m/s) = = 23 m 10 g 10(9.80 m/s 2 )
1 2

Iω 2 = 1 mv 2 , Independent of r. 5

c) All is the same, except there is no rotational kinetic energy term in K : K = 1 mv 2 2 mgh = 1 mv 2 2 h= v2 = 16 m, 0.7 times smaller than the answer in part ( a). 2g

10.74: Break into 2 parts, the rough and smooth sections. Rough : mgh1 = 1 mv 2 + 1 Iω 2 2 2 1 2 12  v  mv +  mR 2   2 25  R  10 v 2 = gh1 7 Smooth: Rotational KE does not change. mgh1 = mgh2 + 1 2 mv + KE Rot = 2 1  10  gh2 +  gh1  = 2 7  vB = = 1 2 mv Bottom + KE Rot 2 1 2 vB 2 10 gh1 + 2 gh2 7
2

10 (9.80 m/s 2 )(25 m) + 2(9.80 m/s 2 )(25 m) 7 = 29.0 m/s

10.75: a) Use conservation of energy to find the speed v 2 of the ball just before it leaves the top of the cliff. Let point 1 be at the bottom of the hill and point 2 be at the top of the hill. Take y = 0 at the bottom of the hill, so y1 = 0 and y 2 = 28.0 m. K1 + U 1 = K 2 + U 2
1 2 2 2 mv12 + 1 Iω 12 = mgy 2 + 1 mv 2 + 1 Iω 2 2 2 2

Rolling without slipping means ω = v r and 1 Iω 2 = 2
7 10 2 7 mv12 = mgy 2 + 10 mv 2

1 2 2 5

(

mr 2 (v / r ) 2 = 1 mv 2 5

)

v 2 = v12 − 10 gy 2 = 15.26 m s 7 Consider the projectile motion of the ball, from just after it leaves the top of the cliff until just before it lands. Take + y to be downward. Use the vertical motion to find the time in the air: v0 y = 0, a y = 9.80 m s 2 , y − y0 = 28.0 m, t = ? During this time the ball travels horizontally x − x0 = v0 xt = (15.26 m s ) ( 2.39 s ) = 36.5 m. b) At the bottom of the hill, ω = v r = ( 25.0 m s ) r. The rotation rate doesn't change while the ball is in the air, after it leaves the top of the cliff, so just before it lands ω = (15.3 s) r. The total kinetic energy is the same at the bottom of the hill and just before it lands, but just before it lands less of this energy is rotational kinetic energy, so the translational kinetic energy is greater.
2 2 Just before it lands, v y = v0 y + a y t = 23.4 s and vx = v0 x = 15.3 s v = vx + v y = 28.0 m s

y − y0 = v0 y t + 1 a y t 2 gives t = 2.39 s 2

10.76: (a) mgh =

1 2 1 2 mv + Iω (1) 2 2

1  I = I rim + I spokes = M r R 2 + 6 m s R 2  3  Uniform density means: m r = λ 2πR and m s = λR. No slipping means that ω = v R. Also, m = mr + ms = 2πRλ + 6 Rλ = 2 Rλ( π + 3) substituting into (1) gives 2 Rλ( π + 3) gh = ω=

1 ( 2 Rλ ) ( π + 3)( Rω) 2 + 1 2πRλR2 + 6 1 πRR 2  ω2   2 2 3  

( π + 3) gh R 2 ( π + 2)

=

( π + 3) (9.80 m s2 )( 58.0 m ) ( 0.210 m ) 2 ( π + 2)

= 124 rad s

and v = Rω = 26.0 m s (b) Doubling the density would have no effect because it does not appear in the answer. 1 ω α R , so doubling the diameter would double the radius which would reduce ω by half, but v = Rω would be unchanged. a) The front wheel is turning at ω = 1.00 rev s = 2π rad s. υ = rω = (0.330 m)(2π rad s) = 2.07 s b) ω = v r = (2.07 m s) (0.655 m = 3.16 rad s = 0.503 rev s c) ω = v r = ( 2.07 m s) (0.220 m) = 9.41 rad s = 1.50 rev s

10.77:

10.78: a) The kinetic energy of the ball when it leaves the tract (when it is still rolling without slipping) is (7 10)mv 2 and this must be the work done by gravity, W = mgh, so v = 10gh 7. The ball is in the air for a time t = 2 y g , so x = vt = 20hy 7. b) The answer does not depend on g, so the result should be the same on the moon. c) The presence of rolling friction would decrease the distance. d) For the dollar coin, modeled as a uniform disc, K = (3 4)mv 2 , and so x = 8hy 3.

10.79:

10 K a) v = = 7m

(10)( 0.800) (1 2) ( 400 N m ) ( 0.15 m ) 2 7( 0.0590 kg )

= 9.34 m s.

b) Twice the speed found in part (a), 18.7 m s. c) If the ball is rolling without slipping, the speed of a point at the bottom of the ball is zero. d) Rather than use the intermediate calculation of the speed, the fraction of the initial energy that was converted to gravitational potential energy is ( 0.800) ( 0.900) , so ( 0.720) (1 2) kx 2 = mgh and solving for h gives 5.60 m. 10.80: a)

b) R is the radius of the wheel (y varies from 0 to 2R) and T is the period of the wheel’s rotation. c) Differentiating, 2πR   2πt  vx = 1 − cos T  T    2πR  2πt  vy = sin   T  T   2π   2πt  ax =   R sin   T   T   2π   2πt  a y =   R cos . T   T 
2 2

 2πt  d) v x = v y = 0 when   = 2π or any multiple of 2π , so the times are integer  T  multiples of the period T. The acceleration components at these times are 4π 2 R a x = 0, a y = . T2 2 4π 2 R  2π  2 2 2  2πt  2  2πt  e) a x + a y =   R cos  ,  + sin  = T2 T   T   T  independent of time. This is the magnitude of the radial acceleration for a point moving on a circle of radius R with constant angular velocity 2Tπ . For motion that consists of this  circular motion superimposed on motion with constant velocity ( a = 0) , the acceleration due to the circular motion will be the total acceleration.

10.81: For rolling without slipping, the kinetic energy is (1 2) m + I R 2 v 2 = ( 5 6 ) mv 2 ; initially, this is 32.0 J and at the return to the bottom it is 8.0 J. Friction has done − 24.0 J of work, − 12.0 J each going up and down. The potential energy at the highest point was 20 0 20.0 J, so the height above the ground was ( 0.600 kg ) (.9.J m s 2 ) = 3.40 m. 80 10.82: Differentiating , and obtaining the answer to part (b), dθ θ  ω= = 3bt 2 = 3b  dt b
23

(

)

= 3b1 3θ 2 3 ,
13

dω θ  α =− = 6bt = 6b  dt b a)

= 6b 2 3θ 1 3 . 9 I cm b 2 3θ 4 3 . 2

W = ∫ I cm α dθ = 6b 2 3 I cm ∫ θ 1 3dθ =

c) The kinetic energy is 1 9 K = I cm ω 2 = I cm b 2 3θ 4 3 , 2 2 in agreement with Eq. (10.25); the total work done is the change in kinetic energy. 10.83: Doing this problem using kinematics involves four unknowns (six, counting the two angular accelerations), while using energy considerations simplifies the calculations greatly. If the block and the cylinder both have speed v, the pulley has angular velocity v/R and the cylinder has angular velocity v/2R, the total kinetic energy is  3 1  2 M (2 R) 2 MR 2 Mv + (v 2 R ) 2 + (v R ) 2 + Mv 2  = Mv 2 .  2 2 2  2 This kinetic energy must be the work done by gravity; if the hanging mass descends a distance y, K = Mgy, or v 2 = (2 3) gy. For constant acceleration, v 2 = 2ay , and comparison of the two expressions gives a = g 3. K=

10.84: (a)

Στ = Iα 1 1 mg cosθ = mL2α 2 3 3 cosθ  3  (9.80 m s 2 ) cos 60° α= g =  = 0.92 rad s 2 2 L 8.00 m  2 (b) As the bridge lowers, θ changes, so α is not constant. Therefore Eq. (9.17) is not valid. (c) Conservation of energy: PEi = KEf → mgh = 11  mg L sin θ =  mL2 ω2 2 23  ω= = 3g sin θ L 3(9.8 m s 2 ) sin 60° = 1.78 rad s 8.00 m 1 2 Iω 2

10.85: The speed of the ball just before it hits the bar is v = 2 gy = 15.34 m s. Use conservation of angular momentum to find the angular velocity ω of the bar just after the collision. Take the axis at the center of the bar. L1 = mvr = ( 5.00 kg ) (15.34 m s ) ( 2.00 m ) = 153.4 kg ⋅ m 2 Immediately after the collsion the bar and both balls are rotating together. L2 = I tot ω 1 1 2 2 I tot = Ml 2 + 2mr 2 = ( 8.00 kg ) ( 4.00 m ) + 2( 5.00 kg ) ( 2.00 m ) = 50.67 kg ⋅ m 2 12 12 L2 = L1 = 153.4 kg ⋅ m 2 ω = L2 I tot = 3.027 rad s Just after the collision the second ball has linear speed v = rw = ( 2.00 m ) ( 3.027 rad s ) = 6.055 m s and is moving upward. 1 2 mv = mgy gives y = 1.87 m for the height the second ball goes. 2

10.86: a) The rings and the rod exert forces on each other, but there is no net force or torque on the system, and so the angular momentum will be constant. As the rings slide toward the ends, the moment of inertia changes, and the final angular velocity is given by Eq. (10.33) ,
1  12 ML2 + 2mr12  I1 5.00 × 10 −4 kg ⋅ m 2 ω1 ω2 = ω1 = ω1  1 = ω1 = , 2 2 I2 2.00 × 10 − 3 kg ⋅ m 2 4  12 ML + 2mr2 

ans so ω 2 = 7.5 rev min. Note that conversion from rev/min to rad/s is not necessary. b) The forces and torques that the rings and the rod exert on each other will vanish, but the common angular velocity will be the same, 7.5 rev/min. 10.87: The intial angular momentum of the bullet is ( m 4 ) ( v ) ( L 2 ) , and the final moment of intertia of the rod and bullet is ( m 3) L2 + ( m 4 )( L 2 ) = (19 48) mL2 . Setting the initial mvL 8 6 angular moment equal to ωI and solving for ω gives ω = = v L. 2 (19 48) mL 19
2

b)

(1 2) Iω2 = (19 48) mL2 ( ( 6 19) ( v L ) ) 2 (1 2 ) ( m 4 ) v 2 ( m 4) v 2

=

3 . 19

10.88: Assuming the blow to be concentrated at a point (or using a suitably chosen “average” point) at a distance r from the hinge, Στ ave = rFave , and ∆L = rFave ∆t = rJ . The angular velocity ω is then ∆L rFave ∆t ( l 2 ) Fave ∆t 3 Fave ∆t ω= = = 1 2 = , I I ml 2 ml 3 Where l is the width of the door. Substitution of the given numeral values gives ω = 0.514 rad s. a) The initial angular momentum is L = mv( l 2 ) and the final moment of inertia
2

10.89:

is I = I 0 + m( l 2 ) , so ω=

mv( l 2 ) = 5.46 rad s. ( M 3) l 2 + m( l 2) 2

b) ( M + m ) gh = (1 2) ω2 I , and after solving for h and substitution of numerical values, h = 3.16 × 10 −2 m. c) Rather than recalculate the needed value of ω , note that ω will be proportional to v and hence h will be proportional to v 2 ; for the board to swing all the way over, h = 0.250 m. and so v = ( 360 m s )
0.250 m 0.0316 m

= 1012 m s.

10.90: Angular momentum is conserved, so I 0ω 0 = I 2ω 2 , or, using the fact that for a common mass the moment of inertia is proportional to the square of the radius, 2 2 R02ω0 = R2 ω2 , or R02ω 0 = ( R0 + ∆R ) ( ω 0 + ∆ω ) ~ R02ω 0 + 2 R0 ∆Rω 0 + R02 ∆ω , where the terms in ∆R∆ω and ∆ω 2 have been omitted. Canceling the R02ω 0 term gives R ∆ω ∆R = − 0 = −1.1 cm. 2 ω0 10.91: The initial angular momentum is L1 = ω0 I A and the initial kinetic energy is is (1 4 ) ω0 and the final kinetic energy is (1 2) 4 I A ( ω0 4 ) = (1 4 ) K1. (This result may be
2 2 K1 = I Aω0 2. The final total moment of inertia is 4 I A , so the final angular velocity

obtained more directly from K = L2 I . Thus, ∆K = −( 3 4 ) K1 and K1 = −( 4 3) ( − 2400 J ) = 3200 J. 10.92: The tension is related to the block’s mass and speed, and the radius of the circle, v2 by T = m . The block’s angular momentum with respect to the hole is L = mvr , so in r terms of the angular momentum, 1 m 2v 2 r 2 ( mvr ) L2 T = mv = = = . r m r3 mr 3 mr 3
2 2

The radius at which the string breaks can be related to the initial angular momentum by L2 ( mv1r1 ) 2 = ( ( 0.250 kg ) ( 4.00 m s ) ( 0.800 m ) ) 2 , r3 = = mTmax mTmax ( 0.250 kg ) ( 30.0 N ) from which r = 0.440 m. 10.93: The train’s speed relative to the earth is 0.600 m s + ω ( 0.475 m ) , so the total angular momentum is

( ( 0.600 m s ) + ω( 0.475 m ) ) (1.20 kg )( 0.475 m ) + ω(1 2) ( 7.00 kg )  1.00 m  = 0,    2 
from which ω = −0.298 rad s ,with the minus sign indicating that the turntable moves clockwise, as expected.

2

10.94:

a), g)

    b) Using the vector product form for the angular momentum, v1 = −v 2 and r1 = − r2 , so     mr2 × v 2 = mr1 × v1 ,  so the angular momenta are the same. c) Let ω = ωˆ. Then, j    ˆ ˆ v1 = ω × r1 = ω zi − xk , and    ˆ ˆ L1 = mr1 × v1 = mω ( − xR ) i + x 2 + y 2 ˆ + ( xR ) k . j

(

(

)

(

)

)

   With x 2 + y 2 = R 2 , the magnitude of L is 2mωR 2 , and L1 ⋅ ω = mω 2 R 2 , and so  mω 2 2 cosθ = ( 2 mωR 2R)(ω ) = 1 , and θ = π . This is true for L2 as well, so the total angular 2 6 momentum makes an angle of
π 6

with the +y-axis.
2

d) From the intermediate

calculation of part (c), Ly1 = mωR = mvR, so the total y-component of angular momentum is Ly = 2mvR. e) Ly is constant, so the net y-component of torque is zero. f) Each particle moves in a circle of radius R with speed v, and so is subject to an inward force of magnitude mv 2 R . The lever arm of this force is R, so the torque on each has magnitude mv 2 . These forces are directed in opposite directions for the two particles, and the position vectors are opposite each other, so the torques have the same magnitude and direction, and the net torque has magnitude 2mv 2 .

10.95:

a) The initial angular momentum with respect to the pivot is mvr , and the

final total moment of inertia is I + mr 2 , so the final angular velocity is ω = mvr mr 2 + I . b) The kinetic energy after the collision is 1 K = ω2 mr 2 + I = ( M + m ) gh, or 2 2( M + m ) gh ω= . mr 2 + I

(

)

(

)

(

)

 m  c) Substitution of Ι = Μr 2 into either of the result of part (a) gives ω =  ( v r ) , m+M  and into the result of part (b), ω = 2 gh (1 r ), which are consistent with the forms for v. 10.96: The initial angular momentum is Ιω1 − mRv1 , with the minus sign indicating that runner’s motion is opposite the motion of the part of the turntable under his feet. The final angular momentum is ω 2 (Ι + mR 2 ), so Ιω − mRv1 ω2 = 1 Ι + mR 2 (80 kg ⋅ m 2 )(0.200 rad s) − (55.0 kg)(3.00 m)( 2.8 m s) = (80 kg ⋅ m 2 ) + (55.0 kg)(3.00 m) 2 = −0.776 rad s, where the minus sign indicates that the turntable has reversed its direction of motion (i.e., the man had the larger magnitude of angular momentum initially). 10.97: From Eq. (10.36), ωr (50.0 kg)(9.80 m s 2 )(0.040 m) Ω= = = 12.7 rad s, Ιω (0.085 kg ⋅ m 2 )((6.0 m s) (0.33 m)) or 13 rad s to two figures, which is quite large.

J 10.98: The velocity of the center of mass will change by ∆vcm = m , and the angular

velocity will change by ∆ω =

⋅ The change is velocity of the end of the bat will J J ( x − xcm ) xcm then be ∆vend = ∆vcm _ ∆ωxcm = − ⋅ Setting m Ι ∆vend = 0 allows cancellati on of J , and gives Ι = ( x − xcm ) xcm m, which when solved for x is Ι (5.30 × 10 −2 kg ⋅ m 2 ) x= + xcm = + (0.600 m) = 0.710 m. xcm m (0.600 m)(0.800 kg) 10.99: In Fig. (10.34(a)), if the vector r , and hence the vector L are not horizontal but make an angle β with the horizontal, the torque will still be horizontal (the torque must be perpendicular to the vertical weight). The magnitude of the torque will be ω r cos β , and this torque will change the direction of the horizontal component of the angular momentum, which has magnitude L cos β . Thus, the situation of Fig. (10.36) is reproduced, but with L horiz instead of L . Then, the expression found in Eq. (10.36) becomes Ω= dφ d L L horiz τ mgr cos β ωr = = = = ⋅ dt dt L cos β Ιω Lhoriz

J ( x − x cm ) I

10.100:

a)

The distance from the center of the ball to the midpoint of the line joining the points where the ball is in contact with the rails is R 2 − ( d 2 ) , so vcm = ω R 2 − d 2 4. when d = 0, this reduces to v cm = ωR, the same as rolling on a flat surface. When
2

d = 2R, the rolling radius approaches zero, and v cm → 0 for any ω .
b) K= 1 2 1 2 mv + Iω 2 2   1 2 = mvcm + ( 2 5) mR 2   2   mv 2  2 = cm 5 + 2 10  1 − d 4R 2

c)

  2 2 R − d 4   vcm

2

(

)

   

 .  Setting this equal to mgh and solving for v cm gives the desired result. c) The denominator in the square root in the expression for vcm is larger than for the case d = 0, so v cm is smaller. For a given speed, ω is large than the d = 0 case, so a larger fraction of the kinetic energy is rotational, and the translational kinetic energy, and hence v cm , is smaller. d) Setting the expression in part (b) equal to 0.95 of that of the d = 0 case and solving for the ratio d R gives d R = 1.05. Setting the ratio equal to 0.995 gives d R = 0.37.

(

)

10.101:

a)

The friction force is f = μk n = μk Mg , so a = μk g . The magnitude of the angular acceleration is gives t= and
2  Rω0  1 1 R 2ω0  = d = at 2 = ( μk g )   3 μ g  18 μ g . 2 2 k  k  2 fR I

=

( 1 2 ) MR 2

μ k MgR

=

2 μk g R

. b) Setting v = at = ωR = ( ω 0 − ωt ) R and solving for t

Rω0 Rω0 Rω0 = = , a + Rα μk g + 2 μk g 3 μk g

c) The final kinetic energy is ( 3 4 ) Mv 2 = ( 3 4) M ( at ) , so the change in kinetic energy
2

is 3  Rω0  1 1 2 2  − MR 2 ω0 = MR 2 ω0 . M  μk g   4 4  3 μk g  6
2

10.102: Denoting the upward forces that the hands exert as FL and FR , the conditions that FL and FR must satisfy are FL + FR = w Iω , r where the second equation is τ = ΩL, divided by r. These two equations can be solved for the forces by first adding and then subtracting, yielding 1 Iω  FL =  ω + Ω  2 r  1 Iω  FR =  ω − Ω . 2 r  Using the values ω = mg = (8.00 kg)(9.80 m s 2 ) = 78.4 N and FL − FR = Ω Iω (8.00 kg )(0.325 m) 2 (5.00 rev s × 2π rad rev) = = 132.7 kg ⋅ m s r (0.200 m) gives FL = 39.2 N + Ω(66.4 N ⋅ s), FR = 39.2 N − Ω(66.4 N ⋅ s). a) Ω = 0, FL = FR = 39.2 N. b) Ω = 0.05 rev s = 0.314 rad s, FL = 60.0 N, FR = 18.4 N. c) Ω = 0.3 rev s = 1.89 rad s, FL = 165 N, FR = −86.2 N, with the minus sign indicating a downward force. d) FR = 0 gives Ω =
39.2 N 66.4 N ⋅s

= 0.575 rad s , which is 0.0916 rev s .

10.103:

  a) See Problem 10.92; T = mv12 r12 r 3 . b) T and dr are always antiparallel, so W = − ∫ T dr = mv12 r12 ∫
r1 r2

dr mv12 2  1 1  = r1  2 − 2 . r2 r3 2  r2 r1 
r1

c) v2 = v1 (r1 r2 ), so
2  1 mv12  r1  2 2   − 1 ' ∆K = m(v2 − v1 ) = 2 2  r2      which is the same as the work found in part (b).

Chapter 11

11.1:

Take the origin to be at the center of the small ball; then, (1.00 kg)(0) + (2.00 kg )(0.580 m) xcm = = 0.387 m 3.00 kg from the center of the small ball. 11.2: The calculation of Exercise 11.1 becomes (1.00 kg)(0) + (1.50 kg )(0.280 m) + (2.00 kg )(0.580 m) xcm = = 0.351 m 4.50 kg This result is smaller than the one obtained in Exercise 11.1. 11.3: In the notation of Example 11.1, take the origin to be the point S , and let the child’s distance from this point be x. Then, M (− D 2) + mx MD scm = = 0, x = = 1.125 m, M +m 2m which is ( L 2 − D 2) 2, halfway between the point S and the end of the plank. 11.4: a) The force is applied at the center of mass, so the applied force must have the same magnitude as the weight of the door, or 300 N. In this case, the hinge exerts no force. b) With respect to the hinge, the moment arm of the applied force is twice the distance to the center of mass, so the force has half the magnitude of the weight, or 150 N . The hinge supplies an upward force of 300 N − 150 N = 150 N. 11.5:

F (8.0 m) sin 40° = (2800 N)(10.0 m), so F = 5.45 kN, keeping an extra figure.

11.6: The other person lifts with a force of 160 N − 60 N = 100 N. Taking torques about the point where the 60 - N force is applied,  160 N  (100 N) x = (160 N)(1.50 m), or x = (1.50 m)  100 N  = 2.40 m.   

11.7: If the board is taken to be massless, the weight of the motor is the sum of the m )( applied forces, 1000 N. The motor is a distance ( 2.001000 600 N ) = 1.200 m from the end where ( N) the 400-N force is applied.

11.8: The weight of the motor is 400 N + 600 N − 200 N = 800 N. Of the myriad ways to do this problem, a sneaky way is to say that the lifters each exert 100 N to the lift the board, leaving 500 N and 300 N to the lift the motor. Then, the distance of the motor m from the end where the 600-N force is applied is ( 2.00(800)(300 N ) = 0.75 m .The center of N) gravity is located at is applied. 11.9: The torque due to Tx is − Tx h = − Lw cot θ h, and the torque due to Ty is Ty D = Lw . D
( 200 N )(1.0 m ) + (800 N )( 0.75 m ) (1000 N )

= 0.80 m from the end where the 600 N force

h The sum of these torques is Lw(1 − D cot θ ). From Figure (11.9(b)), h = D tan θ , so the net torque due to the tension in the tendon is zero.

11.10: a) Since the wall is frictionless, the only vertical forces are the weights of the man and the ladder, and the normal force. For the vertical forces to balance, n2 = w1 + wm = 160 N + 740 N = 900 N, and the maximum frictional forces is μs n2 = (0.40)(900 N) = 360 N (see Figure 11.7(b)). b) Note that the ladder makes contact with the wall at a height of 4.0 m above the ground. Balancing torques about the point of contact with the ground, (4.0 m)n1 = (1.5 m)(160 N) + (1.0 m)(3 5))(740 N) = 684 N ⋅ m, so n1 = 171.0 N, keeping extra figures. This horizontal force about must be balanced by the frictional force, which must then be 170 N to two figures. c) Setting the frictional force, and hence n1 , equal to the maximum of 360 N and solving for the distance x along the ladder, (4.0 m)(360 N) = (1.50 m)(160 N) + x(3 5)(740 N), so x = 2.70 m, or 2.7 m to two figures. 11.11: Take torques about the left end of the board in Figure (11.21). a) The force F at the support point is found from F (1.00 m) = +(280 N)(1.50 m) + (500 N)(3.00 m), or F = 1920 N. b) The net force must be zero, so the force at the left end is (1920 N) − (500 N) − (280 N) = 1140 N, downward.

11.12:

a)

b) x = 6.25 m when FA = 0, which is 1.25 m beyond point B. c) Take torques about the right end. When the beam is just balanced, FA = 0, so FB = 900 N. The distance that point B must be from the right end is then
( 300 N )( 4.50 m ) ( 900 N )

= 1.50 m.

11.13: In both cases, the tension in the vertical cable is the weight ω. a) Denote the length of the horizontal part of the cable by L. Taking torques about the pivot point, TL tan 30.0° = wL + w( L 2), from which T = 2.60 w. The pivot exerts an upward vertical force of 2 w and a horizontal force of 2.60 w , so the magnitude of this force is 3.28w , directed 37.6° from the horizontal. b) Denote the length of the strut by L , and note that the angle between the diagonal part of the cable and the strut is 15.0°. Taking torques about the pivot point, TL sin 15.0° = wL sin 45.0° + ( w 2) L sin 45°, so T = 4.10 w. The horizontal force exerted by the pivot on the strut is then T cos 30.0° = 3.55ω and the vertical force is (2 w) + T sin 30° = 4.05w, for a magnitude of 5.38w, directed 48.8°. 11.14: a) Taking torques about the pivot, and using the 3-4-5 geometry, (4.00 m) (3 5)T = (4.00 m)(300 N) + (2.00 m)(150 N), so T = 625 N. b) The horizontal force must balance the horizontal component of the force exerted by the rope, or T (4 5) = 500 N. The vertical force is 300 N + 150 N − T (3 5) = 75 N, upwards.

11.15: To find the horizontal force that one hinge exerts, take the torques about the other hinge; then, the vertical forces that the hinges exert have no torque. The horizontal force is found from FH (1.00 m) = (280 N)(0.50 m), from which FH = 140 N. The top hinge exerts a force away from the door, and the bottom hinge exerts a force toward the door. Note that the magnitudes of the forces must be the same, since they are the only horizontal forces. 11.16: (a) Free body diagram of wheelbarrow:

Στ wheel = 0 − (450 N)(2.0 m) + (80 N)(0.70 m) + WL (0.70 m) = 0 WL = 1200 N (b) From the ground. 11.17: Consider the forces on Clea.

nr = 89 N, nf = 157 N nr + nf = w so w = 246 N

11.18: a) Denote the length of the boom by L, and take torques about the pivot point. The tension in the guy wire is found from

TL sin 60° = (5000 N) L cos 60.0° + (2600 N)(0.35 L) cos 60.0°,
so T = 3.14 kN. The vertical force exerted on the boom by the pivot is the sum of the F  weights, 7.06 kN and the horizontal force is the tension, 3.14 kN. b) No; tan  v  ≠ 0. F   H 11.19: To find the tension TL in the left rope, take torques about the point where the rope at the right is connected to the bar. Then, TL (3.00 m) sin 150° = (240 N)(1.50 m) + (90 N)(0.50 m), so TL = 270 N. The vertical component of the force that the rope at the end exerts must be (330 N) − (270 N) sin 150° = 195 N, and the horizontal component of the force is − (270 N) cos 150°, so the tension is the rope at the right is TR = 304 N. and θ = 39.9°. 11.20: The cable is given as perpendicular to the beam, so the tension is found by taking torques about the pivot point; T (3.00 m) = (1.00 kN)(2.00 m) cos 25.0° + (5.00 kN)(4.50 m) cos 25.0°, or T = 7.40 kN. The vertical component of the force exerted on the beam by the pivot is the net weight minus the upward component of T , 6.00 kN − T cos 25.0° = 0.17 kN. The horizontal force is T sin 25.0° = 3.13 kN. 11.21: a) F1 (3.00 m) − F2 (3.00 m + l ) = (8.00 N)(−l ). This is given to have a magnitude of 6.40 N.m, so l = 0.80m. b) The net torque is clockwise, either by considering the figure or noting the torque found in part (a) was negative. c) About the point of contact of F 2 , the torque due to F 1 is − F1l , and setting the magnitude of this torque to 6.40 N ⋅ m gives l = 0.80 m, and the direction is again clockwise. 11.22: From Eq. (11.10), l (0.200 m) Y=F 0 =F = F (1333 m −2 ). −2 −4 2 ∆lΑ (3.0 × 10 m)(50.0 × 10 m )

Then, F = 25.0 N corresponds to a Young’s modulus of 3.3 × 10 4 Pa, and F = 500 N corresponds to a Young’s modulus of 6.7 × 10 5 Pa.

11.23:

A=

Fl 0 (400 N)(2.00 m) = = 1.60 × 10 −6 m 2 , 10 −2 Y∆l (20 × 10 Pa )(0.25 × 10 m)

and so d = 4 A π = 1.43 × 10 −3 m, or 1.4 mm to two figures. =

11.24:

a) The strain, from Eq. (11.12), is
d2 4

∆l l0

F YA

. For steel, using Y from Table (11.1)

and A = π

= 1.77 × 10 −4 m 2 , (4000 N) ∆l = = 1.1 × 10 − 4. 11 l 0 (2.0 × 10 Pa )(1.77 × 10 − 4 m 2 )

Similarly, the strain for copper (Y = 1.10 × 1011 Pa ) is 2.1 × 10−4. b) Steel: (1.1 × 10−4 ) × (0.750 m) = 8.3 × 10−5 m . Copper: (2.1 × 10−4 )(0.750 m) = 1.6 × 10−4 m ⋅ 11.25: From Eq. (11.10), Y= (5000 N)(4.00 m) = 2.0 × 1011 Pa. −4 2 −2 (0.50 × 10 m )(0.20 × 10 m)

11.26:

From Eq. (11.10), (65.0 kg )(9.80 m s 2 )(45.0 m) Y= = 6.8 × 10 8 Pa. −3 2 (π (3.5 × 10 m) )(1.10 m)

11.27:

a) The top wire is subject to a tension of (16.0 kg )(9.80 m s 2 ) = 157 N and
(157 N ) ( 20×1010 Pa )( 2.5×10 − 7 m 2 )

hence a tensile strain of

= 3.14 × 10 −3 , or 3.1 × 10 −3 to two figures. The

bottom wire is subject to a tension of 98.0 N, and a tensile strain of 1.96 × 10 −3 , or 2.0 × 10 −3 to two figures. b) (3.14 × 10−3 )(0.500 m) = 1.57 mm, (1.96 × 10 −3 )(0.500 m) = 0.98 mm.

11.28:

a)

( 8000 kg )(9.80 m 2 )
s

π (12.5×10 − 2 m ) 2

= 1.6 × 106 Pa. b)

1.6 ×10 6 Pa 2.0 ×1010 Pa

= 0.8 × 10 −5.

c) (0.8 × 10 −5 ) × (2.50 m) = 2 × 10−5 m. (2.8 − 1)(1.013 × 10 5 Pa )(50.0 m 2 ) = 9.1 × 10 6 N.

11.29:

11.30: a) The volume would increase slightly. b) The volume change would be twice as great. c) The volume is inversely proportional to the bulk modulus for a given pressure change, so the volume change of the lead ingot would be four times that of the gold.

11.31:

a)

250 N 0.75 ×10 -4 m 2

= 3.33 × 106 Pa. b) (3.33 × 106 Pa)(2)(200 × 10−4 m 2 ) = 133 kN.

11.32:

a) Solving Eq. (11.14) for the volume change, ΔV = −kV∆P = −(45.8 × 10 −11 Pa −1 )(1.00 m3 )(1.16 × 108 Pa − 1.0 × 105 Pa ) = −0.0531 m 3 .

b) The mass of this amount of water not changed, but its volume has decreased to 03×103 1.000 m 3 − 0.053 m 3 = 0.947 m 3 , and the density is now 1.0.947 mkg = 1.09 ×10 3 kg m 3 . 3

11.33:

B=

(600 cm 3 )(3.6 × 106 Pa ) 1 = 4.8 × 109 Pa , k = = 2.1 × 10 −10 Pa −1. 3 (0.45 cm ) B

11.34:

a) Using Equation (11.17), Shear strain = F|| (9 × 105 N) = = 2.4 × 10− 2. AS [(.10 m)(.005 m)][7.5 × 1010 Pa ]

b) Using Equation (11.16), x = Shear stain ⋅ h = (.024)(.1 m) = 2.4 ×10 −3 m. 11.35: The area A in Eq. (11.17) has increased by a factor of 9, so the shear strain for the larger object would be 1 9 that of the smaller.

11.36:

Each rivet bears one-quarter of the force, so Shear stress =
1 F|| (1.20 × 104 N) = 4 = 6.11 × 108 Pa. A π(.125 × 10− 2 m) 2

11.37:

F A

=

( 90.8 N ) π ( 0.92 ×10 −3 m ) 2

= 3.41 × 107 Pa , or 3.4 × 107 Pa to two figures.

11.38:

a) (1.6 × 10−3 )(20 × 1010 Pa )(5 × 10−6 m 2 ) = 1.60 × 103 N. b) If this were the case, the wire would stretch 6.4 mm.

c) (6.5 × 10−3 )(20 × 1010 Pa )(5 × 10−6 m 2 ) = 6.5 × 103 N.

11.39:

Ftot (2.40 × 108 Pa ) (3.00 × 10 −4 m 2 ) 3 a= = − 9.80 m s 2 = 10.2 m s 2 . m (1200 kg)

11.40:

A=

350 N 4.7 ×10 8 Pa

= 7.45 × 10−7 m 2 , so d = 4 A π = 0.97 mm.

11.41: a) Take torques about the rear wheel, so that fωd = ωxcm , or xcm = fd . b) (0.53)(2.46 m) = 1.30 m to three figures. 11.42: If Lancelot were at the end of the bridge, the tension in the cable would be (from taking torques about the hinge of the bridge) obtained from T (12.0 N) = (600 kg )(9.80 m s 2 )(12.0 m) + (200 kg )(9.80 m s 2 )(6.0 m), so T = 6860 N. This exceeds the maximum tension that the cable can have, so Lancelot is going into the drink. To find the distance x Lancelot can ride, replace the 12.0 m multiplying Lancelot’s weight by x and the tension T by Tmax = 5.80 × 10 3 N and solve for x; x= (5.80 × 103 N)(12.0 m) − (200 kg)(9.80 m s 2 )(6.0 m) = 9.84 m. (600 kg)(9.80 m s 2 )

11.43:

For the airplane to remain in level flight, both ∑ F = 0 and ∑ τ = 0 .

Taking the clockwise direction as positive, and taking torques about the center of mass, Forces: − Ftail − W + Fwing = 0 Torques: − (3.66 m) Ftail + (.3 m) Fwing = 0 A shortcut method is to write a second torque equation for torques about the tail, and solve for the Fwing : −(3.66 m)(6700 N) + (3.36 m) Fwing = 0. This gives Fwing = 7300 N(up), and Ftail = 6700 N − 7300 N = −600 N(down). Note that the rear stabilizer provides a downward force, does not hold up the tail of the aircraft, but serves to counter the torque produced by the wing. Thus balance, along with weight, is a crucial factor in airplane loading. 11.44: The simplest way to do this is to consider the changes in the forces due to the extra weight of the box. Taking torques about the rear axle, the force on the front wheels is decreased by 3600 N 1.00 m = 1200 N, so the net force on the front wheels 3.00 m is 10,780 N − 1200 N = 9.58 × 10 3 N to three figures. The weight added to the rear wheels is then 3600 N + 1200 N = 4800 N, so the net force on the rear wheels is 8820 N + 4800 N = 1.36 × 10 4 N, again to three figures. b) Now we want a shift of 10,780 N away from the front axle. Therefore, W
1.00 m 3.00 m

= 10,780 N and so w = 32,340 N.

11.45: Take torques about the pivot point, which is 2.20 m from Karen and 1.65 m from Elwood. Then wElwood (1.65 m) = (420 N)(2.20 m) + (240 N)(0.20 m), so Elwood weighs 589 N. b) Equilibrium is neutral.

11.46: a) Denote the weight per unit length as α, so w1 = α (10.0 cm), w2 = α (8.0 cm), and w3 = αl . The center of gravity is a distance x cm to the right of point O where xcm = w1 (5.0 cm) + w2 (9.5 cm) + w3 (10.0 cm − l 2) w1 + w2 + w3

=

(10.0 cm)(5.0 cm) + (8.0 cm)(9.5 cm) + l (10.0 cm − l 2) . (10.0 cm) + (8.0 cm) + l

Setting x cm = 0 gives a quadratic in l , which has as its positive root l = 28.8 cm. b) Changing the material from steel to copper would have no effect on the length l since the weight of each piece would change by the same amount.

    Let ri′ = ri − R ,where R is the vector from the point O to the point P.    The torque for each force with respect to point P is then τi′ = ri′ × Fi , and so the net torque is
11.47:

In the last expression, the first term is the sum of the torques about point O, and the second term is given to be zero, so the net torques are the same.  11.48: From the figure (and from common sense), the force F1 is directed along the length of the nail, and so has a moment arm of (0.0800 m) sin 60° . The moment arm of  F2 is 0.300 m, so F2 = F1 (0.0800 m) sin 60° = (500 N)(0.231) = 116 N. (0.300 m)

∑ τ =∑ ( r − R ) × F     = ∑r × F − ∑ R× F     = ∑r × F − R × ∑F .
i i i
i i i i i i









11.49: The horizontal component of the force exerted on the bar by the hinge must  balance the applied force F , and so has magnitude 120.0 N and is to the left. Taking torques about point A, (120.0 N)(4.00 m) + FV (3.00 m), so the vertical component is − 160 N , with the minus sign indicating a downward component, exerting a torque in a direction opposite that of the horizontal component. The force exerted by the bar on the hinge is equal in magnitude and opposite in direction to the force exerted by the hinge on the bar. 11.50: a) The tension in the string is w2 = 50 N, and the horizontal force on the bar must balance the horizontal component of the force that the string exerts on the bar, and is equal to (50 N) sin 37° = 30 N, to the left in the figure. The vertical force must be  50 N  2 2 (50 N) cos 37° + 10 N = 50 N, up. b) arctan   30 N  = 59°. c) (30 N) + (50 N) = 58 N.    d) Taking torques about (and measuring the distance from) the left end, (50 N) x = (40 N )(5.0 m ) , so x = 4.0 m , where only the vertical components of the forces exert torques. 11.51: a) Take torques about her hind feet. Her fore feet are 0.72 m from her hind feet, N) ( .28 and so her fore feet together exert a force of (190( 0.720m) m) = 73.9 N, so each foot exerts a force of 36.9 N, keeping an extra figure. Each hind foot then exerts a force of 58.1 N. b) Again taking torques about the hind feet, the force exerted by the fore feet is (190 N) ( 0.28 m) + ( 25 N) ( 0.09 m) = 105.1 N, so each fore foot exerts a force of 52.6 N and each hind 0.72 m foot exerts a force of 54.9 N. 11.52: a) Finding torques about the hinge, and using L as the length of the bridge and wT and wB for the weights of the truck and the raised section of the bridge, TL sin 70° = wT ( 3 L ) cos 30° + wB ( 1 L ) cos 30° , so 4 2

T=

( 3 mT + 12 mB ) (9.80 m s 2 )cos 30° = 2.57 × 105 N. 4
sin 70°

b) Horizontal: T cos( 70° − 30°) = 1.97 × 105 N. Vertical: wT + wB − T sin 40° = 2.46 × 105 N.

 11.53: a) Take the torque exerted by F2 to be positive; the net torque is then − F1 ( x)sin φ + F2 ( x + l ) sin φ = Fl sin φ, where F is the common magnitude of the forces. b) τ1 = −(14.0 N)(3.0 m) sin 37° = −25.3 N ⋅ m, keeping an extra figure, and τ2 = (14.0 N)(4.5 m) sin 37° = 37.9 N ⋅ m, and the net torque is 12.6 N ⋅ m. About point P, τ1 = (14.0 N)(3.0 m)(sin 37°) = 25.3 N ⋅ m, and τ2 = (−14.0 N)(1.5 m)(sin 37°) = −12.6 N ⋅ m, and the net torque is 12.6 N ⋅ m. The result of part (a) predicts (14.0 N)(1.5 m) sin 37°, the same result. 11.54: a) Take torques about the pivot. The force that the ground exerts on the ladder is given to be vertical, and FV (6.0 m) sin θ = (250 N )(4.0 m) sin θ + (750 N)(1.50 m) sin θ , so FV = 354 N. b) There are no other horizontal forces on the ladder, so the horizontal pivot force is zero. The vertical force that the pivot exerts on the ladder must be (750 N) + (250 N) − (354 N) = 646 N, up, so the ladder exerts a downward force of 646 N on the pivot. c) The results in parts (a) and (b) are independent of θ. 11.55: a) V = mg + w and H = T . To find the tension, take torques about the pivot point. Then, denoting the length of the strut by L, 2  2   L T  L  sin θ = w L  cos θ + mg   cos θ , or 3  3  6 mg   T = w +  cot θ. 4   b) Solving the above for w , and using the maximum tension for T , mg w = T tan θ − = (700 N) tan 55.0° − (5.0 kg) (9.80 m s 2 ) = 951 N. 4 c) Solving the expression obtained in part (a) for tan θ and letting ω → 0, tan θ = mg = 0.700, so θ = 4.00°. 4T

11.56:

(a) and (b)

Lower rod:

Στ p = 0 : (6.0 N)(4.0 cm) = A(8.0 cm) A = 3 .0 N ΣF = 0 : T3 = 6.0 N + A = 6.0 N + 3.0 N = 9.0 N Middle rod:

Στp = 0 : B (3.0 cm) = (9.0 N)(5.0 cm)

B = 15 N ΣF = 0 : T2 = B + T3 = 15 N + 9.0 N = 24 N

Upper rod:

Στp = 0 : (24 N)(2.0 cm) = C (6.0 cm)

C = 8.0 N ΣF = 0 : T1 = T2 + C = 24 N + 8.0 N = 32 N

11.57:

Στ = 0, axis at hinge T (6.0 m)(sin 40°) − w(3.75 m)(cos 30°) = 0 T = 760 N
11.58: (a)

ΣτHinge = 0 T (3.5 m)sin 37° = (45,000 N)(7.0 m) cos 37° T = 120,000 N (b) ∑ Fx = 0 : H = T = 120,000 N ∑ Fx = 0 : V = 45,000 N The resultant force exerted by the hinge has magnitude 1.28 × 10 5 N and direction 20.6° above the horizontal.

11.59:

a) ∑ τ = 0, axis at lower end of beam Let the length of the beam be L. L T (sin 20°) L = −mg   cos 40° = 0 2 1 mg cos 40° T= 2 = 2700 N sin 20° b) Take +y upward. ∑ Fy = 0 gives n − w + T sin 60° = 0 so n = 73.6 N ∑ Fx = 0 gives f s = T cos 60° = 1372 N f 1372 N f s = μs n, μs = s = = 19 n 73.6 N The floor must be very rough for the beam not to slip. 11.60: a) The center of mass of the beam is 1.0 m from the suspension point. Taking torques about the suspension point,

w(4.00 m) + (140.0 N)(1.00 m) = (100 N)(2.00 m)
(note that the common factor of sin 30° has been factored out), from which w = 15.0 N. b) In this case, a common factor of sin 45° would be factored out, and the result would be the same.

11.61: a) Taking torques about the hinged end of the pole (200 N)(2.50 m) + (600 N) × (5.00 m) − Ty (5.00 m) = 0 . Therefore the y-component of the tension is Ty = 700 N . The x-component of the tension is then Tx = (1000 N) 2 − (700 N) 2 = 714 N . The height above the pole that the wire must be attached is (5.00 m) 700 = 4.90 m . b) The y-component of the tension remains 700 N and 714
4.90 m the x-component becomes (714 N) 4.40 m = 795 N , leading to a total tension of

(795 N) 2 + (700 N) 2 = 1059 N, an increase of 59 N. 11.62: A and B are straightforward, the tensions being the weights suspended; Τ A = (0.0360 kg)(9.80 m s 2 ) = 0.353 N, TB = (0.0240 kg + 0.0360 kg )(9.80 m s 2 ) = 0.588 N. To find TC and TD , a trick making use of the right angle where the strings join is available; use a coordinate system with axes parallel to the strings. Then, TC = TB cos 36.9° = 0.470 N, TD = TB cos 53.1° = 0.353 N, To find TE , take torques about the point where string F is attached; TE (1.000 m) = TD sin 36.9° (0.800 m) + TC sin 53.1° (0.200 m) + (0.120 kg)(9.80 m s 2 )(0.500 m) = 0.833 N ⋅ m, so TE = 0.833 N. TF may be found similarly, or from the fact that TE + TF must be the total weight of the ornament. (0.180 kg )(9.80 m s 2 ) = 1.76 N, from which TF = 0.931 N. 11.63: a) The force will be vertical, and must support the weight of the sign, and is 300 N. Similarly, the torque must be that which balances the torque due to the sign’s weight about the pivot, (300 N)(0.75 m) = 225 N ⋅ m . b) The torque due to the wire must balance the torque due to the weight, again taking torques about the pivot. The minimum tension occurs when the wire is perpendicular to the lever arm, from one corner of the sign to the other. Thus, T (1.50 m) 2 + (0.80 m) 2 = 225 N ⋅ m, or T = 132 N. The angle that the wire 0.80 makes with the horizontal is 90° − arctan ( 1.50 ) = 62.0°. Thus, the vertical component of the force that the pivot exerts is (300 N) –(132 N) sin 62.0° = 183 N and the horizontal force is (132 N) cos 62.0° = 62 N , for a magnitude of 193 N and an angle of 71° above the horizontal.

11.64: b)

a) ∆w = −σ (∆l l ) w0 = −(0.23)(9.0 × 10 −4 ) 4(0.30 × 10−4 m 2 ) π = 1.3 μm. ∆l 1 ∆w = AY l σ w 11 (2.1 × 10 Pa) (π (2.0 × 10 − 2 m) 2 ) 0.10 × 10 −3 m = = 3.1 × 106 N, −2 0.42 2.0 × 10 m

F⊥ = AY

where the Young’s modulus for nickel has been used. 11.65: a) The tension in the horizontal part of the wire will be 240 N. Taking torques about the center of the disk, (240 N)(0.250 m) − w(1.00 m)) = 0, or w = 60 N. b) Balancing torques about the center of the disk in this case, (240 N) (0.250 m) − ((60 N)(1.00 m) + (20 N)(2.00 m)) cos θ = 0, so θ = 53.1° . 11.66: a) Taking torques about the right end of the stick, the friction force is half the weight of the stick, f = w ⋅ Taking torques about the point where the cord is attached to 2 the wall (the tension in the cord and the friction force exert no torque about this point),and noting that the moment arm of the normal force is f l tan θ, n tan θ = w ⋅ Then, n = tan θ < 0.40, so θ < arctan (0.40) = 22°. 2 b) Taking torques as in part (a), and denoting the length of the meter stick as l , l l fl = w + w(l − x) and nl tan θ = w + wx . 2 2 In terms of the coefficient of friction µ s ,

µs >
Solving for x,

l + (l − x ) f 3l − 2 x = 2 l tan θ = tan θ. n +x l + 2x 2

l 3 tan θ − μs = 30.2 cm. 2 μs + tan θ c) In the above expression, setting x = 10 cm and solving for µ s gives (3 − 20 l ) tan θ μs > = 0.625. 1 + 20 l x>

11.67: Consider torques around the point where the person on the bottom is lifting. The center of mass is displaced horizontally by a distance (0.625 m − 0.25 m) sin 45° and the horizontal distance to the point where the upper person is lifting is (1.25 m) sin 45° , and sin 45 so the upper lifts with a force of w 01..375sin 45°° = (0.300) w = 588 N. The person on the 25 bottom lifts with a force that is the difference between this force and the weight, 1.37 kN. The person above is lifting less.

11.68: (a)

ΣτElbow = 0 FB (3.80 cm) = (15.0 N)(15.0 cm) FB = 59.2 N

(b)

Στ E = 0 FB (3.80 cm) = (15.0 N)(15.0 cm) + (80.0 N)(33.0 cm) FB = 754 N

11.69:

a) The force diagram is given in Fig. 11.9.

Στ = 0, axis at elbow wL − ( T sin θ ) D = 0 sin θ = h
2 2

wmax = Tmax b)

h +D hD

so w = T

hD L h2 + D2

L h2 + D 2

dwmax Tmax h  D2  1 − 2 ; the derivative is positive = 2   dD L h2 + D2  h + D  c) The result of part (b) shows that wmax increases when D increases. 11.70:

By symmetry, A=B and C=D. Redraw the table as viewed from the AC side. Στ (about right end ) = 0 : 2 A(3.6 m) = ( 90.0 N ) (1.8 m) + (1500 N ) (0.50 m) A = 130 N = B ΣF = 0 : A + B + C + D = 1590 N Use A = B = 130 N and C = D C = D = 670 N By Newton’s third law of motion, the forces A, B, C, and D on the table are the same as the forces the table exerts on the floor.

11.71:

a) Consider the forces on the roof

V and H are the vertical and horizontal forces each wall exerts on the roof. w = 20 ,000 N is the total weight of the roof. 2V = w so V = w 2 Apply Στ = 0 to one half of the roof, with the axis along the line where the two halves join. Let each half have length L. ( w 2)( L 2)(cos 35.0°) + HL sin 35.0° − VL cos 35.0° = 0 L divides out, and use V = w 2 H sin 35.0° = 1 w cos 35.0° 4 w = 7140 N 4 tan 35.0° By Newton’s 3rd law, the roof exerts a horizontal, outward force on the wall. For torque about an axis at the lower end of the wall, at the ground, this force has a larger moment arm and hence larger torque the taller the walls. b) H=

Consider the torques on one of the walls.

 11.72: a) Take torques about the upper corner of the curb. The force F acts at a perpendicular distance R − h and the weight acts at a perpendicular distance
R 2 − ( R − h ) = 2 Rh − h 2 . Setting the torques equal for the minimum necessary force,
2

2 Rh − h 2 . R−h  b) The torque due to gravity is the same, but the force F acts at a perpendicular distance 2 R − h, so the minimum force is ( mg ) 2 Rh − hv /2R–h. c) Less force is required when the force is applied at the top of the wheel. F = mg

11.73: a) There are several ways to find the tension. Taking torques about point B (the force of the hinge at A is given as being vertical, and exerts no torque about B), the tension acts at distance r = (4.00 m) 2 + (2.00 m) 2 = 4.47 m and at an angle of  2.00  φ = 30° + arctan   = 56.6°. Setting  4.00  Tr sin φ = (500 N) (2.00 m) and solving for T gives T = 268 N . b) The hinge at A is given as exerting no horizontal force, so taking torques about point D, the lever arm for the vertical force at point B is (2.00 m) + (4.00 m) tan 30.0° = 4.31 m, so the horizontal force (500 N) (2.00 m) at B is = 232 N. Using the result of part (a), 4.31 m however, (268 N) cos 30.0° = 232 N In fact, finding the horizontal force at B first simplifies the calculation of the tension slightly. c) (500 N) − (268 N) sin 30.0° = 366 N. Equivalently, the result of part (b) could be used, taking torques about point C , to get the same result.

11.74: a) The center of gravity of top block can be as far out as the edge of the lower block. The center of gravity of this combination is then 3L 4 from the right edge of the upper block, so the overhang is 3L 4. b) Take the two-block combination from part (a), and place it on the third block such that the overhang of 3L 4 is from the right edge of the third block; that is, the center of gravity of the first two blocks is above the right edge of the third block. The center of mass of the three-block combination, measured from the right end of the bottom block, is − L 6 and so the largest possible overhang is (3L 4) + ( L 6) = 11 L 12. Similarly, placing this three-block combination with its center of gravity over the right edge of the fourth block allows an extra overhang of L 8, for a total of 25L 24. c) As the result of part (b) shows, with only four blocks, the overhang can be larger than the length of a single block. 11.75: a)

FB = 2 w = 1.47 N sin θ = R 2 R so θ = 30° τ = 0, axis at P FC (2 R cos θ ) − wR = 0 mg = 0.424 N 2 cos 30° FA = FC = 0.424 N b) Consider the forces on the bottom marble. The horizontal forces must sum to zero, so FA = n sin θ FC = FA = 0.848 N sin 30° Could use instead that the vertical forces sum to zero FB − mg − n cos θ = 0 n= n= FB − mg = 0.848 N, which checks. cos 30°

11.76: (a) Writing an equation for the torque on the right-hand beam, using the hinge as an axis and taking counterclockwise rotation as positive: θ L θ L θ Fwire L sin − Fc cos − w sin = 0 2 2 2 2 2 where θ is the angle between the beams, Fc is the force exerted by the cross bar, and w is the weight of one beam. The length drops out, and all other quantities except Fc are known, so Fc = Therefore Fwire sin θ − 1 w sin 2 2 1 cos θ 2 2
θ 2

= (2 Fwire − w) tan

θ 2

53° = 130 N 2 b) The cross bar is under compression, as can be seen by imagining the behavior of the two beams if the cross bar were removed. It is the cross bar that holds them apart. F = 260 tan c) The upward pull of the wire on each beam is balanced by the downward pull of gravity, due to the symmentry of the arrangement. The hinge therefore exerts no vertical force. It must, however, balance the outward push of the cross bar: 130 N horizontally to the left for the right-hand beam and 130 N to the right for the left-hand beam. Again, it’s instructive to visualize what the beams would do if the hinge were removed. 11.77: a) The angle at which the bale would slip is that for which f = μs Ν = μs w cos β = w sin β , or β = arctan( μs ) = 31.0°. The angle at which the bale would tip is that for which the center of gravity is over the lower contact point, or 0 arctan ( 0..25 m) = 26.6°, or 27° to two figures. The bale tips before it slips. b) The angle for 50 m) tipping is unchanged, but the angle for slipping is arctan (0.40) = 21.8°, or 22° to two figures. The bale now slips before it tips. 11.78: a) F = f = μk Ν = μk mg = (0.35)(30.0 kg)(9.80 m s 2 ) = 103 N b) With respect to the forward edge of the bale, the lever arm of the weight is 0.250 m = 0.125 m and the lever arm h of the applied force is then h 2 = (0.125 m)
mg F

= (0.125 m) µ1k =

0.125 m 0.35

= 0.36 m.

11.79: a) Take torques about the point where wheel B is in contact with the track. With respect to this point, the weight exerts a counterclockwise torque and the applied force and the force of wheel A both exert clockwise torques. Balancing torques, FA (2.00 m) + ( F )(1.60 m) = (950 N)(1.00 m). Using F = µk w = 494 N, FA = 80 N, and FB = w − FA = 870 N. b) Again taking torques about the point where wheel B is in contact with the tract, and using F = 494 N as in part (a), (494 N) h = (950 N)(1.00 N), so h = 1.92 m. 11.80: a) The torque exerted by the cable about the left end is TL sin θ . For any angle θ, sin (180° − θ ) = sin θ , so the tension T will be the same for either angle. The horizontal component of the force that the pivot exerts on the boom will be 1 T cos θ or Tcos (180° − θ) = −T cos θ . b) From the result of part (a), T α sin θ , and this becomes infinite as θ → 0 or → 180°. Also, c), the tension is a minimum when sin θ is a maximum, or θ = 90°, a vertical string. d) There are no other horizontal forces, so for the boom to be in equilibrium, the pivot exerts zero horizontal force on the boom. 11.81: a) Taking torques about the contact point on the ground, T (7.0 m) sin θ = w(4.5 m) sin θ , so T = (0.64) w = 3664 N. The ground exerts a vertical force on the pole, of magnitude w − T = 2052 N . b) The factor of sin θ appears in both terms of the equation representing the balancing of torques, and cancels. 11.82: a) Identifying x with ∆l in Eq. (11.10), k = Y A l0 .

b) (1 2)kx 2 = Y Ax 2 2l0 . 11.83: a) At the bottom of the path the wire exerts a force equal in magnitude to the centripetal acceleration plus the weight, F = m(((2.00 rev s)(2π rad rev)) 2 (0.50 m) + 9.80 m s 2 ) =1.07 × 10 3 N. From Eq. (11.10), the elongation is (1.07 × 103 N)(0.50 m) = 5.5 mm. (0.7 × 1011 Pa )(0.014 × 10− 4 m 2 ) b) Using the same equations, at the top the force is 830 N, and the elongation is 0.0042 m.

11.84:

a)

b) The ratio of the added force to the elongation, found from taking the slope of the graph, doing a least-squares fit to the linear part of the data, or from a casual glance at the F data gives ∆l = 2.00 × 10 4 N m . From Eq. (11.10), (3.50 m) F l0 = (2.00 × 10 4 N m) = 1.8 × 1011 Pa. −3 2 ∆l A (π (0.35 × 10 m) ) c) The total force at the proportional limit is 20.0 N + 60 N = 80 N, and the stress at ( this limit is π ( 0.3580 N−)3 m ) 2 = 2.1 × 10 8 Pa. ×10 Y= 11.85: a) For the same stress, the tension in wire B must be two times in wire A, and so the weight must be suspended at a distance (2 3)(1.05 m) = 0.70 m from wire A.. b) The product Y A for wire B is (4 3) that of wire B, so for the same strain, the tension in wire B must be (4 3) that in wire A, and the weight must be 0.45 m from wire B. 11.86: a) Solving Eq. (11.10) for ∆l and using the weight for F, Fl (1900 N)(15.0 m) ∆l = 0 = = 1.8 × 10 − 4 m. 11 −4 2 YA (2.0 × 10 Pa )(8.00 × 10 m ) b) From Example 5.21, the force that each car exerts on the cable is F = mω 2l0 = w w2l0 , and so g ∆l = Fl0 wω2l02 (1900 N)(0.84 rad s) 2 (15.0 m) 2 = = = 1.9 × 10− 4 m. 2 11 −4 2 YA gYA (9.80 m s )(2.0 × 10 Pa )(8.00 × 10 m )

11.87: Use subscripts 1 to denote the copper and 2 to denote the steel. a) From Eq. (11.10), with ∆l1 = ∆l 2 and F1 = F2 ,  A2Y2   (1.00 cm 2 )(21 × 1010 Pa )  L2 = L1   A Y  = (1.40 m) (2.00 cm 2 )(9 × 1010 Pa )  = 1.63 m.     1 1   8 F F b) For nickel, A1 = 4.00 × 10 Pa and for brass, A2 = 2.00 × 10 8 Pa. c) For nickel,
4.00×108 Pa 21×1010 Pa

= 1.9 × 10 −3 and for brass,

2.00×108 Pa 9×1010 Pa

= 2.2 × 10 −3.

11.88:

 ∆l  a) Fmax = YA  = (1.4 × 1010 Pa )(3.0 × 10 −4 m 2 )(0.010) = 4.2 × 10 4 N. l   0  max

b) Neglect the mass of the shins (actually the lower legs and feet) compared to the rest of the body. This allows the approximation that the compressive stress in the shin bones is uniform. The maximum height will be that for which the force exerted on each lower leg by the ground is Fmax found in part (a), minus the person’s weight. The impulse that the ground exerts is J = (4.2 × 10 4 N − (70 kg )(9.80 m s 2 ))(0.030 s) = 1.2 × 10 3 kg ⋅ m s. The speed at the ground is 2 gh, so 2 J = m 2 gh and solving for h, 1  2J  h=   = 64 m, 2g  m 
2

but this is not recommended.

11.89: a) Two times as much, 0.36 mm, b) One-fourth (which is (1 2) 2 ) as much, 0.045 mm.c) The Young’s modulus for copper is approximately one-half that for steel, so 20×1010 Pa the wire would stretch about twice as much. (0.18 mm ) 11×1010 Pa = 0.33 mm. 11.90: Solving Eq. (11.14) for ∆V , mg ∆V = −kV0 ∆P = − kV0 A (1420 kg )(9.80 m s 2 ) −11 −1 = −(110 × 10 Pa )(250 L) π (0.150 m) 2 = −0.0541 L.

The minus sign indicates that this is the volume by which the original hooch has shrunk, and is the extra volume that can be stored.

11.91: The normal component of the force is F cos θ and the area (the intersection of the red plane and the bar in Figure (11.52)) is A / cos θ , so the normal stress is ( F A) cos2θ. b) The tangential component of the force is F sin θ , so the shear stress is
(F A) sin θ cos θ.

c) cos2θ is a maximum when cos θ = 1, or θ = 0. d) The shear stress can be expressed as ( F 2 A) sin (2θ ), which is maximized when
sin (2θ ) = 1, or θ = 90° = 45°. Differentiation of the original expression with respect 2

to θ and setting the derivative equal to zero gives the same result. 11.92:

a) Taking torques about the pivot, the tension T in the cable is related to
mg . The horizontal component of the 2 sin θ mg mg cot θ and the stress is cot θ. 2 2A

the weight by T sin θ l0 = mgl0 2, so T =

force that the cable exerts on the rod, and hence the horizontal component of the force that the pivot exerts on the rod, is b)
l0 F mgl0 cot θ = . AY 2 AY c) In terms of the density and length, (m A) = ρ l0 , so the stress is ∆l = ( ρl0 g 2) cot θ and the change in length is ( ρl02 g 2Υ ) cot θ. d) Using the numerical

values, the stress is 1.4 × 10 5 Pa and the change in length is 2.2 × 10 −6 m. e) The stress is proportional to the length and the change in length is proportional to the square of the length, and so the quantities change by factors of 2 and 4.

11.93: a) Taking torques about the left edge of the left leg, the bookcase would Ν )( . tip when F = (1500(1.80 0m90 m ) = 750 Ν, and would slip when F = (µs )(1500 Ν ) = 600 Ν, so ) the bookcase slides before tipping. b) If F is vertical, there will be no net horizontal force and the bookcase could not slide. Again taking torques about the )( . left edge of the left leg, the force necessary to tip the case is (1500( 0Ν10 0m90 m ) = 13.5 kN. . ) c) To slide, the friction force is f = μs ( w + F cos θ ), and setting this equal to F sin θ and solving for F gives
F= μs w . sin θ − μs cos θ

To tip, the condition is that the normal force exerted by the right leg is zero, and taking torques about the left edge of the left leg, F sin θ (1.80 m) + F cos θ (0.10 m) = w(0.90 m), and solving for F gives
F= w . (1 9) cos θ + 2 sin θ

Setting the expression equal gives
μs ((1 9) cos θ + 2 sin θ ) = sin θ − μs cos θ ,

and solving for θ gives
 (10 9) μs  θ = arctan   (1 - 2 μ )  = 66°.  s  

11.94: a) Taking torques about the point where the rope is fastened to the ground, the lever arm of the applied force is h and the lever arm of both the 2 weight and the normal force is h tan θ, and so F h = (n − w)h tan θ. Taking torques 2 about the upper point (where the rope is attached to the post), f h = F h . Using 2 f ≤ µs n and solving for F,
1 1  1  1  F ≤ 2w −  µ tan θ  = 2(400 nN ) 0.30 − tan 36.9°  = 400 nN ,     s  b) The above relations between F , n and f become 3 2 F h = (n − w)h tan θ, f = F , 5 5 and eliminating f and n and solving for F gives 2 5 3 5  F ≤ w  µ − tan θ  ,   s 
−1 −1 −1

and substitution of numerical values gives 750 N to two figures. c) If the force is applied a distance y above the ground, the above relations become
Fy = (n − w)h tan θ, F (h − y ) = fh, which become, on eliminating n and f ,
y y  1− h  w ≥ F − h . tan θ  µs

(

) ()

As the term in square brackets approaches zero, the necessary force becomes unboundedly large. The limiting value of y is found by setting the term in square brackets equal to zero. Solving for y gives y tan θ tan 36.9° = = = 0.71. h µ s + tan θ 0.30 + tan 36.9°

11.95: Assume that the center of gravity of the loaded girder is at L 2, and that the cable is attached a distance x to the right of the pivot. The sine of the angle between the lever arm and the cable is then h h 2 + (( L 2) − x) 2 , and the tension is obtained from balancing torques about the pivot;
  hx T  = w L 2,  h 2 + (( L 2) − x) 2    where w is the total load (the exact value of w and the position of the center of

gravity do not matter for the purposes of this problem). The minimum tension will occur when the term in square brackets is a maximum; differentiating and setting the derviative equal to zero gives a maximum, and hence a minimum tension, at xmin = (h 2 L) + ( L 2). However, if x min > L, which occurs if h > L 2 , the cable must be attached at L, the furthest point to the right. 11.96: The geometry of the 3-4-5 right triangle simplifies some of the intermediate algebra. Denote the forces on the ends of the ladders by FL and FR (left and right). The contact forces at the ground will be vertical, since the floor is assumed to be frictionless. a) Taking torques about the right end, FL (5.00 m) = (480 N)(3.40 m) + (360 N)(0.90 m), so FL = 391 N. FR may be found in a similar manner, or from FR = 840 N − FL = 449 N. b) The tension in the rope may be found by finding the torque on each ladder, using the point A as the origin. The lever arm of the rope is 1.50 m. For the left ladder, T (1.50 m) = FL (3.20 m) − (480 N)(1.60 m), so T = 322.1 N (322 N to three figures). As a check, using the torques on the right ladder, T (1.50 m) = FR (1.80 m) − (360 N)(0.90 m) gives the same result. c) The horizontal component of the force at A must be equal to the tension found in part (b). The vertical force must be equal in magnitude to the difference between the weight of each ladder and the force on the bottom of each ladder, 480 N−391 N = 449 N−360 N = 89 N. The magnitude of the force at A is then
(322.1 N) 2 + (89 N) 2 = 334 N.

d) The easiest way to do this is to see that the added load will be distributed at the floor in such a way that ′ FL′ = FL + (0.36)(800 N) = 679 N, and FR = FR + (0.64)(800 N) = 961 N. Using these forces in the form for the tension found in part (b) gives
T= ′ FL′ (3.20 m) − (480 N)(1.60 m) FR (1.80 m) − (360 N)(0.90 m) = = 936.53 N, (1.50 m) (1.50 m)

which is 937 N to three figures.

11.97: The change in the volume of the oil is = kOvO ∆p and the change in the volume of the sodium is = ksvs ∆p. Setting the total volume change equal to Ax (x is positive) and using ∆p = F A,
Ax = ( kOVO + ksVs )( F A),

and solving for ks gives
 A2 x 1 ks =  − kOVO  ⋅  F V   s

11.98:

a) For constant temperature ( ∆T = 0 ) ,
∆( pV ) = ( ∆p)V + p (∆ V ) = 0 and B = −

(∆p )V = p. ( ∆V ) ∆V = 0, V

b) In this situation,
(∆p )V γ + γp( ∆V ) V γ −1 = 0, (∆p ) + γp

and
B=− (∆p )V = γp. ∆V

11.99:

a) From Eq.(11.10), ∆l =

( 4.50 kg )(9.80 m s 2 )(1.50 m ) ( 20 ×1010 Pa )( 5.00×10 −7 m 2 )

= 6.62 × 10− 4 m, or 0.66 mm

to two figures. b) (4.50 kg )(9.80 m s 2 )(0.0500 × 10 −2 m) = 0.022 J. c) The magnitude F will be vary with distance; the average force is Y A(0.0250 cm l0 ) = 16.7 N, and so the work done by the applied force is (16.7 N)(0.0500 × 10−2 m) = 8.35 × 10−3 J. d) The wire is initially stretched a distance 6.62 × 10 −4 m ( the result of part (a)), and so the average elongation during the additional stretching is 9.12 × 10 −4 m , and the average force the wire exerts is 60.8 N. The work done is negative, and equal to − (60.8 N)(0.0500 × 10 −2 m) = −3.04 × 10−2 J. e) See problem 11.82. The change in elastic potential energy is
(20 × 1010 Pa )(5.00 × 10 −7 m 2 ) ((11.62 × 10 − 4 m 2 ) − (6.62 × 10 − 4 m) 2 ) = 3.04 × 10− 2 J, 2(1.50 m )

the negative of the result of part (d). (If more figures are kept in the intermediate calculations, the agreement is exact.)

Chapter 12

Note: to obtain the numerical results given in this chapter, the following numerical values of certain physical quantities have been used; G = 6.673 × 10−11 N ⋅ m 2 kg 2 , g = 9.80 m s 2 and mE = 5.97 × 1024 kg. Use of other tabulated values for these quantities may result in an answer that differs in the third significant figure. 12.1: The ratio will be the product of the ratio of the mass of the sun to the mass of the earth and the square of the ratio of the earth-moon radius to the sun-moon radius. Using the earth-sun radius as an average for the sun-moon radius, the ratio of the forces is  3.84 × 10 8 m    1.50 × 1011 m     12.2: Use of Eq. (12.1) gives
2

 1.99 × 10 30 kg    5.97 × 10 24 kg  = 2.18.   

m1 m2 (5.97 × 10 24 kg)(2150 kg) −11 2 2 Fg = G 2 = (6.673 × 10 N ⋅ m kg ) = 1.67 × 10 4 N. 5 6 2 r (7.8 × 10 m + 6.38 × 10 m) The ratio of this force to the satellite’s weight at the surface of the earth is (1.67 × 10 4 N) = 0.79 = 79%. (2150 kg )(9.80 m s 2 ) (This numerical result requires keeping one extra significant figure in the intermediate calculation.) The ratio, which is independent of the satellite mass, can be obtained directly as GmE m r 2 GmE  RE  = 2 =  , mg r g  r 
2

yielding the same result.

12.3:

G

(nm1 )(nm2 ) mm = G 1 2 2 = F12 . 2 ( nr12 ) r12

12.4: The separation of the centers of the spheres is 2R, so the magnitude of the gravitational attraction is GM 2 (2 R ) 2 = GM 2 4 R 2 .

12.5: a) Denoting the earth-sun separation as R and the distance from the earth as x, the distance for which the forces balance is obtained from GM S m GM E m = , ( R − x) 2 x2 which is solved for R x= = 2.59 × 108 m. MS 1+ ME b) The ship could not be at equilibrium for long, in that the point where the forces balance is moving in a circle, and to move in that circle requires some force. The spaceship could continue toward the sun with a good navigator on board. 12.6: gives a) Taking force components to be positive to the right, use of Eq. (12.1) twice

 ( 5.00 kg ) (10.0 kg ) , Fg = 6.673 × 10−11 Ν ⋅ m 2 kg 2 ( 0.100 kg ) − + 2 ( 0.600 m ) 2   ( 0.400 m )  −11 = −2.32 × 10 Ν with the minus sign indicating a net force to the left.

(

)

b) No, the force found in part (a) is the net force due to the other two spheres.

12.7:

( 6.673 × 10

−11

Ν. m kg
2

2

( 70kg ) ( 7.35 × 1022 kg ) = 2.4 × 10−3 Ν. )

(3.78 × 10 m)
8

2

12.8:

( 333, 000) = 6.03 × 10− 4 ( 23, 500) 2

12.9: a)

Denote the earth-sun separation as r1 and the earth-moon separation as r2 .

mS mE  + 2  = 6.30 × 1020 Ν, 2 r2   (r1 + r2 ) toward the sun. b)The earth-moon distance is sufficiently small compared to the earthsun distance (r2 << r2) that the vector from the earth to the moon can be taken to be perpendicular to the vector from the sun to the moon. The components of the gravitational force are then

( GmM ) 



GmM mS GmM mE = 4.34 × 1020 Ν, = 1.99 × 10 20 Ν, 2 2 r1 r2 and so the force has magnitude 4.77 × 10 20 Ν and is directed 24.6° from the direction toward the sun. c) toward the sun.

( GmM ) 



mS mE  − 2  = 2.37 × 1020 Ν, 2 r2   ( r1 − r2 )

12.10:

FonA = 2 FB cos 45° + FD =2 = GmA mB cos 45° GmA mD + 2 2 rAB rAD

2(6.67 × 10−11 Nm 2 kg 2 )(800 kg) 2 cos 45° (0.10 m) 2 + (6.67 × 10 −11 Nm 2 kg 2 ) (800 kg) 2 (0.10 m) 2

= 8.2 × 10− 3 N, toward the center of the square

12.11:

m1 = m2 = m3 = 500 kg r12 = 0.10 m; r23 = 0.40 m F1 = G F3 = G m1m2 = 1.668 × 10 −3 N 2 r12 m2 m3 = 1.043 × 10 −4 N 2 r23

F = F1 − F3 = 1.6 × 10 −3 N, to the left 12.12: will be The direction of the force will be toward the larger mass, and the magnitude Gm2 m Gm1m 4Gm(m2 − m1 ) − = . (d 2) 2 (d 2) 2 d2

12.13: For convenience of calculation, recognize that the mass of the small sphere will cancel. The acceleration is then 2G (0.260 kg) 6.0 × = 2.1 × 10− 9 m s 2 , −2 2 (10.0 × 10 m) 10.0 directed down. 12.14: Equation (12.4) gives g=

( 6.763 × 10

−11

)( (1.15 × 10 m)
6 2

N ⋅ m 2 kg 2 1.5 × 1022 kg

) = 0.757 m s .
2

12.15: To decrease the acceleration due to gravity by one-tenth, the distance from the earth must be increased by a factor of 10 , and so the distance above the surface of the earth is

(
12.16:

10 − 1 R E = 1.38 × 107 m.

)

a) Using g E = 9.80 m s 2 , Eq (12.4) gives m  R  Gm g v = 2 v = G  v m E  E  m  R  Rv  E  v
2 2

 1   2 R   E
2

Gm  m   R   1  = 2E  v   E  = g E (.815)  m  R  RE  E   v   .949  = (9.80 m s 2 )(.905) = 8.87 m s 2 ,

where the subscripts v refer to the quantities pertinent to Venus. b) (8.87 m s 2 ) (5.00 kg) = 44.3N. 12.17: a) See Exercise 12.16;  ( 8) 2  2  g Titania = (9.80 m s 2 )   1700  = 0.369 m s .   b) ρT = ρE
mT mE

512 (5500 kg m 3 ) ( 1700 ) = 1656 kg m3 , or about 0.39 ρE.

. rE3 , or rearranging and solving for density, ρT = ρE . (1 1700) m E . (1 8rE mE r r
T

3

3 E

)3

=

12.18:

M=

gR 2 G

= 2.44 × 1021 kg and ρ =

( 4π 3) R 3

M

= 1.30 × 103 Kg m 3 .

12.19:

mmE r2 r = 600 × 10 3 m + RE so F = 610 N F =G

At the surface of the earth,

w = mg = 735 N.

The gravity force is not zero in orbit. The satellite and the astronaut have the same acceleration so the astronaut’s apparent weight is zero. 12.20: Get g on the neutron star mg ns = GmM ns R2 GM ns g ns = R2

Your weight would be wns = mg ns = mGM ns R2

 675N =  9.8 m s 2 

 (6.67 × 1011 Nm 2 kg 2 )(1.99 × 10 30 kg)   (10 4 m) 2 

= 9.1 × 1013 N
2 12.21: From eq. (12.1), G = Fr 2 m1m 2 , and from Eq. (12.4), g = GmE RE ; combining and solving for RE ,

mE =

2 gm1m2 RE = 5.98 × 1024 kg. 2 Fr

12.22: a) From Example 12.4 the mass of the lander is 4000 kg. Assuming Phobos to be spherical, its mass in terms of its density ρ and radius R is (4π 3) ρR 3 , and so the gravitational force is G (4π 3)(4000 kg ) ρR 3 = G (4π 3)(4000 kg )(2000 kg m 3 )(12 × 10 3 m) = 27 N. 2 R b) The force calculated in part (a) is much less than the force exerted by Mars in Example 12.4.

12.23:

2 GM R = 2(6.673 × 10−11 N ⋅ m 2 kg 2 ) (3.6 × 1012 kg) (700 m) = 0.83 m s.

One could certainly walk that fast. a) F = GmE m r 2 and U = GmE m r , so the altitude above the surface of the
U F

12.24: earth is

− RE = 9.36 × 105 m. b) Either of Eq. (12.1) or Eq. (12.9) can be used with the
2

result of part (a ) to find m, or noting that U 2 = G 2 M E m 2 r 2 , m = U 2 FGM E = 2.55 × 103 kg.

12.25:

The escape speed, from the results of Example 12.5, is

2GM R.

a) 2(6.673 × 10 −11 N ⋅ m 2 kg 2 ) (6.42 × 10 23 kg ) (3.40 × 10 6 m) = 5.02 × 10 3 m s . b) 2(6.673 × 10 −11 N ⋅ m 2 kg 2 ) (1.90 × 10 27 kg ) (6.91 × 10 7 m) = 6.06 × 10 4 m s. c) Both the kinetic energy and the gravitational potential energy are proportional to the mass.

12.26:

a) The kinetic energy is K = 1 mv 2 , or K = 1 (629 kg )(3.33 × 103 m s) 2 , 2 2

or KE = 3.49 × 10 9 J.
b) U = − GMm (6.673 × 10 −11 N ⋅ m 2 kg 2 )(5.97 × 1024 kg)(629 kg ) = , r 2.87 × 109 m

or U = −8.73 × 10 7 J. 12.27: a) Eliminating the orbit radius r between Equations (12.12) and (12.14) gives T= 2πGmE 2π 6.673 × 10−11 N ⋅ m 2 kg 2 5.97 × 1024 kg = v3 ( 6200 m s ) 3

(

)(

)

= 1.05 × 104 s = 175 min. b) 2πv = 3.71 m s 2 . T Substitution into Eq. (12.14) gives T = 6.96 × 10 3 s, or 116 minutes.

12.28:

12.29:

Using Eq. (12.12), v=

( 6.673 × 10 N ⋅ m kg )(5.97 × 10 ( 6.38 × 10 m + 7.80 × 10 m )
−11 2 2 6 5

24

kg

) = 7.46 × 10

3

m s.

12.30:

Applying Newton’s second law to the Earth ∑ F = ma; GmE ms v2 = mE r2 r 2 rv 2πr ms = and v = G TEarth ms = = r

( )
G

2 πr 2 TE

=

4π 2 r 3 GTE2

4π 2 (1.50 × 1011 m) 3 4 (6.67 × 10 −11 Nm 2 kg 2 ) [(365.3d )( 8.64×10 s )] 2 d

= 2.01 × 1030 kg

12.31: ∑ F = ma c for the baseball. The net force is the gravity force exerted on the baseball by Deimos, so G mmD v2 =m 2 RD RD

v = GmD RD = (6.67 × 10−11 N ⋅ m 2 kg 2 ) (2.0 × 1015 kg) (6.0 × 103 m) = 4.7 m s A world-class sprinter runs 100 m in 10 s so have v = 10 m s; v = 4.7 m s for a thrown baseball is very achieveable.

12.32:

Apply Newton’s second law to Vulcan. Gms mv v2 ∑ F = ma : = mv r2 r 2πr v= T Gms  2πr  =  r  T  T=
2

4π 2 r 3 Gms

4π 2 2 (5.79 × 1010 m) 3 3 = −11 (6.67 × 10 Nm 2 kg 2 )(1.99 × 1030 kg)  1d  = 4.14 × 106 s  = 47.9 days  86,400 s 

[

]

12.33:

a) v = Gm r = (6.673 × 10 −11 N ⋅ m 2 kg 2 )(0.85 × 1.99 × 10 30 kg) ((1.50 × 1011 m)(0.11)) = 8.27 × 10 4 m s.

b) 2πr v = 1.25 × 106 s (about two weeks). 12.34: From either Eq. (12.14) or Eq. (12.19), mS = 4π 2 r 3 4π 2 (1.08 × 1011 m) 3 = GT 2 (6.673 × 10 −11 N ⋅ m 2 kg 2 ) ((224.7 d)(8.64 × 10 4 s d)) 2

= 1.98 × 1030 kg.

12.35: a) The result follows directly from Fig. 12.18. b) (1 − 0.248)(5.92 × 1012 m) = 4.45 × 1012 m, (1 + 0.010)(4.50 × 1012 m) = 4.55 × 1012 m. c) T = 248 y.

12.36:

a)

r=

Gm1m2 = 7.07 × 1010 m. F

b) From Eq. (12.19), using the result of part (a), 2π (7.07 × 1010 m) 3 2 T= = 1.05 × 107 s = 121 days. −11 2 2 30 (6.673 × 10 N ⋅ m kg )(1.90 × 10 kg) c) From Eq. (12.14) the radius is (8) 2 3 = four times that of the large planet’s orbit, or 2.83 × 1011 m. 12.37: a) For a circular orbit, Eq. (12.12) predicts a speed of (6.673 × 10−11 N ⋅ m 2 kg 2 )(1.99 × 1030 kg) (43 × 109 m) = 56 km s. The speed doesn’t have this value, so the orbit is not circular. b) The escape speed for any object at this radius is 2 (56 km s) = 79 km s , so the spacecraft must be in a closed elliptical orbit.

12.38: a) Divide the rod into differential masses dm at position l, measured from the right end of the rod. Then , dm = dl ( M L ), and dU = − Integrating, U= − GmM L Gm dm GmM dl =− . l+x L l+x dl GmM  L  =− ln 1 + . l+x L x 

∫

L

O

For x >> L, the natural logarithm is ~ ( L x ) , and U → −Gm M x. b) The x-component of the gravitational force on the sphere is Fx = − δU GmM (− L x 2 ) GmM = =− 2 , δx L (1 + ( L x)) ( x + Lx )

with the minus sign indicating an attractive force. As x >> L, the denominator in the above expression approaches x 2 , and Fx → Gm M x 2 , as expected. The derivative may also be taken by expressing  L ln 1 +  = ln( x + L) − ln x x  at the cost of a little more algebra.

12.39:

a) Refer to the derivation of Eq. (12.26) and Fig. (12.22). In this case, the red

ring in Fig. (12.22) has mass M and the common distance s is x 2 + a 2 . Then,

U = − GMm x 2 + a 2 . b) When x >> a, the term in the square root approaches x 2 and U → − GMm x , as expected. GMmx δU =− 2 c) Fx = − , ( x + a 2 )3 2 , δx

with the minus sign indicating an attractive force. d) when x >> a, the term inside the parentheses in the above expression approaches x 2 and Fx → − GMmx (x 2 )3 2 − GMm = − GMm x 2 , as expected. e) The result of part (a) indicates that U = when a x = 0. This makes sense because the mass at the center is a constant distance a from the mass in the ring. The result of part (c) indicates that Fx = 0 when x = 0. At the center of the ring, all mass elements that comprise the ring attract the particle toward the respective parts of the ring, and the net force is zero. 12.40: At the equator, the gravitational field and the radial acceleration are parallel, and taking the magnitude of the weight as given in Eq. (12.30) gives w = mg 0 − marad . The difference between the measured weight and the force of gravitational attraction is ω the term marad . The mass m is found by solving the first relation for m, m = g 0 − arad . Then, marad = w arad w = . g 0 − arad ( g 0 arad ) − 1

Using either g 0 = 9.80 m s 2 or calculating g 0 from Eq. (12.4) gives ma rad = 2.40 N. 12.41: a) GmN m R 2 = 10.7 m s 2 ( 5.00 kg ) = 53.5 N, or 54 N to two figures.  4π 2 2.5 × 10 7 m   = 52.0 N. b) m( g 0 − arad ) = ( 5.00 kg ) 10.7 m s 2 −  [ (16 h ) ( 3600 s h ) ] 2   

(

)

(

)

12.42:

a)

GMm RSc 2 2 mc 2 RS = = . r2 r2 2r 2

(

)

b)

( 5.00 kg ) ( 3.00 × 10 8 m s ) 2 (1.4 × 10 −2 m ) = 350 N.
2 3.00 × 10 6 m

(

)

2

c) Solving Eq. (12.32) for M , R c 2 14.00 × 10− 3 m 3.00 × 108 m s M = S = 2G 2 6.673 × 10 −11 N ⋅ m 2 kg 2

(

(

)(

)

)

2

= 9.44 × 10 24 kg.

12.43:

a) From Eq. (12.12), Rv 2 ( 7.5 ly ) 9.461 × 1015 m ly 200 × 103 m s M= = G 6.673 × 10−11 N ⋅ m 2 kg 2

(

(

)(

)

)

2

= 4.3 × 1037 kg = 2.1 × 107 M S . b) It would seem not. c) which does fit. 12.44: Using the mass of the sun for M in Eq. (12.32) gives RS = 2 6.673 × 10 −11 N ⋅ m 2 kg 2 1.99 × 1030 kg RS = 2GM 2v 2 R = 2 = 6.32 × 1010 m, 2 c c

(

(3.00 × 10

8

)( m s)

2

) = 2.95 km.

That is, Eq. (12.32) may be rewritten 2Gmsun RS = c2

 M   M    m  = 2.95 km ×  m .     sun   sun 

Using 3.0 km instead of 2.95 km is accurate to 1.7%.

12.45:

RS 2 6.67 × 10 −11 N m 2 kg 2 5.97 × 10 24 kg = = 1.4 × 10− 9. 2 8 6 RE 3 × 10 m s 6.38 × 10 m

(

(

)(

)(

)

)

12.46: a) From symmetry, the net gravitational force will be in the direction 45 ° from the + x -axis (bisecting the x and y axes), with magnitude  (2.0 kg)  (1.0 kg ) (6.673 × 10−11 N ⋅ m 2 kg 2 )(0.0150 kg )  +2 sin 45° 2 2 (0.50 m)  (2(0.50 m) )  −12 = 9.67 × 10 N. b) The initial displacement is so large that the initial potential may be taken to be zero. From the work-energy theorem,  (2.0 kg) 1 2 (1.0 kg)  mv = Gm  +2 . 2 (0.50 m)   2 (0.50 m) Canceling the factor of m and solving for v, and using the numerical values gives υ = 3.02 × 10 −5 m s. 12.47: The geometry of the 3-4-5 triangle is available to simplify some of the algebra, The components of the gravitational force are Fy = (6.673 × 10 −11 N ⋅ m 2 kg 2 )(0.500 kg )(80.0 kg ) 3 (5.000 m) 2 5

= 6.406 × 10 −11 N
 (60.0 kg ) (80.0 kg) Fx = −(6.673 × 10−11 N ⋅ m 2 kg 2 )(0.500 kg )  + 2 (5.000 m) 2  ( 4.000 m) = −2.105 × 10 −10 N, so the magnitude is 2.20 × 10 −10 N and the direction of the net gravitational force is 163 ° counterclockwise from the + x - axis. b) A at x = 0, y = 1.39 m. 4 5 

12.48: a) The direction from the origin to the point midway between the two large masses is arctan ( 0.100 m ) = 26.6°, which is not the angle(14.6°) found in the example. 0.200 m b) The common lever arm is 0.100 m, and the force on the upper mass is at an angle of 45° from the lever arm. The net torque is  (0.100 m)sin 45° (0.100 m)  (6.673 × 10−11 N ⋅ m 2 kg 2 )(0.0100 kg)(0.500 kg)  − 2 (0.200 m) 2   2(0.200 m)  −13 = −5.39 × 10 N ⋅ m, with the minus sign indicating a clockwise torque. c) There can be no net torque due to gravitational fields with respect to the center of gravity, and so the center of gravity in this case is not at the center of mass. 12.49: a) The simplest way to approach this problem is to find the force between the spacecraft and the center of mass of the earth-moon system, which is 4.67 × 10 6 m from the center of the earth.

The distance from the spacecraft to the center of mass of the earth-moon system is 3.82 × 10 8 m. Using the Law of Gravitation, the force on the spacecraft is 3.4 N, an angle of 0.61° from the earth-spacecraft line. This equilateral triangle arrangement of the earth, moon and spacecraft is a solution of the Lagrange Circular Restricted Three-Body Problem. The spacecraft is at one of the earth-moon system Lagrange points. The Trojan asteriods are found at the corresponding Jovian Lagrange points.
b) The work is W = − GMm = − 6.673×10 r
−11

N⋅m 2 / kg 2 )( 5.97×10 24 kg + 7.35×10 22 kg)(1250 kg) 3.84×108 m

, or

W = −1.31 × 109 J.

12.50: Denote the 25-kg sphere by a subscript 1 and the 100-kg sphere by a subscript 2. a) Linear momentum is conserved because we are ignoring all other forces, that is, the net external force on the system is zero. Hence, m1v1 = m2 v 2. This relationship is useful in solving part (b) of this problem. b)From the workenergy theorem,
1 1 1 2 Gm1m2  −  = m1m12 + m2v2  rf ri  2

(

)

and from conservation of momentum the speeds are related by m1v1 = m2 v 2 . Using the conservation of momentum relation to eliminate v 2 in favor of v1 and simplifying yields
v12 =
2 2Gm2  1 1  − , m1 + m2  rf ri   

with a similar expression for v 2 . Substitution of numerical values gives v1 = 1.63 × 10 −5 m s, v 2 = 4.08 × 10 −6 m s. The magnitude of the relative velocity is the sum of the speeds, 2.04 × 10 −5 m s. c) The distance the centers of the spheres travel ( x1 and x 2 ) is proportional to x a their acceleration, and x = a = m , or x1 = 4 x2 . When the spheres finally make m
1 1 2 1 2 2

contact, their centers will be a distance of 2 R apart, or x1 + x2 + 2 R = 40 m, or x2 + 4 x2 + 2 R = 40 m. Thus, x 2 = 8 m − 0.4 R, and x1 = 32 m − 1.6 R. 12.51: Solving Eq. (12.14) for r,
 T  R = GmE    2π 
3 2

 (27.3 d)(86,400 s d)  = (6.673 × 10 N ⋅ m kg )(5.97 × 10 kg)   2π   = 5.614 × 10 25 m 3 , from which r = 3.83 × 10 8 m.
−11 2 2 24

2

12.52:

 g= g =

( 6.673×10 −11 N ⋅ m 2 kg 2 )( 20.0 kg) (1.50 m) 2

= 5.93 × 10 −10 N kg, directed toward the

center of the sphere.

12.53:
3

a) From Eq. (12.14),
2 2

T   86,164 s  r = GmE   = (6.673 × 10−11 N ⋅ m 2 kg 2 ) (5.97 × 10 24 kg)    2π   2π  = 7.492 × 1022 m 3 ,

and so h = r − RE = 3.58 × 107 m. Note that the period to use for the earth’s rotation is the siderial day, not the solar day (see Section 12.7). b) For these observers, the satellite is below the horizon.

12.54:

Equation 12.14 in the text will give us the planet’s mass: 2πr 3 2
T= GM P 4π 2 r 3 GM P 4π 2 r 3 4π 2 (5.75 × 105 m + 4.80 × 106 m) 3 = GT 2 (6.673 × 10−11 N ⋅ m 2 kg 2 )(5.8 × 103 s) 2 T2 = MP =

= 2.731 × 1024 kg , or about half earth’s mass.

Now we can find the astronaut’s weight on the surface (The landing on the north pole removes any need to account for centripetal acceleration):
w= GM p ma r
2 p

=

( 6.673 × 10

−11

N ⋅ m 2 kg 2 2.731 × 1024 kg ( 85.6 kg )

( 4.80 × 10 m )
6

)(

2

)

= 677 N

12.55: In terms of the density ρ , the ratio M R is ( 4π 3) ρR 2 , and so the escape speed is
v=

( 8π 3) ( 6.673 × 10−11 N ⋅ m 2

kg 2 2500 kg m 3 150 × 103 m

)(

)(

)

2

= 177 m s.

12.56: a) Following the hint, use as the escape velocity v = 2gh, where h is the height one can jump from the surface of the earth. Equating this to the expression for the escape speed found in Problem 12.55,
2 gh = 8π 3 gh ρGR 2 , or R 2 = , 3 4π ρG

where g = 9.80 m s 2 is for the surface of the earth, not the asteroid. Using h = 1 m (variable for different people, of course), R = 3.7 km. As an alternative, if one’s jump speed is known, the analysis of Problem 12.55 shows that for the same density, the escape speed is proportional to the radius, and one’s jump speed as a fraction of 60 m s gives the largest radius as a fraction of 50 km. b) With 3a a = v 2 R, ρ = 4πGR = 3.03 × 10 3 kg m 3 . 12.57: a) The satellite is revolving west to east, in the same direction the earth is rotating. If the angular speed of the satellite is ωs and the angular speed of the earth is ωE , the angular speed ωrel of the satellite relative to you is ωrel = ωs − ωE .
1 ωrel = (1 rev ) (12 h ) = ( 12 ) rev h 1 ωE = ( 12 ) rev h

ωs = ωrel + ωE = ( 1 ) rev h = 2.18 × 10-4 rad s 8   mm v2 ∑ F = ma says G 2 E = m r r GmE Gm v2 = and with v = rω this gives r 3 = 2E ; r = 2.03 × 107 m r ω This is the radius of the satellite’s orbit. Its height h above the surface of the earth is h = r − RE = 1.39 × 107 m.

b) Now the satellite is revolving opposite to the rotation of the earth. 1 If west to east is positive, then ωrel = ( − 12 ) rev h
1 ωs = ωrel + ωE = ( − 24 ) rev h = −7.27 × 10 -5 rad s Gm r 3 = 2E gives r = 4.22 × 107 m and h = 3.59 × 107 m ω

12.58:

(a) Get radius of X : 1 ( 2πR ) = 18,850 km 4
R = 1.20 × 107 m

943 Astronant mass: m = ω = 9.80 mNs = 96.2 kg Use astronant at north pole to get mass of g
2

X:

∑ F = ma :

GmM x = mg x R2 mg x R 2 (915 N)(1.20 × 107 m) 2 Mx = = = 2.05 × 10 25 kg Gm (6.67 × 10−11 Nm 2 kg 2 )(96.2 kg ) − Fscale mv 2 = R

Apply Newton’s second law to astronant on a scale at the equator of X.

∑ F = ma : F

grav

(b) For satellite:

∑ F = ma →

m( 2πR ) 2πR 4π 2 mR R → Fgrav − Fscale = = T R T2 4π 2 (96.2 kg )(1.20 × 10 7 m) 915.0 N − 850.0 N = T2  1 hr  T = 2.65 × 10 4 s  = 7.36 hr, which is one day  3600 s 
2

v=

Gms mx r
2

=

ms v 2 r

where v =

2πr T

π . Gm x = ( 2T r ) r

2

T=

4π 2 r 3 4π 2 (1.20 × 107 m + 2 × 106 m)3 = Gmx (6.67 × 10 −11 Nm 2 kg 2 )(2.05 × 10 25 kg )

T = 8.90 × 103 s = 2.47 hours

12.59:

The fractional error is
1− mgh g = 1− ( RE + h)( RE ). 1 1 GmmE RE − RE +h GmE

(

)

At this point, it is advantageous to use the algebraic expression for g as given in Eq. (12.4) instead of numerical values to obtain the fractional difference as 1 − ( RE + h) RE = − h RE , so if the fractional difference is
− 1%, h = (0.01) RE = 6.4 × 104 m.

If the algebraic form for g in terms of the other parameters is not used, and the numerical values from Appendix F are used along with g = 9.80 m s 2 , h RE = 8.7 × 10 −3 , which is qualitatively the same.

12.60: (a) Get g on Mongo: It takes 4.00 s to reach the maximum height, where v = 0 then v0 − gt → 0 = 12.0 m s − g (4.00 s)
g = 3.00 m s 2

Apply Newton’s second law to a falling object:
∑ F = ma : mg = GmM → M = gR 2 G 2 R 2πR = C → R = C 2π
2

× (3.00 m s 2 ) 2.00210 m π M = gR G = = 4.56 × 10 25 kg 6.67 × 10 −11 N m 2 kg 2

(

8

)

2

b) Apply Newton’s second law to the orbiting starship.
GmM mv 2 ΣF = ma : = r2 r v= 4π 2 r 3 GM C r = R + 30,000 km = + 3.0 × 107 m 2π
8

2πr →T = T

× 4π 2 ( 2.00210 m + 3.0 × 10 7 m) 3 π T= −11 (6.67 × 10 Nm 2 kg 2 )(4.56 × 10 25 kg)

 1h  = 5.54 × 10 4 s   = 15.4 h  3600 s 

12.61:

At Sacramento, the gravity force on you is F1 = G

mmE ⋅ 2 RE

At the top of Mount Everest, a height of h = 8800 m above sea level, the gravity force on you is
F2 = G = mmE mmE =G 2 2 2 ( RE + h) R E (1 + h RE )  1 − 2h  2h , F2 = F1   R   RE  E 

(1 + h RE ) − 2 ≈ 1 −

F1 − F2 2h = = 0.28 % F1 rE

12.62:

a) The total gravitational potential energy in this model is
m mM  U = −Gm  E + ⋅ REM − r   r 

b) See Exercise 12.5. The point where the net gravitational field vanishes is
r= REM = 3.46 × 108 m. 1 + mM mE

Using this value for r in the expression in part (a) and the work-energy theorem, including the initial potential energy of − Gm(mE RE + mM ( REM − RE )) gives 11.1 km s. c) The final distance from the earth is not RM , but the Earth-moon distance minus the radius of the moon, or 3.823 × 10 8 m. From the work-energy
theorem, the rocket impacts the moon with a speed of 2.9 km s. 12.63: One can solve this problem using energy conservation, units of J/kg for energy, and basic concepts of orbits. E = K + U , or − GM = 1 v 2 − GM , where E , K and U are the 2a 2 r energies per unit mass, v is the circular orbital velocity of 1655 m/s at the lunicentric distance of 1.79 × 10 6 m. The total energy at this distance is − 1.37 × 10 6 J Kg. When the velocity of the spacecraft is reduced by 20 m/s, the total energy becomes E= 1 (6.673 × 10 −11 N ⋅ m 2 / kg 2 ) (7.35 × 10 22 kg) (1655 m / s − 20 m / s) 2 − , 2 (1.79 × 10 6 m)

or E = −1.40 × 10 6 J kg. Since E = − GM , we can solve for a, a = 1.748 × 10 6 m, the semi 2a –major axis of the new elliptical orbit. The old distance of 1.79 × 10 6 m is now the apolune distance, and the perilune can be found from r +r a = a 2 p , rp = 1.706 × 106 m. Obviously this is less than the radius of the moon, so the spacecraft crashes! At the surface, U = − GM or, U = −2.818 × 106 J kg. Rm Since the total energy at the surface is − 1.40 × 10 6 J kg, the kinetic energy at the surface is 1.415 × 10 6 J kg. So, 1 v 2 = 1.415 × 10 6 J kg, or v = 1.682 × 10 3 m s = 6057 km h. 2

12.64: Combining Equations (12.13) and (3.28) and setting arad = 9.80 m s 2 ( so that ω = 0 in Eq. (12.30) ) , T = 2π R = 5.07 × 10 3 s, a rad

which is 84.5 min, or about an hour and a half. 12.65: The change in gravitational potential energy is ∆U =

GmE m GmE m h − = −GmE m , ( RE + h ) RE RE ( RR + h ) so the speed of the hammer is, from the work-energy theorem, 2GmE h . ( R E + h ) RE 12.66: a) The energy the satellite has as it sits on the surface of the Earth is Ei =
− GmM E 2 RE GmM E 2 RE − GmM E RE

.

The energy it has when it is in orbit at a radius R ≈ RE is Ef = put it in orbit is the difference between these: W = Ef - Ei =

. The work needed to

.

b) The total energy of the satellite far away from the Earth is zero, so the additional − GmM GmM work needed is 0 − 2 RE E = 2 RE E .

(

)

c) The work needed to put the satellite into orbit was the same as the work needed to put the satellite from orbit to the edge of the universe.

12.67:

The escape speed will be

m m  v = 2G  E + s  = 4.35 × 104 m s.  RE RES  a) Making the simplifying assumption that the direction of launch is the direction of the earth’s motion in its orbit, the speed relative the earth is v− 2πRES 2π(1.50 × 1011 m) = 4.35 × 10 4 m s − = 1.37 × 104 m s. 7 T (3.156 × 10 s)
2 π ( 6.38×10 6 m) cos 28.5 ° 86164 s 4

b) The rotational at Cape Canaveral is

= 4.09 × 102 m s, so the speed

relative to the surface of the earth is 1.33 × 10 m s. c) In French Guiana, the rotational speed is 4.63 × 10 2 m s, so the speed relative to the surface of the earth is 1.32 × 10 4 m s. 12.68: a) The SI units of energy are kg ⋅ m 2 s 2 , so the SI units for φ are m 2 s 2 . Also, it is known from kinetic energy considerations that the dimensions of energy, kinetic or potential, are mass × speed 2 , so the dimensions of gravitational potential must be the same as speed 2 . b) φ = − U = m c)
GmE r

.

1 1 ∆φ = GmE  −  = 3.68 × 10 6 J kg.  RE rf  d) m∆φ = 5.53 × 1010 J. (An extra figure was kept in the intermediate calculations.) 12.69: a) The period of the asteroid is T =
2 πa 3 2 GM

. Inserting 3 × 1011 m for a gives

2.84 y and 5 × 1011 m gives a period of 6.11 y. b) If the period is 5.93 y, then a = 4.90 × 1011 m. c) This happens because 0.4 = 2 5, another ratio of integers. So once every 5 orbits of the asteroid and 2 orbits of Jupiter, the asteroid is at its perijove distance. Solving when T = 4.74 y, a = 4.22 × 1011 m.

12.70:

a) In moving to a lower orbit by whatever means, gravity does positive work, v 2 Gm 1/ 2 and so the speed does increase. b) From = 2 E , v = ( GmE ) r −1 / 2 , so r r − ∆r  − 3 / 2  ∆r  GmE 1/ 2  ∆v = ( GmE )  − =  ⋅ r 3 2    2  r Note that a positive ∆r is given as a decrease in radius. Similarly, the kinetic energy is K = (1 2 ) mv 2 = (1 2) GmE m / r , and so ∆K = (1 2 ) GmE m / r 2 ∆r , ∆U = − GmE m / r 2 ∆r and W = ∆U + ∆K = − GmE m / 2r 2 ∆r , is agreement with part (a). c)v = GmE r = 7.72 × 103 m/s, ∆v = ( ∆r 2) GmE r = 28.9 m/s, E = −GmE m / 2r =
3

(

)

(

)

(

)

− 8.95 × 1010 J (from Eq. (12.15)), ∆K = GmE m / 2r 2 ( ∆r ) = 6.70 × 108 J, ∆U = −2∆K =

(

)

− 1.34 × 109 J and W = −∆K = −6.70 × 108 J. d)As the term “burns up” suggests, the energy is converted to heat or is dissipated in the collisions of the debris with the grounds.
12.71: a) The stars are separated by the diameter of the circle d =2R, so the 2 gravitational force is GM2 . 4R b) The gravitational force found in part (b) is related to the radial acceleration by Fg = Marad = Mv 2 R for each star, and substituting the expression for the force from part (a) and solving for v gives v = GM 4R. The period is Τ= − GM 2 R and the initial kinetic energy is 2(1 2) Mv 2 = .GM 2 4 R, so the total mechanical energy is − GM 2 2 R. If the stars have zero speed when they are very far apart, the energy needed to separate them is GM 2 4 R.
2πR υ 2

= 16π 2 R 3 GM = 4πR 3 2

GM . c) The initial gravitational potential energy is

12.72: a) The radii R1 and R2 are measured with respect to the center of mass, and so M 1 R1 = M 2 R2 , and R1 R2 = M 2 M 1 . b) If the periods were different, the stars would move around the circle with respect to one another, and their separations would not be constant; the orbits would not remain circular. Employing qualitative physical principles, the forces on each star are equal in magnitude, and in terms of the periods, the product of the mass and the radial accelerations are 4π2 M 1R1 4π2 M 2 R2 = . T12 T22 From the result of part (a), the numerators of these expressions are equal, and so the denominators are equal, and the periods are the same. To find the period in the symmetric from desired, there are many possible routes. An elegant method, using a bit of hindsight, M is to use the above expressions to relate the periods to the force Fg = ( GM 1R )22 , so that R +
1 2

equivalent expressions for the period are M 2T 2 = 4π 2 R1(R1 + R2 )2 G

4π2 R2 ( R1 + R2 ) 2 M 1T = . G
2

Adding the expressions gives ( M 1 + M 2 )T 2 = 4π2 ( R1 + R2 )3 2π( R1 + R2 )3 2 or T = . G G(M1 + M 2 )

c) First we must find the radii of each orbit given the speed and period data. In a π circular orbit, v = 2TR , or R = vT 2π Thus , R α =
( 36×103 m s )(137 d ×86,400 s d ) 2π

= 6.78 × 1010 m, and R β =

(12×103 m s )(137 d×86, 400 s d) 2π

= 2.26 × 1010 m. Now find the sum of the masses and use M α Rα = M β Rβ , and the fact
that Rα = 3Rβ .( M α + M β ) = (M α + M β ) =
4 π 2. ( Rα + Rβ )3 T 2G

, inserting the values of T, and the radii, .M α + M β = 3.12 × 1030 kg. Since

4π 2 ( 6.78×1010 m + 2.26×1010 m)3 (137 d ×86,400 s d ) 2 ( 6.673×10 −11 N⋅m 2 kg 2 )

M β = M α Rα R β = 3M α ,4 M α = 3.12 × 10 30 kg, or M α = 7.80 × 10 29 kg, and M β = 2.34 × 10 30 kg. d) Let α refer to the star and β refer to the black hole. Use the relationships derived in parts (a) and (b): R β = ( M α M β ) Rα = (0.67 3.8) Rα = (0.176) Rα , ( Rα +

Rβ) = 3

( M α + M β ) T 2G 4π 2

, and v =

2 πR T

. For Rα , inserting the values for M and T and

12.73:

From conservation of energy, the speed at the closer distance is  1 1 2 v = v0 + 2Gms  −  = 6.8 × 10 4 m s . r r  i   f

12.74:

Using conservation of energy, GM S mM 1 GM S mM 1 2 mM va2 − = mM v p − , or 2 ra 2 rp 1 1 vp = va2 − 2GM S  −  = 2.650 × 104 m s.  ra rp   

The subscripts a and p denote aphelion and perihelion. To use conservation of angular momentum, note that at the extremes of distance (periheleion and aphelion), Mars’ velocity vector must be perpendicular to its radius vector, and so the magnitude of the angular momentum is L = mrv . Since L is constant, the product rv must be a constant, and so v p = va ra (2.492 × 1011 m) = (2.198 × 10 4 m s) = 2.650 × 10 4 m s, rp (2.067 × 1011 m)

a confirmation of Kepler’s Laws.

12.75: a) The semimajor axis is the average of the perigee and apogee distances, a = 1 (( RE + hp ) + ( RE + ha )) = 8.58 × 106 m. From Eq. (12.19) with the mass of the earth, 2 the period of the orbit is T= 2πa 3 2 = 7.91 × 103 s, GM E
υp υa

a little more than two hours. b) See Problem 12.74;

=

ra rp

= 1.53. c) The equation that

represents conservation of energy (apart from a common factor of the mass of the spacecraft) is 1 2 GmE 1 2 GmE 1  rp  2 GmE vp − = va − =   vp − , 2 rp 2 ra 2  ra  ra   where conservation of angular momentum has been used to eliminate va is favor of vp .
2 Solving for vp and simplifying, 2

2 vp =

2GmE ra = 7.71 × 107 m 2 s 2 , rp ( rp + ra )

from which vp = 8.43 × 103 m s and va = 5.51 × 103 m s. d) The escape speed for a given distance is ve = 2GM r , and so the difference between escape speed and v p is, after some algebra, ve − v p = 2GmE 1 − 1 / 1 + (rp ra ) ⋅ rp

[

]

Using the given values for the radii gives ve − vp = 2.41 × 103 m s. The similar calculation at apogee give ve − va = 3.26 × 103 m s, so it is more efficient to fire the rockets at perigee. Note that in the above, the escape speed ve is different at the two points, vpe = 1.09 × 10 4 m s and vae = 8.77 × 103 m s.

12.76:

a) From the value of g at the poles, g R2 11.1m s 2 2.556 × 107 m mU = U U = = 1.09 × 1026 kg. −11 2 2 G 6.673 × 10 N ⋅ m kg

(

2 b) GmU r 2 = g U ( RU r ) = 0.432 m s . c) GmM RM = 0.080 m s 2 . d) No; Miranda’s gravity is sufficient to retain objects released near its surface. 2 2

(

)(

) )

2

12.77: Using Eq. (12.15), with the mass M m instead of the mass of the earth, the energy needed is ∆E = Gmm m  1 1  − 2  ri rf    −11 6.673 × 10 N ⋅ m 2 kg 2 6.42 × 10 23 kg ( 3000 kg ) = 2   1 1 × − 6 6 6 6 4.00 × 10 m + 3.40 × 10 m   2.00 × 10 m + 3.40 × 10 m  9 = 3.22 × 10 J.

(

)(

)

(

) (

)

12.78:

a) The semimajor axis is 4 × 1015 m and so the period is

( 6.673 × 10

2π 4 × 1015 m
−11

(

N⋅m

2

) kg ) (1.99 × 10
32 2

30

kg

)

= 1.38 × 1014 s,

which is about 4 million years. b) Using the earth-sun distance as an estimate for the distance of closest approach, v = 2GmS RES = 4 × 104

m s. c) (1 2) mv 2 = GmS m R = 10 24 J. This is far larger than the energy of a volcanic eruption and is comparable to the energy of burning the fossil fuel.

12.79:

a) From Eq. (12.14) with the mass of the sun,  6.673 × 10 −11 N ⋅ m 2 kg 2 1.99 × 1030 kg  r=  2  × 3 × 104 y 3.156 × 107 s y 4π2   

(

((

)(

)(

))

)

13

= 1.4 × 1014 m.

This is about 24 times the orbit radius of Pluto and about 1 250 of the way to Alpha Centauri. 12.80: Outside the planet it behaves like a point mass, so at the surface: ∑ F = ma : GmM = mg → g = GM R 2 2 R

Get M : M = ∫ dm = ∫ ρdV = ∫ ρ 4πr 2 dr. The density is ρ = ρ0 − br , where

ρ 0 = 15.0 × 10 3 kg m 3 at the center at the surface, ρS = 2.0 × 10 3 kg m 3 , so b = 4π R M = ∫ 0 ( ρ0 − br ) 4πr 2 dr = ρ0 R 3 − πbR 4 3 4  ρ − ρs   3 1 = πR 3 ρ0 − πR 4  0  = πR  ρ0 + ρs  3  R  3 
GM Gπ R 3 ( 1 ρ0 + ρs ) 1  3 g= 2 = = πRG ρ0 + ρs  2 R R 3 

ρ0 − ρ s R

 15.0 × 10 3 kg m 3  π 6.38 × 10 6 m 6.67 × 10 −11 Nm 2 kg 2  + 2.0 × 10 3 kg m 3    3   2 = 9.36 m s

(

)(

)

12.81: The radius of the semicircle is R = L π Divide the semicircle up into small segments of length R dθ

dM = ( M L ) R dθ = ( M π ) dθ  dF is the gravity force on m exerted by dM ∫ dFy = 0; the y-components from the upper half of the semicircle cancel the ycomponents from the lower half. The x-components are all in the +x-direction and all add. mdM R2 mdM Gmπm dFx = G cos θ = cosθ dθ 2 R L2 π 2 GmπM π 2 Gmπm Fx = ∫ dFx = ∫−π 2 cos θdθ = L2 (2) −π 2 L2 2πGmM F= L2 dF = G

12.82: The direct calculation of the force that the sphere exerts on the ring is slightly more involved than the calculation of the force that the ring exerts on the ball. These forces are equal in magnitude but opposite in direction, so it will suffice to do the latter calculation. By symmetry, the force on the sphere will be along the axis of the ring in Fig. (12.34), toward the ring. Each mass element dM of the ring exerts a force of magnitude GmdM on the sphere, and the x-component of this force is a2 + x2 GmdM a2 + x2 x a +x
2 2

=

(a

GmdMx
2

+ x2

)

3/ 2

.

As x >> a the denominator approaches x 3 and F → GMm , as expected, and so the force on x2 the sphere is GmMx / a 2 + x 2 , in the − x - direction. The sphere attracts the ring with a force of the same magnitude. (This is an alternative but equivalent way of obtaining the result of parts (c) and (d) of Exercise 12.39.)2 12.83: Divide the rod into differential masses dm at position l, measured from the right end of the rod. Then, dm = dl ( M L ) , and the contribution GmMdl dFx from each piece is dFx = − . Integrating from l = 0 to l = L gives ( l + x) 2 L F =− GmM L

(

)

3/ 2

∫ ( l + x)
0

L

dl

2

=

GmM  1 1 GmM  x + L − x  = − x( x + L ) , L  
GmM x( x + L )

with the negative sign indicating a force to the left. The magnitude is F = equivalent way of obtaining the result of part (b) Exercise 12.39.)

. As x >>

L, the denominator approaches x 2 and F → GmM , as expected. (This is an alternative but x2

12.84: a) From the result shown in Example 12.10, the force is attractive and its magnitude is proportional to the distance the object is from the center of the earth. Comparison with equations (6.8) and (7.9) show that the gravitational potential energy is given by GmE m 2 r . 3 2 RE This is also given by the integral of Fg from 0 to r with respect to distance. b) From part U (r) = GmE m . Equating initial potential energy 1RE and final kinetic energy (initial kinetic energy and final potential energy are both zero) GmE gives v 2 = , so v = 7.90 × 103 m/s. RE (a), the initial gravitational potential energy is

12.85: T + ∆T = Since v = Since T =

a) T =
2π GM E GM E r

2π r 3 2 GM E

, therefore
2π r 3 2 GM E r r (1 + ∆rr ) 3 2 ≈ 2πGM (1 + 32∆rr ) = T + 3π GM∆ r .
32 E 12 E

(r + ∆r )3 2 = , ∆T = , ∆v =
3π ∆ r v

. v = GM E r

−1 2

, therefore
∆ r −1 2 r

v − ∆ v = GM E (r + ∆ r ) −1 2 = GM E r −1 2 (1 +
2π r 3 2 GM E π∆r T

)

≈ GM E r −1 2 (1 − ∆ rr ) = v − 2

GM E 2r 3 2

∆ r.

.

b) Note: Because of the small change in r, several significant figures are needed to r3 2 see the results. Starting with T = 2πGM (Eq.(12.14)), T = 2π r v , and v = GM r (Eq.(12.12)) find the velocity and period of the initial orbit: (6.673 × 10−11 N ⋅ m 2 kg 2 )(5.97 × 10 24 kg ) = 7.672 × 103 m s, and 6.776 × 106 m T = 2π r v = 5549 s = 92.5 min. We then can use the two derived equations to approximate the 3 π 100 ∆ T and ∆ v, ∆ T = 3 πv∆ r and ∆v = πT∆r .∆T = 7.672(×10 3m) s = 0.1228 s, and ∆v m v= =
π ∆r T

=

π (100 m ) ( 5549 s )

= .05662 m s.

Before the cable breaks, the shuttle will have traveled a distance d, d = (125 m 2 ) − (100 m 2 ) = 75 m. So, (75 m) (.05662 m s) = 1324.7 s = 22 min. It will take 22 minutes for the cable to break. c) The ISS is moving faster than the space shuttle, so the total angle it covers in an orbit must be 2π radians more than the angle that the space shuttle covers before they are once again in line. Mathematically, vt − ((vr− ∆vr))t = 2π . Using the binomial theorem and r +∆ neglecting terms of order ∆v∆r , vt − ( v −r∆v ) t (1 + r t=
2π r v∆r    ∆v +  r   ∆r −1 r v∆T 3π

)

≈t

(

∆v r

+ vr∆2r = 2π . Therefore, =
T2 ∆T

)

=

vT π ∆r v∆r + T r

. Since 2π r = vT and ∆r =

,t =

vT π  v∆T  2 π  v∆T   +   t  3π  T  3π 

, as was to be

shown. t =

T2 ∆T

=

( 5549 s ) 2 ( 0.1228 s )

= 2.5 × 108 s = 2900 d = 7.9 y. It is highly doubtful the shuttle

crew would survive the congressional hearings if they miss!

12.86: a) To get from the circular orbit of the earth to the transfer orbit, the spacecraft’s energy must increase, and the rockets are fired in the direction opposite that of the motion, that is, in the direction that increases the speed. Once at the orbit of Mars, the energy needs to be increased again, and so the rockets need to be fired in the direction opposite that of the motion. From Fig. (12.37), the semimajor axis of the transfer orbit is the arithmetic average of the orbit radii of the earth and Mars, and so from Eq. (12.19), the energy of spacecraft while in the transfer orbit is intermediate between the energies of the circular orbits. Returning from Mars to the earth, the procedure is reversed, and the rockets are fired against the direction of motion. b) The time will be half the period as given in Eq. (12.19), with the semimajor axis a being the average of the orbit radii, a = 1.89 × 1011 m, so t= T π (1.89 × 1011 m)3 2 = = 2.24 × 107 s, −11 2 2 30 2 (6.673 × 10 N ⋅ m kg )(1.99 × 10 kg)

which is more than 8 1 months. c) During this time, Mars will pass through an angle of 2
( × (360°) ( 6872d.248610 s )s d ) = 135.9° , and the spacecraft passes through an angle of 180° , so the )( , 400
7

angle between the earth-sun line and the Mars-sun line must be 44.1° .

12.87:

a) There are many ways of approaching this problem; two will be given here.

I) Denote the orbit radius as r and the distance from this radius to either ear as δ . Each ear, of mass m , can be modeled as subject to two forces, the gravitational force from the black hole and the tension force (actually the force from the body tissues), denoted by F . Then, the force equations for the two ears are GMm − F = mω 2 (r − δ ) (r − δ ) 2 GMm + F = mω 2 (r + δ ), 2 (r + δ ) where ω is the common angular frequency. The first equation reflects the fact that one ear is closer to the black hole, is subject to a larger gravitational force, has a smaller acceleration, and needs the force F to keep it in the circle of radius r − δ . The second equation reflects the fact that the outer ear is further from the black hole and is moving in a circle of larger radius and needs the force F to keep in in the circle of radius r + δ . Dividing the first equation by r − δ and the second by r + δ and equating the resulting expressions eliminates ω , and after a good deal of algebra, (r + δ ) F = (3GMmδ ) 2 . (r − δ 2 ) 2 At this point it is prudent to neglect δ in the sum and difference, but recognize that F is proportional to δ , and numerically F = 3GMmδ = 2.1 kN. (Using the result of Exercise r3 12.39 to express the gravitational force in terms of the Schwartzschild radius gives the same result to two figures.) II) Using the same notation, GMm − F = mω 2 (r + δ ), 2 (r + δ ) where δ can be of either sign. Replace the product mω 2 with the value for δ = 0, mω 2 = GMm r 3 and solve for r + δ 1  GMm −2 F = (GMm)  3 − = r + δ − r (1 + (δ r)) ⋅ 2 3 (r + δ)  r  r Using the binomial theorem to expand the term in square brackets in powers of δ r , GMm GMm F = 3 [ r + δ − r (1 − 2(δ r ) ) ] = 3 (3δ), r r the same result as above. Method (I) avoids using the binomial theorem or Taylor series expansions; the approximations are made only when numerical values are inserted and higher powers of δ are found to be numerically insignificant.

[

]

12.88: As suggested in the problem, divide the disk into rings of radius r and thickness dr. Each ring has an area dA = 2πr dr and mass dM = πM 2 dA = 2aM r dr. The magnitude of 2 a the force that this small ring exerts on the mass m is then (G m dM )( x (r 2 + x 2 )3 2 ), the expression found in Problem 12.82, with dM instead of M and the variable r instead of a. 2GMmx rdr Thus, the contribution dF to the force is dF = . 2 2 a (x + r 2 )3 2 The total force F is then the integral over the range of r; 2GMmx a r F = ∫ dF = 2 ∫0 ( x 2 + r 2 ) 3 2 dr. a The integral (either by looking in a table or making the substitution u = r 2 + a 2 ) is 1  1  a r 1 x ∫0 ( x 2 + r 2 ) 3 2 dr =  x − a 2 + x 2  = x 1 − a 2 + x 2 .     Substitution yields the result  2GMm  x F= 1 − . a2  a2 + x2  The second term in brackets can be written as = (1 + ( a x) ) 1 + (a x) 2 if x >> a, where the binomial approximation (or first-order Taylor series expansion) has been used. Substitution of this into the above form gives F≈ as it should. GMm , x2 1
2 −1 2 2

1a ≈1−   2 x

12.89: From symmetry, the component of the gravitational force parallel to the rod is zero. To find the perpendicular component, divide the rod into segments of length dx and M mass dm = dx 2L , positioned at a distance x from the center of the rod. The magnitude of the gravitational force from each segment is Gm dM GmM dx dF = 2 = . 2L x 2 + a 2 x + a2 The component of dF perpendicular to the rod is dF 2a 2 , and so the net gravitational
x +a

force is GmMa dx F = ∫ dF = ∫L ( x 2 + a 2 )3 2 . 2L − −L The integral can be found in a table, or found by making the substitution x = a tanθ. Then, dx = a sec 2θ dθ , ( x 2 + a 2 ) = a 2 sec 2θ , and so dx a sec 2θ dθ 1 1 x ∫ ( x 2 + a 2 ) 3 2 = ∫ a 3 sec 3θ = a 2 ∫ cosθ dθ = a 2 sin θ = a 2 x 2 + a 2 , and the definite integral is GmM F= . a a 2 + L2 When a >> L, the term in the square root approaches a 2 and F → GmM , as expected. a2
L L

Chapter 13

13.1: a) b)

T=
1 4 ( 220 Hz)

1 f

= 4.55 × 10−3 s, ω =

2π T

= 2πf = 1.38 × 103 rad s.

= 1.14 × 10−3 s, ω = 2πf = 5.53 × 103 rad s.

13.2: a) Since the glider is released form rest, its initial displacement (0.120 m) is the amplitude. b) The glider will return to its original position after another 0.80 s, so the period is 1.60 s. c) The frequency is the reciprocal of the period (Eq. (13.2)), 1 f = 1.60 s = 0.625 Hz. 13.3: The period is ω=
2π T 0.50 s 440

= 1.14 × 10 −3 s and the angular frequency is

= 5.53 × 103 rad s.

13.4: (a) From the graph of its motion, the object completes one full cycle in 2.0 s; its period is thus 2.0 s and its frequency = 1 period = 0.5 s −1 . (b) The displacement varies from − 0.20 m to + 0.20 m, so the amplitude is 0.20 m. (c) 2.0 s (see part a) 13.5: This displacement is
1 4

of a period.

T = 1 f = 0.200 s, so t = 0.0500 s. 13.6: The period will be twice the time given as being between the times at which the glider is at the equilibrium position (see Fig. (13.8));  2π  2π  k =ω m=  m=  2(2.60 s) T  
2 2

  (0.200 kg) = 0.292 N m.  
k ω2

2

13.7: a) T =

1 f

= 0.167 s. b) ω = 2πf = 37.7 rad s. c) m =

= 0.084 kg.

13.8: Solving Eq. (13.12) for k,  2π   2π  3 k = m  = (0.600 kg)   = 1.05 × 10 N m. T   0.150 s 
2 2

13.9: From Eq. (13.12) and Eq. (13.10), T = 2π ω = 2πf = 16.7 rad s. 13.10: a) ax = ax =
dv x dt d2x dt 2

0.500 kg 140 N m

= 0.375 s, f =

1 T

= 2.66 Hz,

= −ω2 A sin( ωt + β ) = −ω2 x, so x(t ) is a solution to Eq. (13.4) if
dx dt

k ω2 = m . b) a = 2 Aω a constant, so Eq. (13.4) is not satisfied. c) vx =

= iωi ( ωt + β ) ,

= (iω) 2 Aei ( ωt + β ) = −ω2 x, so x(t ) is a solution t o Eq. (13.4) if ω2 = k m ⋅

13.11: a) x = (3.0 mm) cos ((2π )(440 Hz)t ) b) (3.0 × 10−3 m)(2π )(440 Hz) = 8.29 m s, (3.0 mm)(2π ) 2 (440 Hz) 2 = 2.29 × 10 4 m s 2 . c) j (t ) = (6.34 × 107 m s 3 ) sin((2 π )(440 Hz)t ), jmax = 6.34 × 107 m s3 . 13.12: a) From Eq. (13.19), A =
v0 ω

=

v0 k m

= 0.98 m. b) Equation (13.18) is

13.16: Empty chair: T = 2π

m k

k= With person in chair:

4π 2 m 4π 2 (42.5 kg) = = 993 N/m T2 (1.30 s) 2

T = 2π m k T 2 k (2.54 s) 2 (993 N/m) = = 162 kg 4π 2 4π 2 mperson = 162 kg − 42.5 kg = 120 kg m=

13.17:

T = 2π m k , m = 0.400 kg

Use ax = −2.70 m/s 2 to calculate k : − kx = ma x gives k = − max (0.400 kg)(−2.70 m/s 2 ) =− = +3.60 N/m x 0.300 m T = 2π m k = 2.09 s

13.18: We have vx (t ) = (3.60 cm/s)sin(( 4.71 s −1 ) t − π 2). Comparing this to the general form of the velocity for SHM: − ωA = 3.60 cm/s ω = 4.71 s −1 φ = −π 2 (a) (b) (c ) amax T = 2π ω = 2π 4.71 s −1 = 1.33 s 3.60 cm s 3.60 cm s A= = = 0.764 cm ω 4.71 s −1 = ω2 A = (4.71 s −1 ) 2 (0.764 cm) = 16.9 cm s 2

13.19: a) x (t ) = (7.40 cm) cos((4.16 rad s)t − 2.42 rad) When t = T , (4.16 rad s)T = 2π so T = 1.51 s b) T = 2π m k so k = m(2π T ) 2 = 26.0 N m c) A = 7.40 cm = 0.0740 m
1 2

mv 2 + 1 kx 2 = 1 kA2 gives vmax = A k m = 0.308 m s 2 2

d) F = −kx so Fmax = kA = 1.92 N e) x(t ) evaluated at t = 1.00 s gives x = −0.0125 m v = ± k m A2 − x 2 = ± 26.0 1.50 (0.0740) 2 − (0.0125) 2 m s = ±0.303 m s Speed is 0.303 m s . a = − kx m = −(26.0 1.50)(−0.0125) m s 2 = +0.216 m s 2 13.20: See Exercise 13.15; t = (arccos(− 1.5 6))(0.3 (2π )) = 0.0871 s. 13.21: a) Dividing Eq. (13.17) by ω , x0 = A cos φ, Squaring and adding,
2 v0 x + 2 = A2 , ω which is the same as Eq. (13.19). b) At time t = 0, Eq. (13.21) becomes 1 2 1 2 1 2 1 k 2 1 2 kA = mv0 + kx0 = v0 + kx0 , 2 2 2 2 ω2 2 2 where m = kω (Eq. (13.10)) has been used. Dividing by k 2 gives Eq. (13.19). 2 0

v0 = − A sin φ. ω

13.22: a) vmax = (2πf ) A = (2π (392 Hz))(0.60 × 10−3 m) = 1.48 m s. 1 1 b) K max = m(Vmax ) 2 = (2.7 × 10 −5 kg)(1.48 m s) 2 = 2.96 × 10 −5 J. 2 2 13.23: a) Setting
1 2

mv 2 = 1 kx 2 in Eq. (13.21) and solving for x gives x = ± 2

A 2

.

Eliminating x in favor of v with the same relation gives vx = ± kA2 2m = ± ωA . b) This 2 happens four times each cycle, corresponding the four possible combinations of + and – in the results of part (a). The time between the occurrences is one-fourth of a period or 2 2 2π T / 4 = 4 ω = 2πω . c) U = 1 E , K = 3 E U = kA , K = 3kA 4 4 8 8

(

)

13.24: a) From Eq. (13.23), vmax = k A= m 450 N m (0.040 m) = 1.20 m/s. 0.500 kg

b) From Eq. (13.22), 450 N v= (0.040 m) 2 − (−0.015 m) 2 = 1.11 m/s. 0.500 kg c) The extremes of acceleration occur at the extremes of motion, when x = ± A, and kA (450 N/m)(0.040 m) a max = = = 36 m/s 2 m (0.500 kg)
0.015 d) From Eq. (13.4), a x = − ( 450 N/m)( −kg) m) = 13.5 m/s 2 . (0.500

e) From Eq. (13.31), E = 1 (450 N/m)(0.040 m) 2 = 0.36 J. 2 13.25: a) amax = ω2 A = (2πf ) 2 A = ( 2π(0.85 Hz) ) (18.0 × 10 −2 m) = 5.13 m/s 2 . vmax =
2

ωA = 2πfA = 0.961 m/s . b) a x = −(2πf ) 2 x = −2.57 m/s 2 ,
v = (2πf ) A2 − x 2 = ( 2π (0.85 Hz) ) (18.0 × 10− 2 m) 2 − (9.0 × 10 − 2 m) 2 = 0.833 m/s. c) The fraction of one period is (1 2π ) arcsin (12.0 18.0), and so the time is (T 2π ) × arcsin (12.0 18.0) = 1.37 × 10 −1 s. Note that this is also arcsin ( x A) ω . d) The conservation of energy equation can be written 1 kA 2 = 1 mv 2 + 1 kx 2 . We are 2 2 2 given amplitude, frequency in Hz, and various values of x . We could calculate velocity from this information if we use the relationship k m = ω2 = 4π 2 f 2 and rewrite the conservation equation as
1 2

A2 =

1 v2 2 4π 2 f

2

+ 1 x 2 . Using energy principles is generally a good 2

approach when we are dealing with velocities and positions as opposed to accelerations and time when using dynamics is often easier. 13.26: In the example, A2 = A1
M M +m

and now we want A2 = 1 A1 . So 1 = 2 2

M M +m

, or

2 m = 3M . For the energy, E 2 = 1 kA2 , but since A2 = 1 A1 , E 2 = 1 E1 , or 3 E1 is lost to 2 2 4 4 heat.

13.27: a) b)
2 x0 +

1 2

mv 2 + 1 kx 2 = 0.0284 J . 2

2 v0 (0.300 m/s) 2 = (0.012 m) 2 + = 0.014 m. ω2 (300 N/m) (0.150 kg)

c) ωA = k mA = 0.615 m s ⋅

13.28: At the time in question we have x = A cos (ωt + φ) = 0.600 m v = −ωA sin( ωt + φ) = 2.20 m s a = −ω2 A cos (ωt + φ) = −8.40 m s 2 Using the displacement and acceleration equations: − ω2 A cos (ωt + φ) = −ω2 (0.600 m) = −8.40 m s 2 ω2 = 14.0 and ω = 3.742 s −1 To find A, multiply the velocity equation by ω : − ω2 A sin (ωt + φ) = (3.742 s −1 ) (2.20 m s) = 8.232 m s 2 Next square both this new equation and the acceleration equation and add them: ω4 A2 sin 2 (ωt + φ) + ω4 A2 cos2 (ωt + φ) = (8.232 m s 2 ) 2 + (−8.40 m s 2 ) 2 = ω4 A2 sin 2 (ωt + φ) + cos2 (ωt + φ) ω4 A2 = 67.77 m 2 s 4 + 70.56 m 2 s 4 = 138.3 m 2 s 4 A2 = 138.3 m 2 s 4 138.3 m 2 s 4 = = 0.7054 m 2 4 −1 4 ω (3.742 s ) A = 0.840 m

The object will therefore travel 0.840 m − 0.600 m = 0.240 m to the right before stopping at its maximum amplitude. 13.29: vmax = A k m Use T to find k m : T = 2π m k so k m = (2π T ) 2 = 158 s − 2 Use amax to find A : amax = kA m so A = amax (k m) = 0.0405 m. Then vmax = A k m = 0.509 m s

13.30:

Using k =

F0 L0

from the calibration data, ( F0 L0 ) (200 N) (1.25 × 10 −1 m) = = 6.00 kg. (2πf ) 2 (2π (2.60 Hz))2

m=

13.31: a) k =

mg (650 kg) (9.80 m s 2 ) = = 531 × 103 N m. Δl (0.120 m) m ∆l 0.120 m = 2π = 2π = 0.695 s. k g 9.80 m s 2

b) T = 2π =

13.32: a) At the top of the motion, the spring is unstretched and so has no potential energy, the cat is not moving and so has no kinetic energy, and the gravitational potential energy relative to the bottom is 2mgA = 2(4.00 kg)(9.80 m/s 2 ) × (0.050 m) = 3.92 J . This is the total energy, and is the same total for each part. b) U grav = 0, K = 0, so U spring = 3.92 J . c) At equilibrium the spring is stretched half as much as it was for part (a), and so U spring = 1 (3.92 J) = 0.98 J, U grav = 1 (3.92 J) = 1.96 J, and so K = 0.98 J . 4 2 13.33: The elongation is the weight divided by the spring constant, w mg gT 2 ∆ l = = 2 = 2 = 3.97 cm . k ω m 4π 13.34: See Exercise 9.40. a) The mass would decrease by a factor of (1 3) 3 = 1 27 and so the moment of inertia would decrease by a factor of (1 27)(1 3) = (1 243) , and for the same spring constant, the frequency and angular frequency would increase by a factor of 243 = 15.6 . b) The torsion constant would need to be decreased by a factor of 243, or changed by a factor of 0.00412 (approximately). 13.35: a) With the approximations given, I = mR 2 = 2.72 × 10 −8 kg ⋅ m 2 , or 2.7 × 10 −8 kg ⋅ m 2 to two figures. b) κ = (2πf ) 2 I = (2π 2 Hz) 2 (2.72 × 10−8 kg ⋅ m 2 ) = 4.3 × 10 −6 N ⋅ m rad . 13.36: Solving Eq. (13.24) for κ in terms of the period, 2  2π  κ=  I T   2π  −3 −2 2 =  ((1 2)(2.00 × 10 kg)(2.20 × 10 m) )  1.00 s  = 1.91 × 10− 5 N ⋅ m/rad.
2 2

13.37: I=

κ 0.450 N ⋅ m/rad = = 0.0152 kg ⋅ m 2 . 2 2 (2πf ) ( 2π (125) (265 s) )

13.38: The equation θ = Θcos (ωt + φ) describes angular SHM. In this problem, φ = 0. a)
dθ dt

= −ω Θ sin( ω t ) and

d 2θ dt 2

= −ω2 Θ cos(ω t ).

b) When the angular displacement is Θ, Θ = Θ cos(ω t ) , and this occurs at t = 0, so dθ d 2θ = 0 since sin(0) = 0, and 2 = −ω2Θ, since cos(0) = 1. dt dt When the angular displacement is Θ 2, Θ = Θ cos(ω t ), or 1 = cos(ω t ). 2 2 dθ − ωΘ 3 3 d 2θ − ω 2 Θ = since sin( ω t ) = , and 2 = , since cos(ω t ) = 1 2. dt 2 2 dt 2 This corresponds to a displacement of 60° . 13.39: Using the same procedure used to obtain Eq. (13.29), the potential may be expressed as U = U 0 [(1 + x R0 ) −12 − 2(1 + x R0 ) −6 ]. Note that at r = R0 , U = −U 0 . Using the appropriate forms of the binomial theorem for | x R0 | << 1,  ( − 12)( − 13) ( x R ) 2   0  1 − 12( x R0 ) + 2   U ≈ U0   ( − 6)( − 7 ) ( x R ) 2  0  − 21 − 6( x R0 ) + 2   

 36  = U 0 − 1 + 2 x 2  R0   1 = kx 2 − U 0 . 2 where k = 72U 0 / R 2 has been used. Note that terms in u 2 from Eq. (13.28) must be kept ; the fact that the first-order terms vanish is another indication that R0 is an extreme (in this case a minimum) of U.

13.40: f =

1 2π

k 1 = ( m 2 ) 2π

2(580 N/m) = 1.33 × 1014 Hz. (1.008) (1.66 × 10− 27 kg)

13.41: T = 2π L g , so for a different acceleration due to gravity g ′, T ′ = T g g ′ = (1.60 s ) 9.80 m s 2 3.71 m s 2 = 2.60 s.

13.42: a) To the given precision, the small-angle approximation is valid. The highest speed is at the bottom of the arc, which occurs after a quarter period, T = π L = 0.25 s. 4 2 g b) The same as calculated in (a), 0.25 s. The period is independent of amplitude. 13.43: Besides approximating the pendulum motion as SHM, assume that the angle is sufficiently small that the length of the spring does not change while swinging in the arc. Denote the angular frequency of the vertical motion as ω0 = = 1 ω0 = 2
kg 4w k m

=

kg ω

and ω′ =

g L

spring; the unstretched length is L0 = L − w k = 3 w k = 3(1.00 N ) (1.50 N/m ) = 2.00 m. 13.44:

, which is solved for L = 4w k . But L is the length of the stretched

13.45: The period of the pendulum is T = (136 s ) 100 = 1.36 s. Then, g= 4π 2 L 4π 2 ( .5 m ) = = 10.67 m s 2 . 2 2 T (1.36 s )

13.46: From the parallel axis theorem, the moment of inertia of the hoop about the nail is I = MR 2 + MR 2 = 2 MR 2 , so T = 2π 2 R g , with d = R in Eq.(13.39). Solving for R,

R = gT 2 8π = 0.496 m.
13.47: For the situation described, I = mL2 and d = L in Eq. (13.39); canceling the factor of m and one factor of L in the square root gives Eq. (13.34).

2

13.48:

a) Solving Eq. (13.39) for I,
2 2

T   0.940 s  2 2 I =   mgd =   (1.80 kg ) 9.80 m s ( 0.250 m ) = 0.0987 kg ⋅ m .  2π   2π  b) The small-angle approximation will not give three-figure accuracy for Θ = 0.400 rad. From energy considerations, 1 mgd (1 − cos Θ ) = IΩ 2 . max 2 Expressing Ω max in terms of the period of small-angle oscillations, this becomes Ω max  2π   2π  = 2  (1 − cos Θ ) = 2  (1 − cos( 0.40 rad ) ) = 2.66 rad s . T   0.940 s 
2 2

(

)

13.49: Using the given expression for I in Eq. (13.39), with d=R (and of course m=M), T = 2π 5 R 3g = 0.58 s. 13.50: From Eq. (13.39), T   120 s 100  2 I = mgd   = (1.80 kg ) 9.80 m s 2 ( 0.200 m )   = 0.129 kg.m . 2π  2π    13.51: a) From Eq. (13.43), ω′ = b) b = 2
2

(

)

2

( 2.50 N m ) − ( 0.90 kg s ) 2 = 2.47 rad s , so ( 0.300 kg ) 4( 0.300 kg ) 2 km = 2 ( 2.50 N m ) ( 0.300 kg ) = 1.73 kg s .

f′=

ω′ = 0.393 Hz. 2π

13.52: From Eq. (13.42) A2 = A1 exp ( − 2bm t ). Solving for b, 2m  A1  2(0.050 kg )  0.300 m  ln   = ln   0.100 m  = 0.0220 kg s.  A  t (5.00 s)    2 As a check, note that the oscillation frequency is the same as the undamped frequency to 4.8 × 10 −3%, so Eq. (13.42) is valid. b=

13.53: a) With φ = 0, x(0) = A. dx  b  vx = = Ae− ( b 2 m ) t − cos ω′t − ω′ sin ω′t , b) dt  2m  and at t = 0, v = − Ab 2m ; the graph of x versus t near t = 0 slopes down. c) and at t = 0, ax =  b 2   dvx ω′b = Ae − ( b 2 m ) t  2 − ω′2  cos ω ' t + sin ω′t ,  4m  dt 2m   

 b2   b2 k ax = A 2 − ω′2  = A 2 − .  4m   2m m     (Note that this is (− bv0 − kx0 ) m.) This will be negative if b < 2km , zero if b = 2km and positive if b > 2km . The graph in the three cases will be curved down, not curved, or curved up, respectively. 13.54: At resonance, Eq. (13.46) reduces to A = Fmax bωd . a) A1 . b) 2 A1. Note that 3 the resonance frequency is independent of the value of b (see Fig. (13.27)). 13.55: a) The damping constant has the same units as force divided by speed, or kg ⋅ m s 2 [ m s] = [ kg s]. ⋅ b)The units of km are the same as [[kg s 2 ][kg]]1 2 = [ kg s],
2 the same as those for b. c) ωd = k m. (i) bωd = 0.2 k , so A = Fmax ( 0.2k ) = 5 Fmax k . (ii) bωd = 0.4k , so A = Fmax (0.4 k ) = 2.5Fmax k , as shown in Fig.(13.27).

[

]

13.56: The resonant frequency is k m = (2.1 × 106 N m) 108 kg ) = 139 rad s = 22.2 Hz, and this package does not meet the criterion.

13.57: a)  π rad s    0.100 m   2 3 a = Aω =    = 6.72 × 10 m s .   (3500 rev min )    2  30 rev min  
2 2

 π rad s  b) ma = 3.02 × 10 3 N. c) ωA = (3500 rev min ) (.05 m)  = 18.3 m s .  30 rev min  K = 1 mv 2 = ( 1 )(.45 kg)(18.3 m s) 2 = 75.6 J. d) At the midpoint of the stroke, cos( ω t)=0 2 2
π rad s and so ωt = π 2, thus t = π 2ω. ω = (3500 rev min )( 30 rev min ) = 350 π 3

t=

3 2 ( 350 )

3 s. Then P = ∆K ∆t , or P = 75.6 J ( 2(350) s) = 1.76 × 10 4 W.

e) If the frequency doubles, the acceleration and hence the needed force will quadruple (12.1 × 10 3 N). The maximum speed increases by a factor of 2 since v α ω, so the speed will be 36.7 m/s. Because the kinetic energy depends on the square of the velocity, the kinetic energy will increase by a factor of four (302 J). But, because the time to reach the midpoint is halved, due to the doubled velocity, the power increases by a factor of eight (141 kW). 13.58: Denote the mass of the passengers by m and the (unknown) mass of the car by M. The spring cosntant is then k = mg ∆l . The period of oscillation of the empty car is TE = 2π M k and the period of the loaded car is TL = 2π M +m 2 ∆l = TE2 + ( 2π ) , so k g
2

TE = TL2 − ( 2π )

∆l = 1.003 s. g

13.59: a) For SHM, the period, frequency and angular frequency are independent of amplitude, and are not changed. b) From Eq. (13.31), the energy is decreased by a factor of 1 . c) From Eq. (13.23), the maximum speed is decreased by a factor of 1 d) Initially, 4 2 the speed at A1 4 was ω
15 4 3 4

ωA1; after the amplitude is reduced, the speed is
1 5

( A1 2) 2 − ( A1 4) 2

=

ωA1 , so the speed is decreased by a factor of

(this result is

valid at x = − A1 4 as well). e) The potential energy depends on position and is unchanged. From the result of part (d), the kinetic energy is decreased by a factor of 1 . 5 13.60: This distance L is L = mg k ; the period of the oscillatory motion is m L = 2π , k g which is the period of oscillation of a simple pendulum of lentgh L. T = 2π

13.61: a) Rewriting Eq. (13.22) in terms of the period and solving, 2π A2 − x 2 T= = 1.68 s. v b) Using the result of part (a),  vT  A 2 −   = 0.0904 m.  2π  c) If the block is just on the verge of slipping, the friction force is its maximum, 2 2 f = μs n = μs mg. Setting this equal to ma = mA( 2π T ) gives μs = A( 2π T ) g = 0.143. x= 13.62: a) The normal force on the cowboy must always be upward if he is not holding on. He leaves the saddle when the normal force goes to zero (that is, when he is no longer in contact with the saddle, and the contact force vanishes). At this point the cowboy is in free fall, and so his acceleration is − g ; this must have been the acceleration just before he left contact with the saddle, and so this is also the saddle’s acceleration. b) x = + a (2π f ) 2 = +(9.80 m s 2 ) 2π (1.50 Hz ))2 = 0.110 m. c) The cowboy’s speed will be the saddle’s speed, v = (2πf ) A2 − x 2 = 2.11 m s. d) Taking t = 0 at the time when the cowboy leaves, the position of the saddle as a function of time is given by Eq. g (13.13), with cos φ = − 2 ; this is checked by setting t = 0 and finding that ω A g a x = ω 2 = − ω 2 . The cowboy’s position is xc = x0 + v0t − ( g 2)t 2 . Finding the time at which the cowboy and the saddle are again in contact involves a transcendental equation which must be solved numerically; specifically, (0.110 m) + (2.11m s)t − (4.90 m s 2 )t 2 = (0.25 m) cos ((9.42 rad s)t − 1.11 rad), which has as its least non-zero solution t = 0.538 s. e) The speed of the saddle is (−2.36 m s) sin (ω t + φ) = 1.72 m s , and the cowboy’s speed is (2.11 m s) − (9.80 m s 2 ) × (0.538 s) = −3.16 m s, giving a relative speed of 4.87 m s (extra figures were kept in the intermediate calculations). 13.63: The maximum acceleration of both blocks, assuming that the top block does not slip, is amax = kA (m + M ), and so the maximum force on the top block is m ( m + M ) kA = μs mg , and so the maximum amplitude is Amax = μs (m + M ) g k.
2

13.64: (a) Momentum conservation during the collision: mv0 = (2m)V 1 1 V = v0 = (2.00 m s) = 1.00 m s 2 2 Energy conservation after the collision: 1 1 MV 2 = kx 2 2 2 x= MV 2 (20.0 kg)(1.00 m s) 2 = = 0.500 m (amplitude ) k 80.0 N m

ω = 2πf = k M f = 1 1 80.0 N m k M = = 0.318 Hz 2π 2π 20.0 kg 1 1 = = 3.14 s f 0.318 Hz

T=

(b) It takes 1 2 period to first return: 1 (3.14 s) = 1.57 s 2

13.65:

a) m → m 2 Splits at x = 0 where energy is all kinetic energy, E = 1 mv 2 , so E → E 2 2 k stays same

E = 1 kA2 so A = 2 E k 2 Then E → E 2 means A → A 2

T = 2π m k so m → m 2 means T → T 2 b) m → m 2 Splits at x = A where all the energy is potential energy in the spring, so E doesn’t change. E = 1 kA2 so A stays the same. 2 T = 2π m k so T → T 2, as in part (a). c) In example 13.5, the mass increased. This means that T increases rather than decreases. When the mass is added at x = 0, the energy and amplitude change. When the mass is added at x = ± A, the energy and amplitude remain the same. This is the same as in this problem. 13.66: a)

For space considerations, this figure is not precisely to the scale suggested in the problem. The following answers are found algebraically, to be used as a check on the graphical method. 2E 2(0.200 J) b) A= = = 0.200 m. k (10.0 N/m) c) v0 = −
E 4

= 0.050 J. d) If U = 1 E , x = 2 and x 0 =
2U 0 k

A 2

= 0.141 m. e) From Eq. (13.18), using

2 K0 m

, v − 0 = ω x0
2K0 m k m 2U 0 k

=

and φ = arctan

(

0.429 = 0.580 rad .

)

K0 = 0.429 U0

13.67: a) The quantity ∆l is the amount that the origin of coordinates has been moved from the unstretched length of the spring, so the spring is stretched a distance ∆l − x (see Fig. (13.16 ( c ))) and the elastic potential energy is U el = (1 2)k (∆l − x) 2 . 1 2 1 2 kx + ( ∆ l ) − k∆ lx + mgx − mgx0 . 2 2 Since ∆l = mg k , the two terms proportional to x cancel, and b) U = U el + mg ( x − x0 ) = U= 1 2 1 2 kx + k ( ∆l ) − mgx0 . 2 2

c) An additive constant to the mechanical energy does not change the dependence of the force on x, Fx = − dU , and so the relations expressing Newton’s laws and the dx resulting equations of motion are unchanged. The “spring constant” for this wire is k = f = 1 2π k 1 = m 2π g 1 = ∆l 2π
mg ∆l

13.68:

, so

9.80 m s 2 = 11.1 Hz. 2.00 × 10− 3 m

πA 13.69: a) 2T = 0.150 m s. b) a = −( 2π T ) x = −0.112 m s 2 . The time to go from equilibrium to half the amplitude is sin ωt = (1 2) , or ωt = π 6 rad, or one-twelfth of a period. The needed time is twice this, or one-sixth of a period, 0.70 s. d) ∆l = mg = ωg2 = ( 2 πgr ) 2 = 4.38 m. k 2

13.70: Expressing Eq. (13.13) in terms of the frequency, and with φ = 0, and taking two derivatives,  2πt  x = ( 0.240 m ) cos   1.50 s      2π ( 0.240 m )   2πt   2πt  v x = −  (1.50 s )  sin  1.50 s  = −(1.00530 m s ) sin  1.50 s            
2

 2π   2πt   2πt  2 ax = −  1.50 s  ( 0.240 m ) cos 1.50 s  = − 4.2110 m s cos 1.50 s .           

(

)

a) Substitution gives x = −0.120 m, or using t = T gives x = A cos 120° = 3 b) Substitution gives ma x = +( 0.0200 kg ) 2.106 m s 2 = 4.21 × 10 −2 N, in the + x - direction.

−A 2

.

A c) t = 2Tπ arccos −3A 4 = 0.577 s. d) Using the time found in part (c) , v = 0.665 m s (Eq.(13.22) of course gives the same result).

(

)

(

)

13.71: a) For the totally inelastic collision, the final speed v in terms of the initial speed V = 2 gh is v =V
2 v0

M m+M

= 2 9.80 m s 2 ( 0.40 m ) ( 2..2 ) = 2.57 m s, or 2.6 m s to two figures. b) When 24
Mg k

(

)

the steak hits, the pan is
ω2

above the new equilibrium position. The ratio and so the amplitude of oscillation is

is v

2

( k ( m + M ) ) = 2 ghM 2 k ( m + M ) ,
2

2 ghM 2  Mg  A=   +  k  k(m + M )  ( 2.2 kg ) 9.80 m/s 2 =   ( 400 N m )  = 0.206 m.

(

) 

2

 + 

2(9.80 m s 2 )(0.40 m)(2.2 kg) 2 (400 N m)(2.4 kg)

(This avoids the intermediate calculation of the speed.) c) Using the total mass, T = 2π (m + M ) k = 0.487 s.

k 13.72: f = 0.600 Hz, m = 400 kg; f = 1 m gives k = 5685 N/m. 2 This is the effective force constant of the two springs. a) After the gravel sack falls off, the remaining mass attached to the springs is 225 kg. The force constant of the springs is unaffected, so f = 0.800 Hz. To find the new amplitude use energy considerations to find the distance downward that the beam travels after the gravel falls off. Before the sack falls off, the amount x0 that the spring is stretched at equilibrium is

given by mg − kx 0 , so x 0 = mg k = ( 400 kg ) 9.80 m/s 2 ( 5685 N/m ) = 0.6895 m. The maximum upward displacement of the beam is A = 0.400 m. above this point, so at this point the spring is stretched 0.2895 m.

(

)

With the new mass, the mass 225 kg of the beam alone, at equilibrium the spring is stretched mg k = (225 kg) (9.80 m s 2 ) (5685 N m) = 0.6895 m. The new amplitude is therefore 0.3879 m − 0.2895 m = 0.098 m. The beam moves 0.098 m above and below the new equilibrium position. Energy calculations show that v = 0 when the beam is 0.098 m above and below the equilibrium point. b) The remaining mass and the spring constant is the same in part (a), so the new frequency is again 0.800 Hz. The sack falls off when the spring is stretched 0.6895 m. And the speed of the beam at this point is v = A k m = ( 0.400 m ) = ( 5685 N/m ) ( 400 kg ) = 1.508 m/s. . Take y = 0 at this point. The total energy of the beam at this point, just after the sack falls off, is 2 E = K + U el + U g = 1 ( 225 kg ) 1.508 m/s 2 + 1 ( 5695 N/m )( 0.6895 m ) + 0 = 1608 J. Let this 2 2 be point 1. Let point 2 be where the beam has moved upward a distance d and where 2 v = 0 . E2 = 1 k ( 0.6985 m − d ) + mgd . E1 = E2 gives d = 0.7275 m . At this end point of 2 motion the spring is compressed 0.7275 m – 0.6895 m =0.0380 m. At the new equilibrium position the spring is stretched 0.3879 m, so the new amplitude is 0.3789 m + 0.0380 m = 0.426 m. Energy calculations show that v is also zero when the beam is 0.426 m below the equilibrium position.

(

)

13.73: The pendulum swings through 1 cycle in 1.42 s, so T = 2.84 s. L = 1.85 m. 2 Use T to find g: 2 T = 2π L g so g = L( 2π T ) = 9.055 m/s 2 Use g to find the mass M p of Newtonia:
2 g = GM p / Rp

2πRp = 5.14 × 107 m, so Rp = 8.18 × 106 m mp =
2 gRp

G

= 9.08 × 1024 kg

13.74:

a) Solving Eq. (13.12) for m , and using k =
2 2

F ∆l

 T  F  1  40.0 N m=  =  = 4.05 kg.  2π  ∆l  2π  0.250 m b) t = (0.35)T , and so x = − Asin2 π (0.35) = −0.0405 m. Since t > T , the mass has 4 already passed the lowest point of its motion, and is on the way up. c) Taking upward forces to be positive, Fspring − mg = −kx, where x is the displacement from equilibrium , so Fspring = −(160 N m)( −0.030 m) + (4.05 kg)(9.80 m/s 2 ) = 44.5 N. 13.75: Of the many ways to find the time interval, a convenient method is to take 1 φ = 0 in Eq. (13.13) and find that for x = A 2, cos ω t = cos(2πt / T ) = 2 and so t = T / 6 . The time interval available is from − t to t , and T / 3 = 1.17 s. 13.76: See Problem 12.84; using x as the variable instead of r , dU GM E m GM E g F ( x) = − =− x, so ω2 = = . 3 3 dx RE RE RE The period is then T= or 84.5 min. 2π R 6.38 × 106 m = 2π E = 2π = 5070 s, ω g 9.80 m/s 2

13.77:

Take only the positive root (to get the least time), so that dx = dt dx A −x
2 2

k A2 − x 2 , or m k dt m k t1 dt = m ∫0 k t1 m k (t1 ) m

= =

∫

A

dx A2 − x 2

0

arcsin(1) =

π k = t1 , 2 m where the integral was taken from Appendix C. The above may be rearranged to show that t1 = π k = T , which is expected. 4
2 m

13.78:

a)

U = − ∫ F dx = c ∫ x 3 dx =
0 0

x

x

c 4 x. 4

a) From conservation of energy, 1 mv 2 = c ( A4 − x 4 ) , and using the technique of Problem 2 4 13.77, the separated equation is dx c = dt. 2m A4 − x 4 Integrating from 0 to A with respect to x and from 0 to T 4 with respect to t ,

∫
0

A

dx A −x
4 4

=

c T . 2m 4

x To use the hint, let u = A , so that dx = a du and the upper limit of the u − integral is

u = 1. Factoring A 2 out of the square root,

1 du 1.31 c ∫ 1 − u 4 = A = 32m T , A0 which may be expressed as T = motion is not simple harmonic.
7.41 A m c

1

. c) The period does depend on amplitude, and the

13.79: As shown in Fig. (13.5( b ) ) , v = −vtan sin θ. With vtan = Aω and θ = ωt + φ, this is Eq. (13.15). 13.80: a) Taking positive displacements and forces to be upwad, 2 n − mg = ma, a = −( 2πf ) x, so n = m g − ( 2πf ) A cos( ( 2πf ) t + φ ) .
2

(

)

a) The fact that the ball bounces means that the ball is no longer in contact with the lens, and that the normal force goes to zero periodically. This occurs when the amplitude of the acceleration is equal to g , or when g = ( 2πf b ) A.
2

13.81: a) For the center of mass to be at rest, the total momentum must be zero, so the momentum vectors must be of equal magnitude but opposite directions, and the momenta   can be represented as p and − p. b) K tot = 2 × p2 p2 = . 2m 2( m 2)

c) The argument of part (a) is valid for any masses. The kinetic energy is K tot = p2 p2 p 2  m1 + m2  + = 2m1 2m2 2  m1 m2   p2 =  2( m m ( m + m ) ) . 1 2 1 2 

13.82: a)

Fr = −

 R07  1  dU = Α  9  − 2 . dr  r  r 

b) Setting the above expression for Fr equal to zero, the term in square brackets 7 1 7 vanishes, so that R90 = 2 , or R0 = r 7 , and r = R0 . r r c) U ( R0 ) = − 7Α = −7.57 × 10 −19 J. 8 R0

d) The above expression for Fr can be expressed as A Fr = 2 R0 = ≈ =  r  −9  r  −2    −      R   R0   0   

A (1 + ( x R0 ) ) −9 − (1 + ( x R0 ) ) − 2 2 R0 A [ (1 − 9( x R0 ) ) − (1 − 2( x R0 ) ) ] R02 A ( − 7 x R0 ) R02

[

]

 7A = − 3  x. R   0

e)

f =

1 1 k m= 2π 2π

7A = 8.39 × 1012 Ηz. 3 R0 m

13.83: a)

1  dU 1 = A 2 − . 2  dx ( r − 2 R0 )  r   b) Setting the term in square brackets equal to zero, and ignoring solutions with r < 0 or r > 2 R0 , r = 2 R0 − r , or r = R0 . c) The above expression for Fr may be written as Fr = − A Fr = 2 R0 = ≈
−2  r  −2  r     −  − 2     R   R0   0    

A (1 + ( x R0 ) ) − 2 − (1 − ( x R0 ) ) − 2 2 R0

[

]

A [ (1 − 2( x R0 ) ) − (1 − ( − 2) ( x R0 ) ) ] R02

 4A  = −  3  x, R   0 corresponding to a force constant of k = 4 A R0 . d) The frequency of small oscillations would be f = (1 2π ) k m = (1 π ) A mR0 . 13.84: a) As the mass approaches the origin, the motion is that of a mass attached to a spring of spring constant k, and the time to reach the origin is π m k . After passing 2 through the origin, the motion is that of a mass attached to a spring of spring constant 2k and the time it takes to reach the other extreme of the motions is π m 2k . The period is 2 twice the sum of these times, or T = π
m k 3 3

(1 + ) . The period does not depend on the
1 2

amplitude, but the motion is not simple harmonic. B) From conservation of energy, if the negative extreme is A′, 1 kA2 = 1 (2k ) A′2 , so A′ = − A2 ; the motion is not symmetric about 2 2 the origin.

13.85: There are many equivalent ways to find the period of this oscillation. Energy considerations give an elegant result. Using the force and torque equations, taking torques about the contact point, saves a few intermediate steps. Following the hint, take torques about the cylinder axis, with positive torques counterclockwise; the direction of positive rotation is then such that α = Ra , and the friction force f that causes this torque acts in the –x-direction. The equations to solve are then Max = − f − kx, Which are solved for ax = kx k =− x, 2 M +I R (3 2) M fR = I cm α, a = Rα,

where I = I cm = (1 2) MR 2 has been used for the combination of cylinders. Comparison with Eq. (13.8) gives T =
2π ω

= 2π 3M 2k .

13.86: Energy conservation during downward swing: m2 gh0 = 1 m2v 2 2 v = 2 gh0 = 2(9.8 m s )(0.100 m) = 1.40 m s Momentum conservation during collision: m2v = (m2 + m3 )V V= m2v (2.00 kg )(1.40 m s) = = 0.560 m s m2 + m3 5.00 kg
2

Energy conservation during upward swing: 1 MV 2 2 (0.560 m s) 2 hf = V 2 2 g = = 0.0160 m = 1.60 cm 2(9.80 m s 2 ) Mghf =

cos θ =

48.4 cm 50.0 cm θ = 14.5°

1 g 1 9.80 m s 2 f = = = 0.705 Hz 2π l 2π 0.500 m

13.87:

T = 2π I mgd , m = 3M d = ycg = m1 y1 + m2 y2 m1 + m2

d=

2M ( [1.55 m] 2 ) + M (1.55 m + [1.55 m] 2 ) = 1.292 m 3M

I + I1 + I 2 I1 =
1 3

( 2M )(1.55 m ) 2 = (1.602 m 2 ) M
2

1 I 2, cm = 12 M (1.55 m )

The parallel-axis theorem (Eq. 9.19) gives 2 I 2 = I 2, cm + M (1.55 m + [1.55 m ] 2) = 5.06 m 2 M I = I1 + I 2 = 7.208 m 2 M Then T = 2π I mgd = 2π

(

)

(

)

( 7.208 m ) M = 2.74 s. ( 3M ) (9.80 m s )(1.292 m )
2 2

This is smaller than T = 2.9 s found in Example 13.10.

13.88: The torque on the rod about the pivot (with angles positive in the direction indicated in the figure) is τ = −( k L θ ) L . Setting this equal to the rate of change of angular 2 2 momentum, Iα = I
d 2θ dt 2

, d 2θ L2 4 3k = −k θ = − θ, 2 dt I M

1 where the moment of inertia for a slender rod about its center, I = 12 ML2 has been used.

It follows that ω2 =

3K M

, and T =

2π ω

= 2π

M 3k

.

13.89: The period of the simple pendulum (the clapper) must be the same as that of the bell; equating the expression in Eq. (13.34) to that in Eq. (13.39) and solving for L gives L = Ι md = (18.0 kg ⋅ m 2 ) ((34.0 kg)(0.60 m)) = 0.882 m. Note that the mass of the bell, not the clapper, is used. As with any simple pendulum, the period of small oscillations of the clapper is independent of its mass. 13.90: The moment of inertia about the pivot is 2(1 3) ML2 = (2 3) ML2 , and the center of gravity when balanced is a distance d = L (2 2 ) below the pivot (see Problem 8.95). From Eq. (13.39), the frequency is f = 1 1 = T 2π 3g 1 = 4 2 L 4π 3g ⋅ 2L

13.91: a) L = g (T 2π ) 2 = 3.97 m. b)There are many possibilities. One is to have a uniform thin rod pivoted about an axis perpendicular to the rod a distance d from its center. Using the desired period in Eq. (13.39) gives a quadratic in d, and using the maximum size for the length of the rod gives a pivot point a distance of 5.25 mm, which is on the edge of practicality. Using a “dumbbell,” two spheres separated by a light rod of length L gives a slight improvement to d=1.6 cm (neglecting the radii of the spheres in comparison to the length of the rod; see Problem 13.94).

k 13.92: Using the notation 2bm = γ, m = ω2 and taking derivatives of Eq. (13.42) (setting the phase angle φ = 0 does not affect the result),

x = Αe − γt cosω′ t vx = − Αe − γt (ω′ sin ω′ t + γ cos ω′t ) ax = − Αe − γt ((ω′ 2 − γ 2 ) cos ω′ t − 2ω′ γ sin ω′ t ). Using these expression in the left side of Eq. (13.41), − kx − bvx = Αe − γt ( − k cos ω′ t + (2γ m)ω′ sin ω′t + 2mγ 2 cos ω′ t ) = mΑ e − γt ((2γ 2 − ω2 ) cos ω′ t + 2γω′ sin ω′t ). The factor ( 2γ 2 − ω2 ) is γ 2 − ω′2 (this is Eq. (13.43)), and so − kx − bvx = mΑ e − γt ((γ 2 − ω′ 2 ) cos ω′ t + 2γω′ sin ω′ t ) = max . 13.93: a) In Eq. (13.38), d=x and from the parallel axis theorem, gx I = m( L2 12 + x 2 ) , so ω2 = ( L2 12 )+ x 2 . b) Differentiating the ratio ω2 g = respect to x and setting the result equal to zero gives 1 2x2 = , or 2 x 2 = x 2 + L2 12, 2 2 2 2 2 ( L 12) + x (( L 12) + x ) Which is solved for x = L 12. ω2 L 12 6 3 = = = 2 g 2 L 12 L 12 L,

x ( L2 12 ) + x 2

with

c) When x is the value that maximizes ω the ratio so the length is L = 3g = 0.430 m. ω2

(

)

13.94: a) From the parellel axis theorem, the moment of inertia about the pivot point is M L2 + ( 2 5) R 2 .

(

)

Using this in Eq. (13.39), With d = L gives. L2 + ( 2 5) R 2 L T = 2π = 2π 1 + 2 R 2 5 L2 = Tsp 1 + 2 R 2 5 L2 . gL g b) Letting 1 + 2 R 2 5 L2 = 1.001 and solving for the ratio L R (or approximating the
L R

square root as 1 + R 2 5L2 ) gives c) (14.1) (1.270 cm ) = 18.0 cm.

= 14.1.

13.95: a) The net force on the block at equilibrium is zero, and so one spring (the one with k1 = 2.00 Ν m ) must be stretched three times as much as the one with k 2 = 6.00 Ν m . The sum of the elongations is 0.200 m, and so one spring stretches 0.150 m and the other stretches 0.050 m, and so the equilibrium lengths are 0.350 m and 0.250 m. b) There are many ways to approach this problem, all of which of course lead to the result of Problem 13.96(b). The most direct way is to let ∆x1 = 0.150 m and x2 = 0.050 m, the results of part (a). When the block in Fig.(13.35) is displaced a distance x to the right, the net force on the block is − k1 ( ∆x1 + x ) + k 2 ( ∆x 2 − x ) = [ k1 ∆x1 − k 2 ∆x 2 ] − ( k1 + k 2 ) x. From the result of part (a), the term in square brackets is zero, and so the net force is − ( k1 + k 2 ) x, the effective spring constant is k eff = k1 + k 2 and the period of vibration is T = 2π
0.100 kg 8.00 Ν m

= 0.702 s.

13.96: In each situation, imagine the mass moves a distance ∆x, the springs move distances ∆x1 and ∆x 2 , with forces F1 = −k1∆x1 , F2 = − k2 ∆x2 . a) ∆x = ∆x1 = ∆x2 , F = F1 + F2 = −( k1 + k2 ) ∆x, so keff = k1 + k 2 . b) Despite the orientation of the springs, and the fact that one will be compressed when the other is extended, ∆x = ∆x1 + ∆x2 , and the above result is still valid; k eff = k1 + k 2 . c) For massless springs, the force on the block must be equal to the tension in any point F F of the spring combination, and F = F1 = F2 , and so ∆x1 = − , ∆x2 = − , and k1 k2 1 1 k + k2 ∆x = − +  F = − 1 F k k  k1k 2  1 2  = κκ1+κκ22 . d) The result of part (c) shows that when a spring is cut in half, the 1 2.

and κeff

effective spring constant doubles, and so the frequency increases by a factor of 13.97: a) Using the hint, 1 ∆g   T + ∆ T ≈ 2π L  g −1 2 − g −3 2∆ g  = T − T ,   2 2g

so ∆T = −(1 2 )( T g ) ∆g . This result can also be obtained from T 2 g = 4π 2 L, from which

( 2T∆T ) g + T 2 ∆g = 0.

Therefore,

∆T T

= − 1 ∆gg . b) The clock runs slow; ∆T > 0, ∆g < 0 2

 2∆T  2 and g + ∆g = g 1 −  = 9.80 m s T  

(

)

 2( 4.00 s )  2 1 −  ( 86,400 s )  = 9.7991 m s .   

13.98: Denote the position of a piece of the spring by l ; l = 0 is the fixed point and l = L is the moving end of the spring. Then the velocity of the point corresponding to l , l denoted u , is u ( l ) = v (when the spring is moving, l will be a function of time, and so L u is an implicit function of time). a) dm = M dl , and so L 1 1 Mv 2 2 2 dK = dm u = l dl , 2 2 L3 and K =∫ Mv 2 2 Mv 2 dK = l dl = . 2 L3 ∫ 6 0
M 3 L

b) mv dv + kx dx = 0, or ma + kx = 0, which is Eq. (13.4). c) m is replaced by dt dt ω=
3k M

, so

and M ′ =

M 3

.

13.99: a) With I = (1 3) ML2 and d = L 2 in Eq. (13.39 ) , T0 = 2π 2 L 3g . With the addedmass, I = M L2 3 + y 2 , m = 2 M and d = ( L 4 ) + y 2, T = 2π ×

(L

2

3 + y2

) ( g( L 2 + y) )

((

)

)

and r= T = T0 L2 + 3 y 2 . L2 + 2 yL

b) From the expression found in part a), T = T0 when y = 2 L. At this point, a simple 3 pendulum with length y would have the same period as the meter stick without the added mass; the two bodies oscillate with the same period and do not affect the other’s motion.

13.100: Let the two distances from the center of mass be d1 and d 2 . There are then two 2 relations of the form of Eq. (13.39); with I1 = I cm + md12 and I 2 = I cm + md 2 , these relations may be rewritten as mgd1T 2 = 4π 2 I cm + md12 mgd 2T Subtracting the expressions gives
2 mg ( d1 − d 2 )T 2 = 4π 2 m d12 − d 2 = 4π 2 m( d1 − d 2 ) ( d1 + d 2 ) , and dividing by the common factor of m( d1 − d 2 ) and letting d 1 + d 2 = L gives the desired result. 2 2 cm

( = 4π ( I )

) + md ).
2 2

(

13.101: a) The spring, when stretched, provides an inward force; using ω′2 l for the magnitude of the inward radial acceleration, kl0 mω′l = k ( l − l0 ) , or l = . k − mω′2 b) The spring will tend to become unboundedly long. 13.102: Let r = R0 + x, so that r − R0 = x and F = A[e −2bx − e − bx ]. When x is small compared to b −1 , expanding the exponential function gives F ≈ A [ (1 − 2bx ) − (1 − bx ) ] = − Abx, corresponding to a force constant of Ab = 579.2 N m or 579 N m to three figures. This is close to the value given in Exercise 13.40.

Chapter 14

14.1:

w = mg = ρVg

= 7.8 × 103 kg m 3 ( 0.858 m ) π 1.43 × 10−2 m 9.80 m s 2 = 41.8 N or 42 N to two places. A cart is not necessary.
2

(

)

(

)(

)

14.2:

ρ=

m m 7.35 × 1022 kg = 4 3= = 3.33 × 103 kg m 3 . 3 6 4 V 3 πr π 1.74 × 10 m 3

(

(

) )

14.3:

m ρ=V =

( 0.0158 kg ) ( 5.0×15.0×30.0 ) mm 3

= 7.02 × 103 kg m . You were cheated.

3

14.4:

The length L of a side of the cube is 3  m 3  40.0 kg L =V =   =  ρ  21.4 × 10 3 kg m 3  = 12.3 cm.     
1 3 1 1

14.5:

Same mass means ra3 ρa = r13 ρ1 ( a = aluminum, l = lead ) ra  ρ1  =  r1  ρa   
13

m = ρV = 4 πr 3 ρ 3

 11.3 × 103  =  2.7 × 103    

13

= 1.6

14.6:

a)

D=

M sun 1.99 × 1030 kg 1.99 × 1030 kg = = 3 4 Vsun 1.412 × 1027 m 3 π 6.96 × 108 m 3

(

)

= 1.409 × 10 kg m 3
3

b)

1.99 × 1030 kg D= = = 0.594 × 1017 kg m 3 3 13 3 4 4 3.351 × 10 m π 2.00 × 10 m 3

1.99 × 1030 kg

(

)

= 5.94 × 1016 kg m 3 p − p0 = ρgh

14.7:

p − p0 1.00 × 105 Pa h= = = 9.91m ρg (1030 kg m 3 ) (9.80 m s 2 )

14.8: The pressure difference between the top and bottom of the tube must be at least 5980 Pa in order to force fluid into the vein: ρgh = 5980 Pa 5980 Pa 5980 N m 2 h= = = 0.581m gh (1050 kg m 3 ) (9.80 m s 2 ) a) ρgh = 600 kg m 3 9.80 m s 2 ( 0.12 m ) = 706 Pa.

14.9:

(

)(

)

b) 706 Pa + 1000 kg m 3 9.80 m s 2 ( 0.250 m ) = 3.16 × 103 Pa. 14.10: is a) The pressure used to find the area is the gauge pressure, and so the total area (16.5 × 10 3 N) = 805 cm 2 ⋅ 3 (205 × 10 Pa ) b) With the extra weight, repeating the above calculation gives 1250 cm 2 .

(

)(

)

14.11: a) ρgh = (1.03 × 103 kg m 3 )(9.80 m s )(250 m) = 2.52 × 106 Pa. b) The pressure difference is the gauge pressure, and the net force due to the water and the air is (2.52 × 106 Pa )(π (0.15 m) 2 ) = 1.78 × 105 N. p = ρgh = (1.00 × 103 kg m 3 )(9.80 m s 2 )(640 m) = 6.27 × 106 Pa = 61.9 atm. a) pa + ρgy2 = 980 × 102 Pa + (13.6 × 103 kg m3 )(9.80 m s 2 )(7.00 × 10−2 m) =

2

14.12:

14.13:

1.07 × 10 5 Pa. b) Repeating the calcultion with y = y 2 − y1 = 4.00 cm instead of y 2 gives 1.03 × 10 5 Pa. c) The absolute pressure is that found in part (b), 1.03 × 10 5 Pa. d) ( y2 − y1 ) ρg = 5.33 × 103 Pa (this is not the same as the difference between the results of parts (a) and (b) due to roundoff error).
14.14: ρgh = (1.00 × 103kg m3 )(9.80 m s 2 )(6.1 m) = 6.0 × 104 Pa.

14.15: With just the mercury, the gauge pressure at the bottom of the cylinder is p = p 0 + p m ghm⋅ With the water to a depth hw , the gauge pressure at the bottom of the cylinder is p = p0 + ρm ghm + pw ghw . If this is to be double the first value, then ρw ghw = ρm ghm. hw = hm ( ρm ρw ) = (0.0500 m)(13.6 ×103 1.00 × 103 ) = 0.680 m The volume of water is V = hA = (0.680 m)(12.0 × 10−4 m 2 ) = 8.16 × 10−4 m 3 = 816 cm3 14.16: a) Gauge pressure is the excess pressure above atmospheric pressure. The pressure difference between the surface of the water and the bottom is due to the weight of the water and is still 2500 Pa after the pressure increase above the surface. But the surface pressure increase is also transmitted to the fluid, making the total difference from atmospheric 2500 Pa+1500 Pa = 4000 Pa. b) The pressure due to the water alone is 2500 Pa = ρgh. Thus h= 2500 N m 2 = 0.255 m (1000 kg m3 ) (9.80 m s 2 )

To keep the bottom gauge pressure at 2500 Pa after the 1500 Pa increase at the surface, the pressure due to the water’s weight must be reduced to 1000 Pa: 1000 N m 2 h= = 0.102 m (1000 kg m 3 )(9.80 m s 2 ) Thus the water must be lowered by 0.255 m − 0.102 m = 0.153 m 14.17: The force is the difference between the upward force of the water and the downward forces of the air and the weight. The difference between the pressure inside and out is the gauge pressure, so F = ( ρgh) A − w = (1.03 × 103 ) (9.80 m s 2 ) (30 m) (0.75 m 2 ) − 300 N = 2.27 × 105 N.

14.18: 130 × 103 Pa + (1.00 × 103 kg m 3 )(3.71 m s 2 )(14.2 m) − 93 × 103 Pa (2.00 m 2 ) = 1.79 × 10 5 N.

[

]

14.19: The depth of the kerosene is the difference in pressure, divided by the product ρg = mg , V (16.4 × 103 N) (0.0700 m 2 ) − 2.01 × 105 Pa h= = 4.14 m. (205 kg)(9.80 m s 2 ) (0.250 m3 )

14.20:

F mg (1200 kg)(9.80 m s 2 ) p= = = = 1.66 × 105 Pa = 1.64 atm. 2 2 A π (d 2) π (0.15 m)

14.21:

The buoyant force must be equal to the total weight; ρwaterVg = ρiceVg + mg , so V= m 45.0 kg = = 0.563 m 3 , ρwater − ρice 1000 kg m 3 − 920 kg m3

or 0.56 m 3 to two figures. 14.22: The buoyant force is B = 17.50 N − 11.20 N = 6.30 N, and V= B ρwater g = (6.30 N) = 6.43 × 10− 4 m3 . 3 2 (1.00 × 10 kg m )(9.80 m s )
3

The density is ρ= m w g w  17.50  3 3 = = ρwater = (1.00 × 103 kg m3 )   = 2.78 × 10 kg m . V B ρwater g B  6.30 

14.23: a) The displaced fluid must weigh more than the object, so ρ < ρ fluid . b) If the ship does not leak, much of the water will be displaced by air or cargo, and the average density of the floating ship is less than that of water. c) Let the portion submerged have ρ V volume V, and the total volume be V0 . Then, ρVo = ρfluid V , so V0 = ρfluid ⋅ The fraction above the fluid is then 1 −
P Pfluid

. If p → 0, the entire object floats, and if ρ → ρ fluid , none

of the object is above the surface. d) Using the result of part (c), 1−

ρ ρ fluid

(0.042 kg) (5.0 × 4.0 × 3.0 × 10 -6 m 3 ) = 1− = 0.32 = 32%. 1030 kg m 3

14.24: a) B = ρwater gV = 1.00 × 103 kg m 3 9.80 m s 2 0.650 m 3 = 6370 N. − b) m = w = B g T = 6370 Nm900 N = 558 kg. g 9.80 s 2 c) (See Exercise 14.23.) If the submerged volume is V ′, V′ = w ρwater g and V′ w 5470 N = = = 0.859 = 85.9%. V ρwater gV 6370 N

(

)(

)(

)

14.25:

a) ρoil ghoil = 116 Pa.

b) 790 kg m3 ( 0.100 m ) + 1000 kg m3 ( 0.0150 m ) 9.80 m s 2 = 921 Pa. c) m=
2 w ( pbottom − ptop ) A ( 805 Pa ) ( 0.100 m ) = = = 0.822 kg. g g 9.80 m s 2

((

)

(

)

)(

)

(

)

The density of the block is p =

0.822 kg

density of the fluid displaced, ( 0.85) 790 kg m3 + ( 0.15) (1000 kg m3 ) .

( 0.10 m ) 3

(

kg = 822 m 3 . Note that is the same as the average

)

14.26:

a) Neglecting the density of the air, V= m w g w ( 89 N ) = = = = 3.36 × 10 − 3 m 3 , 2 3 3 ρ ρ gρ 9.80 m s 2.7 × 10 kg m

(

)(

)

or 3.4 × 10−3 m 3 to two figures.
 ρ b) T = w −B = w − gρwaterV = ω1 − water  ρaluminum   1.00   = ( 89 N ) 1 −   = 56.0 N.  2.7   

14.27: a) The pressure at the top of the block is p = p 0 + ρgh, where h is the depth of the top of the block below the surface. h is greater for block Β , so the pressure is greater at the top of block Β . b) B = ρ flVobj g . The blocks have the same volume Vobj so experience the same buoyant force. c) T − w + B = 0 so T = w −B.

w = ρVg . The object have the same V but ρ is larger for brass than for aluminum so w is larger for the brass block. B is the same for both, so T is larger for the brass block, block B.
14.28: The rock displaces a volume of water whose weight is 39.2 N - 28.4 N = 10.8 N.

The mass of this much water is thus 10.8 N 9.80 m s 2 = 1.102 kg and its volume, equal to the rock’s volume, is 1.102 kg = 1.102 × 10− 3 m 3 3 3 1.00 × 10 kg m The weight of unknown liquid displaced is 39.2 N − 18.6 N = 20.6 N, and its mass is 20.6 N 9.80 m s 2 = 2.102 kg. The liquid’s density is thus 2.102 kg 1.102 × 10 −3 m 3 = 1.91 × 103 kg m 3 , or roughly twice the density of water. 14.29: v1 Α1 = v2 Α2 , v2 = v1 ( Α1 Α2 ) Α1 = π (0.80 cm) 2 , Α2 = 20π (0.10 cm) 2 v2 = (3.0 m s) π (0.80)2 = 9.6 m s 20π (0.10) 2

14.30:

v2 = v1

Α1 (3.50 m s)(0.0700 m 2 ) 0.245 m 3 s = = ⋅ Α2 Α2 Α2

a) (i) Α2 = 0.1050 m 2 , v2 = 2.33 m s. (ii) Α2 = 0.047 m 2 , v2 = 5.21 m s. b) v1 Α1t = υ2 Α2t = (0.245 m 3 s) (3600 s) = 882 m 3 . dV dt (1.20 m 3 s) a) v = = = 16.98. A π (0.150 m) 2

14.31:

b) r2 = r1 v1 v2 = (dV dt ) πv2 = 0.317 m. 14.32: a) From the equation preceding Eq. (14.10), dividing by the time interval dt gives Eq. (14.12). b) The volume flow rate decreases by 1.50% (to two figures). 14.33: The hole is given as being “small,”and this may be taken to mean that the velocity of the seawater at the top of the tank is zero, and Eq. (14.18) gives v = 2( gy + ( p ρ)) = 2((9.80 m s 2 )(11.0 m) + (3.00)(1.013 × 10 5 Pa) (1.03 × 10 3 kg m )) = 28.4 m s. Note that y = 0 and p = pa were used at the bottom of the tank, so that p was the given gauge pressure at the top of the tank.
3

14.34:

a) From Eq. (14.18), v = 2 gh = 2(9.80 m s 2 )(14.0 m) = 16.6 m s.

b) vΑ = (16.57 m s)(π (0.30 × 10−2 m) 2 ) = 4.69 × 10−4 m3 s. Note that an extra figure was kept in the intermediate calculation. 14.35: The assumption may be taken to mean that v1 = 0 in Eq. (14.17). At the maximum height, v2 = 0, and using gauge pressure for p1 and p2 , p2 = 0 (the water is open to the atmosphere), p1 = ρgy2 = 1.47 × 105 Pa.

14.36:

Using v2 = 1 v1 in Eq. (14.17), 4  15   1 2 ρ(v12 − v2 ) + ρg ( y1 − y2 ) = p1 + ρ  υ12 + g ( y1 − y2 ) 2  32  

p2 = p1 +

 15  = 5.00 × 104 Pa + (1.00 × 103 kg m 3 )  (3.00 m s) 2 + (9.80 m s 2 )(11.0 m)   32  5 = 1.62 × 10 Pa. 14.37: Neglecting the thickness of the wing (so that y1 = y2 in Eq. (14.17)), the pressure 2 difference is ∆p = (1 2) ρ(v2 − v12 ) = 780 Pa. The net upward force is then (780 Pa) × (16.2 m 2 ) − (1340 kg)(9.80 m s 2 ) = −496 N.

14.38:
0.355 kg 1.30 kg s 1000 kg m 3

( 0 355 a) ( 220 )60..0 s kg ) = 1.30 kg s. b) The density of the liquid is

0.355 ×10 −3 m 3

= 1000 kg m 3 , and so the volume flow rate is This result may also be obtained
−3 3

= 1.30 × 10−3 m 3 s = 1.30 L s.
60.0 s

from

( 220 )( 0.355 L )

= 1.30 L s. c) v1 = 1.230×10 −4m 2 s .00 ×10 m

= 6.50 m s, v2 = v1 4 = 1.63 m s. 1 2 d) p1 = p2 + ρ v2 − v12 + ρg ( y2 − y1 ) 2 2 2 = 152 kPa + (1 2) 1000 kg m 3 (1.63 m s ) − ( 6.50 m s )

(

+ 1000 kg = 119 kPa 14.39:

(

)( m ) ( 9.80 m s ) ( − 1.35 m ) (
3 2

)

)

The water is discharged at a rate of v1 =

4.65×10 −4 m 3 s 1.32×10 −3 m 2

= 0.352 m s. The pipe is

given as horizonatal, so the speed at the constriction is v2 = v12 + 2∆ p ρ = 8.95 m s, keeping an extra figure, so the cross-section are at the constriction is 4.65×10 −4 m 3 s = 5.19 × 10− 5 m 2 , and the radius is r = A π = 0.41 cm. 8.95 m s

14.40:

From Eq. (14.17), with y1 = y 2 , p2 = p1 + 1 1  v2  3 2 ρ v12 − v2 = p1 + ρ v12 − 1  = p1 + ρv12   2 2  4 8

(

)

3 2 1.00 × 10 3 kg m 3 ( 2.50 m s ) = 2.03 × 10 4 Pa, 8 v where the continutity relation v 2 = 1 has been used. 2 = 1.80 × 10 4 Pa + Let point 1 be where r1 = 4.00 cm and point 2 be where r2 = 2.00 cm. The volume flow rate has the value 7200 cm 3 s at all points in the pipe. v1 A1 = v1πr12 = 7200 cm3 , so v1 = 1.43 m s v2 A2 = v2πr22 = 7200 cm 3 , so v2 = 5.73 m s 1 1 2 p1 + ρgy1 + ρv12 = p2 + ρgy2 + ρv2 2 2 1 2 y1 = y2 and p2 = 2.40 × 105 Pa, so p2 = p1 + ρ v12 − v2 = 2.25 × 105 Pa 2 14.41:

(

)

(

)

14.42:

F = ( p0 − p ) π D4 . b) The force on each hemisphere due to the atmosphere is
2

a) The cross-sectional area presented by a sphere is π D4 , therefore

2

π 5.00 × 10− 2 m

(

) (1.013 × 10
2

5

Pa ( 0.975) = 776 Ν.

)

14.43: a) ρgh = 1.03 × 103 kg m3 9.80 × m s 2 10.92 × 103 m = 1.10 × 108 Pa. b) The fractional change in volume is the negative of the fractional change in density. The density at that depth is then ρ = ρ0 (1 + k∆p ) = 1.03 × 103 kg m3 1 + 1.16 × 108 Pa 45.8 × 10−11 Pa −1 = 1.08 × 103 kg m 3 , A fractional increase of 5.0%. Note that to three figures, the gauge pressure and absolute pressure are the same.

(

)(

)(

)

(

)( (

)(

))

14.44:

a) The weight of the water is ρgV = 1.00 × 103 kg m 3 9.80 m s 2

(

)(

) ( ( 5.00 m) ( 4.0 m) ( 3.0 m) ) = 5.88 × 10

5

N,

or 5.9 × 10 5 N to two figures. b) Integration gives the expected result the force is what it would be if the pressure were uniform and equal to the pressure at the midpoint; F = ρgA d 2 = 1.00 × 103 kg m 3 9.80 m s 2

(

)(

) ( ( 4.0 m ) ( 3.0 m ) ) (1.50 m) = 1.76 × 10

5

N,

or 1.8 × 10 5 N to two figures.

14.45: Let the width be w and the depth at the bottom of the gate be H . The force on a strip of vertical thickness dh at a depth h is then dF = ρgh( wdh ) and the torque about the hinge is dτ = ρgwh ( h − H 2 ) dh; integrating from h = 0 to h = H gives τ = ρgωH 3 12 = 2.61 × 10 4 N ⋅ m. 14.46: a) See problem 14.45; the net force is ∫ dF from h = 0 to h = H , F = ρgωH 2 2 = ρgAH 2, where A = ωH . b) The torque on a strip of vertical thickness dh about the bottom is dτ = dF ( H − h ) = ρgwh ( H − h ) dh, and integrating from h = 0 to h = H gives τ = ρgwH 3 6 = ρgAH 2 6. c) The force depends on the width and the square of the depth, and the torque about the bottom depends on the width and the cube of the depth; the surface area of the lake does not affect either result (for a given width). The acceleration due to gravity on the planet is ∆p ∆p g= = m ρd V d and so the planet’s mass is gR 2 ∆pVR 2 M = = G mGd 14.47:

14.48:

The cylindrical rod has mass M , radius R, and length L with a density that is

proportional to the square of the distance from one end, ρ = Cx 2 . a) M = ∫ ρdV = ∫ Cx 2 dV . The volume element dV = πR 2 dx. Then the integral
L L becomes M = ∫ 0 Cx 2πR 2 dx. Integrating gives M = CπR 2 ∫ 0 x 2 dx = CπR 2

L3 . Solving 3

for C , C = 3M πR 2 L3 . b) The density at the x = L end is ρ = Cx 2 =

(

3M πR 2 L3

)( L ) = ( ). The denominator is
2 3M πR 2 L

just the total volume V , so ρ = 3M V , or three times the average density, M V . So the average density is one-third the density at the x = L end of the rod.

14.49: a) At r = 0, the model predicts ρ = A = 12,700 kg m 3 and at r = R, the model predicts ρ = A − BR = 12,700 kg m 3 − (1.50 × 10 −3 kg m 4 )(6.37 × 10 6 m) = 3.15 × 10 3 kg m 3 . b), c) R  AR3 BR 4   4πR 3   3BR  M = ∫ dm = 4π ∫ [ A − Br ]r 2 dr = 4π  −  =  3  A − 4    4     3  0  4π (6.37 × 106 m) 3   3(1.50 × 10−3 kg m 4 )(6.37 × 106 m)   12,700 kg m 3 − =    3 4    24 = 5.99 × 10 kg, which is within 0.36% of the earth’s mass. d) If m (r ) is used to denote the mass contained in a sphere of radius r , then g = Gm (r ) r 2 . Using the same integration as that in part (b), with an upper limit of r instead of R gives the result. e) g = 0 at r = 0, and g at r = R, g = Gm( R ) R 2 = (6.673 × 10−11 N ⋅ m 2 kg 2 ) (5.99 × 10 24 kg) (6.37 × 106 m) 2 = 9.85 m s 2 . f) dg  4πG  d  3Br 2   4πG   3Br  = Ar −    =  3  A − 2  ; dr  3  dr  4    

setting ths equal to zero gives r = 2 A 3B = 5.64 × 10 6 m , and at this radius  4πG  2 A    3   2 A  g =    A −   B   3  3B    4   3B   4πGA2 = 9B 4π (6.673 × 10−11 N ⋅ m 2 kg 2 ) (12,700 kg m 3 ) 2 = = 10.02 m s 2 . 9(1.50 × 10 − 3 kg m 4 )

14.50: a) Equation (14.4), with the radius r instead of height y, becomes dp = − ρg dr = − ρg s (r R )dr. This form shows that the pressure decreases with increasing radius. Integrating, with p = 0 at r = R, p=− ρg s R

∫ r dr =
R

r

ρg s R

∫

R

r

r dr =

ρgs 2 ( R − r 2 ). 2R
M V

b) Using the above expression with r = 0 and ρ = p (0) =

=

3M 4πR 3

,

3(5.97 × 1024 kg)(9.80 m s 2 ) = 1.71 × 1011 Pa. 8π (6.38 × 106 m) 2

c) While the same order of magnitude, this is not in very good agreement with the estimated value. In more realistic density models (see Problem 14.49 or Problem 9.99), the concentration of mass at lower radii leads to a higher pressure. 14.51: a) ρwater ghwater = (1.00 × 103 kg m 3 )(9.80 m s 2 )(15.0 × 10−2 m) = 1.47 × 103 Pa. b) The gauge pressure at a depth of 15.0 cm − h below the top of the mercury column must be that found in part (a); ρHg g (15.0 cm − h) = ρwater g (15.0 cm), which is solved for h = 13.9 cm. 14.52: Following the hint, F = ∫ ( ρgy )(2πR )dy = ρgπRh2
o h

where R and h are the radius and height of the tank (the fact that 2 R = h is more or less coincidental). Using the given numerical values gives F = 5.07 × 10 8 N. 14.53: For the barge to be completely submerged, the mass of water displaced would need to be ρwaterV = (1.00 × 103 kg m 3 )(22 × 40 × 12 m3 ) = 1.056 × 107 kg. The mass of the barge itself is (7.8 × 103 kg m 3 ) ((2(22 + 40) × 12 + 22 × 40) × 4.0 × 10 −2 m 3 ) = 7.39 × 105 kg, so the barge can hold 9.82 × 10 6 kg of coal. This mass of coal occupies a solid volume of 6.55 × 10 3 m 3 , which is less than the volume of the interior of the barge (1.06 × 10 4 m 3 ), but the coal must not be too loosely packed.

14.54: The difference between the densities must provide the “lift” of 5800 N (see Problem 14.59). The average density of the gases in the balloon is then ρave = 1.23 kg m3 − (5800 N) = 0.96 kg m 3 . 2 3 (9.80 m s )(2200 m )

14.55:

a) The submerged volume V ′ is

w ρ water g

, so

V ′ w ρwater g m (900 kg) = = = = 0.30 = 30% ⋅ V V ρwaterV (1.00 × 103 kg m3 ) (3.0 m3 ) b) As the car is about to sink, the weight of the water displaced is equal to the weight of the car plus the weight of the water inside the car. If the volume of water inside the car is V ′′ , Vρwater g = w + V ′′pwater g , or V ′′ w =1− = 1 − 0.30 = 0.70 = 70% ⋅ V Vpwater g

14.56: a) The volume displaced must be that which has the same weight and mass as 9 the ice, 1.00.70 gm 3 = 9.70 cm3 (note that the choice of the form for the density of water gm cm avoids conversion of units). b) No; when melted, it is as if the volume displaced by the 9.70 gm of melted ice displaces the same volume, and the water level does not change. c)
9.70 gm 1.05 gm cm 3

= 9.24 cm 3 ⋅ d) The melted water takes up more volume than the salt water

displaced, and so 0.46 cm3 flows over. A way of considering this situation (as a thought experiment only) is that the less dense water “floats” on the salt water, and as there is insufficient volume to contain the melted ice, some spills over.

14.57:

The total mass of the lead and wood must be the mass of the water displaced, or VPb ρ Pb + Vwood ρ
wood

= (VPb + Vwood ) ρ

water

;

solving for the volume VPb , ρ − ρ wood VPb = Vwood water ρ Pb − ρ water = (1.2 × 10− 2 m 3 ) 1.00 × 103 kg m 3 − 600 kg m3 11.3 × 103 kg m3 − 1.00 × 103 kg m 3

= 4.66 × 10−4 m 3 , which has a mass of 5.27 kg.

14.58:

, ρ fluid where ρ is the average density of the hydrometer (see Problem 14.23 or Problem 14.55), 1 . Thus, if two fluids are observed to have which can be expressed as ρ fluid = ρ 1− f 1 − f1 floating fraction f1 and f 2 , ρ2 = ρ1 . In this form, it’s clear that a larger f 2 1 − f2 corresponds to a larger density; more of the stem is above the fluid. Using 2 2 f1 = (8.00 (cm)(0.400) cm ) = 0.242, f 2 = ( 3.20 (cm)(0.400) cm ) = 0.097 gives 3 3 13.2 cm 13.2 cm ρalcohol = (0.839) ρwater = 839 kg m3 . 14.59: a) The “lift” is V ( ρair − ρH 2 ) g , from which V = 120,000 N = 11.0 × 10 3 m 3 . 3 2 (1.20 kg m − 0.0899 kg m )(9.80 m s )
3

The fraction f of the volume that floats above the fluid is f = 1 −

ρ

b) For the same volume, the “lift” would be different by the ratio of the density differences,  ρ − ρHe (120,000 N) air  ρair − ρH  2   = 11.2 × 10 4 N.  

This increase in lift is not worth the hazards associated with use of hydrogen.

M . ρA b) The buoyant force is ρgA( L + x) = Mg + F , and using the result of part (a) and F solving for x gives x = ρgA . 14.60: a) Archimedes’ principle states ρgLA = Mg , so L = c) The “spring constant,” that is, the proportionality between the displacement x and the applied force F, is k = ρgA, and the period of oscillation is T = 2π M M = 2π . k ρgA

14.61:

a) x =

w mg m ( 70.0 kg ) = = = = 0.107 m. 3 ρgA ρgA ρA 1.03 × 10 kg m 3 π ( 0.450 m ) 2

(

)

b) Note that in part (c) of Problem 14.60, M is the mass of the buoy, not the mass of the man, and A is the cross-section area of the buoy, not the amplitude. The period is then T = 2π

(1.03 × 10

( 950 kg )
3

kg m

3

) (9.80 m s ) π ( 0.450 m )
2

2

= 2.42 s.

14.62: To save some intermediate calculation, let the density, mass and volume of the life preserver be ρ 0 , m and v, and the same quantities for the person be ρ1 , M and V . Then, equating the buoyant force and the weight, and dividing out the common factor of g, ρwater ( ( 0.80 )V + v ) = ρ0v + ρ1V , Eliminating V in favor of ρ1 and M , and eliminating m in favor of ρ0 and v,   M ρ0v + M = ρwater  ( 0.80) + v .   ρ1  

Solving for ρ 0 ,

   1 M ρ0 =  ρwater  ( 0.80) + v − M      v ρ1     M ρ = ρwater − 1 − ( 0.80) water   v  ρ1   = 1.03 × 103 kg m 3 − = 732 kg m3 ⋅ 14.63: To the given precision, the density of air is negligible compared to that of brass, but not compared to that of the wood. The fact that the density of brass may not be known the three-figure precision does not matter; the mass of the brass is given to three figures. The weight of the brass is the difference between the weight of the wood and the buoyant force of the air on the wood, and canceling a common factor of g , Vwood ( ρwood − ρair ) = M brass, and M wood = ρwoodVwood = M brass  ρwood ρ = M brass 1 − air  ρwood − ρair ρwood 
−1

75.0 kg 0.400 m 3

 1.03 × 103 kg m 3  1 − (0.80)   980 kg m3   

   

−1

 1.20 kg m 3  = (0.0950 kg )1 −  150 kg m 3  = 0.0958 kg.   

14.64: The buoyant force on the mass A, divided by g , must be 7.50 kg − 1.00 kg − 1.80 kg = 4.70 kg (see Example 14.6), so the mass block is 4.70 kg + 3.50 kg = 8.20 kg. a) The mass of the liquid displaced by the block is

4.70 kg, so the density of the liquid is

4.70 kg 3.80×10 - 3 m 3

= 1.24 × 10 3 kg m 3 . b) Scale D will read

the mass of the block, 8.20 kg, as found above. Scale E will read the sum of the masses of the beaker and liquid, 2.80 kg. 14.65: Neglecting the buoyancy of the air, the weight in air is g ( ρAuVAu + ρA1VA1 ) = 45.0 N. and the buoyant force when suspended in water is ρwater (VAu + VA1 ) g = 45.0 N − 39.0 N = 6.0 N. These are two equations in the two unknowns VAu and VA1. Multiplying the second by ρ A1 and the first by ρ water and subtracting to eliminate the VA1 term gives ρwaterVAu g ( ρAu − ρA1 ) = ρwater (45.0 N) − ρA1 (6.0 N) ρAu wAu = ρAu gVAu = ( ρwater ( 45.0 N) − ρAu (6.0)) ρwater ( ρAu − ρA1 ) (19.3) = ((1.00)(45.0 N) − (2.7)(6.0 N)) (1.00)(19.3 − 2.7) = 33.5 N. Note that in the numerical determination of wAu , specific gravities were used instead of densities.

14.66:

The ball’s volume is

4 4 V = πr 3 = π (12.0 cm) 3 = 7238 cm3 3 3 As it floats, it displaces a weight of water equal to its weight. a) By pushing the ball under water, you displace an additional amount of water equal to 84% of the ball’s volume or (0.84)(7238 cm 3 ) = 6080 cm 3 . This much water has a mass of 6080 g = 6.080 kg and weighs (6.080 kg)(9.80 m s 2 ) = 59.6 N, which is how hard you’ll have to push to submerge the ball. b) The upward force on the ball in excess of its own weight was found in part (a): 59.6 N. The ball’s mass is equal to the mass of water displaced when the ball is floating: (0.16)(7238 cm 3 )(1.00 g cm 3 ) = 1158 g = 1.158 kg, and its acceleration upon release is thus a= Fnet 59.6 N = = 51.5 m s 2 m 1.158 kg

14.67: a) The weight of the crown of its volume V is w = ρcrown gV , and when suspended the apparent weight is the difference between the weight and the buoyant force, fw = fρcrown gV = ( ρcrown − ρwater ) gV . Dividing by the common factors leads to − ρwater + ρcrown = fρcrown or ρcrown 1 = . ρwater 1 − f

As f → 0, the apparent weight approaches zero, which means the crown tends to float; from the above result, the specific gravity of the crown tends to 1. As f → 1, the apparent weight is the same as the weight, which means that the buoyant force is negligble compared to the weight, and the specific gravity of the crown is very large, as reflected in the above expression. b) Solving the above equations for f in terms of the ρ water specific gravity, f = 1 − ρcrown , and so the weight of the crown would be c) Approximating the average density by that of lead for a “thin” gold plate, the apparent weight would be (1 − (1 11.3) ) (12.9 N ) = 11.8 N.

(1 − (1 19.3) ) (12.9 N ) = 12.2 N.

14.68:

a) See problem 14.67. Replacing f with, respectively, wwater w and wfluid w gives ρsteel w ρ w = , steel = , ρfluid w - wfluid ρfluid w - wwater

and dividing the second of these by the first gives ρfluid w - wfluid = . ρwater w - wwater b) When wfluid is greater than wwater, the term on the right in the above expression is less than one, indicating that the fluids is less dense than water, and this is consistent with the buoyant force when suspended in liquid being less than that when suspended in water. If the density of the fluid is the same as that of water wfluid = wwater , as expected. Similarly, if wfluid is less than wwater , the term on the right in the above expression is greater than one, indicating the the fluid is denser than water. c) Writing the result of part (a) as ρfluid 1 − f fluid = . ρwater 1 − f water and solving for f fluid , f fluid = 1 − ρfluid (1 − f water ) = 1 − (1.220) ( 0.128) = 0.844 = 84.4%. ρwater

14.69: a) Let the total volume be V; neglecting the density of the air, the buoyant force in terms of the weight is  (w g )  B = ρwater gV = ρwater g  + V0 ,  ρ   m  or

V0 = b)
B ρ water g

B ρwater g

−

w ⋅ ρw g
B ρ water g

−

w ρ Cu g

= 2.52 × 10 m . Since the total volume of the casting is
3

−4

, the

cavities are 12.4% of the total volume.

14.70: a) Let d be the depth of the oil layer, h the depth that the cube is submerged in the water, and L be the length of a side of the cube. Then, setting the buoyant force equal to the weight, canceling the common factors of g and the cross-section area and supressing units, (1000)h + (750)d = (550) L.d , h and L are related by d + h + (0.35)L = L, so h = (0.65)L − d. Substitution into the first relation gives d = L (0.65)(100−0) − (550) = (1000) (750)
2L 5.00

= 0.040 m. b) The

gauge pressure at the lower face must be sufficient to support the block (the oil exerts only sideways forces directly on the block), and p = ρ wood gL = (550 kg m 3 )(9.80 m s 2 )(0.100 m) = 539 Pa. As a check, the gauge pressure, found from the depths and densities of the fluids, is ((0.040 m)(750 kg m 3 ) + (0.025 m)(1000 kg m 3 ))(9.80 m s 2 ) = 539 Pa. 14.71: The ship will rise; the total mass of water displaced by the barge-anchor combination must be the same, and when the anchor is dropped overboard, it displaces some water and so the barge itself displaces less water, and so rises. To find the amount the barge rises, let the original depth of the barge in the water be h0 = ( mb + ma ) ( ρ water A) , where mb and ma are the masses of the barge and the anchor, and A is the area of the bottom of the barge. When the anchor is dropped, the buoyant force on the barge is less than what it was by an amount equal to the buoyant force on the anchor; symbolically, h′ρ water Ag = h0 ρ water Ag − ( ma ρ steel ) ρ water g , which is solved for ∆h = h0 − h′ = or about 0.56 mm. ma =

ρ steel A

( 7860 kg

( 35.0 kg )
m 3 8.00 m 2

)(

) = 5.57 × 10

−4

m,

14.72: a) The average density of a filled barrel is 15.0 m ρ oil + V = 750 kg m 3 + 0.120 kg3 = 875 kg m 3 , which is less than the density of seawater, m so the barrel floats. b) The fraction that floats (see Problem 14.23) is 1−

ρ ave 875 kg m 3 = 1− = 0.150 = 15.0%. ρ water 1030 kg m 3

32.0 kg kg c) The average density is 910 m 3 + 0.120 kg3 = 1172 m 3 which means the barrel sinks. m

In order to lift it, a tension kg kg T = (1177 m 3 )(0.120 m 3 )(9.80 sm ) − (1030 m 3 )(0.120 m3 )(9.80 sm ) = 173 N is required. 2 2 14.73: a) See Exercise 14.23; the fraction of the volume that remains unsubmerged is 1 − ρB . b) Let the depth of the liquid be x and the depth of the water be y. Then ρL

ρLgx + ρwgy = ρ B gL and x + y = L. Therefore x = L − y and y =
78 y = 13..6 −1..0 (0.10 m) = 0.046 m. 13 6 −

( ρL − ρB ) L ρ L − ρω

. c)

∆V , A where A is the surface area of the water in the lock. ∆V is the volume of water that has the same weight as the metal, so 14.74: a) The change is height ∆y is related to the displaced volume ∆V by ∆y = ∆y = = ∆V w ρwater g w = = A A ρwater gA (2.50 × 106 N) = 0.213 m. (1.00x103 kg m3 )(9.80 m s 2 )((60.0 m)(20.0 m))

b) In this case, ∆V is the volume of the metal; in the above expression, ρ water is replaced by ρ metal = 9.00 ρ water , which gives ∆y′ = water sinks by this amount.
∆y 9

, and ∆y − ∆y′ = 8 ∆y = 0.189 m; the 9

14.75: a) Consider the fluid in the horizontal part of the tube. This fluid, with mass ρAl , is subject to a net force due to the pressure difference between the ends of the tube, which is the difference between the gauge pressures at the bottoms of the ends of the tubes. This difference is ρg ( yL − yR ), and the net force on the horizontal part of the fluid is

ρg ( yL − yR ) A = ρAla ,
or ( yL − yR ) = a l. g

b) Again consider the fluid in the horizontal part of the tube. As in part (a), the fluid is accelerating; the center of mass has a radial acceleration of magnitude a rad = ω 2 l 2, and so the difference in heights between the columns is (ω 2 l 2)(l g ) = ω 2 l 2 2 g . Anticipating Problem, 14.77, an equivalent way to do part (b) is to break the fluid in the horizontal part of the tube into elements of thickness dr; the pressure difference between the sides of this piece is dp = ρ (ω 2 r )dr (see Problem 14.78), and integrating from r = 0 to r = l gives ∆p = ρω 2 l 2 2, giving the same result. c) At any point, Newton’s second law gives dpA = pAdla from which the area A cancels out. Therefore the cross-sectional area does not affect the result, even if it varies. Integrating the above result from 0 to l gives ∆p = pal between the ends. This is related to the height of the columns through ∆p = pg∆y from which p cancels out.

14.76: a) The change in pressure with respect to the vertical distance supplies the force necessary to keep a fluid element in vertical equilibrium (opposing the weight). For the rotating fluid, the change in pressure with respect to radius supplies the force necessary to ∂ keep a fluid element accelerating toward the axis; specifically, dp = ∂ p dr = ρa dr , and p using a = ω2 r gives
∂p ∂p

= ρω2 r. b) Let the pressure at y = 0, r = 0 be p a (atmospheric
∂p ∂p

pressure); integrating the expression for

from part (a) gives

ρω2 2. p( r , y = 0 ) = pa + r 2

c) In Eq. (14.5), p2 = pa , p1 = p(r , y = 0) as found in part (b), y1 = 0 and y 2 = h(r ), the height of the liquid above the y = 0 plane. Using the result of part (b) gives h(r ) = ω 2 r 2 2 g . 14.77: a) The net inward force is ( p + dp ) A − pA = Adp, and the mass of the fluid element is ρAdr ′. Using Newton’s second law, with the inward radial acceleration of ω 2 r ' , gives dp = ρω2 r ′dr ′. b) Integrating the above expression,

∫

p

p0

dp = ∫ ρω2 r ′dr ′
r0

r

 ρω2  2 2 p − p0 =   2  r −r 0 ,    which is the desired result. c) Using the same reasoning as in Section 14.3 (and Problem 14.78), the net force on the object must be the same as that on a fluid element of the same shape. Such a fluid element is accelerating inward with an acceleration of magnitude ω2 Rcm , and so the force on the object is ρVω 2 Rcm . d) If ρR cm > ρob Rcmob, the inward

(

)

force is greater than that needed to keep the object moving in a circle with radius Rcmob at angular frequency ω , and the object moves inward. If ρRcm < ρob Rcmob, , the net force is insufficient to keep the object in the circular motion at that radius, and the object moves outward. e) Objects with lower densities will tend to move toward the center, and objects with higher densities will tend to move away from the center.

14.78: (Note that increasing x corresponds to moving toward the back of the car.) a) The mass of air in the volume element is ρdV = ρAdx , and the net force on the element in the forward direction is ( p + dp ) A − pA = Adp. From Newton’s second law, Adp = ( ρA dx)a, from which dp = ρadx. b) With ρ given to be constant, and with p = p0 at x = 0, p = p0 + ρax. c) Using ρ = 1.2 kg/m 3 in the result of part (b) gives

m3 5.0 m s 2 ( 2.5 m ) = 15.0 Pa ~ 15 × 10-5 patm , so the fractional pressure difference is negligble. d) Following the argument in Section 14-4, the force on the balloon must be the same as the force on the same volume of air; this force is the product of the mass ρV and the acceleration, or ρVa. e) The acceleration of the balloon is the force found in part (d) divided by the mass ρbalV , or ( ρ ρbal ) a. The acceleration relative to the car is the difference between this acceleration and the car’s acceleration, arel = [ ( ρ ρbal ) − 1] a. f) For a balloon filled with air, ( ρ ρbal ) < 1 (air balloons tend to sink in still air), and so the quantity in square brackets in the result of part (e) is negative; the balloon moves to the back of the car. For a helium balloon, the quantity in square brackets is positive, and the balloon moves to the front of the car. 14.79: If the block were uniform, the buoyant force would be along a line directed through its geometric center, and the fact that the center of gravity is not at the geometric center does not affect the buoyant force. This means that the torque about the geometric center is due to the offset of the center of gravity, and is equal to the product of the block’s weight and the horizontal displacement of the center of gravity from the geometric center, (0.075 m) 2. The block’s mass is half of its volume times the density of water, so the net torque is (0.30 m)3 (1000 kg m3 ) 0.075 m (9.80 m s 2 ) = 7.02 N ⋅ m, 2 2 or 7.0 N ⋅ m to two figures. Note that the buoyant force and the block’s weight form a couple, and the torque is the same about any axis. 14.80: a) As in Example 14.8, the speed of efflux is 2gh. After leaving the tank, the water is in free fall, and the time it takes any portion of the water to reach the ground is t =
2( H − h) g

(1.2 kg

)(

)

, in which time the water travels a horizontal distance

R = vt = 2 h( H − h). b) Note that if h′ = H − h, h′( H − h′ ) = ( H − h)h, and so h′ = H − h gives the same range. A hole H − h below the water surface is a distance h above the bottom of the tank.

14.81: The water will rise until the rate at which the water flows out of the hole is the rate at which water is added; A 2 gh = which is solved for
−4 3 1  dV dt  1  2.40 × 10 m s  h= =   1.50 × 10 −4 m 2  2(9.80 m s 2 ) = 13.1 cm.   A  2g   2 2

dV , dt

Note that the result is independent of the diameter of the bucket.

14.82: a) v3 A3 = 2 g ( y1 − y3 ) A3 = 2(9.80 m s 2 )(8.00 m) (0.0160 m 2 ) = 0.200 m 3 s . b) Since p3 is atmospheric, the gauge pressure at point 2 is
2 1 1 2   A3   8 2 2 1 −    = ρg ( y − y ), p2 = ρ (v 3 − v2 ) = ρv3 1 3   A2   9 2 2     using the expression for υ3 found above. Subsititution of numerical values gives

p2 = 6.97 × 10 4 Pa. 14.83: The pressure difference, neglecting the thickness of the wing, is 2 2 ∆p = (1 2) ρ(vtop − vbottom ), and solving for the speed on the top of the wing gives vtop = (120 m s) 2 + 2(2000 Pa) (1.20 kg m 3 ) = 133 m s . The pressure difference is comparable to that due to an altitude change of about 200 m, so ignoring the thickness of the wing is valid.

14.84:

a) Using the constancy of angular momentum, the product of the radius and  30  speed is constant, so the speed at the rim is about (200 km h)   = 17 km h. b) The  350  pressure is lower at the eye, by an amount  1m s  1 3 ∆p = (1.2 kg m 3 ) ((200 km h ) 2 − (17 km h) 2 )  3.6 km h  = 1.8 × 10 Pa.  2   = 160 m to two figures. d) The pressure at higher altitudes is even lower.
2

c)

v2 2g

14.85: The speed of efflux at point D is 2 gh1 , and so is 8gh1 at C. The gauge pressure at C is then ρgh1 − 4 ρgh1 = −3 ρgh1, and this is the gauge pressure at E. The height of the fluid in the column is 3h1 . a) v =

14.86:

dV dt A

, so the speeds are

6.00 × 10−3 m3 s 6.00 × 10−3 m 3 s = 6.00 m s and = 1.50 m s . 10.0 × 10− 4 m 2 40.0 × 10− 4 m 2
2 b) ∆p = 1 ρ (v12 − v 2 ) = 1.688 × 10 4 Pa, or 1.69 × 10 4 Pa to three figures. 2

c) ∆h =

∆p ρH g g

=

(1.688 ×10 4 Pa) (13.6×10 3 kg m 3 )( 9.80 m s 2 )

= 12.7 cm.

14.87:

a) The speed of the liquid as a function of the distance y that it has fallen is

2 v = v 0 + 2 gy , and the cross-section area of the flow is inversely proportional to this speed. The radius is then inversely proportional to the square root of the speed, and if the radius of the pipe is r0 , the radius r of the stream a distance y below the pipe is −1 4

r=

r0 v0
2 (v 0 + 2 gy )1 4

 2 gy  = r0 1 + 2   v0   

.

2 b) From the result of part (a), the height is found from (1 + 2 gy v0 )1 4 = 2, or 2 15v0 15(1.2 m s) 2 = = 1.10 m. 2g 2(9.80 m s 2 )

y=

14.88:

a) The volume V of the rock is V= B = w − T ((3.00 kg)(9.80 m s 2 ) − 21.0 N) = = 8.57 × 10 − 4 m 3 . ρ water g (1.00 × 103 kg m3 )(9.80 m s 2 )

ρ water g

In the accelerated frames, all of the quantities that depend on g (weights, buoyant forces, gauge pressures and hence tensions) may be replaced by g ′ = g + a, with the positive direction taken upward. Thus, the tension is T = mg ′ − B′ = (m − ρV ) g ′ = T0 g , g where T0 = 21.0 N. b) g ′ = g + a; for a = 2.50 m s 2 , T = (21.0 N) 9.80 + 2.50 = 26.4 N. 9.80 c) For a = −2.50 m s 2 , T = (21.0 N) 9.80.− 2.50 = 15.6 N. 9 80 d) If a = − g , g ′ = 0 and T = 0.
′

14.89:

a) The tension in the cord plus the weight must be equal to the buoyant force, so T = Vg ( ρ water − ρ foam ) = (1 2)(0.20 m) 2 (0.50 m)(9.80 m s 2 )(1000 kg m 3 − 180 kg m 3 ) = 80.4 N.

b) The depth of the bottom of the styrofoam is not given; let this depth be h0 . Denote the length of the piece of foam by L and the length of the two sides by l. The pressure force on the bottom of the foam is then ( p0 + ρgh0 ) L 2l and is directed up. The pressure on each side is not constant; the force can be found by integrating, or using the result of Problem 14.44 or Problem 14.46. Although these problems found forces on vertical surfaces, the result that the force is the product of the average pressure and the area is valid. The average pressure is p0 + ρg (h0 − (l (2 2 ))), and the force on one side has magnitude ( p0 + ρg (h0 − l (2 2 ))) Ll

( )

and is directed perpendicular to the side, at an angle of 45.0° from the vertical. The force on the other side has the same magnitude, but has a horizontal component that is opposite that of the other side. The horizontal component of the net buoyant force is zero, and the vertical component is B = ( p 0 + ρgh0 ) Ll 2 − 2(cos 45.0°)( p 0 + ρg (h0 − l (2 2 ))) Ll = ρg the weight of the water displaced. Ll 2 , 2

14.90: When the level of the water is a height y above the opening, the efflux speed is 2gy , and dV = π (d 2) 2 2 gy . As the tank drains, the height decreases, and dt

π (d 2) 2 2 gy dV dt dy d =− =− = −  2 gy . 2 dt A π ( D 2) D This is a separable differential equation, and the time T to drain the tank is found from
2

dy d = −  y  D which integrates to

2

2 g dt ,
2

[2 y ]
or

0 H

d = −   D

2 gT ,
2

D 2 H D T =  =  2g  d  d

2

2H . g

14.91: a) The fact that the water first moves upwards before leaving the siphon does not change the efflux speed, 2gh . b) Water will not flow if the absolute (not gauge) pressure would be negative. The hose is open to the atmosphere at the bottom, so the pressure at the top of the siphon is pa − ρg ( H + h), where the assumption that the crosssection area is constant has been used to equate the speed of the liquid at the top and bottom. Setting p = 0 and solving for H gives H = ( pa ρg ) − h. 14.92: Any bubbles will cause inaccuracies. At the bubble, the pressure at the surfaces of the water will be the same, but the levels need not be the same. The use of a hose as a level assumes that pressure is the same at all point that are at the same level, an assumption that is invalidated by the bubble.

Chapter 15

15.1: a) The period is twice the time to go from one extreme to the other, and v = f λ = λ T = (6.00 m) (5.0 s) = 1.20 m s, or 1.2 m s to two figures. b) The amplitude is half the total vertical distance, 0.310 m. c) The amplitude does not affect the wave speed; the new amplitude is 0.150 m. d) For the waves to exist, the water level cannot be level (horizontal), and the boat would tend to move along a wave toward the lower level, alternately in the direction of and opposed to the direction of the wave motion. 15.2: f = 15.3:

fλ=v
v 1500 m s = = 1.5 × 106 Hz λ 0.001 m

a) λ = v f = (344 m s) (784 Hz) = 0.439 m.

b) f = v λ = (344 m s) (6.55 × 10 −5 m) = 5.25 × 106 Hz. Denoting the speed of light by c, λ = c , and f
3.00 ×10 8 m s 540 ×10 Hz
3

15.4: a)

= 556 m.

3.00 ×10 b) 104.5×10 6 m s = 2.87 m. Hz

8

15.5: a) λ max = (344 m s) (20.0 Hz) = 17.2 m, λ min = (344 m s) (20,000 Hz) = 1.72 cm. b) λ max = (1480 m s) (20.0 Hz) = 74.0 m, λ min = (1480 m s) (20,000 Hz) = 74.0 mm. 15.6: c) f = Comparison with Eq. (15.4) gives a) 6.50 mm, b) 28.0 cm, 1 = 0.0360 s = 27.8 Hz and from Eq. (15.1), d) v = (0.280 m)(27.8 Hz) = 7.78 m s ,

1 T

e) + x direction.

15.7: a) f = v λ = (8.00 m s) (0.320 m) = 25.0 Hz, T = 1 f = 1 (25.0 Hz) = 4.00 × 10 −2 s, k = 2π λ = (2π ) (0.320 m) = 19.6 rad m.   x . b) y ( x, t ) = (0.0700 m) cos 2π  t (25.0 Hz) +  0.320 m    c) (0.0700 m) cos [ 2π ((0.150 s)(25.0 Hz) + (0.360 m) (0.320 m))] = −4.95 cm. d) The argument in the square brackets in the expression used in part (c) is 2π (4.875), and the displacement will next be zero when the argument is 10π ; the time is then T (5 − x λ ) = (1 25.0 Hz)(5 − (0.360 m) (0.320 m)) = 0.1550 s and the elapsed time is 0.0050 s, e) T 2 = 0.02 s.

15.8:

a)

b)

15.9:

a)

∂y = − Ak sin( kx + ωt ) ∂x ∂y = − Aω sin (kx + ωt ) ∂t

∂2 y = − Ak 2 cos(kx + ωt ) 2 ∂x ∂2 y = − Aω2 cos(kx + ωt ), 2 ∂t

and so b)

∂2 y ∂x
2

=

2 k2 ∂ y 2 ω ∂t 2

, and y ( x, t ) is a solution of Eq. (15.12) with v = ω k . ∂2 y = − Ak 2 sin( kx + ωt ) 2 ∂x ∂2 y = − Aω2 sin (kx + ωt ), ∂t 2 c) Both waves

∂y = + Ak cos(kx + ωt ) ∂x ∂y = + Aω cos(kx + ωt ) ∂t

and so

∂2 y ∂x
2

=

are moving in the − x -direction, as explained in the discussion preceding Eq. (15.8). d) Taking derivatives yields v y ( x, t ) = −ωA cos (kx + ωt ) and a y ( x, t ) = −ω 2 A sin (kx + ωt ).

2 k2 ∂ y 2 ω ∂t 2

, and y ( x, t ) is a solution of Eq. (15.12) with v = ω k .

15.10:

a) The relevant expressions are y ( x, t ) = A cos(kx − ωt ) ∂y vy = = ωA sin (kx − ωt ) ∂t ∂ 2 y ∂v a y = 2 = y = −ω2 A cos (ωt − kx). ∂t ∂t

b) (Take A, k and ω to be positive. At x t = 0, the wave is represented by (19.7(a)); point (i) in the problem corresponds to the origin, and points (ii)-(vii) correspond to the points in the figure labeled 1-7.) (i) v y = ωA cos(0) = ωA, and the particle is moving upward (in the positive y-direction). a y = −ω2 A sin( 0) = 0, and the particle is instantaneously not accelerating. (ii) v y = ωA cos(− π 4) = ωA

2 , and the particle is moving up.

a y = −ω2 A sin( − π 4) = ω2 A

2 , and the particle is speeding up.

(iii) v y = ωA cos(− π 2) = 0, and the particle is instantaneously at rest. a y = −ω2 A sin( − π 2) = ω2 A, and the particle is speeding up. (iv) v y = ωA cos(− 3π 4) = − ωA

2 , and the particle is moving down. 2 , and the particle is slowing down ( v y is becoming less

a y = −ω2 A sin( − 3π 4) = ω2 A

negative). (v) v y = ωA cos(−π ) = −ωA and the particle is moving down. a y = −ω2 A sin( −π ) = 0, and the particle is instantaneously not accelerating. (vi) v y = ωA cos(− 5π 4) = − ωA

2 and the particle is moving down. 2 and the particle is speeding up ( v y and a y have the

a y = −ω 2 A sin( − 5π 4) = −ω 2 A

same sign). (vii) v y = ωA cos(− 3π 2) = 0, and the particle is instantaneously at rest. a y = −ω2 A sin( − 3π 2) = −ω2 A and the particle is speeding up. (viii) v y = ωA cos(− 7π 4) = ωA

2 , and the particle is moving upward. 2 and the particle is slowing down ( v y and a y have

a y = −ω A sin( − 7π 4) = − ω A opposite signs).
2 2

15.11: Reading from the graph, a) A = 4.0 mm , b) T = 0.040 s. c) A displacement of 0.090 m corresponds to a time interval of 0.025 s; that is, the part of the wave represented by the point where the red curve crosses the origin corresponds to the point where the blue curve crosses the t-axis ( y = 0) at t = 0.025 s, and in this time the wave has traveled 0.090 m, and so the wave speed is 3.6 m s and the wavelength is vT = (3.6 m s)(0.040 s) = 0.14 m . d) 0.090 m 0.015 s = 6.0 m s and the wavelength is 0.24 m. d) No; there could be many wavelengths between the places where y (t ) is measured. 2π  λ  x t  A cos 2π  −  = + A cos  x − t  λ  T  λ T  = + A cos λ = λf = v has been used. T 2π ( x − vt ) , λ

15.12:

a)

where b)

∂y 2πv 2π ( x − vt ). = A sin ∂t λ λ c) The speed is the greatest when the cosine is 1, and that speed is 2πvA λ . This will be equal to v if A = λ 2π , less than v if A < λ 2π and greater than v if A > λ 2π . vy =

15.13: a) t = 0 : ____________________________________________________________ x(cm) 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 ____________________________________________________________ y(cm) 0.000 −0.212 −0.300 −0.212 0.000 0.212 0.300 0.212 0.000 ____________________________________________________________

b) i) t = 0.400 s: _______________________________________________________________ x(cm) 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 _______________________________________________________________ y(cm) 0.285 0.136 −0.093 −0.267 −0.285 −0.136 0.093 0.267 0.285 _______________________________________________________________

ii ) t = 0.800 s : ________________________________________________________________ x(cm) 0.00 1.50 3.00 4.50 6.00 7.50 9.00 10.50 12.00 ________________________________________________________________ y(cm) 0.176 0.296 0.243 0.047 −0.176 −0.296 −0.243 −0.047 0.176 ________________________________________________________________

15.14: Solving Eq. (15.13) for the force F ,  0.120 kg  2 2 F = μv 2 = μ( f λ ) =   ((40.0 Hz ) (0.750 m)) = 43.2 Ν. 2.50 m  

15.15: a) Neglecting the mass of the string, the tension in the string is the weight of the pulley, and the speed of a transverse wave on the string is v= F (1.50 kg )(9.80 m s 2 ) = = 16.3 m s. μ (0.0550 kg m)

b) λ = v f = (16.3 m s) (120 Hz) = 0.136 m. c) The speed is proportional to the square root of the tension, and hence to the square root of the suspended mass; the answers change by a factor of 2 , to 23.1 m s and 0.192 m.

15.16: a) v = F μ = (140.0 Ν ) (10.0 m) (0.800 kg ) = 41.8 m s. b) λ = v f = (41.8 m s) (1.20 Hz) = 34.9 m. c) The speed is larger by a factor of 2 , and so for the same wavelength, the frequency must be multiplied by 2 , or 1.70 Hz.

15.17: Denoting the suspended mass by M and the string mass by m, the time for the pulse to reach the other end is t= L = v L = Mg (m L) mL (0.800 kg)(14.0 m) = = 0.390 s. Mg (7.50 kg )(9.80 m s 2 )

15.18: a) The tension at the bottom of the rope is due to the weight of the load, and the speed is the same 88.5 m s as found in Example 15.4 b) The tension at the middle of the rope is (21.0 kg ) (9.80 m s 2 ) = 205.8 N (keeping an extra figure) and the speed of the rope is 90.7 m s. c) The tension at the top of the rope is (22.0 kg )(9.80 m s 2 ) = 215.6 m s and the speed is 92.9 m s . (See Challenge Problem (15.80) for the effects of varying tension on the time it takes to send signals.)

15.19: a) v = F μ = (5.00 N) (0.0500 kg m) = 10.0 m s b) λ = v f = (10.0 m s) (40.0 Hz) = 0.250 m c) y ( x, t ) = A cos(kx − ωt ) (Note : y (0.0) = A, as specified. ) k = 2π λ = 8.00π rad m; ω = 2πf = 80.0π rad s y ( x, t ) = (3.00 cm)cos[π (8.00 rad m) x − (80.0π rad s)t ] d) v y = + Aω sin( kx − ωt ) and a y = − Aω2cos(kx − ωt ) a y , max = Aω2 = A(2πf ) 2 = 1890 m s 2 e) a y , max is much larger than g, so ok to ignore gravity.

15.20:

a) Using Eq.(15.25), 1 Pave = μF ω 2 A2 2  3.00 × 10− 3 kg   (25.0 N) (2π(120.0 Hz))2 (1.6 × 10− 3 m) 2   0.80 m   = 0.223 W, = 1 2

or 0.22 W to two figures. W.

b) Halving the amplitude quarters the average power, to 0.056

15.21:

Fig. 15.13 plots P ( x, t ) = μF ω2 A2 sin 2 (kx − ωt ) at x = 0. For x = 0, P ( x, t ) = μF ω2 A2 sin 2 (ωt ) = Pmax sin 2 (ωt ) When x = λ 4, kx = (2π λ) (λ 4) = π 2. sin (π 2 − ωt ) = cos ωt , so P (λ 4, t ) = Pmax cos2 ωt

The graph is shifted by T 4 but is otherwise the same. The instantaneous power is still never negative and Pav = 1 Pmax , the same as at x = 0. 2

15.22:

r2 = r1

Ι1 Ι2

= (7.5 m)

0.11 W m 2 1.0 W m 2

= 2.5 m, so it is possible to

move r1 − r2 = 7.5 m − 2.5 m = 5.0 m closer to the source.

15.23:

2 a) Ι1r 1 = Ι 2 r

2 2

Ι 2 = Ι1 (r1 r2 ) 2 = (0.026 W m 2 )(4.3 m 3.1 m) 2 = 0.050 W m 2 b) P = 4πr 2 Ι = 4π (4.3m ) 2 (0.026 W m 2 ) = 6.04 W Energy = Pt = (6.04 W )(3600 s) = 2.2 × 10 4 J

15.24: v=
ω k

(a) A = 2.30 mm. (b) f =

ω 2π

=

742 rad s 2π

118 Hz. (c) λ =

2π k

=

2π 6.98 rad m

= 0.90 m. (d)

=

= 106 m s. (e) The wave is traveling in the –x direction because the

phase of y (x,t) has the form kx + ωt. (f) The linear mass density is μ = (3.38 × 10 −3 kg ) (1.35 m) = 2.504 × 10 −3 kg m , so the tension is F = μv 2 = (2.504 × 10 −3 kg m)(106.3 m s) 2 = 28.3 N (keeping an extra figure in v for accuracy). (g ) Pav =
1 2

μF ω2 A2 =

1 2

(2.50 × 10−3 kg m)(28.3 N ) (742 rad s) 2

(2.30 × 10−3 m) 2 = 0.39 W. 15.25: I = 0.250 W m 2 at r = 15.0 m P = 4πr 2 I = 4π (15.0 m) 2 (0.250 W m 2 ) = 707 W

15.26:

a) The wave form for the given times, respectively, is shown.

b)

15.27:

a) The wave form for the given times, respectively, is shown.

b)

15.28:

15.29:

15.30: Let the wave traveling in the + x direction be y1 ( x, t ) = A cos (kx − ωt). The wave traveling in the − x direction is inverted due to reflection from the fixed end of the string at x = 0, so it has the form y2 ( x, t ) = − A cos(kx + ωt ). The wave function of the resulting standing wave is then y ( x, t ) = y1 ( x, t ) + y2 ( x, t ) , where A = 2.46 mm, ω = 2π T = 2π (3.65 × 10−3 s) = 1.72 × 103 rad s, k = ω v = (1.72 × 103 rad s)(111 m s) = 15.5 rad m. 15.31: a) The nodes correspond to the places where y = 0 for all t in Eq. (15.1); that is, sin kxnode = 0 or kxnode = nπ , n an integer . With k = 0.75π rad m, xnode = (1.333 m) n and for n = 0, 1, 2, ..., xnode = 0, 1.333 m, 2.67 m, 4.00 m, 5.33 m, 6.67 m,... b) The antinodes correspond to the points where cos kx = 0, which are halfway between any two adjacent nodes, at 0.667 m, 2.00 m, 3.33 m, 4.67 m, 6.00 m, ... ∂2 y ∂2 y = −k 2 [ Asw sin ωt ] sin kx, = −ω 2 [ Asw sin ωt ] sin kx, 2 2 ∂x ∂t − ω2 ω so for y ( x, t ) to be a solution of Eq. (15.12), − k 2 = 2 , and v = . v k b) A standing wave is built up by the superposition of traveling waves, to which the relationship v = ω k applies. 15.32: a)

15.33: a) The amplitude of the standing wave is Asw = 0.85 cm, the wavelength is twice the distance between adjacent antinodes, and so Eq. (15.28) is y ( x, t ) = (0.85 cm) sin(( 2π 0.075 s)t ) sin( 2πx 30.0 cm). b) c) v = λ f = λ T = (30.0 cm) (0.0750 s) = 4.00 m/s. (0.850 cm) sin( 2π (10.5 cm) (30.0 cm)) = 0.688 cm. y1 + y2 = A [− cos(kx + ωt ) + cos(kx − ωt )] = A [− cos kx cos ωt + sin kx sin ωt + cos kx cos ωt + sin kx sin ωt ] = 2 A sin kx sin ωt.

15.34:

15.35: The wave equation is a linear equation, as it is linear in the derivatives, and differentiation is a linear operation. Specifically, ∂y ∂ ( y1 + y2 ) ∂y1 ∂y2 = = + . ∂x ∂x ∂x ∂x Repeating the differentiation to second order in both x and t, ∂ 2 y1 ∂ 2 y2 ∂2 y = + 2 , ∂x 2 ∂x ∂x 2 ∂ 2 y1 ∂ 2 y 2 ∂2 y = + . ∂t 2 ∂t 2 ∂t 2

The functions y1 and y2 are given as being solutions to the wave equation; that is, ∂2 y ∂ 2 y1 ∂ 2 y2  1  ∂ 2 y1  1  ∂ 2 y 2 = + 2 = 2 + 2 2 2 ∂x 2 ∂x ∂x 2  v  ∂t  v  ∂t
2 2  1   ∂ y ∂ y2  =  2   21 +  ∂t 2   v   ∂t 2  1 ∂ y = 2 2  v  ∂t

and so y = y1 + y 2 is a solution of Eq. (15.12).

15.36:

a) From Eq. (15.35), f1 = 1 2L FL 1 = 2(0.400 m) m (800 N)(0.400 m) =408 Hz. (3.00 × 10 −3 kg )

b)

10 , 000 Hz 408 Hz

= 24.5, so the 24 th harmonic may be heard, but not the 25 th .

15.37: a) In the fundamental mode, λ = 2 L = 1.60 m and so v = f λ = (60.0 Hz)(1.60 m) = 96.0 m s. b) F = v 2 μ = v 2 m L = (96.0 m s) 2 (0.0400 kg ) (0.800 m) = 461 Ν.

15.38: The ends of the stick are free, so they must be displacement antinodes. 1st harmonic:

L= 2nd harmonic:

1 λ1 → λ1 = 2 L = 4.0 m 2

3 harmonic:

rd

L = 1λ 2 → λ 2 = L = 2.0 m

3 2L L = λ3 → λ3 = = 1.33 m 2 3

15.39:

a)

b) Eq. (15.28) gives the general equation for a standing wave on a string: y ( x, t ) = ( Asw sin kx ) sin ωt Asw = 2 A, so A = ASW 2 = (5.60 cm) 2 = 2.80 cm c) The sketch in part (a) shows that L = 3(λ 2) k = 2π λ, λ = 2π k Comparison of y ( x, t ) given in the problem to Eq.(15.28) gives k = 0.0340 rad cm. So, λ = 2π (0.0340 rad cm) = 184 .8 cm L = 3(λ 3) = 277 cm d) λ = 185 cm, from part (c) ω = 50.0 rad s so f = ω 2π = 7.96 Hz period T = 1 f = 0.126 s v = f λ = 1470 cm s e) v y = dy dt = ωAsw sin kx cos ωt v y , max = ωASW = (50.0 rad s)(5.60 cm) = 280 cm s f) f 3 = 7.96 Hz = 3 f 1 , so f 1 = 2.65 Hz is the fundamental f 8 = 8 f1 = 21.2 Hz; ω8 = 2πf 8 = 133 rad s λ = v f = (1470 cm s) (21.2 Hz) = 69.3 cm and k = 2π λ = 0.0906 rad cm. y ( x, t ) = (5.60 cm) sin ([0.0906 rad cm] x) sin ([133 rad s]t )

15.40: ( c) f =

(a) A = 1 ASW = 1 (4.44 mm) = 2.22 mm. (b) λ = 2 2
ω 2π

2π k

=

2π 32.5 rad m

= 0.193 m.

=

754 rad m 2π

= 120 Hz. (d) v =

ω k

=

= 23.2 m s. (e) If the wave traveling in

the + x direction is written as y1 ( x, t ) = A cos(kx − ωt ), then the wave traveling in the − x direction is y2 ( x, t ) = − A cos(kx + ωt ), where A = 2.22 mm from (a), and k = 32.5 rad m and ω = 754 rad s. (f) The harmonic cannot be determined because the length of the string is not specified. 15.41: a) The traveling wave is y ( x, t ) = (2.30 m) cos ([6.98 rad m] x ) + [742 rad s]t ) A = 2.30 mm so ASW = 4.60 mm; k = 6.98 rad m and ω = 742 rad s The general equation for a standing wave is y ( x, t ) = ( ASW sin kx) sin ωt , so y ( x, t ) = (4.60 mm) sin([6.98 rad m] x)sin([742 rad s]t ) b) L = 1.35 m (from Exercise 15.24) λ = 2π k = 0.900 m L = 3(λ 2), so this is the 3rd harmonic c) For this 3rd harmonic, f = ω 2π = 118 Hz f 3 = 3 f 1 so f 1 = (118 Hz) 3 = 39.3 Hz 15.42: The condition that x = L is a node becomes kn L = nπ. The wave number and wavelength are related by kn λ n = 2π , and so λ n = 2 L n. 15.43: a) The product of the frequency and the string length is a constant for a given string, equal to half of the wave speed, so to play a note with frequency 587 Hz, x = (60.0 cm) (440 Hz) (587 Hz) = 45.0 cm. b) Lower frequency requires longer length of string free to vibrate. Full length of string gives 440 Hz, so this is the lowest note possible. 15.44: a) (i) x =
λ 2

is a node, and there is no motion. (ii) x =

λ 4

is an antinode,
1 2

and vmax = A(2πf ) = 2πfA, amax = (2πf )vmax = 4π 2 f 2 A. (iii ) cos π = 4

, and this factor

multiplies the results of (ii), so vmax = 2π fA, amax = 2 2π 2 f 2 A . b) The amplitude is A sin kx, or (i)0, (ii) A, (iii) A 2. c) The time between the extremes of the motion is the same for any point on the string (although the period of the zero motion at a node might be considered indeterminate) and is 21f .

15.45:

a) λ1 = 2 L = 3.00 m, f1 =

v 2L

=

( 48.0 m s ) 2 (1.50 m )

= 16.0 Hz.

b) λ 3 = λ1 3 = 1.00 m, f 2 = 3 f1 = 48.0 Hz. c) λ 4 = λ1 4 = 0.75 m, f 3 = 4 f1 = 64.0 Hz. 15.46: a) For the fundamental mode, the wavelength is twice the length of the string, and v = f λ = 2 fL = 2(245 Hz)(0.635 m) = 311 m s. b) The frequency of the fundamental mode is proportional to the speed and hence to the square root of the tension; (245 Hz) 1.01 = 246 Hz. c) The frequency will be the same, 245 Hz. The wavelength will be λ air = vair f = (344 m s) (245 Ηz) = 1.40 m, which is larger than the wavelength of standing wave on the string by a factor of the ration of the speeds.

15.47: a) f = v λ = (36.0 m s) (1.80 m) = 20.0 Hz, ω = 2πf = 126 rad s, k = ω v = 2π λ = 3.49 rad m. b) y ( x, t ) = A cos (kx − ωt ) = (2.50 mm )cos [ (3.49 rad m) x − (126 rad s)t ]. c)At x = 0, y (0, t ) = A cos ωt = (2.50 mm ) cos [ (126 rad s)t ]. From this form it may be seen that at x = 0, t = 0, ∂y > 0. d) At x = 1.35 m = 3λ 4, kx = 3π 2 and ∂t y ( 3λ 4, t ) = A cos [ 3π 2 − ωt ]. e) See Exercise 15.12; ωA = 0.315 m s. f) From the result of part ( d ) , y = 0 mm. v y = − 0.315 m s.

15.48: b)

a) From comparison with Eq. (15.4 ) , A = 0.75 cm, λ =
1 f

2 0.400 cm

= 5.00 cm,

f = 125 Hz, T =

= 0.00800 s and v − λf = 6.25 m s.

c) To stay with a wavefront as t increases, x and so the wave is moving in the − x direction. d) From Eq. (15.13) , the tension is F = μv 2 = (0.50 kg m) (6.25 m s) 2 = 19.5 N. e) Pav =
1 2

μF ω2 A2 = 54.2 W.

15.49:

a) Speed in each segment is v = F μ. The time to travel through a segment is
μ1 F

t = L v. The travel times then, are t1 = L t total = L
μ1 F

, t2 = L

4 μ1 F

, and t3 = L

μ1 4F

+ 2L

μ1 F

+1L 2

μ1 F

=7L 2

μ1 F

.

b) No, because the tension is uniform throughout each piece.

15.50: The amplitude given is not needed, it just ensures that the wave disturbance is small. Both strings have the same tension F , and the same length L = 1.5 m. The wave takes different times t1 and t2 to travel along each string, so the design requirements is t1 + t 2 = 0.20 s. Using t = L v and v = F μ = FL m gives m1 + m2 L F = 0.20 s, with m1 = 90 × 10 −3 kg and m1 = 10 × 10 −3 kg . Solving for F gives F = 6.0 N. 15.51: a) y ( x, t ) = A cos(kx − ωt ) v y = dy dt = + Aω sin( kx − ωt ) v y , max = Aω = 2πfA f = v and v = λ F  1  FL , so f =   ( m L) λ M

(

)

 2πA  FL v y , max =    λ  M b) To double v y , max increase F by a factor of 4

15.52:
2

The maximum vertical acceleration must be at least g. Because
gλ2 μ 4π 2 F

a = ω A, g = ω2 Amin and thus Amin = g ω2. Using ω = 2πf = 2πv λ and v = F μ, this becomes Amin = .

15.53:

a) See Exercise 15.10; a y =

∂2 y ∂t 2

= −ω2 y, and so k ′ = ∆mω2 = ∆xμ ω2 .
2

b)

4π 2 F  2πv  2 ω2 = ( 2πf ) =  =  μλ 2  λ 

and so k ′ = (4π 2 F λ2 )∆x. The effective force constant k ′ is independent of amplitude, as for a simple harmonic oscillator, and is proportional to the tension that provides the restoring force. The factor of 1 λ2 indicates that the curvature of the string creates the restoring force on a segment of the string. More specifically, one factor of 1 λ is due to the curvature, and a factor of 1 (λμ ) represents the mass in one wavelength, which determines the frequency of the overall oscillation of the string. The mass ∆m = μ∆x also contains a factor of μ, and so the effective spring constant per unit length is independent of μ.

15.54: a), b)

c) The displacement is a maximum when the term in parentheses in the denominator is zero; the denominator is the sum of two squares and is minimized when x = vt , and the maximum displacement is A. At x = 4.50 cm, the displacement is a maximum at t = (4.50 × 10 −2 m) (20.0 m s) = 2.25 × 10−3 s. The displacement will be half of the maximum when ( x − vt ) 2 = A2 , or t = ( x ± A) v = 1.75 × 10−3 s and 2.75 × 10−3 s. d) Of the many ways to obtain the result, the method presented saves some algebra and minor calculus, relying on the chain rule for partial derivatives. Specifically, let ∂f dg ∂u dg ∂f dg ∂u dg u = u ( x, t ) = x − vt , so that if f ( x, t ) = g(u ), = = and = = − v. ∂x du ∂t du ∂t du ∂t du (In this form it may be seen that any function of this form satisfies the wave equation; see Problem 15.59.) In this case, y ( x, t ) = A3 ( A2 + u 2 )−1 , and so ∂y − 2 A3u = 2 , ∂x ( A + u 2 ) 2 ∂y 2 A3u =v 2 , ∂t ( A + u 2 )2 ∂2 y 2 A3 ( A2 − 3u 2 ) =− ∂x 2 ( A2 + u 2 ) 3
3 2 2 ∂2 y 2 2 A ( A − 3u ) = −v , ∂t 2 ( A2 + u 2 ) 2

and so the given form for y ( x, t ) is a solution to the wave equation with speed v.

15.55: a) and b) (1): The curve appears to be horizontal, and v y = 0. As the wave moves, the point will begin to move downward, and a y < 0. (2): As the wave moves in the + x -direction (to the right in Fig. (15.34)), the particle will move upward so v y > 0. The portion of the curve to the left of the point is steeper, so a y > 0. (3) The point is moving down, and will increase its speed as the wave moves; v y < 0, a y < 0. (4) The curve appears to be horizontal, and v y = 0. As the wave moves, the point will move away from the x -axis, and a y >0. (5) The point is moving downward, and will increase its speed as the wave moves; v y < 0, a y < 0. (6) The particle is moving upward, but the curve that represents the wave appears to have no curvature, so v y > 0 and a y = 0. c) The accelerations, which are related to the curvatures, will not change. The transverse velocities will all change sign. 15.56: (a ) The wave travels a horizontal distance d in a time t= d d 8.00 m = = = 0.333 s. v λ f ( 0.600 m ) ( 40.0 Hz )

(b) A point on the string will travel a vertical distance of 4 A each cycle. Although the transverse velocity v y ( x, t ) is not constant, a distance of h = 8.00 m corresponds to a whole number of cycles, n = h (4 A) = (8.00 m) ((4(5.00 × 10−3 m)) = 400, so the amount of time is t = nT = n f = (400) (40.0 Hz) = 10.0 s. (c ) The answer for (a ) is independent of amplitude. For (b), the time is halved if the amplitude is doubled.

15.57: a) y 2 ( x, y ) + z 2 ( x, y ) = A2 The trajectory is a circle of radius A.

At t = 0, y (0,0) = A, z (0,0) = 0.
At t = π 2ω, y (0, π 2ω) = 0, z (0, π 2ω) = − A. At t = π ω, y (0, π ω) = − A, z (0, π 2ω) = 0. At t = 3π 2ω, y (0, 3π 2ω) = 0, z (0, 3π 2ω) = + A b) v y = dy dt = + Aω sin( kx − ωt ), vz = dz dt = − Aω cos(kx − ωt )
2 v = v y + vz2 = Aω, so the speed is constant.

 ˆ r = yˆ + zk j   r ⋅ v = yv y + zv z = A2ω sin (kx − ωt ) cos(kx − ωt ) − A2ω cos(kx − ωt ) sin( kx − ωt )

   r ⋅ v = 0, so v is tangent to the circular path.
c) a y = dv y dt = − Aω2 cos(kx − ωt ), a z = dvz dt = − Aω2 sin( kx − ωt )   r ⋅ a = ya y + za z = − A2ω2 [cos2 (kx − ωt ) + sin 2 (kx − ωt )] = − A2ω2   r = A, a = Aω2 , so r ⋅ a = −ra

     r ⋅ a = ra cos φ so φ = 180° and a is opposite in direction to r ; a is radially
inward. y 2 + z 2 = A 2 , so the path is again circular, but the particle rotates in the opposite sense compared to part (a ).

15.58: The speed of light is so large compared to the speed of sound that the travel time of the light from the lightning or the radio signal may be neglected. Them, the distance from the storm to the dorm is (344 m s)(4.43 s) = 1523.92 m and the distance from the storm to the ballpark is (344 m s)(3.00 s) = 1032 m. The angle that the direction from the storm to the ballpark makes with the north direction is found from these distances using the law of cosines;  (1523.92 m) 2 − (1032 m) 2 − (1120 m) 2  θ = arccos   = 90.07°, − 2(1032 m) (1120 m)   so the storm can be considered to be due west of the park. 15.59: a) As time goes on, someone moving with the wave would need to move in such a way that the wave appears to have the same shape. If this motion can be described by x = vt + c, with c a constant (not the speed light),then y ( x, t ) = f (c), and the waveform is the same to such observer. b) See Problem 15.54. The derivation is completed by taking the second partials, ∂2 y 1 d 2 f = , ∂x 2 v 2 du 2 ∂2 y d 2 f = , ∂t 2 du 2 c) This is

so y ( x, t ) = f (t − x / v ) is a solution to the wave equation with wave speed v . of the form y ( x, t ) = f (u ), with u = t − x v and
f (u ) = De − C
2

(t − ( B c ) x ) 2

,

and the result of part (b) may be used to determine the speed v = C B immediately.

15.60: a)

b) ∂y = ωΑ sin( kx − ωt + φ ). c) No; φ = π 4 or φ = 3π 4 would both give Α 2. If the ∂t particle is known to be moving downward, the result of part b) shows that cos φ < 0, and so φ = 3π 4. d) To identify φ uniquely, the quadrant in which φ is must be known. In physical terms, the signs of both the position and velocity, and the magnitude of either, are necessary to determine φ (within additive multiples of 2π ).

15.61: a) μF = F μ F = F v = F k ω and substituting this into Eq. (15.33) gives the result. b) Quadrupling the tension for F to F ′ = 4 F increases the speed v = F μ by a factor of 2, so the new frequency ω′ and new wave number k ′ are related to ω and k by (ω′ k ′) = 2(ω k ). For the average power to be the same, we must have Fkω = F ′k ′ω′, so kω = 4k ′ω′ and k ′ω′ = kω 4 . Multiplying the first and second equations together gives ω′2 = ω2 2, so ω′ = ω 2. 2. Dividing the second equation by

Thus, the frequency must decrease by a factor of the first equation gives k ′2 = k 2 8, so k ′ = k 8.

15.62: (a)

(b) The wave moves in the + x direction with speed v, so in the expression for y (x,0) replace x with x − vt : for ( x − vt ) < − L 0 h ( L + x − vt ) L for − L < ( x − vt ) < 0  y ( x, t ) =  h ( L − x + vt ) L for 0 < ( x − vt ) < L 0 for ( x − vt ) > L  (c) From Eq. (15.21): for ( x − vt ) < − L − F (0)(0) = 0  2 ∂y ( x, t ) ∂y ( x, t ) − F (h L)(− hv L) = Fv (h L) for − L < ( x − vt ) < 0 P ( x, t ) = − F = 2 ∂x ∂t − F (− h L)(hv L) = Fv (h L) for 0 < ( x − vt ) < L − F (0)(0) = 0 for ( x − vt ) > L  Thus the instantaneous power is zero except for − L < ( x − vt ) < L, where it has the constant value Fv (h L) 2 .

15.63: a) Pav = v= F μ

1 2

μF ω2 A2 so F =v μ

ω = 2πf = 2π (v λ ) Using these two expressions to replace F and ω gives Pav = 2 μπ 2v 3 A2 λ2 ; μ = (6.00 × 10 −3 kg) (8.00 m)  2λ2 P  A =  2 3av  = 7.07 cm  4π v μ    3 b) Pav ~ v so doubling v increases Pav by a factor of 8.
2 1

Pav = 8(50.0 W) = 400.0 W

15.64: a) , d)

b)The power is a maximum where the displacement is zero, and the power is a minimum of zero when the magnitude of the displacement is a maximum. c) The direction of the energy flow is always in the same direction. d) In this case, ∂y = − kΑ sin( kx + ωt ), and so Eq. (15.22) becomes ∂x P ( x, t ) = − FkωA2sin 2 (kx + ωt ). The power is now negative (energy flows in the − x -direction), but the qualitative relations of part (b) are unchanged. F1 2 F2 F1 − YΑαΔΤ , v2 = = ⋅ μ μ μ

15.65: Solving for α,

v12 =

α=

2 2 v12 − v2 v 2 − v2 = 1 ⋅ Y ( Α μ)∆Τ (Y ρ)∆Τ

1 15.66: (a) The string vibrates through 1 2 cycle in 4 × 5000 min, so

1 4 T= min → T = 1.6 × 10 − 3 min = 9.6 × 10− 2 s 2 5000 f = 1 T = 1 9.6 × 10−2 s = 10.4 Hz

λ = L = 50.0 cm = 0.50 m
(b) Second harmonic (c) v = fλ = (10.4 Hz)(0.50 m) = 5.2 m s (d) (i)Maximum displacement, so v = 0 (ii) v y = Speed = v y = ω(1.5 cm)sin kx sin ωt at maximum speed, sin kx = sin ωt = 1 v y = ω(1.5 cm) = 2πf (1.5 cm) = 2π (10.4 Hz)(1.5 cm) = 98 cm s = 0.98 m s (e) v = F μ → μ = F v 2 M = μL = F (1.00 N)(0.500 m) L= = 1.85 × 10− 2 kg 2 v (5.2 m s) 2 =18.5 g
∂y ∂t

= ∂∂t (1.5 cm sin kx sin ωt )

15.67: There is a node at the post and there must be a node at the clothespin. There could be additional nodes in between. The distance between adjacent nodes is λ 2, so the distance between any two nodes is n (λ 2) for n = 1, 2, 3, ... 45.0 cm = n(λ 2), λ = v f , so f = n[v (90.0 cm)] = (0.800 Hz)n, n = 1, 2, 3, ...

15.68: (a) The displacement of the string at any point is y ( x, t ) = ( ASW sin kx) sin ωt. For the fundamental mode λ = 2L, so at the midpoint of the string sin kx = sin( 2π λ )( L 2) = 1, and y = ASW sin ωt. Taking derivatives gives v y =
∂v

∂y ∂t

= ωASW cos ωt , with maximum value

v y max = ωASW , and a y = ∂ty = −ω2 ASW sin ωt , with maximum value a y max = ω2 ASW . Dividing these gives ω = a y max v y max = (8.40 × 103 m s 2 ) (3.80 m s) = 2.21 × 103 rad s, and then ASW = v y max ω = (3.80 m s) (2.21 × 103 rad s) = 1.72 × 10−3 m. (b) v = λf = (2 L)(ω 2π ) = Lω π = (0.386 m) (2.21× 103 rad s) π = 272 m s.

15.69: a) To show this relationship is valid, take the second time derivative: ∂ 2 y ( x, t ) ∂ 2 = 2 [( ASW sin kx) cos ωt ], ∂t 2 ∂t 2 ∂ y ( x, t ) ∂ = −ω [( ASW sin kx) sin ωt ] 2 ∂t ∂t 2 ∂ y ( x, t) = −ω2 [ ( Αsw sin kx) cos ωt ], 2 ∂t ∂ 2 y ( x, t ) = −ω2 y ( x, t ), Q.E.D. 2 ∂t The displacement of the harmonic oscillator is periodic in both time and space. b) Yes, the travelling wave is also a solution of the wave equation.

15.70: a) The wave moving to the left is inverted and reflected; the reflection means that the wave moving to the left is the same function of − x, and the inversion means that the function is − f (− x ). More rigorously, the wave moving to the left in Fig. (15.17) is obtained from the wave moving to the right by a rotation of 180° , so both the coordinates ( f and x ) have their signs changed. b). The wave that is the sum is f ( x) − f (− x) (an inherently odd function), and for any f , f (0) − f (−0) = 0. c) The wave is reflected but not inverted (see the discussion in part (a) above), so the wave moving to the left in Fig. (15.18) is + f (− x ). d) dy d df ( x) df (− x) df ( x) df (− x) d (− x) = ( f ( x) + f (− x)) = + = + dx dx dx dx dx d (− x) dx = df df − dx dx .
x=−x

At x = 0 , the terms are the same and the derivatives is zero. (See Exercise 20-2 for a situation where the derivative of f is not finite, so the string is not always horizontal at the boundary.) y ( x, t ) = y1 ( x, t ) + y2 ( x, t ) = Α[ cos(kx + ωt ) + cos(kx − ωt )] = Α[ cosωt cos kx − sin ωt sin kx + cos ωt cos kx + sin ωt sin kx ] = (2 Α) cosωt cos kx. b) At x = 0, y (0, t ) = (2 Α)cosωt , and so x = 0 is an antinode. c) The maximum displacement is, front part (b), ΑSW = 2 Α, the maximum speed is ωΑSW = 2ωΑ and the 15.71: a) magnitude of the maximum acceleration is ω2 Αsw = 2ω2 Α.

15.72: a) λ = v f = (192.0 m s) (240.0 Hz) = 0.800 m , and the wave amplitude is ΑSW = 0.400 cm. The amplitude of the motion at the given points is (i) (0.400 cm)sin (π ) = 0 (a node), (ii) (0.400 cm) sin( π 2) = 0.004 cm (an antinode) and (iii) (0.400 cm) sin( π 4) = 0.283 cm. b). The time is half of the period, or 1 (2 f ) = 2.08 × 10 −3 s. c) In each case, the maximum velocity is the amplitude multiplied by ω = 2πf and the maximum acceleration is the amplitude multiplied by ω2 = 4π 2 f 2 , or (i) 0, 0; (ii) 6.03 m s, 9.10 × 103 m s 2 ; (iii) 4.27 m s , 6.43 × 103 m s 2 .

15.73: The plank is oscillating in its fundamental mode, so λ = 2 L = 10.0 m, with a frequency of 2.00 Hz. a) v = fλ = 20.0 m s. b) The plank would be its first overtone, with twice the frequency, or 4 jumps s.

15.74: (a) The breaking stress is

F πr 2

= 7.0 × 108 N m 2 , and the maximum tension is
900 N π (7.0×10 8 N m 2 )

F = 900 N, so solving for r gives the minimum radius r =
The mass and density are fixed, ρ = length L =
M πr 2 ρ M πr 2 L

= 6.4 × 10−4 m.

, so the minimum radius gives the maximum

=

4.0×10 −3 kg π (6.4×10
−4

m) 2 ( 7800 kg m 3 )

= 0.40 m.
1 2L F µ

(b) The fundamental frequency is f 1 = when F = 900 N, f1 =
1 2 900 N (4.0×10 −3 kg)(0.40 m)

=

1 2L

F M L

=

1 2

F ML

. Assuming the

maximum length of the string is free to vibrate, the highest fundamental frequency occurs = 376 Hz.

15.75: a) The fundamental has nodes only at the ends, x = 0 and x = L. b) For the second harmonic, the wavelength is the length of the string, and the nodes are at x = 0, x = L 2 and x = L. b)

d) No; no part of the string except for x = L 2, oscillates with a single frequency.

15.76: a) The new tension F ′ in the wire is F′ = F − B = w −  1 ρwater (1 3w) ρwater = w1 −  3 ρ ρA1  A1    

 (1.00 × 103 kg m3 )  = w 1 −  3(2.7 × 103 kg m 3 )  = (0.8765) w = (0.87645) F .    The frequency will be proportional to the square root of the tension, and so f ′ = (200 Hz) 0.8765 = 187 Hz. b) The water does not offer much resistance to the transverse waves in the wire, and hencethe node will be located a the point where the wire attaches to the sculpture and not at the surface of the water. 15.77: a) Solving Eq. (15.35) for the tension F, F = 4 L2 f12 μ = 4mLf12 = 4(14.4 × 10−3 kg)(0.600 m)(65.4 Hz) 2 = 148 N. b) The tension must increase by a factor of ( 73..4 ) 2 , and the percent increase is 65 4 (73.4 65.4) 2 − 1 = 26.0%. 15.78: a) Consider the derivation of the speed of a longitudinal wave in Section 16.2. Instead of the bulk modulus B, the quantity of interest is the change in force per fractional length change. The force constant k ′ is the change in force force per length change, so the force change per fractional length change is k ′L, the applied force at one end is F = (k ′L)(v y v) and the longitudinal impulse when this force is applied for a time t is k ′Lt v y v . The change in longitudinal momentum is ((vt ) m L)v y and equating the expressions, canceling a factor of t and solving for v gives v 2 = L2 k ′ m. An equivalent method is to use the result of Problem 11.82(a), which relates the force constant k ′ and the “Young’s modulus” of the Slinky TM , k ′ = YA L , or Y = k ′ L A. The mass density is ρ = m ( AL), and Eq. (16.8) gives the result immediately. b) ( 2.00 m) (1.50 N m) (0.250 kg) = 4.90 m s.

15.79: a) uk = b)
∂y ∂t

2 2 ∆K (1 2)∆mv y 1  ∂y  = = μ  . ∆x ∆m μ 2  ∂t 

= ωA sin( kx − ωt ) and so uk = 1 μω2 A2 sin 2 (kx − ωt ). 2

c) The piece has width ∆x and height ∆x ∂y , and so the length of the piece is ∂x
2   ∂y   2  (∆x) +  ∆x     ∂x     12

  ∂y  2  = ∆x1 +      ∂x    

12

 1  ∂y  2  ≈ ∆x 1 +   ,  2  ∂x    

where the relation given in the hint has been used. d) e) ∆x 1 + 1 ( ∂y ) 2 − ∆x 1  ∂y  2 ∂x up = F = F  . ∆x 2  ∂x  ∂y = −kA sin( kx − ωt ), and so ∂x up = 1 2 2 Fk A sin 2 (kx − ωt ) 2

[

]

2

and f) comparison with the result of part (c)with k 2 = ω2 v 2 = ω2 μ F , shows that for a

v sinusoidal wave u k = u p . g) In this graph, uk and up coincide, as shown in part (f).

15.80: a) The tension is the difference between the diver’s weight and the buoyant force, F = (m − ρwaterV ) g = (120 kg − (1000 kg m 3 )(0.0800 m 3 )(9.80 m s 2 )) = 392 N. b) The increase in tension will be the weight of the cable between the diver and the point at x, minus the buoyant force. This increase in tension is then m − (1000 kg m3 )π (1.00 × 10−2 m) 2 )(9.80 m s ) x = (7.70 N m) x The tension as a function of x is then F ( x) = (392 N) + (7.70 N m) x. c) Denote the tension as F ( x) = F0 + ax, where F0 = 392 N and a = 7.70 N m. Then, the speed of transverse waves as a function of x is v = dx = ( F0 + a ) μ and the time t x dt needed for a wave to reach the surface is found from t = ∫ dt = ∫ Let the length of the cable be L, so t = μ∫ = 2 μ a
L

( µx − ρ( Ax) ) g = (1.10 kg

2

μ dx =∫ dx. dx dt F0 + ax

0

dx 2 = μ F0 + ax a F0 + ax F0 + aL − F0

L 0

(

)

=

2 1.10 kg m ( 392 N + (7.70 N m)(100 m ) − 392 N ) = 3.98 s. 7.70 N m

15.81: The tension in the rope will vary with radius r. The tension at a distance r from the center must supply the force to keep the mass of the rope that is further out than r − accelerating inward. The mass of this piece in m LL r , and its center of mass moves in a circle of radius L+ r , and so 2
2  L − r  2  L + r  mω 2 T (r ) = m ω  = ( L − r 2 ). L   L  2L    An equivalent method is to consider the net force on a piece of the rope with length dr and mass dm = dr m L . The tension must vary in such a way that T (r ) − T (r + dr ) = −ω2 r dm, or dT = −(mω2 L)r dr. This is integrated to obtained dr

T (r ) = −(mω2 2 L)r 2 + C , where C is a constant of integration. The tension must vanish at r = L, from which C = (mω2 L 2) and the previous result is obtained. The speed of propagation as a function of distance is dr T (r ) TL ω v( r ) = = = = L2 − r 2 , dt μ m 2 dr where dt > 0 has been chosen for a wave traveling from the center to the edge. Separating variables and integrating, the time t is 2 L dr t = ∫ dt = . ω ∫0 L2 − r 2 The integral may be found in a table, or in Appendix B. The integral is done explicitly by letting r = L sin θ , dr = L cos θ dθ , L2 − r 2 = L cos θ , so that dr r ∫ L2 − r 2 = θ = arcsin L , and t= 2 π arcsin(1) = . ω ω 2

15.82: a) ∂y = kAS W coskx sin ωt, ∂y = −ωASW ωsin kx cosωt , and so the instantaneous ∂x ∂t power is P = FA2 SW ωk (sin kx cos kx)(sin ωt cos ωt ) 1 = FA2 SW ωk sin( 2kx) sin( 2ωt ). 4 b) The average value of P is proportional to the average value of sin( 2ωt ), and the average of the sine function is zero; Pav = 0. c) The waveform is the solid line, and the power is the dashed line. At time t = π 2ω , y = 0 and P = 0 and the graphs coincide. d) When the standing wave is at its maximum displacement at all points, all of the energy is potential, and is concentrated at the places where the slope is steepest (the nodes). When the standing wave has zero displacement, all of the energy is kinetic, concentrated where the particles are moving the fastest (the antinodes). Thus, the energy must be transferred from the nodes to the antinodes, and back again, twice in each cycle. Note that P is greatest midway between adjacent nodes and antinodes, and that P vanishes at the nodes and antinodes.

15.83: a) For a string, f n =

n 2L

F μ

and in this case, n = 1. Rearranging this and solving

for F gives F = μ 4 L2 f 2 . Note that μ = πr 2 ρ, so μ = π (.203 × 10−3 m) 2 (7800 kg m 3 ) = 1.01 × 10−3 kg m. Substituting values, F = (1.01 × 10 −3 kg m)4(.635 m) 2 (247.0 Hz) 2 = 99.4 N. b) To find the fractional change in the frequency we must take the ration of ∆f to f : f = 1 2L F , μ

  1 F 1   = ∆ 1 F 2 , ∆ ( f ) = ∆  2L μ   2L μ      1 1 ∆f = ∆ F2 2L μ 1 1 ∆F ∆f = . 2L μ 2 F

( )

Now divide both sides by the original equation for f and cancel terms: ∆f = f
1 1 ∆F 2L μ 2 F 1 2L F μ

∆f 1 ΔF = f 2 F c) From Section 17.4, ∆F = − YαA∆T , so ∆F = − 2.00 × 1011 Pa 1.20 × 10−5 C° × (π (.203 × 10−3 m) 2 )(11°C) = 3.4 N. Then, ∆ F F = − 0.034, ∆ f f = −0.017, and finally, ∆f = −4.2 Hz, or the pitch falls. This also explains the constant the constant tuning in the string sections of symphonic orchestras.

(

)(

)

Chapter 16

16.1: a) λ = v f = (344 m s) (100 Hz) = 0.344m. b) if p → 1000 p 0 , then A → 1000A0 Therefore, the amplitude is 1.2 × 10 −5 m. c) Since pmax = BkA, increasing pmax while keeping A constant requires decreasing k, and increasing π , by the same factor. Therefore the new wavelength is (0.688 m)(20) = 6.9 m, f new = 3449m s = 50 Hz. 6. m 16.2: A=
p max v ( 3.0×10 −2 Pa) (1480 m s ) = 2 πBf 2 π ( 2.2×10 9 Pa) (1000 Hz)

, or A = 3.21 × 10 −12 m. The much higher bulk modulus

increases both the needed pressure amplitude and the speed, but the speed is proportional to the square root of the bulk modulus. The overall effect is that for such a large bulk modulus, large pressure amplitudes are needed to produce a given displacement. 16.3: From Eq. (16.5), pmax = BkA = 2π BA λ = 2πBA f v.

a) 2π (1.42 × 105 Pa) (2.00 × 10−5 m) (150 Hz) (344 m s) = 7.78 Pa. b) 10 × 7.78 Pa = 77.8 Pa. c) 100 × 7.78 Pa = 778 Pa. The amplitude at 1500 Hz exceeds the pain threshold, and at 15,000 Hz the sound would be unbearable. The values from Example 16.8 are B = 3.16 × 104 Pa, f = 1000 Hz,
216 K 293K

16.4:

A = 1.2 × 10−8 m. Using Example 16.5, v = 344 m s amplitude of this wave is pmax = BkA = B

= 295 m s , so the pressure

2πf A = (3.16 × 104 Pa). v

2π (1000 Hz) (1.2 × 10 −8 m) = 8.1 × 10− 3 Pa. This is (8.1 × 10 −3 Pa) (3.0 × 10−2 Pa) = 0.27 295 m s times smaller than the pressure amplitude at sea level (Example 16-1), so pressure amplitude decreases with altitude for constant frequency and displacement amplitude. a) Using Equation (16.7), B = v 2 ρ = (λf ) 2 , so B = [ (8 m)( 400 s) ] × (1300 kg m 3 ) = 1.33 × 1010 Pa.
2

16.5:

b) Using Equation (16.8), Y = v 2 ρ = ( L t ) 2 ρ = (1.5 m) (3.9 × 10− 4 s × (6400 kg m 3 ) = 9.47 × 1010 Pa.

[

]

2

16.6: a) The time for the wave to travel to Caracas was 9 min 39 s = 579 s and the speed was 1.085 × 10 4 m s (keeping an extra figure). Similarly, the time for the wave to travel to Kevo was 680 s for a speed of 1.278 ×10 4 m s, and the time to travel to Vienna was 767 s for a speed of 1.258 ×10 4 m s. The average speed for these three measurements is 1.21 × 10 4 m s. Due to variations in density, or reflections (a subject addressed in later chapters), not all waves travel in straight lines with constant speeds. b) From Eq. (16.7), B = v 2 ρ , and using the given value of ρ = 3.3 × 10 3 kg m 3 and the speeds found in part (a), the values for the bulk modulus are, respectively, 3.9 × 1011 Pa, 5.4 × 1011 Pa and 5.2 × 1011 Pa. These are larger, by a factor of 2 or 3, than the largest values in Table (11-1). 16.7: Use vwater = 1482 m s at 20°C, as given in Table (16.1) The sound wave travels in water for the same time as the wave travels a distance 22.0 m − 1.20 m = 20.8 m in air, and so the depth of the diver is

( 20.8 m ) v water
vair

= ( 20.8 m )

1482 m s = 89.6 m. 344 m s

This is the depth of the diver; the distance from the horn is 90.8 m. 16.8: a), b), c) Using Eq. (16.10 ) , vH 2 = vH e = vAr =

(1.41) ( 8.3145 J mol ⋅ K ) ( 300.15 K )

( 2.02 × 10

−3

kg mol kg mol

)

= 1.32 × 103 m s = 1.02 × 103 m s = 323 m s .

(1.67 ) ( 8.3145 J mol ⋅ K ) ( 300.15 K ) (1.67 ) ( 8.3145 J mol ⋅ K ) ( 300.15 K )

( 4.00 × 10

−3

)

(39.9 × 10

−3

kg mol

)

d) Repeating the calculation of Example 16.5 at T = 300.15 K gives vair = 348 m s , and so vH 2 = 3.80 vair , vHe = 2.94 vair and vAr = 0.928 vair .

16.9:

Solving Eq. (16.10) for the temperature, (28.8 × 10
−3

Mv 2 T = = γR

  850 km h   1 m s   kg mol)       0.85   3.6 km hr  

2

(1.40)( 8.3145 J mol ⋅ K )

= 191 K,

or − 82°C. b) See the results of Problem 18.88, the variation of atmospheric pressure with altitude, assuming a non-constant temperature. If we know the altitude we can use  α y   Rα   the result of Problem 18.88, p = p0 1 − . Since T = To − αy,  T0    for T = 191 K, α = .6 × 10−2 °C m, and T0 = 273 K, y = 13,667 m (44,840 ft.). Although a very high altitude for commercial aircraft, some military aircraft fly this high. This result assumes a uniform decrease in temperature that is solely due to the increasing altitude. Then, if we use this altitude, the pressure can be found:  (.6 × 10 ° C m) (13,667 m)   p = p o 1 −   273 K  
−2  ( 28.8×10 −3 kg mol)(9.8 m s 2 )     (8.315 J mol ⋅ K)(.6×10 − 2 ° C m)     Mg   

,

p = p o (.70) 5.66 = .13p o , or about .13 atm. Using an altitude of 13,667 m in the equation derived in Example 18.4 gives p = .18p o , which overestimates the pressure due to the assumption of an isothermal atmosphere. and 16.10: As in Example 16 - 5, with T = 21°C = 294.15 K, v= γRT (1.04)(8.3145 J mol ⋅ K)(294.15 K) = = 344.80 m s. M 28.8 × 10− 3 kg mol

The same calculation with T = 283.15 Κ gives 344.22 m s, so the increase is 0.58 m s. 16.11: Table 16.1 suggests that the speed of longitudinal waves in brass is much higher than in air, and so the sound that travels through the metal arrives first. The time difference is ∆t = L L 80.0 m 80.0 m − = − = 0.208 s. vair vBrass 344 m s (0.90 × 1011 Pa) (8600 kg m 3 )

16.12:

(1.40)(8.3145 J mol ⋅ K)(300.15 K) (1.40)(8.3145 J mol ⋅ K)(260.15 K) − −3 (28.8 × 10 kg mol) (28.8 × 10− 3 kg mol) = 24 m s.

(The result is known to only two figures, being the difference of quantities known to three figures.) 16.13: The mass per unit length µ is related to the density (assumed uniform) and the cross-section area A by μ = Aρ, so combining Eq. (15.13) and Eq. (16.8) with the given relations between the speeds, Υ F Υ = 900 so F A = ⋅ ρ Αρ 900

16.14:

Υ ρ (11.0 × 1010 Pa) (8.9 × 103 kg m 3 ) v a) λ = = = = 16.0 m. f f 220 Hz

b) Solving for the amplitude A (as opposed to the area a = πr 2 ) in terms of the average power Pav = Ιa, A= = (2 Pav a ) ρΥω2 2(6.50 × 10−6 ) W) (π (0.800 × 10-2 m) 2 ) (8.9 × 10 kg m )(11.0 × 10 Pa) (2π (220 Hz))
3 10 3 2

= 3.29 × 10−8 m.

c) ωΑ = 2π f Α = 2π (220 Hz)(3.289 × 10−8 m) = 4.55 × 10−5 m s.

16.15:

a) See Exercise 16.14. The amplitude is 2Ι Α= ρΒω2 = 2(3.00 × 10−6 W m 2 ) (1000 kg m )(2.18 × 10 Pa) (2π (3400 Hz))
3 9 2

= 9.44 × 10−11 m.

The wavelength is v λ= = f B ρ ( 2.18 × 109 Pa) (1000 kg m 3 ) = = 0.434 m. f 3400 Hz

b) Repeating the above with B = γp = 1.40 × 105 Pa and the density of air gives A = 5.66 × 10−9 m and λ = 0.100 m. c) The amplitude is larger in air, by a factor of about 60. For a given frequency, the much less dense air molecules must have a larger amplitude to transfer the same amount of energy.
2 From Eq. (16.13), I = vp max 2 B, and from Eq. (19.21), v 2 = B ρ . Using

16.16:

Eq. (16.7) to eliminate v, I = eliminate B, I = vp
2 max 2

(

2 2 B ρ pmax 2 B = pmax 2 ρB . Using Eq. (16.7) to 2 max

)

2(v ρ) = p

2 ρv.

16.17:

a) pmax = BkA =

2 πBfA v

=

2 π (1.42×10 5 Pa) (150 Hz) (5.00 ×10 −6 m) (344 m s)

= 1.95 Pa.

b) From Eq. (16.14), 2 I = pmax 2 ρv = (1.95 Pa) 2 (2 × (1.2 kg m 3 )(344 m s)) = 4.58 × 10−3 W m 2 . c) 10 × log

(

4.58 ×10 −3 10 −12

) = 96.6 dB.

16.18: (a) The sound level is β = (10 dB) log II0 , so β = (10 dB) log

0.500 μ W m 2 10 −12 W m 2

, or β = 57 dB.

b) First find v, the speed of sound at 20.0 °C, from Table 16.1, v = 344 m s. The density of air at that temperature is 1.20 kg m 3 . Using Equation (16.14), I=
2 pmax (0.150 N m 2 ) 2 = , or I = 2.73 × 10 −5 W m 2 . Using this in Equation 2 ρv 2(1.20 kg m3 )(344 m s)

(16.15), β = (10 dB) log

2.73 × 10−5 W m 2 , or β = 74.4 dB. 10−12 W m 2

16.19:

a) As in Example 16.6, I =

( 6.0×10 −5 Pa) 2 2 (1.20 kg m 3 )( 344 m s)

= 4.4 × 10−12 W m 2 . β = 6.40 dB.

16.20: a) 10 × log ( 4II ) = 6.0 dB. b) The number must be multiplied by four, for an increase of 12 kids. 16.21: Mom is five times further away than Dad, and so the intensity she hears is −2 1 of the intensity that he hears, and the difference in sound intensity levels is 25 = 5 10 × log(25) = 14 dB. 16.22: ∆ (Sound level) = 75 dB − 90 dB = −25 dB ∆(Sound level) = 10 log I 0f − 10 log II0i = 10 log IIfi Therefore I − 25 dB = 10 log Ifi
If Ii I

= 10−2.5 = 3.2 × 10−3

16.23:

β = (10 dB)log II0 , or 13 dB = (10 dB)log II0 . Thus, I I 0 = 20.0, or the intensity has

increased a factor of 20.0.

16.24:

Open Pipe: λ1 = 2 L = v v = f1 594 Hz v f

Closed at one end: λ1 = 4 L = Taking ratios: 2 L v 594 Hz = 4L v f 594 Hz f = = 297 Hz 2 16.25: a) Refer to Fig. (16.18). i) The fundamental has a displacement node at L = 0.600 m , the first overtone mode has displacement nodes at L = 0.300 m 2 4 and 34L = 0.900 m and the second overtone mode has displacement nodes at L = 0.200 m, L = 0.600 m and 56L = 1.000 m . ii) Fundamental: 0, L = 1.200 m. First : 0, 6 2 L = 0.600 m, L = 1.200 m. Second : 0, L = 0.400 m, 23L = 0.800 m, L = 1.200 m. 3 2 b) Refer to Fig. (16.19); distances are measured from the right end of the pipe in the figure. Pressure nodes at: Fundamental: L = 1.200 m . First overtone: L 3 = 0.400 m, L = 1.200 m. Second overtone: L 5 = 0.240 m, 3L 5 = 0.720 m , L = 1.200 m. Displacement nodes at Fundamental: 0. First overtone: 0, 2 L 3 = 0.800 m. Second overtone: 0, 2 L 5 = 0.480 m, 4 L 5 = 0.960 m

16.26:

a) f1 =

v 2L

=

( 344 m s ) 2 ( 0.450 m)

= 382 Hz, 2 f1 = 764 Hz, f 3 = 3 f1 = 1147 Hz,

f 4 = 4 f1 = 1529 Hz. b) f1 = 4vL = 191 Hz, f 3 = 3 f1 = 573 Hz, f 5 = 5 f1 = 956 Hz, f 7 = 7 f1 = 1338 Hz. Note that the symbol “ f 1 ” denotes different frequencies in the two parts. The frequencies are not always exact multiples of the fundamental, due to rounding. c) Open: 20 ,f000 = 52.3, so the 52nd harmonic is heard. Stopped; 20 ,f000 = 104.7, so 103 rd 1 1 highest harmonic heard. f1 =
( 344 m/s) 4(0.17 m)

16.27:

= 506 Hz, f 2 = 3 f1 = 1517 Hz, f 3 = 5 f1 = 2529 Hz.

16.28: a) The fundamental frequency is proportional to the square root of the ratio (see Eq. (16.10)), so f He = f air = γ He M air (5 3) 28.8 ⋅ = (262 Hz) ⋅ = 767 Hz, γ air M He (7 5) 4.00

γ M

b) No; for a fixed wavelength , the frequency is proportional to the speed of sound in the gas. 16.29: a) For a stopped pipe, the wavelength of the fundamental standing wave is 4 L = 0.56 m, and so the frequency is f1 = ( 344 m s ) (0.56 m) = 0.614 kHz. b) The length of the column is half of the original length, and so the frequency of the fundamental mode is twice the result of part (a), or 1.23 kHz. 16.30: For a string fixed at both ends, Equation (15.33) , f n =
( 2 )( .635 m )( 588 /s )
3 nv 2L

, is useful. It is
2 fnL n

important to remember the second overtone is the third harmonic. Solving for v, v = and inserting the data, v = , and v = 249 m s .

,

16.31: a) For constructive interference, the path difference d = 2.00 m must be equal to an integer multiple of the wavelength, so λ n = d n, fn = v vn 344 m s v = = n  = n = n(172 Hz ). λn d 2.00 m d 

Therefore, the lowest frequency is 172 Hz. b) Repeating the above with the path difference an odd multiple of half a wavelength, f n = ( n + 1 ) (172 Hz ). Therefore, the lowest frequency is 86 Hz ( n = 0 ). 2

16.32: The difference in path length is ∆x = ( L − x ) − x = L − 2 x, or x = ( L − ∆x ) 2 . For destructive interference, ∆x = (n + (1 2))λ ,and for constructive interference, ∆x = nλ. The wavelength is λ = v f = ( 344 m s ) (206 Hz) = 1.670 m (keeping an extra figure), and so to have 0 ≤ x ≤ L, − 4 ≤ n ≤ 3 for destructive interference and − 4 ≤ n ≤ 4 for constructive interference. Note that neither speaker is at a point of constructive or destructive interference. a) The points of destructive interference would be at x = 0.58 m, 1.42 m. b) Constructive interference would be at the points x = 0.17 m, 1.00 m, 1.83 m. c) The positions are very sensitive to frequency, the amplitudes of the waves will not be the same (except possibly at the middle), and exact cancellation at any frequency is not likely. Also, treating the speakers as point sources is a poor approximation for these dimensions, and sound reaches these points after reflecting from the walls , ceiling, and floor. 16.33: λ = v f = ( 344 m s ) ( 688 Hz ) = 0.500 m To move from constructive interference to destructive interference, the path difference must change by λ 2. If you move a distance x toward speaker B, the distance to B gets shorter by x and the difference to A gets longer by x so the path difference changes by 2x. 2 x = λ 2 and x = λ 4 = 0.125 m 16.34: We are to assume v = 344 m s , so λ = v f = ( 344 m/s ) (172 Hz ) = 2.00 m. If rA = 8.00 m and rB are the distances of the person from each speaker, the condition for destructive interference is rB − rA = ( n + 1 ) λ, where n is any integer. Requiring 2 1 1 rB = rA + ( n + 2 ) λ > 0 gives n + 2 > − rA λ = − ( 8.00 m ) ( 2.00 m ) = −4, so the smallest value of rB occurs when n = −4, and the closest distance to B is rB = 8.00 m + ( - 4 + 1 ) ( 2.00 m ) = 1.00 m. 2

16.35: λ = v f = ( 344 m s ) ( 860 Hz ) = 0.400 m The path difference is 13.4 m − 12.0 m = 1.4 m. path difference = 3.5 λ The path difference is a half-integer number of wavelengths, so the interference is destructive. 16.36: a) Since f beat = f a − f b , the possible frequencies are 440.0 Hz ± 1.5 Hz = 438.5 Hz or 441.5 Hz b) The tension is proportional to the square of the frequency. ( 1.5 Hz Therefore T ∝ f 2 and ∆T ∝ 2 f∆f . So ∆TT = 2 ∆f . i) ∆TT = 2440 Hz ) = 6.82 × 10 −3. f ii)
∆T T

=

2 ( −1.5 Hz ) 440 Hz

= −6.82 × 10 −3.

16.37: a) A frequency of 1 (108 Hz + 112 Hz ) = 110 Hz will be heard, with a beat 2 frequency of 112 Hz–108 Hz = 4 beats per second. b) The maximum amplitude is the sum of the amplitudes of the individual waves, 2 1.5 × 10 −8 m = 3.0 × 10 −8 m. The minimum amplitude is the difference, zero.

(

)

Solving Eq. (16.17) for v, with vL = 0, gives fL 1240 Hz   v= vS =  ( − 25.0 m s ) = 775 m s , fS − f L  1200 Hz − 1240Hz  or 780 m s to two figures (the difference in frequency is known to only two figures). Note that vS < 0, since the source is moving toward the listener. Redoing the calculation with +20.0 m s for vS and − 20.0 m/s for vL gives 267

16.38:

16.39: Hz.

′ 16.40: a) From Eq. (16.17 ) , with vS = 0, vL = −15.0 m s , f A = 375 Hz. ′ b) With vS = 35.0 m s , vL = 15.0 m s , f B = 371 Hz. ′ ′ ′ c) f A − f B = 4 Hz (keeping an extra figure in f A ) . The difference between the frequencies is known to only one figure.

16.41:

In terms of wavelength, Eq. (16.29) is v + vs λL = λS ⋅ v + vL a) vL = 0, vS = −25.0 m and λ L = ( 319 ) ( 344 m s ) ( 400 Hz ) = 0.798 m. This is, of 344 course, the same result as obtained directly from Eq. (16.27). vS = 25.0 m s and vL = ( 369 m s ) ( 400 Hz ) = 0.922 m. The frequencies corresponding to these wavelengths are c) 431 Hz and d) 373 Hz.

16.42:

a) In terms of the period of the source, Eq. (16.27) becomes vS = v − λ 0.12 m = 0.32 m s − = 0.25 m s . TS 1.6 s

b) Using the result of part (a) in Eq. (16.18), or solving Eq. (16.27) for v S and substituting into Eq. (16.28) (making sure to distinguish the symbols for the different wavelengths) gives λ = 0.91 m.  v + vL  fL =   v + v  fS  S   a) The direction from the listener to source is positive, so vS = − v 2 and vL = 0.  v  fL =   v − v 2  f S = 2 fS = 2.00 kHz    b) vS = 0, vL = + v 2 v +v 2 fL =   fS = 3 fS = 1.50 kHz 2 v   This is less than the answer in part (a). The waves travel in air and what matters is the velocity of the listener or source relative to the air, not relative to each other. For a stationary source, vS = 0, so f L =
v + vL v + vS

16.43:

16.44:

which gives vL = v

(

fL fS

490 Hz − 1 = ( 344 m s ) ( 520 Hz − 1) = −19.8 m/s.

)

f S = (1 + vL v ) f S ,

This is negative because the listener is moving away from the source.

16.45: a) vL = 18.0 m/s, vS = −30.0 m s , and Eq. (16.29) gives f L = ( 362 ) ( 262 Hz ) 314 = 302 Hz. b) vL = −18.0 m s , vS = 30.0 m/s and f L = 228 Hz. 16.46: a) In Eq. (16.31), v vS = 1 1.70 = 0.588 and α = arcsin(0.5 88) = 36.0°. b) As in Example 16.20, ( 950 m ) t= = 2.23 s. (1.70) (344 m s) ( tan(36.0°))

16.47: a) Mathematically, the waves given by Eq. (16.1) and Eq. (16.4) are out of phase. Physically, at a displacement node, the air is most compressed or rarefied on either side of the node, and the pressure gradient is zero. Thus, displacement nodes are pressure antinodes. b) (This is the same as Fig. (16.3).) The solid curve is the pressure and the dashed curve is the displacement.

c)

The pressure amplitude is not the same. The pressure gradient is either zero or undefined. At the places where the pressure gradient is undefined mathematically (the “cusps” of the y - x plot), the particles go from moving at uniform speed in one direction to moving at the same speed in the other direction. In the limit that Fig. (16.43) is an accurate depiction, this would happen in a vanishing small time, hence requiring a very large force, which would result from a very large pressure gradient. d) The statement is true, but incomplete. The pressure is indeed greatest where the displacement is zero, but the pressure is equal to its largest value at points other than those where the displacement is zero. 16.48: The altitude of the plane when it passes over the end of the runway is (1740 m − 1200 m)tan 15° = 145 m , and so the sound intensity is 1 (1.45) 2 of what the intensity would be at 100 m. The intensity level is then 100.0 dB − 10 × log (1.45) 2 = 96.8 dB, so the airliner is not in violation of the ordinance.

[

]

16.49:

a) Combining Eq. (16.14) and Eq. (16.15), pmax = 2 ρvI 010( β 10 ) = 2(1.20 kg m 3 )(344 m s)(10−12 W m )105.20 = 1.144 × 10 −2 Pa,
2

or = 1.14 × 10 −2 Pa, to three figures.

b) From Eq. (16.5), and as in Example 16.1,

pmax pmax v (1.144 × 10−2 Pa) (344 m s) A= = = = 7.51 × 10 − 9 m. 5 Bk B 2πf 2π (1.42 × 10 Pa)( 587 Hz ) c) The distance is proportional to the reciprocal of the square root of the intensity, and hence to 10 raised to half of the sound intensity levels divided by 10. Specifically, (5.00 m)10(5.20 − 3.00) 2 = 62.9 m. a) p = IA = I 010( β 10 dB) A. b) 1.00 × 10−12 W m 2 (105.50 )(1.20 m 2 )
−7

16.50:

(

)

= 3.79 × 10 W.
16.51: For the flute, the fundamental frequency is 344.0 m s v f 1f = = = 800.0 Hz 4 L 4(0.1075 m) For the flute and string to be in resonance, nf f1f = ns f1s , where f1s = 600.0 Hz is the fundamental frequency for the string. ns = nf ( f 1 f f 1s ) = 4 nf 3 ns is an integer when nf = 3N , N = 1, 3, 5... (the flute has only odd harmonics) nf = 3 N gives ns = 4 N Flute harmonic 3 N resonates with string harmonic 4 N , N = 1,3,5,...

16.52: f
string 3

(a) The length of the string is d = L 10, so its third harmonic has frequency

= 3 21d F μ. The stopped pipe has length L, so its first harmonic has frequency v 1 f1pipe = s . Equating these and using d = L 10 gives F = μvs2 . 4L 3600 (b) If the tension is doubled, all the frequencies of the string will increase by a factor of 2 . In particular, the third harmonic of the string will no longer be in resonance with the first harmonic of the pipe because the frequencies will no longer match, so the sound produced by the instrument will be diminished. (c) The string will be in resonance with a standing wave in the pipe when their frequencies are equal. Using f1pipe = 3 f1string , the frequencies of the pipe are nf1pipe = 3nf1string , (where n=1, 3, 5…). Setting this equal to the frequencies of the string n′f1string , the harmonics of the string are n′ = 3n = 3, 9, 15,... 16.53: a) For an open pipe, the difference between successive frequencies is the fundamental, in this case 392 Hz, and all frequencies would be integer multiples of this frequency. This is not the case, so the pipe cannot be an open pipe. For a stopped pipe, the difference between successive frequencies is twice the fundamental, and each frequency is an odd integer multiple of the fundamental. In this case, f1 = 196 Hz, and 1372 Hz = 7 f1 , 1764 Hz = 9 f1. b) n = 7 for 1372 Hz, n = 9 for 1764 Hz. c) f1 = v 4 L, so L = v 4 f1 = ( 344 m/s ) ( 784 Hz ) = 0.439 m. 16.54: The steel rod has standing waves much like a pipe open at both ends, as shown in Figure (16.18). An integral number of half wavelengths must fit on the rod, that is, nv fn = . 2L a) The ends of the rod are antinodes because the ends of the rod are free to ocsillate. b) The fundamental can be produced when the rod is held at the middle because a node is located there. (1) ( 5941 m s ) = 1980 Hz. c) f1 = 2(1.50 m ) d) The next harmonic is n = 2, or f 2 = 3961 Hz. We would need to hold the rod at an n = 2 node, which is located at L 4 from either end, or at 0.375 m from either end.

16.55: The shower stall can be modeled as a pipe closed at both ends, and hence there are nodes at the two end walls. Figure (15.23) shows standing waves on a string fixed at both ends but the sequence of harmonics is the same, namely that an integral number of half wavelengths must fit in the stall. nv a) The condition for standing waves is f n = 2 L , so the first three harmonics are n = 1, 2, 3. nv b) A particular physics professor’s shower has a length of L = 1.48 m. Using f n = 2L , the resonant frequencies can be found when v = 344 m s . n f( Hz ) 1 116 2 232 3 349 Note that the fundamental and second harmonic, which would have the greatest amplitude, are frequencies typically in the normal range of male singers. Hence, men do sing better in the shower! (For a further discussion of resonance and the human voice, see Thomas D. Rossing , The Science of Sound, Second Edition, Addison-Wesley, 1990, especially Chapters 4 and 17.) 16.56: a) The cross-section area of the string would be a = (900 N) (7.0 × 108 Pa) = 1.29 × 10 −6 m 2 , corresponding to a radius of 0.640 mm (keeping extra figures). The length is the volume divided by the area, L= V m ρ (4.00 × 10 −3 kg) = = = 0.40 m. a a (7.8 × 103 kg m 3 )(1.29 × 10− 6 m 2 )

b) Using the above result in Eq. (16.35) gives f1 = 377 Hz, or 380 Hz to two figures.

16.57: a) The second distance is midway between the first and third, and if there are no other distances for which resonance occurs, the difference between the first and third positions is the wavelength λ = 0.750 m. (This would give the first distance as λ 4 = 18.75 cm, but at the end of the pipe, where the air is not longer constrained to move along the tube axis, the pressure node and displacement antinode will not coincide exactly with the end). The speed of sound in the air is then v = fλ = (500 Hz)(0.750 m) = 375 m s. b) Solving Eq. (16.10) for γ , γ= Mv 2 (28.8 × 10−3 kg mol)(375 m s) 2 = = 1.39. RT (8.3145 J mol ⋅ K)(350.15 K)

c) Since the first resonance should occur at τ 4 = 0.875 m but actually occurs at 0.18 m, the difference is 0.0075 m. 16.58: a) Considering the ear as a stopped pipe with the given length, the frequency of the fundamental is f1 = v 4 L = (344 m s) (0.10 m) = 3440 Hz; 3500 Hz is near the resonant frequency, and the ear will be sensitive to this frequency. b) The next resonant frequency would be 10,500 Hz and the ear would be sensitive to sounds with frequencies close to this value. But 7000 Hz is not a resonant frequency for an open pipe and the ear is not sensitive at this frequency. 16.59: a) From Eq. (15.35), with m the mass of the string and M the suspended mass, f1 = F Mg = = 4mL πd 2 L2 ρ (420.0 × 10−3 kg)(9.80 m s 2 ) = 77.3 Hz π (225 × 10 − 6 m) 2 (0.45 m) 2 (21.4 × 103 kg m 3 ) and the tuning fork frequencies for which the fork would vibrate are integer multiples of 77.3 Hz. b) The ratio m M ≈ 9 × 10 −4 , so the tension does not vary appreciably along the string.

16.60: a) L = λ 4 = v 4 f = (344 m s) (4(349 Hz)) = 0.246 m. b) The frequency will be proportional to the speed, and hence to the square root of the Kelvin temperature. The temperature necessary to have the frequency be higher is (293.15 K)(1.060) 2 = 329.5 K, which is 56.3° C.

The wavelength is twice the separation of the nodes, so γRT v = λf = 2 Lf = . M Solving for γ, γ= M (16.0 × 10−3 kg) ( 2(0.200 m)(1100 Hz)) 2 = 1.27. (2 Lf ) 2 = RT (8.3145 J mol ⋅ K ) (293.15 K)

16.61:

16.62: If the separation of the speakers is denoted h, the condition for destructive interference is

x 2 + h 2 − x = βλ ,
where β is an odd multiple of one-half. Adding x to both sides, squaring, cancelling the x 2 term from both sides and solving for x gives h2 β x= − λ. 2 βλ 2 Using λ = and 0.026 m ( β = 9 ) . These are the only allowable values of β 2 that give positive solutions for x . (Negative values of x may be physical, depending on speaker design, but in that case the difference between path lengths is x 2 + h 2 + x.) b) Repeating the above for integral values of β , constructive interference occurs at 4.34 m, 1.84 m, 0.86 m, 0.26 m . Note that these are between, but not midway between, the answers to part (a). c) If h = λ 2 , there will be destructive interference at speaker B. If λ 2 > h, the path difference can never be as large as λ 2 . (This is also obtained from the above expression for x , with x = 0 and β = 1 .) The minimum frequency is then 2 v 2h = (344 m s) (4.0 m) = 86 Hz. a) The wall serves as the listener, want f L = 600 Hz.  v + vS  fS =   v + v  fL   L  vL = 0, vS = −30 m s, v = 344 m s fS = 548 Hz b) Now the wall serves as a stationary source with f s = 600 Hz  v + vL  fL =   v + v  fS  S   vS = 0, vL = +30 m s, v = 344 m s f L = 652 Hz

( β = 52 ), 0.53 m ( β = 7 ) 2

v f

and h from the given data yields 9.01 m ( β =

1 2

), 2.71 m( β = 3 ), 1.27 m 2

16.63:

16.64: To produce a 10.0 Hz beat, the bat hears 2000 Hz from its own sound plus 2010 Hz coming from the wall. Call v the magnitude of the bat’s speed, f w the frequency the wall receives (and reflects), and V the speed of sound. Bat is moving source and wall is stationary observer: V V −v = f w 2000 Hz Bat is moving observer and wall is stationary source: V +v V = 2010 Hz f w Solve (1) and (2) together: v = 0.858 m s (2) (1)

16.65: a) A = ∆R ⋅ pmax = BkA = pmax = 2π ρB f∆R, I =

2πBA 2πBAf = . In air v = λ v

B . Therefore ρ

p 2 max = 2π 2 ρB f 2 ( ∆R ) 2 . 2 ρB ,

b) PTot = 4πR 2 I = 8π3 ρB f 2 R 2 (∆R ) 2 c) I =
PTot 4 πd 2

=

2π 2

ρB f 2 R 2 ( ∆R ) 2 d2

p max = (2 ρB I )1 2 =

2π ρB fR( ∆R ) p max R (∆R ) ,A= = . d d 2π ρB f

16.66: (See also Problems 16.70 and 16.74). Let f 0 = 2.00 MHz be the frequency of the generated wave. The frequency with which the heart wall receives this wave is f H = v +vv H f 0 , and this is also the frequency with which the heart wall re-emits the wave. The detected frequency of this reflected wave is f ′ v −vv H , f H , with the minus sign indicating that the heart wall, acting now as a source of
H waves, is moving toward the receiver. Combining, f ′ v − v H f 0 , and the beat frequency is

v+v

 v + vH  2vH f beat = f ′ − f 0 =  − 1 f 0 = f0 . v − vH  v − vH 

Solving for vH ,     f beat 85 Hz vH = v    2 f + f  = (1500 m s )  2(2.00 × 106 Hz) + (85 Hz)     beat   0 −2 = 3.19 × 10 m s. Note that in the denominator in the final calculation, f beat is negligible compared to f 0 . 16.67: a) λ = v f = (1482 m s) (22.0 × 103 Hz) = 6.74 × 10 −2 m. or Problem 16.70; the difference in frequencies is

b) See Problem 16.66

 2vW  2( 4.95 m s ) 3 ∆f = f S   v − v  = 22.0 × 10 Hz (1482 m s ) − ( 4.95 m s ) = 147 Hz.  W  

(

)

The reflected waves have higher frequency. 16.68: a) The maximum velocity of the siren is ωP AP = 2πf P AP . You hear a sound with frequency f L = f siren v ( v + vS ) , where vS varies between + 2πf P AP and − 2πf P AP . So f L − max = f siren v ( v − 2πf P AP ) and f L − min = f siren v ( v + 2πf P AP ) . b) The maximum (minimum) frequency is heard when the platform is passing through equilibrium and moving up (down).

16.69: a) Let vb be the speed of the bat, vi the speed of the insect and f i the frequency with which the sound waves both strike and are reflected from the insect. The frequencies at which the bat sends and receives the signals are related by  v + vb   v + vi  v + vb   = fs  fL = fi   v−v   v − v  v − v .    i  b  i   Solving for vi ,  fS  v + vb    1 −  f L  v − vb     = v  f L ( v − vb ) − fS ( v + vb ) . vi = v   f (v − v ) + f (v + v )   fS  v + vb   b S b   L  1 +  f L  v − vb       Letting f L = f refl and f S = f bat gives the result. b) If f bat = 80.7 kHz, f refl = 83.5 kHz, and vbat = 3.9 m s , vinsect = 2.0 m s . 16.70: (See Problems 16.66, 16.74, 16.67). a) In a time t, the wall has moved a distance v1t and the wavefront that hits the wall at time t has traveled a distance vt, where v = f 0 λ0 , and the number of wavecrests in the total distance is
( v + v1 ) t λ0

.

b) The reflected

wave has traveled vt and the wall has moved v1t , so the wall and the wavefront are separated by (v − v1 )t. c) The distance found in part (b) must contain the number of reflected waves found in part (a), and the ratio of the quantities is the wavelength of the v −v reflected wave, λ 0 v + v1 . d) The speed v divided by the result of part (c), expressed in 1
v−v terms of f 0 is f 0 v + v1 . This is what is predicted by the problem-solving strategy. 1
1 e) f 0 v + v1 − f 0 = f 0 v2−vv11 . v−v

16.71:

a) fR = fL
12 −1 2 1− v c−v  v  v c = fS = f S 1 −  1 +  . c+v 1+ v  c  c c

(1 + x )

b) For small x, the binomial theorem (see Appendix B) gives (1 − x )
−1 2

12

≈ 1 − x 2,

≈ 1 − x 2, so v    v f L ≈ f S 1 −  ≈ f S 1 −   2c   c
2

where the binomial theorem has been used to approximate (1 − x 2 ) ≈ 1 − x.
2

The above result may be obtained without resort to the binomial theorem by expressing f R in terms of f S as f R = fS 1 − (v c ) 1 − (v c ) 1 + (v c ) 1 − (v c ) = fS 1 − (v / c ) 1 − (v c )
2

.

To first order in v c , the square root in the denominator is 1, and the previous result is obtained. c) For an airplane, the approximation v << c is certainly valid, and solving the expression found in part (b) for v, v=c fS − f R f 46.0 Hz = c beat = (3.00 × 108 m s) = 56.8 m s , fS fS 2.43 × 108 Hz

and the approximation v << c is seen to be valid. Note that in this case, the frequency difference is known to three figures, so the speed of the plane is known to three figures.

16.72:

a) As in Problem 16.71, fS − f R − 0.018 × 1014 Hz 8 v=c = (3.00 × 10 m s) = −1.2 × 106 m s , 8 fS 4.568 × 10 Hz

with the minus sign indicating that the gas is approaching the earth, as is expected since f R > f S . b) The radius is (952 yr) (3.156 × 107 s yr )(1.2 × 106 m s) = 3.6 × 1016
m = 3.8 ly . This may also be obtained from (952 yr)
f R − fS fS

.

c) The ratio of the width

of the nebula to 2π times the distance from the earth is the ratio of the angular width (taken as 5 arc minutes) to an entire circle, which is 60 × 360 arc minutes. The distance to the nebula is then (keeping an extra figure in the intermediate calculation) 2 × 3.75 ly (60) × (360) = 5.2 × 10 3 ly, 5

so the explosion actually took place about 4100 B.C 16.73: a) The frequency is greater than 2800 MHz; the thunderclouds, moving toward the installation, encounter more wavefronts per time than would a stationary cloud, and so an observer in the frame of the storm would detect a higher frequency. Using the result of Problem 16.71, with v = −42.0 Km h, f R − fS = f S −v (42.0 km h) (3.6 km h/1 m s) = (2800 × 106 Hz) = 109 Hz. c (3.00 × 108 m s)

b) The waves are being sent at a higher frequency than 2800 MHz from an approaching source, and so are received at a higher frequency. Repeating the above calculation gives the result that the waves are detected at the installation with a frequency 109 Hz greater than the frequency with which the cloud received the waves, or 218 Hz higher than the frequency at which the waves were originally transmitted at the receiver. Note that in doing the second calculation, f S = 2800 MHz + 109 Hz is the same as 2800 MHz to three figures.

16.74:

a) (See also Example 16.19 and Problem 16.66.) The wall will receive and v reflect pulses at a frequency f 0 , and the woman will hear this reflected wave at a v − vw frequency v + vw v v + vw ⋅ f0 = f0 ; v v − vw v − vw The beat frequency is  v + vw   2vw  f beat = f 0   v − v − 1 = f 0  v − v .    w w     b) In this case, the sound reflected from the wall will have a lower frequency, and using f 0 (v − vw ) (v + vw ) as the detected frequency (see Example 21-12; vw is replaced by − vw in the calculation of part (a)),  v − vw   2vw  f beat = f 0 1 −  v + v  = f 0  v + v .    w  w    16.75: Refer to Equation (16.31) and Figure (16.38). The sound travels a distance vT and the plane travels a distance vsT before the boom is found. So, h 2 = (vT ) 2 + (vsT ) 2 , or v hv vsT = h 2 − v 2T 2 . From Equation (16.31), sin α = . Then, vs = . 2 vs h − v 2T 2

16.76:

a)

b) From Eq. (16.4), the function that has the given p( x, 0) at t = 0 is given graphically as shown. Each section is a parabola, not a portion of a sine curve. The period is λ v = (0.200 m) (344 m s) = 5.81 × 10−4 s and the amplitude is equal to the area under the p − x curve between x = 0 and x = 0.0500 m divided by B, or 7.04 × 10 −6 m.

c) Assuming a wave moving in the + x -direction, y (0, t ) is as shown.

d) The maximum velocity of a particle occurs when a particle is moving throughout the origin, and the particle speed is v y = − ∂y v = pv . The maximum velocity is found from ∂x B the maximum pressure, and v ymax = (40 Pa)(344 m s) (1.42 × 105 Pa) = 9.69 cm s . The maximum acceleration is the maximum pressure gradient divided by the density, a max = (80.0 Pa) (0.100 m) = 6.67 × 10 2 m s 2 . 3 (1.20 kg m )

e) The speaker cone moves with the displacement as found in part (c ); the speaker cone alternates between moving forward and backward with constant magnitude of acceleration (but changing sign). The acceleration as a function of time is a square wave with amplitude 667 m s 2 and frequency f = v λ = (344 m s) (0.200 m) = 1.72 kHz.

16.77: Taking the speed of sound to be 344 m s, the wavelength of the waves emitted by each speaker is 2.00 m. a) Point C is two wavelengths from speaker A and one and one-half from speaker B , and so the phase difference is 180° = π rad. b) P 8.00 × 10−4 W I= = = 3.98 × 10− 6 W m 2 , 2 2 4πr 4π (4.00 m)

and the sound intensity level is (10 dB) log(3.98 × 106 ) = 66.0 dB. Repeating with P = 6.00 × 10−5 W and r = 3.00 m gives I = 5.31 × 10 −7 W and β = 57.2dB. c) With the result of part (a), the amplitudes, either displacement or pressure, must be subtracted. That is, the intensity is found by taking the square roots of the intensities found in part (b), subtracting, and squaring the difference. The result is that I = 1.60 × 10−6 W and β = 62.1 dB.

Chapter 17

17.1: From Eq. (17.1) , a) ( 9 5)( − 62.8) + 32 = −81.0°F. b) ( 9 5)( 56.7 ) + 32 = 134.1°F. c) ( 9 5)( 31.1) + 32 = 88.0°F. 17.2: From Eq. (17.2 ) , a) ( 5 9)( 41.0 − 32 ) = 5.0°C. b) ( 5 9 )(107 − 32 ) = 41.7°C. c) ( 5 / 9 )( − 18 − 32) = −27.8°C. 17.3: 1 C° = 9 F°, so 40.0 = 72.0 F° 5

T2 = T1 + 70.0 F° = 140.2°F 17.4: a) (5 9) (45.0 − (−4.0)) = 27.2° C. b) (5 9) (−56.0 − 44) = −55.6° C.

17.5: a) From Eq. (17.1), ( 9 5)( 40.2 ) + 32 = 104.4°F, which is cause for worry. b) ( 9 5)(12) + 32 = 53.6°F, or 54°F to two figures. 17.6:

( 9 5)(11.8) = 21.2 F°

17.7: 1 K = 1 C° = 9 F° , so a temperature increase of 10 K corresponds to an increase 5 of 18 F° . Beaker B has the higher temperature. 17.8: For (b), ∆TC = ∆TK = −10.0 C°. Then for (a), ∆TF = 9 ∆TC = 5 = −18.0 F°. 17.9:
9 5

( − 10.0 C°)

Combining Eq. (17.2) and Eq. (17.3), 5 TK = ( TF − 32°) + 273.15, 9 and substitution of the given Fahrenheit temperatures gives a) 216.5 K, b) 325.9 K, c) 205.4 K. 17.10: (In these calculations, extra figures were kept in the intermediate calculations to arrive at the numerical results.) a) TC = 400 − 273.15 = 127°C, TF = (9 / 5)(126.85) + 32 = 260°F. b) TC = 95 − 273.15 = −178°C, TF = (9 / 5)(−178.15) + 32 = −289°F. c) TC = 1.55 × 107 − 273.15 = 1.55 × 107°C, TF = (9 / 5)(1.55 × 107 ) + 32 = 2.79 × 107°F.

17.11:

From Eq. (17.3), TK = (−245.92°C) + 273.15 = 27.23 K. From Eq. (17.4), (7.476)(273.16 K) = 2042.14 K − 273.15 = 1769°C. From Eq. (17.4), (325.0 mm )( 373.15 K ) = 444 mm. 273.16 K

17.12:

17.13:

17.14: On the Kelvin scale, the triple point is 273.16 K, so °R = (9/5)273.15 K = 491.69°R. One could also look at Figure 17.7 and note that the Fahrenheit scale extends from − 460°F to + 32°F and conclude that the triple point is about 492 °R.

17.15: From the point-slope formula for a straight line (or linear regression, which, while perhaps not appropriate, may be convenient for some calculators), 4.80 × 10 4 Pa (0.01°C) − (100.0°C) = −282.33°C, 6.50 × 10 4 Pa − 4.80 × 10 4 Pa which is − 282°C to three figures. b) Equation (17.4) was not obeyed precisely. If it were, the pressure at the triple 4 point would be P = (273.16) 6.50×1015 Pa = 4.76 × 10 4 Pa. 373.

(

)

17.16: ∆T = ( ∆L ) ( αL0 ) = ( 25 × 10−2 m ) so the temperature is 183° C .

(( 2.4 × 10

−5

( C°) −1 ( 62.1 m ) ) = 168° C,

17.17:

αL0 ∆T = (1.2 × 10−5 (C°) −1 )(1410 m)(18.0° C − (−5.0)°C) = +0.39 m.

17.18:

d + ∆d = d (1 + α∆T )
= (0.4500 cm)(1 + (2.4 × 10 −5 (C°) −1 )(23.0° C − (−78.0°C))) = 0.4511 cm = 4.511 mm.

17.19:

a) αD0 ∆T = (2.6 × 10 −5 (C°) −1 )(1.90 cm) (28.0°C) = 1.4 × 10−3 cm, so the

diameter is 1.9014 cm. b) αD0 ∆T = −3.6 × 10 −3 cm, so the diameter is 1.8964 cm.

17.20:

α∆T = (2.0 × 10−5 (C°) −1 )(5.00° C − 19.5° C) = −2.9 × 10−4. α = ( ∆L) ( L0 ∆T ) = 2.3 × 10−4 m
−1

17.21:

= 2.3 × 10−5 ( C°) .

(

) ( ( 40.125 × 10

−2

m ( 25.0 C°)

)

)

17.22: 17.23

From Eq. (17.8), ∆T = β V0 ∆T = 75 × 10−5 ( C°)

(

∆V V0 β −1

) (1700 L) ( − 9.0°C) = −11 L, so there is 11 L of air.

=

1.50 ×10 −3 5.1×10 −5 K −1

= 29.4°C, so T = 49.4°C.

17.24: The temperature change is ∆T = 18.0° C − 32.0° C = −14.0 C°. The volume of ethanol contracts more than the volume of the steel tank does, so the additional amount of ethanol that can be put into the tank is ∆Vsteel − ∆Vethanol = ( βsteel − βethanol ) V0 ∆T = 3.6 × 10−5 ( C°) − 75 × 10−5 (C°) −1 2.80 m 3 ( − 14.0 C°) = 0.0280 m 3
−1

(

)(

)

17.25: The amount of mercury that overflows is the difference between the volume change of the mercury and that of the glass; 8.95 cm3 −1 −5 −1 βglass = 18.0 × 10 K − = 1.7 × 10− 5 ( C°) . 3 1000 cm ( 55.0°C )

(

(

)

)

17.26: a) A = L2 , ∆A = 2 L∆L = 2 ∆LL L2 = 2 ∆LL A0 . But ∆A = 2α∆TA0 = ( 2α ) A0 ∆T . b)

∆L L

= α∆T , and so
2

∆A = ( 2α ) A ∆T = ( 2 ) (2.4 × 10−5 (C°) −1 ) (π × ( .275 m ) ) (12.5°C ) = 1.4 × 10 −4 m 2 . πD 2 π 2 = (1.350 cm ) = 1.431 cm 2 . 4 4

17.27:

a) A0 =

b) A = A0 (1 + 2α ∆T ) = 1.431 cm 2 1 + ( 2 ) 1.20 × 10 −5°C (150°C ) = 1.437 cm 2 .

(

)(

(

)

)

17.28: (a ) No, the brass expands more than the steel. (b) call D the inside diameter of the steel cylinder at 20°C At 150°C : DST = DBR D + ∆DST = 25.000 cm + ∆D BR D = = 25 cm(1 + αBR ∆T ) 1 + αST ∆T

D + αST D° ∆T = 25 cm + αBR (25 cm)∆T

(25 cm) 1 + (2.0 × 10− 5 (C°) −1 )(130C°) 1 + (1.2 × 10− 5 (C°) −1 )(130 C°) = 25.026 cm

[

]

17.29: The aluminum ruler expands to a new length of L = L0 (1 + α∆T ) = (20.0 cm)[1 + (2.4 × 10 −5 (C°) −1 )(100 C°)] = 20.048 cm The brass ruler expands to a new length of L = L0 (1 + α∆T ) = (20.0 cm)[1 + (2.0 × 10 −5 (C°) −1 )(100 C°)] = 20.040 cm The section of the aluminum ruler will be longer by 0.008 cm 17.30: From Eq. (17.12), F = −Yα∆TA = −(0.9 × 1011 Pa)(2.0 × 10 −5 (C°) −1 )(−110°C)(2.01× 10 − 4 m 2 ) = 4.0 × 10 4 N. 17.31: a) α = (∆L L0 ∆T ) = (1.9 × 10−2 m) ( (1.50 m)(400 C°) ) = 3.2 × 10 −5 (C°) −1.

b) Yα∆T = Y∆ L L0 = (2.0 × 1011 Pa)(1.9 × 10−2 m) (1.50 m) = 2.5 × 109 Pa. 17.32: a) ∆L = α∆TL = (1.2 × 10−5 K −1 )(35.0 K)(12.0 m) = 5.0 × 10 −3 m. b) Using absolute values in Eq. (17.12), F = Yα∆T = (2.0 × 1011 Pa)(1.2 × 10−5 K −1 )(35.0 K) = 8.4 × 107 Pa. A

17.33:

a)

(37°C − (−20°C))(0.50 L)(1.3 × 10−3 kg L) (1020 J kg ⋅ K ) = 38 J

b) There will be 1200 breaths per hour, so the heat lost is (1200)(38 J) = 4.6 × ×10 4 J. Q mc∆T (70 kg)(3480 J kg ⋅ K )(7 C°) = = = 1.4 × 103 s, about 24 min. P P (1200 W)

17.34: 17.35:

t=

Using Q=mgh in Eq. (17.13) and solving for ∆Τ gives ∆T = gh (9.80 m s 2 )(225 m) = = 0.53 C°. c (4190 J kg.K )

17.36:

a) The work done by friction is the loss of mechanical energy,

1 1   2 mgh + m(v12 − v2 ) = (35.0 kg) (9.80 m s 2 )(8.00 m) sin 36.9° − (2.50 m s) 2  2 2   3 = 1.54 × 10 J. b) Using the result of part (a) for Q in Eq. (17.13) gives ∆T = 1.54 × 103 J

(

) ( ( 35.0 kg )( 3650 J kg ⋅ K ) ) = 1.21× 10

−2

C°.

17.37:

( 210° C − 20°C) ( (1.60 kg )( 910 J
Assuming Q = ( 0.60) × 10 × K ,

kg ⋅ K ) + ( 0.30 kg )( 470 J kg ⋅ K ) ) = 3.03 × 105 J.

17.38:

1 MV 2 ( 6) 1 (1.80 kg )( 7.80 m s ) = 45.1 C°. K 2 2 ∆T = ( 0.60 ) × 10 × =6 = mc mc 8.00 × 10−3 kg ( 910 J kg ⋅ K ) 2

(

)

17.39:

( 85.0° C − 20.0° C) ( (1.50 kg )( 910 J
= 5.79 × 10 J.
5

kg ⋅ K ) + (1.80 kg )( 4190 J kg ⋅ K ) )

17.40:

a) Q = mc∆T = ( 0.320 kg ) ( 4190 J kg ⋅ K ) ( 60.0 K ) = 8.05 ×10 4 J.
Q P

b) t =

=

8.05×10 4 J 200 W

= 402 s.

17.41:

a) c =

Q (120 s ) ( 65.0 W ) = = 2.51 × 103 J kg ⋅ K. m∆T ( 0.780 kg ) ( 22.54° C − 18.55° C )

b) An overstimate; the heat Q is in reality less than the power times the time interval. 17.42: The temperature change is ∆T = 18.0 K, so c=

Q gQ 9.80 m s 2 1.25 × 104 J = = = 240 J kg ⋅ K. m∆T w ∆T ( 28.4 N )(18.0 K )

(

)(

)

17.43: a) Q = mc∆T , c = 470 J kg ⋅ K We need to find the mass of 3.00 mol: m = nM = ( 3.00 mol ) 55.845 × 10 −3 kg mol = 0.1675 kg ∆T = Q mc = ( 8950 J )

(

)

[ ( 0.1675 kg )( 470 J kg ⋅ K ) ] = 114 K = 114 C°

b) For m = 3.00 kg, ∆T = Q mc = 6.35 C° c) The result of part (a) is much larger; 3.00 kg is more material than 3.00 mol. Qmelt (10,000 J min )(1.5 min ) = = 30,000 J kg m 0.50 kg Q (b) Liquid : Q = mc∆T → c = m∆T (10,000 J min )(1.5 min ) = 1,000 J kg ⋅ C° c= ( 0.50 kg )( 30C°) LF =

17.44: (a)

Solid : c =

(10,000 J min )(1.0 min ) = 1300 J kg ⋅ C° Q = ( 0.50 kg )(15 C°) m∆T

17.45: a) Qwater + Qmetal = 0 mwater c water ∆Twater + mmetal c metal ∆Tmetal = 0 (1.00kg )( 4190 J kg ⋅ K )( 2.0 C°) + ( 0.500 kg )( cmetal )( − 78.0 C°) = 0 cmetal = 215 J/kg ⋅ K b) Water has a larger specific heat capacity so stores more heat per degree of temperature change. c) If some heat went into the styrofoam then Qmetal should actually be larger than in part (a), so the true c metal is larger than we calculated; the value we calculated would be smaller than the true value. 17.46: a) Let the man be designated by the subscript m and the “‘water” by w, and T is the final equilibrium temperature. − mm C m ∆Tm = mw C w ∆Tw − mmCm ( T − Tm ) = mw Cw ( T − Tw ) mm C m ( Tm − T ) = mw C w ( T − Tw ) Or solving for T, T =
m m C m Tm + m w C w T w mm C m + mw C w

. Inserting numbers, and realizing we can change K to

°C , and the mass of water is .355 kg, we get (70.0 kg) (3480 J kg ⋅ K ) (37.0°C ) + (0.355 kg) (4190 J kg ⋅ °C ) (12.0°C ) T= (70.0 kg)(3480 J kg. °C ) + (0.355 kg) (4190 J kg ⋅ °C ) Thus, T = 36.85°C.

b) It is possible a sensitive digital thermometer could measure this change since they can read to .1°C. It is best to refrain from drinking cold fluids prior to orally measuring a body temperature due to cooling of the mouth. The rate of heat loss is ∆Q ∆t. ∆Q = ∆t
7 ×10 6 J day

17.47: ∆t =

( )

mC∆T ∆t

, or ∆t =

mC∆t ∆Q ( ∆t )

. Interesting numbers,

( 70.355 kg)(3480 J kg. ° C)(0.15 ° C)

= 0.005 d, or ∆t = 7.6 minutes. This may acount for mothers

taking the temperature of a sick child several minutes after the child has something to drink. 17.48: Q = m(c∆T + Lf ) = (0.350 kg) (4190 J kg ⋅ K )(18.0 K) + 334 × 103 J kg = 1.43 × 10 J = 34.2 kcal = 136 Btu.
5

(

)

17.49:

Q = m(cice ∆Tice + Lf + cwater ∆Twater + LV )  (2100 J kg ⋅ K)(10.0 C°) + 334 × 103 J kg   = (12.0 × 10 kg)  + (100 C°)(4190 J kg ⋅ K) + 2256 × 103 J kg    4 = 3.64 × 10 J = 8.69 kcal = 34.5 Btu.
−3

17.50:

a) t =
mLf P

Q mc∆T (0.550 kg)(2100 J kg ⋅ K)(15.0 K) = = = 21.7 min . P P (800 J min ) = 230 min, so the time until the ice has melted is

b)

=

( 0.550 kg)(334 ×10 3 J kg ) (800 J min )

21.7 min + 230 min = 252 min.

17.51: 17.52:

( (4000 lb)

2.205 lb kg) ) (334 × 103 J kg) = 7.01 kW = 2.40 × 104 Btu hr. (86,400 s)

a) m(c∆T + Lv ) = (25.0 × 10−3 kg) (4190 J kg ⋅ K)(66.0 K) + 2256 × 103 J kg =

(

)

6.33 × 104 J. b) mc∆T = (25.0 × 10−3 kg)(4190 J kg ⋅ K)(66.0 K) = 6.91 × 103 J. c) Steam burns are far more severe than hot-water burns. With Q = m(c∆T + Lf ) and K = (1 / 2)mv 2 , setting Q = K and solving for v gives v = 2( (130 J Kg ⋅ K)(302.3 C°) + 24.5 × 103 J kg ) = 357 m s .

17.53:

17.54:

a) msweat =

Mc∆T (70.0 kg)(3480 J kg ⋅ K)(1.00 K) = = 101 g. Lv (2.42 × 106 J kg )

b) This much water has a volume of 101 cm 3 , about a third of a can of soda.

17.55:

The mass of water that the camel saves is

Mc∆T (400 kg)(3480 J kg ⋅ K)(6.0 K) = = 3.45 kg, Lv (2.42 × 106 J kg) which is a volume of 3.45 L. 17.56: For this case, the algebra reduces to  ((200)(3.00 × 10−3 kg ))(390 J kg ⋅ K )(100.0 C°)      + (0.240 kg )(4190 J kg )(20.0 C°)  = 35.1°C. T=  ((200)(3.00 × 10− 3 kg )(390 J kg ⋅ K )     + (0.240 kg )(4190 J kg ⋅ .K)    17.57: The algebra reduces to     = 27.5°C

 ((0.500 kg)(390 J kg ⋅ K ) + (0.170 kg)(4190 J kg ⋅ K ))(20.0° C   + (0.250 kg)(470 J kg ⋅ K )(85.0°C) T=  ((0.500 kg)(390 J kg ⋅ K ) + (0.170 kg)(4190 J kg ⋅ K ))      + (0.250 kg)(470 J kg ⋅ K )  

17.58: The heat lost by the sample is the heat gained by the calorimeter and water, and the heat capacity of the sample is c= Q ((0.200 kg)(4190 J kg ⋅ K ) + (0.150 kg)(390 J kg ⋅ K ))(7.1C°) = m∆T (0.0850 kg)(73.9 C°) =1010 J kg ⋅ K, or 1000 J kg ⋅ K to the two figures to which the temperature change is known.

17.59:

The heat lost by the original water is − Q = (0.250 kg )(4190 J kg ⋅ K )(45.0 C°) = 4.714 × 10 4 J,

and the mass of the ice needed is mice = −Q cice ∆Tice + Lf + c water ∆Twater (4.714 × 104 ) J (2100 J kg ⋅ K) (20.0 C°) + (334 × 103 J kg ) + (4190 J kg ⋅ K )(30.0 C°)

=

= 9.40 × 10 −2 kg = 94.0 g. 17.60: The heat lost by the sample (and vial) melts a mass m, where Q ((16.0 g)(2250 J kg ⋅ K ) + (6.0 g)(2800 J kg ⋅ K ))(19.5K) m= = = 3.08 g. Lf (334 × 103 J kg)

Since this is less than the mass of ice, not all of the ice melts, and the sample is indeed cooled to 0°C. Note that conversion from grams to kilograms was not necessary. (4.00 kg)(234 J kg ⋅ K)(750 C°) = 2.10 kg. (334 × 103 J kg)

17.61:

17.62: Equating the heat lost by the lead to the heat gained by the calorimeter (including the water-ice mixtue), mP b c Pb (200°C − T ) = ( mw + mice )c wT + mcu ccu T + mice Lf . Solving for the final temperature T and using numerical values,  (0.750 kg)(130 J kg ⋅ K)(255 C°)     − (0.018 kg)(334 × 103 J kg)   = 21.4°C. T=  (0.750 kg)(130 ⋅ J kgK)     + (0.178 kg)(4190 J kg ⋅ K)   + (0.100 kg)(390 J kg ⋅ K)    (The fact that a positive Celsius temperature was obtained indicates that all of the ice does indeed melt.)

17.63: The steam both condenses and cools, and the ice melts and heats up along with the original water; the mass of steam needed is (0.450 kg)(334 × 103 J kg) + (2.85 kg)(4190 J kg ⋅ K)(28.0 C°) msteam = 2256 × 103 J kg + (4190 J kg )(72.0 C°) = 0.190 kg.

17.64: The SI units of H and dQ are both watts, the units of area are m 2 , temperature dt difference is in K, length in meters, so the SI units for thermal conductivity are [ W][m] W = . 2 [m ][K] m ⋅ K 17.65:
100 a) 0.450Km = 222 K m. b)(385 W m ⋅ K)(1.25 × 10-4 m 2 )(400 K m) = 10.7 W.

c)100.0°C − (222 K m)(12.00 × 10−2 m) = 73.3°C. 17.66: Using the chain rule, H = dQ = Lf dm and solving Eq. (17.21) for k, dt dt dm L k = Lf dt A∆T (8.50 × 10 − 3 kg) (60.0 × 10 − 2 s) = (334 × 103 J kg) (600 s) (1.250 × 10 − 4 m 2 )(100 K) = 227 W m ⋅ K.

17.67: (Although it may be easier for some to solve for the heat flow per unit area, part (b), first the method presented here follows the order in the text.) a) See Example 17.13; as in that example, the area may be divided out, and solving for temperature T at the boundary, T=

( k foam Lfoam )Tin + ( k wood Lwood )Tout N ( k foam Lfoam ) + ( k wood Lwood )

( ( 0.010 W m ⋅ K ) ( 2.2 cm ) )(19.0° C ) + ( ( 0.080 W m ⋅ K ) ( 3.0 cm) )( − 10.0° C) ( ( 0.010 W m ⋅ K ) ( 2.2 cm ) ) + ( ( 0.080 W m ⋅ K ) ( 3.0 cm) )
= −5.8°C.

Note that the conversion of the thickness to meters was not necessary. b) Keeping extra figures for the result of part, (a), and using that result in the temperature difference across either the wood or the foam gives

(19.0° C − ( − 5.767° C) ) H foam H wood = = ( 0.010 W m ⋅ K ) A A 2.2 × 10− 2 m
= ( 0.080 W m ⋅ K ) = 11 W m 2 . 17.68: a) From Eq. (17.21) , H = ( 0.040 W m ⋅ K ) 1040 m 2

( − 5.767°C − ( − 10.0°C ) )
3.0 × 10 −2 m

(

) ( 4.0(140 K ) m) = 196 W, × 10
−2

or 200 W to two figures. b) The result of part (a) is the needed power input. From Eq. (17.23) , the energy that flows in time ∆t is H∆t = A∆T 125 ft 2 ( 34F°) ∆t = ( 5.0 h ) = 708 Btu = 7.5 × 105 J. R 30 ft 2 ⋅ F° ⋅ h Btu

17.69:

(

(

)

)

17.70: a) The heat current will be the same in both metals; since the length of the copper rod is known, H = ( 385.0 W m ⋅ K ) 400 × 10 −4 m 2

(

35.0 K ) ((1.00 m)) = 5.39 W.

b) The length of the steel rod may be found by using the above value of H in Eq. (17.21) and solving for L2 , or, since H and A are the same for the rods, L2 = L

( 50.2 W m ⋅ K )( 65.0 K ) = 0.242 m. k 2 ∆T2 = (1.00 m ) k ∆T ( 385.0 W m ⋅ K )( 35.0 K )

Using H = Lv dm (see Problem 17.66) in Eq. (17.21), dt dm L ∆T = Lv dt kA (0.390 kg) (0.85 × 10−2 m) 3 = (2256 × 10 J kg) = 5.5 C°, (180 s) (50.2 W m ⋅ K )(0.150 m 2 ) and the temperature of the bottom of the pot is 100° C + 6 C° = 106° C. 17.71: 17.72:

∆Q ∆T = kA ∆t L W   300 K   150 J s =  50.2  A  m. K   0.500 m   A = 4.98 × 10− 3 m 2 D A = πR = π   2
2 2

D = 4A π = 4(4.98 × 10 −3 m 2 ) π = 8.0 × 10 − 2 m = 8.0 cm

17.73: H a = H b (a = aluminum, b = brass) A(150.0°C − T ) A(T − 0° C) H a = ka , H b = kb La Lb (It has been assumed that the two sections have the same cross-sectional area.) A(150.0°C − T ) A(T − 0°C) ka = kb La Lb (2050 W m ⋅ K)(150.0° C − T ) (109.0 W m ⋅ K)(T − 0°C) = 0.800 m 0.500 m Solving for T gives T = 90.2°C 17.74: From Eq. (17.25), with e = 1,

a) (5.67 × 10−8 W m 2 ⋅ K 4 )(273 K) 4 = 315 W m 2 . b) A factor of ten increase in temperature results in a factor of 10 4 increase in the output; 3.15 × 10 6 W m 2 . Repeating the calculation with Ts = 273 K + 5.0 C° = 278 K gives H = 167 W.

17.75:

17.76:

The power input will be equal to H net as given in Eq. (17.26);

P = Aeσ (T 4 − Ts4 ) = (4π (1.50 × 10− 2 m) 2 )(0.35)(5.67 × 10−8 W m 2 ⋅ K 4 )((3000 K) 4 − (290 K) 4 ) = 4.54 × 103 W.

17.77:

A=

H 150 W = = 2.10 cm 2 4 4 −8 2 4 eσ T ( 0.35) 5.67 × 10 W m ⋅ K ( 2450K )

(

)

17.78:

The radius is found from R= A = 4π H σT 2 = 4π

(

)

H 1 . 4πσ T 2

Using the numerical values, the radius for parts (a ) and (b ) are Ra = W 1 = 1.61 × 1011 m 2 2 4 4π 5.67 × 10 W m ⋅ K (11,000 K )

(

( 2.7 × 10
−8

32

)

)

Rb =

W 1 = 5.43 × 106 m 2 2 4 4π 5.67 × 10 W m ⋅ K (10,000 K )

(

( 2.10 × 10
−8

23

)

)

c) The radius of Procyon B is comparable to that of the earth, and the radius of Rigel is comparable to the earth-sun distance. 17.79: a) normal melting point of mercury: − 30° C = 0.0°Μ normal boiling point of mercury: 357° C = 100.0°Μ 100 Μ ° = 396 C° so 1 Μ° = 3.96 C° Zero on the M scale is − 39 on the C scale, so to obtain TC multiple TΜ by 3.96 and then subtract 39° : TC = 3.96TM − 39° 1 Solving for TM gives TM = ( TC + 39°) 3.96 1 (100° + 39°) = 35.1°Μ The normal boiling point of water is 100°C; TM = 3.96 b) 10.0 Μ° = 39.6° C° 17.80: All linear dimensions of the hoop are increased by the same factor of α ∆ T , so the increase in the radius of the hoop would be Rα ∆ T = 6.38 × 106 m 1.2 × 10 −5 K −1 ( 0.5 K ) = 38 m.

(

)(

)

17.81: The tube is initially at temperature T0 , has sides of length L0 volume V0 , density ρ0 , and coefficient of volume expansion β. a) When the temperature increase to T0 + ∆T , the volume changes by an amount m m ∆V , where ∆V = β V0 ∆T . Then, ρ = , or eliminating ∆V , ρ = . V0 + ∆V V0 + βV0 ∆T Divide the top and bottom by V0 and substitute ρ0 = m V0 . Then m V0 ρ0 ρ= or ρ = . This can be rewritten as ρ = ρ0 (1 + β ∆T ) −1. Then V0 V0 + βV0 ∆ T V0 1 + β∆T using the expression (1 + x ) ≈ 1 + nx, where n = −1, ρ = ρ0 (1 − β ∆T ).
n

b) The copper cube has sides of length 1.25 cm = .0125 m, and ∆T = 70.0° C − 20.0° C = 50.0° C.

∆V = βV0 ∆T = 5.1 × 10−5 ° C ( .0125 m ) ( 50.0° C ) = 5 × 10−9 m 3 .
3

(

)

Similarly, ρ = 8.9 × 103 kg m 3 (1 − (5.1 × 10 −5 ° C)(50.0° C)), or ρ = 8.877 × 103 kg m 3 ; extra significant figures have been keep. So ∆ρ = −23
kg m3

.

17.82: (a) We can use differentials to find the frequency change because all length changes are small percents . Let m be the mass of the wire v = F μ = F (m L) = FL m v f = and λ = 2 L(fundamenta l) λ FL m 1 F f =v λ= = 2L 2 mL ∂F ∆f ≈ ∆L (only L changes due to heating) ∂L ∆f ≈ f
1 2 F ( 1 )( F mL) −1 2 (− mL2 )∆L 2 1 2 F mL

=

1 ∆L 2 L

1 1 ∆f ≈ (α∆T ) f = (1.7 × 10− 5 (C°) −1 )(40C°)(440 C°)(440 Hz) = 0.15 Hz 2 2 The frequency decreases since the length increases (b) v = F μ = FL m ∆v 1 ( FL m) −1 2 ( F m)∆L ∆L α∆T = 2 = = v 2L 2 FL m 1 = (1.7 × 10 −5 (C°) −1 )(40C°) = 3.4 × 10 − 4 = 0.034% 2 (c) λ = 2 L → ∆λ = 2∆L → ∆λλ = 22∆LL = ∆LL = α∆T ∆λ = (1.7 × 10− 5 C −1 )(40C°) = 6.8 × 10− 4 = 0.068% : λ it increases 17.83: Both the volume of the cup and the volume of the olive oil increase when the temperature increases, but β is larger for the oil so it expands more. When the oil starts to overflow, ∆Voil = ∆Vglass + (1.00 × 10−3 m) A, where A is the cross-sectional area of the cup. ∆Voil = V0, oil βoil ∆T = (9.9 cm) Aβoil ∆T ∆Vglass = V0, glass βglass ∆T = (10.0 cm) Aβglass ∆T (9.9 cm) Aβoil ∆T = (10.0 cm) Aβglass ∆T + (1.00 × 10−3 m) A The area A divides out. Solving for ∆T gives ∆T = 15.5 C° T2 = T1 + ∆T = 37.5°C

Volume expansion: dV = βV dT dV dT Slope of graph β= = V V Construct the tangent to the graph at 2°C and 8°C and measure the slope of this line. 3 At 22°C : Slope ≈ − 0.10Ccm and V ≈ 1000 cm3 3 ° 17.84: 0.10 cm3 3C° ≈ −3 × 10− 5 (C°) −1 3 1000 cm The slope in negative, as the water contracts or it is heated. At 3 8° C : slope ≈ 0.24Ccm and V ≈ 1000 cm 3 4 ° β≈− 0.24 cm3 4C ° ≈ 6 × 10− 5 (C°) −1 1000 cm3 The water now expands when heated. β≈ 17.85: ∆La + ∆Ls = 0.40 cm (a = aluminum, s = steel)

∆L = L0α ∆T , so (24.8 cm)(2.4 × 10 −5 (C °) −1 ))∆T + (34.8 cm)(1.2 × 10−5 (C °) −1 ))∆T = 0.40 cm ∆T = 395 C° T2 = T1 + ∆T = 415°C 17.86: a) The change in height will be the difference between the changes in volume of the liquid and the glass, divided by the area. The liquid is free to expand along the column, but not across the diameter of the tube, so the increase in volume is reflected in the change in the length of the columns of liquid in the stem. b) ∆h = ∆Vliquid − ∆Vglass
−6

A (100 × 10 m 3 ) = (8.00 × 10 − 4 K −1 − 2.00 × 10− 5 K -1 )(30.0 K) (50.0 × 10− 6 m 2 ) = 4.68 × 10 −2 m.

=

V ( βliquid − βglass ) ∆T A

17.87: To save some intermediate calculation, let the third rod be made of fractions f 1 and f 2 of the original rods; then f1 + f 2 = 1 and f1 (0.0650) + f 2 (0.0350) = 0.0580. These two equations in f 1 and f 2 are solved for 0.0580 − 0.0350 f1 = , f 2 = 1 − f1 , 0.0650 − 0.0350 and the lengths are f 1 (30.0 cm) = 23.0 cm and f 2 (30.0 cm) = 7.00 cm 17.88: a) The lost volume, 2.6 L, is the difference between the expanded volume of the fuel and the tanks, and the maximum temperature difference is ∆V ∆T = ( βfuel − βA1 )V0 (2.6 × 10 −3 m 3 ) (9.5 × 10 − 4 (C°) −1 − 7.2 × 10 − 5 (C°) −1 )(106.0 × 10 − 3 m 3 ) = 2.78 C°, or 28°C to two figures; the maximum temperature was 32°C. b) No fuel can spill if the tanks are filled just before takeoff. = 17.89: a) The change in length is due to the tension and heating ∆L F = AY + α∆T . Solving for F A, F = Y ΔL − αΔT . L0 A L0

(

)

b) The brass bar is given as “heavy” and the wires are given as “fine,” so it may be assumed that the stress in the bar due to the fine wires does not affect the amount by which the bar expands due to the temperature increase. This means that in the equation preceding Eq. (17.12), ∆L is not zero, but is the amount αbrass L0 ∆T that the brass expands, and so F = Ysteel (αbrass − αsteel )∆T A = 20 × 1010 Pa)(2.0 × 10−5 (C°) −1 − 1.2 × 10− 5 (C°) −1 )(120°C) = 1.92 × 108 Pa.

17.90: In deriving Eq. (17.12), it was assumed that ∆L = 0; if this is not the case when there are both thermal and tensile stresses, Eq. (17.12) becomes F   ∆L = L0  α∆T + . AY   For the situation in this problem, there are two length changes which must sum to zero, and so Eq. (17.12) may be extended to two materials a and b in the form   F  F   + L0b  αb ∆T +  = 0. L0a  αa ∆T +    AYa  AYb     Note that in the above, ∆T , F and A are the same for the two rods. Solving for the stress F A, F αa L0a + αb L0b =− ∆T A (( Loa Ya ) + ( L0b Yb )) = (1.2 × 10- 5 (C°) −1 )(0.350 m) + (2.4 × 10− 5 (C°) −1 )(0.250 m) (60.0 C°) ((0.350 m) 20 × 1010 Pa) + (0.250 m 7 × 1010 Pa))

= −1.2 × 108 Pa to two figures. a) ∆T =
( 0.0020 in.) ∆R = αR0 (1.2×10 − 5 (C ° ) −1 ( 2.5000 in.)

17.91:

= 67 C° to two figures, so the ring should be

warmed to 87°C. b) the difference in the radii was initially 0.0020 in., and this must be the difference between the amounts the radii have shrunk. Taking R0 to be the same for both rings, the temperature must be lowered by an amount ∆R ∆T = ( αbrass − αsteel ) R0 ( 0.0020 in. ) = = 100 C° −1 −1 2.0 × 10 −5 ( C°) − 1.2 × 10 −5 ( C°) ( 2.50 in. ) to two figures, so the final temperature would be − 80°C.

(

)

17.92: a) The change in volume due to the temperature increase is βV∆T , and the change in volume due to the pressure increase is − V ∆p ( Eq. (11.13) ). Setting the net B change equal to zero, βV∆T = V ∆p = 1.6 × 10 Pa 3.0 × 10
11

(

)(

−5

K

∆p B −1

)(15.0 K ) = 8.64 × 10

, or ∆p = Bβ∆V . b) From the above,
7

Pa.

17.93: As the liquid is compressed, its volume changes by an amount ∆V = −∆pkV0 . When cooled, the difference between the decrease in volume of the liquid and the decrease in volume of the metal must be this change in volume, or ( α1 − αm )V0 ∆T = ∆V . Setting the expressions for ∆V equal and solving for ∆T gives ∆pk 5.065 × 106 Pa 8.50 × 10 −10 Pa −1 ∆T = = = −9.76 C°, α m − α1 3.90 × 10− 5 K −1 − 4.8 × 10 − 4 K −1 so the temperature is 20.2°C.

( (

)(

) )

17.94: Equating the heat lost be the soda and mug to the heat gained by the ice and solving for the final temperature T =  ( ( 2.00 kg )( 4190 J kg ⋅ K ) + ( 0.257 kg )( 910 J kg ⋅ K ) )( 20.0C°)     − ( 0.120 kg ) ( 2100 J kg ⋅ K )(15.0 C°) + 334 × 10 3 J kg    ( 2.00 kg )( 4190 J kg ⋅ K ) + ( 0.257 kg )( 910 J kg ⋅ K ) + ( 0.120 kg )( 4190 J kg ⋅ K ) = 14.1°C. Note that the mass of the ice ( 0.120 kg ) appears in the denominator of this expression multiplied by the heat capacity of water; after the ice melts, the mass of the melted ice must be raised further to T .

(

)

( 7700 m s ) K (1 2 ) mv 2 v2 = = = = 54.3. Q cm∆T 2c∆T 2( 910 J kg ⋅ K )( 600C°) b) Unless the kinetic energy can be converted into forms other than the increased heat of the satellite cannot return intact.
2

17.95:

a)

a) The capstan is doing work on the rope at a rate 2π 2π P = τω = ∆Fr = ( 520 N ) 5.0 × 10− 2 m = 182 W, T ( 0.90 s ) or 180 W to two figures. The net torque that the rope exerts on the capstan, and hence the net torque that the capstan exerts on the rope, is the difference between the forces of the ends times the radius. A larger number of turns might increase the force, but for given forces, the torque is independent of the number of turns. dT dQ dt P (182 W) b) = = = = 0.064 C° s. dt mc mc (6.00 kg)(470 J mol ⋅ K)

17.96:

(

)

17.97:

a) Replacing m with nM and nMc with nC, nk T2 nk Q = ∫ dQ = 3 ∫ T 3 dt = (T24 − T14 ). 3 T1 Θ 4Θ For the given temperatues, (1.50 mol)(1940 J mol ⋅ K) Q= ((40.0 K) 4 − (10.0 K) 4 ) = 83.6 J. 4(281 K) 3 b)
Q (83.6 J) = nΔ T (1.50 mol) (30.0 K)

= 1.86 J mol ⋅ K.

c) C = (1940 J mol ⋅ K) (40.0 K 281 K) 3 = 5.60 J mol ⋅ K. 17.98: Setting the decrease in internal energy of the water equal to the final gravitational potential energy, Lf ρwVw + Cw ρwVw ΔT = mgh. Solving for h, and inserting numbers: ρ V ( L + Cw ΔT ) h= w w f mg = (1000 kg m 3 )(1.9 × .8 × .1 m3 ) 334 × 103 J kg + (4190 J kg ⋅ ° C)(37° C) (70 kg)(9.8 m s 2 )

[

]

= 1.08 × 105 m = 108 km. 17.99: a) (90)(100 W)(3000 s) = 2.7 × 10 7 J.

Q Q 2.7 × 107 J = = = 6.89 C°, cm cρV (1020 J kg ⋅ K)(1.20 kg m 3 )(3200 m 3 ) or 6.9 C° to the more apropriate two figures. c) The answers to both parts (a) and (b) are multiplied by 2.8, and the temperature rises by 19.3 C°. b) ΔT = 17.100: See Problem 17.97. Denoting C by C = a + bT , a and b independent of temperature, integration gives. b   Q = n a(T2 − T1 ) + (T22 − T12 ) . 2   In this form, the temperatures for the linear part may be expressed in terms of Celsius temperatures, but the quadratic must be converted to Kelvin temperatures, T1 = 300 K and T2 = 500 K. Insertion of the given values yields Q = (3.00 mol)(29.5 J mol ⋅ K)(500 K - 300 K) + (4.10 × 10 −3 J mol ⋅ K 2 )((500 K) 2 − (300 K) 2 ))

= 1.97 × 10 4 J.

17.101: a) To heat the ice cube to 0.0°C, heat must be lost by the water, which means that some of the water will freeze. The mass of this water is m C ∆T (0.075 kg)(2100 J kg ⋅ K)(10.0 C°) mwater = ice ice ice = = 4.72 × 10−3 kg = 4.72 g. 3 Lf (334 × 10 J kg) b) In theory, yes, but it takes 16.7 kg of ice to freeze 1 kg of water, so this is impractical. 17.102: The ratio of the masses is ms C w ∆Tw ( 4190 J kg ⋅ K)(42.0 K) = = = 0.0696, m w C w ∆Ts + Lv (4190 J kg ⋅ K)(65.0 K) + 2256 × 10 3 J kg so 0.0696 kg of steam supplies the same heat as 1.00 kg of water. Note the heat capacity of water is used to find the heat lost by the condensed steam. 17.103: a) The possible final states are steam, water and copper at 100°C, water, ice and copper at 0.0°C or water and copper at an intermediate temperature. Assume the last possibility; the final temperature would be  (0.0350 kg)((4190 J kg ⋅ K)(100 C°) + 2256 × 103 J kg)      − (0.0950 kg)(334 × 103 J kg)  = 86.1°C T=  (0.0350 kg)(4190 J kg ⋅ K) + (0.446 kg)(390 J kg ⋅ K)      + (0.0950 kg)(4190 J kg ⋅ K)   This is indeed a temperature intermediate between the freezing and boiling points, so the reasonable assumption was a valid one. b) There are 0.13 kg of water. 17.104: a) The three possible final states are ice at a temperature below 0.0°C, an icewater mixture at 0.0° C or water at a temperature above 0.0°C. To make an educated guess at the final possibility, note that (0.140 kg)(2100 J kg ⋅ K)(15.0 C°) = 4.41 kJ are needed to heat the ice to 0.0°C, and (0.190 kg)(4190 J kg ⋅ K)(35.0 C°) = 27.9 kJ must removed to cool the water to 0.0°C, so the water will not freeze. Melting all of the ice would require an additional (0.140 kg)(334 × 103 J kg) = 46.8 kJ, so some of the ice melts but not all; the final temperature of the system is 0.0°C. Considering the other possibilities would lead to contradictions, as either water at a temperature below freezing or ice at a temperature above freezing. b) The ice will absorb 27.9 kJ of heat energy to cool the water to 0°C. Then, ( 27.9 kJ − 4.41 kJ ) m = 334 ×10 3 J kg = 0.070 kg will be converted to water. There will be 0.070 kg of ice and 0.260 kg of water.

17.105: a) If all of the steam were to condense, the energy available to heat the water would be (0.0400 kg )(2256 × 10 3 J kg ) = 9.02 × 10 4 J. If all of the water were to be heated to 100.0°C, the needed heat would be (0.200 kg )(4190 J kg ⋅ K )(50.0 C°) = 4.19 × 104 J. Thus, the water heats to 100.0°C and some of the steam condenses; the temperature of the final state is 100°C. b) Because the steam has more energy to give up than it takes to raise the water temperature, we can assume that some of the steam is converted to water: 4.19 × 10 4 J m= = 0.019 kg. 2256 × 10 3 J kg Thus in the final state, there are 0.219 kg of water and 0.021 kg of steam. 17.106: The mass of the steam condensed 0.525 kg − 0.490 kg = 0.035 kg. The heat lost by the steam as it condenses and cools is (0.035 kg) Lv + (0.035 kg )(4190 J kg ⋅ K )(29.0 K ), and the heat gained by the original water and calorimeter is ((0.150 kg)(420 J kg ⋅ K) + (0.340 kg)(4190 J kg ⋅ K ))(56.0 K) = 8.33 × 104 J. Setting the heat lost equal to the heat gained and solving for Lv gives 2.26 × 10 6 J kg, or 2.3 × 10 6 J kg to two figures (the mass of steam condensed is known to only two figures).

17.107: a) The possible final states are in ice-water mix at 0.0°C, a water-steam mix at 100.0°C or water at an intermediate temperature. Due to the large latent heat of vaporization, it is reasonable to make an initial guess that the final state is at 100.0°C. To check this, the energy lost by the steam if all of it were to condense would be (0.0950 kg )(2256 × 103 J kg) = 2.14 × 105 J. The energy required to melt the ice and heat it to 100°C is (0.150 kg)(334 × 103 J kg + (4190 J kg ⋅ K) (100 C°)) = 1.13 × 105 J, and the energy required to heat the origianl water to 100°C is (0.200 kg)(4190 J kg.K ) (50.0 C°) = 4.19 × 104 J. Thus, some of the steam will condense, and the final state of the system wil be a water-steam mixture at 100.0°C. b) All of the ice is converted to water, so it adds 0.150 kg to the mass of water. Some of the steam condenses giving up 1.55 × 103 J of energy to melt the ice and raise the temperature. Thus, m =
1.55×10 5 J 2256 ×10 3 J kg

= 0.69 kg and the final mass of steam is 0.026 kg, and

of the water, .150 kg + .069 kg + .20 kg = 0.419 kg. c) Due to the much larger quantity of ice, a reasonable initial guess is an ice-water mix at 0.0° C. The energy required to melt all of the ice would be (0.350 kg) (334 × 103 J kg) = 1.17 × 10 5 J . The maximum energy that could be transferred to the ice would be if all of the steam would condense and cool to 0.0°C and if all of the water would cool to 0.0 C , (0.0120) kg (2256 × 103 J kg + (4190 J kg ⋅ K)(100.0 C°)) + (0.200 kg)(4190 J kg ⋅ K)(40.0 C°) = 6.56 × 104 J. This is insufficient to melt all of the ice, so the final state of the system is an ice-water mixture at 0.0° C. 6.56 × 104 J of energy goes into melting the ice. So, m =
6.56×10 4 J 334×103 J kg

= 0.196 kg. So there is 0.154 kg of ice, and 0.012 kg + 0.196 kg + 0.20 kg = 0.408 kg of water.
17.108: Solving Eq. (17.21) for k, ∆Τ (3.9 × 10−2 m) k=H = (180 W) = 5.0 × 10− 2 W m ⋅ K. 2 Α∆Τ (2.18 m )(65.0 K)

17.109:

  ∆Τ 28.0 C° = (0.120 J mol.K) (2.00 × 0.95 m 2 )  5.0 × 10 − 2 m + 1.8 × 10− 2 m   L   = 93.9 W. b) The flow through the wood part of the door is reduced by a factor of ( 0.50 ) 2 1 − ( 2.00×0.95 ) = 0.868 to 81.5 W . The heat flow through the glass is a) H = kΑ

 28.0 C°  H glass = (0.80 J mol ⋅ K) (0.50 m) 2   12.45 × 10− 2 m  = 45.0 W,    5+ and so the ratio is 81.93.45.0 = 1.35. 9 17.110: R1 =
L k1

, R2 =

L k2

, H1 = H 2 , and so ∆T1 = ∆Τ = ∆Τ1 + ∆T2 =

H A

R1 , ∆Τ 2 =

H A

R2 . The temperature

difference across the combination is H H ( R1 + R2 ) = R, A A

so, R = R1 + R2 . 17.111: The ratio will be the inverse of the ratio of the total thermal resistance, as given by Eq. (17.24). With two panes of glass with the air trapped in between, compared to the single pane, the ratio of the heat flows is ( 2( Lglass kglass ) + R0 + ( Lair kair ) , ( Lglass kglass) + R0 where R0 is the thermal resistance of the air films. Numerically, the ratio is

( 2((4.2 × 10

−3

m) (0.80 W m ⋅ K)) + 0.15 m 2 ⋅ K W + ((7.0 × 10 −3 m) (0.024 W m ⋅ K)) = 2.9. (4.2 × 10− 3 m) (0.80 W m ⋅ K) + 0.15 m 2 ⋅ K W

)

17.112: Denote the quantites for copper, brass and steel by 1, 2 and 3, respectively, and denote the temperature at the junction by T0 . a) H 1 = H 2 + H 3 , and using Eq. (17.21) and dividing by the common area, k1 (100°C − T0 ) = k 2 T0 + k 3 T0 . L1 L2 L3 Solving for T0 gives L1 ) (100°C). ( k1 L2 ) + ( k 3 L3 ) Substitution of numerical values gives T0 = 78.4°C. T0 = b) Using H = kA ∆T for each rod, with ∆T1 = 21.6 C°, ∆T2 = ∆T3 = 78.4°C gives L H1 = 12.8 W, H 2 = 9.50 W and H 3 = 3.30 W . If higher precision is kept, H 1 is seen to be the sum of H 2 and H 3 . 17.113: a) See Figure 17.11. As the temperature approaches 0.0°C, the coldest water rises to the top and begins to freeze while the slightly warmer water, which is more dense, will be beneath the surface. b) (As in part (c), a constant temperature difference is assumed.) Let the thickness of the sheet be x, and the amount the ice thickens in time dt be dx. The mass of ice added per unit area is then ρic e dx, meaning a heat transfer of ρic e Lf dx. This must be the product of the heat flow per unit area times the time, ( H A) dt = ( k∆T x ) dt. Equating these expressions, k∆T k∆T ρic e Lf dx = dt or xdx = dt. x ρice Lf This is a separable differential equation; integrating both sides, setting x = 0 at t = 0, gives 2k∆T x2 = t. ρice Lf The square of the thickness is propotional to the time, so the thickness is propotional to the square root of the time. c) Solving for the time in the above expression, 920 kg m 3 334 × 10 3 J kg ( 0.25 m ) 2 = 6.0 × 10 5 s. t= 2(1.6 J mol ⋅ K )(10°C )

( k1 L1 ) + ( k 2

(

)(

)

d) Using x = 40 m in the above calculation gives t = 1.5 × 1010 s, about 500 y, a very long cold spell.

17.114:

Equation(17.21) becomes H = kA ∂T . ∂x
dQ dt

a) H = (380 J kg ⋅ K)( 2.50 × 10 −4 m 2 )(140 C° m) = 13.3 W. b) Denoting the points as 1 and 2, H 2 − H1 = = mc ∂∂T . Solving for t
∂T ∂x

at 2,

∂T ∂T mc ∂T = + . ∂x 2 ∂x 1 kA ∂t The mass m is ρA∆x, so the factor multiplying ∂∂T in the above expression is t
cρ k

∆x = 137 s m. Then, ∂T ∂x = 140 C° m + (137 s m)(0.250 C° s) = 174 C° m.
2

17.115: The mass of ice per unit area will be the product of the density and the thickness x, and the energy needed per unit area to melt the ice is product of the mass per unit area and the heat of fusion. The time is then ρxLf (920 kg m 3 )(2.50 × 10−2 m)(334 × 103 L kg) t= = P A (0.70)(600 W m 2 ) = 18.3 × 103 s = 305 min. 17.116: a) Assuimg no substantial energy loss in the region between the earth and the sun, the power per unit area will be inversely proportional to the square of the distance from the center of the sun, and so the energy flux at the surface of the sun is

(1.50 × 103 W m 2 )

(

1.50 ×1011 m) 6.96 ×10 m)
8

)

2

= 6.97 × 107 W m 2 . b) Solving Eq. (17.25) with e = 1,
1 1

7 2 4  H 1  4  6.97 × 10 W m T = = = 5920 K. −8 2 4  A σ   5.67 × 10 W m ⋅ K 

17.117: The rate at which the helium evaporates is the heat gained from the surroundings by radiation divided by the heat of vaporization. The heat gained from the surroundings come from both the side and the ends of the cylinder, and so the rate at which the mass is lost is 2 hπ d + 2π ( d 2 ) σe Ts4 − T 4 Lv

(

) (

)

=

  ( 0.250 m ) π ( 0.090 m ) + 2π ( 0.045 m ) 2 ( .200 )    × 5.67 × 10−8 W m 2 ⋅ K 4 ( 77.3 K ) 4 − ( 4.22 K ) 4 

(

( 2.09 × 10

)(

4

J kg

)

)

   

= 1.62 × 10 −6 kg s, which is 5.82 g h. 17.118: a) With ∆p = 0, p∆V = nR∆T = or pV ∆T , T

b)

∆V ∆T 1 = , and β = . V T T βair 1 = = 67. βcopper (293 K)(5.1 × 10-5 K −1 )

17.119: a) At steady state, the input power all goes into heating the water, so P = H = dm c∆T and dt (1800 W) P ∆T = = = 51.6 K, c(dm dt ) (4190 J kg ⋅ K) (0.500 kg min) (60 s min) and the output temperature is 18.0°C + 51.6°C. b) At steady state, the apparatus will neither remove heat from nor add heat to the water. 17.120: a) The heat generated by the hamster is the heat added to the box; dT P = mc = (1.20 kg m 3 )(0.0500 m 3 )(1020 J mol ⋅ K)(1.60 C° h) = 97.9 J h. dt b) Taking the efficiency into account, M P0 P (10%) 979 J h = = = = 40.8 g h. t Lc Lc 24 J g

17.121: For a spherical or cylindrical surface, the area A in Eq. (17.21) is not constant, and the material must be considered to consist of shells with thickness dr and a temperature difference between the inside and outside of the shell dT . The heat current will be a constant, and must be found by integrating a differential equation. a)Equation (17.21) becomes dT H dr H = k (4πr 2 ) or = k dT . dr 4πr 2 Integrating both sides between the appropriate limits, H 1 1  −  = k (T2 − T1 ). 4π  a b  In this case the “appropriate limits” have been chosen so that if the inner temperature T2 is at the higher temperature T1 , the heat flows outward; that is, dT < 0. Solving for the dr heat current, k 4πab(T2 − T1 ) H= . b−a b) Of the many ways to find the temperature, the one presented here avoids some intermediate calculations and avoids (or rather sidesteps) the sign ambiguity mentioned above. From the model of heat conduction used, the rate of changed of temperature with radius is of the form dT = rB2 , with B a constant. Integrating from r = a to r and from dr
r = a to r = b gives

1 1 1 1 T (r ) − T2 = B −  and T1 − R2 = B − . a r a b Using the second of these to eliminate B and solving T(r) (and rearranging to eliminate compound fractions) gives  r − a  b  T (r ) = T2 − (T2 − T1 )  .  b − a  r  There are, of course, many equivalent forms. As a check, note that at r = a, T = T2 and at r = b, T = T1. c) As in part (a), the expression for the heat current is dT H H = k (2πrL) or = kLdT , dr 2πr which integrates, with the same condition on the limits, to H 2πkL(T2 − T1 ) ln( b a ) = kL(T2 − T1 ) or H = . 2π ln( b a ) d) A method similar (but slightly simpler) than that use in part (b) gives ln( r a) T (r ) = T2 + (T1 − T2 ) . ln( b a) e) For the sphere: Let b − a = l , and approximat e b ~ a, with a the common radius. Then the surface area of the sphere is A = 4πa 2 , and the expression for H is that of Eq. (17.21) (with l instead of L, which has another use in this problem). For the cylinder: with the same notation, consider

17.122: From the result of Problem 17.121, the heat current through each of the jackets is related to the temperature difference by H = ln2( πlka ) ∆T , where l is the length of the b cylinder and b and a are the inner and outer radii of the cylinder. Let the temperature across the cork be ∆T1 and the temperature across the styrofoam be ∆T2 , with similar notation for the thermal conductivities and heat currents. Then, ∆T1 + ∆T2 = ∆T = 125 C°. Setting H 1 = H 2 = H and canceling the common factors, ∆T1k1 ∆T2 k 2 = 1n 2 1n 1.5  k ln 1.5  Eliminating ∆T2 and solving for ∆T1 gives ∆T1 = ∆T 1 + 1  k ln 2  .  2   Substitution of numerical values gives ∆T1 = 37 C°, and the temperature at the radius where the layers meet is 140°C − 37°C = 103°C. b) Substitution of this value for ∆T1 into the above expression for H 1 = H gives 2π ( 2.00 m )( 0.04 J mol ⋅ K ) H= ( 37 C°) = 27 W. ln 2
−1

17.123:

a)

b) After a very long time, no heat will flow, and the entire rod will be at a uniform temperature which must be that of the ends, 0°C. c)

d)

temperature gradient is ± (100°C )( π 0.100 m) = ± 3.14 × 103 C° m. e) Taking the phrase “into the rod” to mean an absolute value, the heat current will be kA ∂T = ∂x

∂T ∂x

= (100°C )( π L ) cosπx L. At the ends, x = 0 and x = L, the cosine is ± 1 and the

(385.0 W m ⋅ K) (1.00 × 10-4 m 2 )(3.14 × 103 C° m) = 121 W. f) Either by evaluating ∂T at ∂x the center of the rod, where πx L = π 2 and cos( π 2 ) = 0, or by checking the figure in part (a), the temperature gradient is zero, and no heat flows through the center; this is consistent with the symmetry of the situation. There will not be any heat current at the center of the rod at any later time. g) See Problem 17.114; k (385 W m ⋅ K) = = 1.1 × 10− 4 m 2 s. 3 ρc (8.9 × 10 kg m 3 )(390 J kg ⋅ K) h) Although there is no net heat current, the temperature of the center of the rod is decreasing; by considering the heat current at points just to either side of the center, where there is a non-zero temperature gradient, there must be a net flow of heat out of the region around the center. Specifically, ∂T H (( L 2) + ∆x) − H (( L 2) − ∆x) = ρAΔxc ∂t  ∂T  ∂T  = kA −  ∂x ( L 2 ) + ∆x ∂x ( L 2) − ∆x    2 ∂t = kA 2 ∆x, ∂x

17.124: a) In hot weather, the moment of inertia I and the length d in Eq. (13.39) will both increase by the same factor, and so the period will be longer and the clock will run slow (lose time). Similarly, the clock will run fast (gain time) in cold weather. (An ideal pendulum is a special case of physical pendulum.) b) ∆L = α∆Τ = (1.2 × 10−5 (C°) −1 L0 × (10.0 C°) = 1.2 × 10-4. c) See Problem 13.97; to avoid possible confusion, denote the pendulum period by τ . For this problem, ∆ττ = 1 ∆LL = 6.0 × 10−5 , so in one day the clock 2 will gain (86,400 s)(6.0 × 10−5 ) = 5.2 s so two figures. d) ∆T < 2((1.2 × 10−5 (C°) −1 ) × (86,400))−1 = 1.93 C°. 17.125: rod, The rate at which heat is aborbed at the blackened end is the heat current in the Aσ (TS4 − T24 ) =
Δτ τ

= (1 2)αΔT < (86,400) −1 , so

kA (T2 − T1 ), L where T1 = 20.00 K and T2 is the temperature of the blackened end of the rod. If this were to be solved exactly, the equation would be a quartic, very likely not worth the trouble. Following the hint, approximate T2 on the left side of the above expression as T1 to obtain σL 2 T2 = T1 + (TS − T14 ) = T1 + (6.79 × 10−12 K − 3 )(Ts4 − T14 ) = T1 + 0.424 K. k This approximation for T2 is indeed only slightly than T1, and is a good estimate of the temperature. Using this for T 2 in the original expression to find a better value of ∆T gives the same ∆T to eight figures, and further, and further iterations are not worth – while. A numerical program used to find roots of the quartic equation returns a value for ∆T that differed from that found above in the eighth place; this, of course, is more precision than is warranted in this problem.

17.126: a) The rates are: (i) 280 W, (ii) (54 J h ⋅ C° ⋅ m 2 )(1.5 m 2 )(11 C°) (3600 s h) = 0.248 W, (iii) (1400 W m 2 )(1.5 m 2 ) = 2.10 × 10 3 W, (iv) (5.67 × 10−8 W m 2 ⋅ K 4 )(1.5 m 2 )(320 K) 4 − (309 K 4 ) = 116 W. The total is 2.50 kW, with the largest portion due to radiation from the sun. b) P 2.50 × 103 W = = 1.03 × 10− 6 m 3 s = 3.72 L h. 3 6 ρLv (1000 kg m )(2.42 × 10 J kg ⋅ K)

c) Redoing the above calculations with e = 0 and the decreased area gives a power of 945 W and a corresponding evaporation rate of 1.4 L h. Wearing reflective clothing helps a good deal. Large areas of loose weave clothing also facilitate evaporation.

Chapter 18

In doing the numerical calculations for the exercises and problems for this chapter, the values of the ideal-gas constant have been used with the precision given on page 501 of the text, R = 8.3145 J mol ⋅ K = 0.08206 L ⋅ atm mol ⋅ K. Use of values of these constants with either greater or less precision may introduce differences in the third figures of some answers. 18.1: a) n = mtot M = (0.225 kg) (400 × 10−3 kg mol) = 56.3 mol. b) Of the many ways to find the pressure, Eq. (18.3) gives p= nRT (56.3 mol)(0.08206 L ⋅ atm mol ⋅ K)(291.15 K) = V (20.0 L) = 67.2 atm = 6.81 × 106 Pa. 18.2: a) The final temperature is four times the initial Kelvin temperature, or 4(314.15 K) –273.15= 983°C to the nearest degree. b) mtot = nM = MpV (4.00 × 10−3 kg/mol)(1.30 atm)(2.60 L) = = 5.24 × 10− 4 kg. RT (0.08206 L ⋅ atm/mol ⋅ K )(314.15 K )

18.3: For constant temperature, Eq. (18.6) becomes p2 = p1 (V1 V2 ) = (3.40 atm)(0.110 0.390) = 0.96 atm. 18.4: a) Decreasing the pressure by a factor of one-third decreases the Kelvin temperature by a factor of one-third, so the new Celsius temperatures is 1/3(293.15 K) – 273.15= − 175°C rounded to the nearest degree. b) The net effect of the two changes is to keep the pressure the same while decreasing the Kelvin temperature by a factor of onethird, resulting in a decrease in volume by a factor of one-third, to 1.00 L.

18.5:

Assume a room size of 20 ft X 20 ft X 20 ft V = 4000 ft 3 = 113 m3 . Assume a temperature of 20°C. pV (1.01 × 105 Pa )(113 m3 ) pV = nRT so n = = = 4685 mol RT (8.315 J/mol ⋅ K )(293 K ) N = nN A = 2.8 × 1027 molecules N 2.8 × 10 27 molecules b) = = 2.5 × 1019 molecules/ cm 3 6 3 V 113 × 10 cm

18.6: The temperature is T = 22.0°C = 295.15K. (a) The average molar mass of air is M = 28.8 × 10−3 kg mol, so mtot = nM = pV (1.00 atm)(0.900 L)(28.8 × 10−3 kg mol) M= = 1.07 × 10 − 3 kg. RT (0.08206 L ⋅ atm mol ⋅ K)(295.15 K) pV (1.00 atm)(0.900 L)(4.00 × 10 −3 kg mol) M = = 1.49 × 10− 4 kg. RT (0.08206 L ⋅ atm mol ⋅ K)(295.15 K)

(b) For Helium M = 4.00 × 10−3 kg mol, so mtot = nM =

18.7: From Eq. (18.6), pV   (2.821 × 10 6 Pa)(46.2 cm 3 )  T2 = T1  2 2  = (300.15 K)  pV   (1.01 × 10 5 Pa)(499 cm 3 )  = 776 K = 503°C.   1 1   32.0 × 10 −3 kg mol 4.013 × 10 5 Pa 0.0750 m 3 MpV = = 0.373 kg. RT ( 8.3145 J mol ⋅ K )( 310.15 K )

18.8:

a) mtot =

(

)(

)(

)

b) Using the final pressure of 2.813 × 10 5 Pa and temperature of 295.15 K, m′ = 0.275 kg, so the mass lost is 0.098 kg where extra figures were kept in the intermediate calculation of mtot . 18.9: From Eq. (18.6), T p 2 = p1  2 T  1  V1   V  2
3   430.15 K  0.750 m  5 5  = (1.50 × 10 Pa)   0.48 m 3  = 3.36 × 10 Pa .    300.15 K   

18.10:

a)

n=

pV (1.00 atm)(140 × 10 3 L) = = 5.78 × 10 3 mol. RT (0.08206 L ⋅ atm mol ⋅ K) (295.15 K)

b) (32.0 × 10 −3 kg mol)(5.78 × 10 3 mol) = 185 kg. 18.11: V2 = V1 (T2 T1 ) = (0.600 L)(77.3 292.15) = 0.159 L. 18.12: a) nRT V = 7.28 × 106 Pa while Eq. (18.7) gives 5.87 × 106 Pa. b) The van der Waals equation, which accounts for the attraction between molecules, gives a pressure that is 20% lower. c) 7.28 × 10 5 Pa, 7.13 × 10 5 Pa, 2.1%. d) As n V decreases, the formulas and the numerical values are the same. 18.13: At constant temperature, p 2 = p1 (V1 V2 ) = (1.0 atm)(6.0 5.7) = 1.1 atm.
2 a) V1 = V

18.14:

p1 T2 p2 T1

= (3.50)( 296 K ) = 3.74. b) Lungs cannot withstand such a volume 277K

change; breathing is a good idea. p 2V (100 atm)(3.10 L) = = 343 K = 70.3°C. nR (11.0 mol)( 0.08206 L ⋅ atm mol ⋅ K) b) This is a very small temperature increase and the thermal expansion of the tank may be neglected; in this case, neglecting the expansion means not including expansion in finding the highest safe temperature, and including the expansion would tend to relax safe standards. 18.15: a) T2 =

18.16: (a) The force of any side of the cube is F = pA = (nRT V ) A = (nRT ) L, since the ratio of area to volume is A V = 1 L. For T = 20.0° C = 293.15 K. F= nRT (3 mol) (8.3145 J mol ⋅ K) (293.15 K) = = 3.66 × 104 N. L 0.200 m

b) For T = 100.00°C = 373.15 K, F= nRT (3 mol)(8.3145 J mol ⋅ K)(373.15 K ) = = 4.65 × 104 N. L 0.200 m

18.17: Example 18.4 assumes a temperature of 0° C at all altitudes and neglects the variation of g with elevation. With these approximations, p = p0e − Mgy
/R T

We want y for p = 0.90 p0 so 0.90 = e − Mgy / RT RT y=− ln( 0.90) = 850 m Mg (We have used M = 28.8 × 10−3 kg/mol for air.)

and

18.18: From example 18.4, the pressure at elevation y above sea level is p = p 0 e − Mgy / RT . The average molar mass of air is M = 28.8 × 10 −3 kg/mol, so at an altitude of 100 m, Mgy1 (28.8 × 10 −3 kg mol)(9.80 m s 2 )(100 m) = = 0.01243, RT (8.3145 J mol ⋅ K)(273.15 K) and the percent decrease in pressure is 1 − p p0 = 1 − e −0.01243 = 0.0124 = 1.24%. At an altitude of 1000 m, Mgy2 RT = 0.1243, and the percent decrease in pressure is 1 − e −0.1243 = 0.117 = 11.7%. These answers differ by a factor of 11.7% 1.24% = 9.44, which is less than 10 because the variation of pressure with altitude is exponential rather than linear.

18.19: p = p0e − Myg

RT

from Example 18.4.

Eq. (18.5) says p = ( ρ M ) RT. Example 18.4 assumes a constant T = 273 K, so p and ρ are directly proportional and we can write ρ = ρ0 e − Mgy RT For y = 100 m, Mgy = 0.0124, so ρ = ρ0e − 0.0124 = 0.988 ρ0 RT

The density at sea level is 1.2% larger than the density at 100 m.

18.20: Repeating the calculation of Example 18.4 (and using the same numerical values for R and the temperature gives) p = (0.537) patm = 5.44 × 104 Pa. 18.21: p = ρRT M = 0.364 kg m 3 ( 8.3145 J mol ⋅ K ) ( 273.15K − 56.5K ) (28.8 × 10 −3 kg/mol) = 2.28 × 104 Pa.

(

)

18.22: M = N A m = (6.02 × 1023 molecules mol)(1.41 × 10−21 kg molecule) = 849 kg mol. Find the mass: m = nM = (3.00 mol)(63.546 × 10−3 kg mol) = 0.1906 kg m 0.1906 kg = = 2.14 × 10− 5 m 3 = 21.4 cm3 3 3 ρ 8.9 × 10 kg m

18.23: V=

18.24:

N = nN A =

pV NA RT

=

(9.119 × 10−9 Pa)(1.00 × 10 −6 m3 ) (6.023 × 10 23 molecules mol) (8.3145 J mol ⋅ K)(300 K) = 2.20 × 10 6 molecules.

18.25: p=

a) nRT N RT  molecules  (0.08206 L ⋅ atm mol ⋅ K)(7500 K) = =  80 × 103  23 V V Na  L  (6.023 × 10 molecules mol) = 8.2 × 10 −17 atm,

about 8.2 × 10 −12 Pa. This is much lower, by a factor of a thousand, than the pressures considered in Exercise 18.24. b) Variations in pressure of this size are not likely to affect the motion of a starship. 18.26: Since this gas is at standard conditions, the volume will be N V = (22.4 × 10 −3 m 3 ) = 2.23 × 10 −16 m 3 , and the length of a side of a cube of this NA
1

volume is (2.23 × 10 −16 m 3 ) 3 = 6.1 × 10 −6 m. 1000 g = 55.6 mol, which is (55.6 mol)(6.023 × 1023 molecules mol) = 18.0 g mol

18.27:

3.35 × 1025 molecules. 18.28: a) The volume per molecule is V nRT p RT = = N nN A N Ap = (8.3145 J mol ⋅ K)(300.15 K) (6.023 × 1023 molecules mol)(1.013 × 105 Pa)

= 4.091 × 10 − 26 m3 . If this volume were a cube of side L,

L = ( 4.091 × 10

− 26

m

3

)

1 3

= 3.45 × 10− 9 m,

which is (b) a bit more than ten times the size of a molecule.

18.29: a ) V = m ρ = n M ρ = ( 5.00 mol )(18.0g/mol ) / 1.00 g/cm3 = 90.0 cm 3 = 9.00 × 10−5 m 3 . b) See Excercise 18.28;

(

)

V    N

1/ 3

 V  =  nN    A

1/ 3

 9.00 × 10− 5 m3 =  ( 5.00 mol ) 6.023 × 1023 molecules/ mol  = 3.10 × 10−10 m.

(

)

   

1/ 3

c) This is comparable to the size of a water molecule. 18.30: a) From Eq. (18.16), the average kinetic energy depends only on the temperature, not on the mass of individual molecules, so the average kinetic energy is the same for the molecules of each element. b) Equation (18.19) also shows that the rms speed is proportional to the inverse square root of the mass, and so vrms vrms
Kr Ne

=

20.18 v = 0.491, rms 83.80 vrms vrms vrms
Rn Kr

Rn Ne

=

20.18 = 0.301 222.00

and

=

83.80 = 0.614. 222.00

18.31: a) At the same temperature, the average speeds will be different for the different isotopes; a stream of such isotopes would tend to separate into two groups. b)
0.352 0.349

= 1.004.

18.32: (Many calculators have statistics functions that are preprogrammed for such calculations as part of a statistics application. The results presented here were done on such a calculator.) a) With the multiplicity of each score denoted by  1   1  2 n1 , the average is   ∑ ni xi = 54.6 b)   ∑ ni xi  = 61.1. (Extra significant  150   150   figures are warranted because the sums are known to higher precision.)
1/ 2

18.33: We known that V A = VB and that T A > TB . a) p = nRT / V ; we don’t know n for each box, so either pressure could be higher.  N  pVN A b) pV =   N  RT so N = RT , where N A is Avogadro’s number. We don’t know   A how the pressures compare, so either N could be larger. c) pV = ( m M ) RT . We don’t know the mass of the gas in each box, so they could contain the same gas or different gases. d) 1 m v 2 av = 3 kT 2 2 TA > TB and the average kinetic energy per molecule depends only on T, so the statement must be true. e) v rms = 3kT m We don’t know anything about the masses of the atoms of the gas in each box, so either set of molecules could have a larger vrms . 18.34: Box A has higher pressure than B. This could be due to higher temperature and/or higher particle density in A. Since we know nothing more about these gases, none of the choices is necessarily true, although each of them could be true. a ) m = mP + mn = 3.348 × 10−27 kg; T = 300 × 106 K

( )

18.35:

vrms = 3kT m = 1.9 × 106 m s; vrms c = 0.64%
2 b) T = mvrms 3k

For vmms = 3.0 × 107 m s, T = 7.3 × 1010 K 18.36: From pV = nRT , the temperature increases by a factor of 4 if the pressure and

volume are each doubled. Then the rms speed vrms = 3RT M increases by a factor of 4 = 2, so the final rms speed is 2(250 m s) = 500 m s.

18.37: b)

a)

3 2

kT = (3 2)(1.381 × 10−23 J K)(300 K) = 6.21 × 10−21 J.

2 K ave 2(6.21 × 10−21 J) = = 2.34 × 105 m 2 s 2 . m (32.0 × 10− 3 kg mol) (6.023 × 10 23 molecules mol) vs= 3RT 3(8.3145 J mol ⋅ K)(300 K) = = 4.84 × 102 m s, −3 M (32.0 × 10 kg mol)

c)

which is of course the square root of the result of part (b). M  (32.0 × 10−3 kg mol)  v s = d) mv s =  (4.84 × 102 m s) 23  (6.023 × 10 molecules mol)  NA  = 2.57 × 10−23 kg ⋅ m This may also be obtained from 2mK ave = 2(6.21 × 10 −21 J)(32.0 × 10 −3 kg mol) (6.023 × 10 23 molecules mol)

e) The average force is the change in momentum of the atom, divided by the time between collisions. The magnitude of the momentum change is twice the result of part (d) (assuming an elastic collision), and the time between collisions is twice the length of a side of the cube, divided by the speed. Numerically, Fave = f) g) h) 2mv s mv s 2 2 K s 2(6.21 × 10−21 J) = = = = 1.24 × 10 −19 N. 2L v s L L (0.100 m) pave = Fave L2 = 1.24 × 10 −17 Pa. P Pave = 1.013 × 105 Pa 1.24 × 10−17 Pa = 8.15 × 1021 molecules. pV NA RT   (1.00 atm) (1.00 L ) 23  =  ( 0.08206 L ⋅ atm/mol ⋅ K )( 300 K )  6.023 × 10 molecules mol      N = n NA = = 2.45 × 1022. i) The result of part (g) was obtained by assuming that all of the molecules move in the same direction, and that there was a force on only two of the sides of the cube.

(

)(

)

18.38: This is the same calculation done in Example 16-9, but with p = 3.50 × 10−13 atm, giving λ = 1.6 × 105 m. 18.39: The rms speeds will be the same if the Kelvin temperature is proportional to the molecular mass; TN 2 = TH 2 ( M N 2 M H 2 ) = ( 293.15 K ) (28.0 2.02)

= 4.06 × 103 K = 3.79 × 103 °C.
3kT 3(1.381 × 10−23 J K )(300 K ) = = 6.44 × 10− 3 m s . b) If the particle is m 3.00 × 10−16 kg in thermal equilibrium with its surroundings, its motion will depend only on the surrounding temperature, not the mass of the individual particles. 18.40: a)

(

)

18.41: a) The six degress of freedom would mean a heat capacity at constant volume of . mol ⋅ K 6( 1 ) R = 3R = 24.9 J/mol ⋅ K. 3 R = (3( 8.03145−J kg mol) ) = 1.39 × 103 J/kg ⋅ K , b) vibrations do 2 M 18 ×10 3 contribute to the heat capacity. 18.42: a) Cv = ( C ) ( molar mass ) , so ( 833 J/kg ⋅ °C ) ( 0.018 kg/mol ) = 15.0 J mol ⋅ °C

at − 180°C, (1640 J/kg ⋅ °C ) ( 0.018 kg/mol ) = 29.5 J/kg ⋅ °C at − 60°C, ( 2060 J/kg ⋅ °C ) × ( 0.018 kg/mol ) = 37.1 J/mol ⋅ °C at − 5.0°C. b) Vibrational degrees of freedom become more important. c) CV exceeds 3R because H 2O also has rotational degrees of freedom. 18.43: a) Using Eq. (18.26), Q = ( 2.50 mol )( 20.79 J mol ⋅ K )( 30.0 K ) = 1.56 kJ. b) From Eq. (18.25), 3 of the result of part (a), 936 J. 5

18.44: which is b) from

a)
741 = 0.177 4190

c=

CV 20.76 J/mol ⋅ K = = 741 J/kg ⋅ K, M 28.0 × 10-3 kg/mol

times the specific heat capacity of water. mw Cw . Inserting the given data and the result CN (a) gives

mNCN ∆TN = mw Cw ∆Tw , or mN = part

mN = 5.65 kg. To find and volume, use pV = nRT , or V =

nRT = p [ ( 5.65 kg ) / ( 0.028 kg/mol ) ]( 0.08206 L ⋅ atm/mol ⋅ K )( 293 K ) = 4855 L. 1 atm
From Table (18.2), the speed is (1.60)v s, and so
v s2 = 3kT 3RT v2 = = m M (1.60) 2

18.45:

(see Exercise 18.48), and so the temperature is Mv 2 (28.0 × 10 −3 kg mol) T= = v 2 = (4.385 × 10− 4 K ⋅ s 2 m 2 )v 2 . 2 2 3(1.60) R 3(1.60) (8.3145 J mol ⋅ K) a) (4.385 × 10 −4 K ⋅ s 2 m 2 )(1500 m s) 2 = 987 K b) (4.385 × 10−4 K ⋅ s 2 m 2 )(1000 m s) 2 = 438 K c) (4.385 × 10 −4 K ⋅ s 2 m 2 )(500 m s) 2 = 110 K. 1 2 mv , 2
32

18.46: Making the given substitution ε =  m  f (v ) = 4π    2πkT 
32

2ε − ε kT 8π  m  − ε kT e =  .  εe m m  2πkT 

18.47: Express Eq. (18.33) as f = Aε e − ε kT , with A a constant. Then, df ε − ε kT  ε   − ε kT  = Ae − ε kT − e  = Ae 1 − kT . de kT     Thus, f will be a maximum when the term in square brackets is zero, or ε = which is Eq. (18.34). k R NA R . = = m M NA M 1 2 mv = kT , 2

18.48: Note that

a) 2(8.3145 J mol ⋅ K)(300 K) (44.0 × 10−3 kg mol) = 3.37 × 102 m s. b) 8(8.3145 J mol ⋅ K)(300 K) (π (44.0 × 10−3 kg mol)) = 3.80 × 102 m s.

c) 3(8.3145 J mol ⋅ K)(300 K) (44.0 × 10−3 kg mol) = 4.12 × 102 m s.

18.49: Ice crystals will form if T = 0.0°C; using this in the given relation for temperature as a function of altitude gives y = 2.5 × 10 3 m = 2.5 km. 18.50: a) The pressure must be above the triple point, p1 = 610 Pa. If p < p1 , the water cannot exist in the liquid phase, and the phase transition is from solid to vapor (sublimation). b) p 2 is the critical pressure, p 2 = p c = 221 × 10 5 Pa. For pressures below p 2 but above p1 , the phase transition is the most commonly observed sequence, solid to liquid to vapor, or ice to water to steam. 18.51: The temperature of 0.00° C is just below the triple point of water, and so there will be no liquid. Solid ice and water vapor at 0.00°C will be in equilibrium. 18.52: The atmospheric pressure is below the triple point pressure of water, and there can be no liquid water on Mars. The same holds true for CO 2

18.53: a)

∆V = βVo ∆T = (3.6 × 10−5 °C) (11 L)(21°C) = 0.0083 L ∆V = − kVo ∆p = (6.25 × 10−12 Pa)(11 L) (2.1 × 107 Pa) = −0.0014 L

So the total change in volume is ∆V = 0.0083 L − 0.0014 L = 0.0069 L. b) Yes; ∆V is much less than the original volume of 11.0 L. 18.54: m = nM = MpV RT

=

( 28.0 × 10 −3 kg mol)(2.026 × 10 −8 Pa)(3000 × 10−6 m 3 ) (8.3145 J mol ⋅ K)(295.15 K)

= 6.94 × 10 −16 kg. ∆pVM RT 6 (1.05 × 10 Pa )((1.00 m)π (0.060 m) 2 )(44.10 × 10−3 kg mol) = = 0.213 kg. (8.3145 J mol ⋅ K)(295.15 K)

18.55: ∆m = ∆nM =

18.56: a) The height h′ at this depth will be proportional to the volume, and hence inversely proportional to the pressure and proportional to the Kelvin temperature; h′ = h p T′ patm T′ =h p′ T patm + ρgy T  280.15 K  (1.013 × 10 5 Pa )   5 3 2 (1.013 × 10 Pa) + (1030 kg m )(9.80 m s )(73.0 m)  300.15 K   

= (2.30 m)

= 0.26 m,
so ∆h = h − h′ = 2.04 m. b) The necessary gauge pressure is the term ρgy from the above calculation, pg = 7.37 × 105 Pa.

18.57: The change in the height of the column of mercury is due to the pressure of the air. The mass of the air is mair = nM = ρ g∆hV PV M = Hg M RT RT

 (13.6 × 103 kg m 3 )(9.80 m s 2 )(0.060 m)     × ((0.900 m − 0.690 m))(0.620 × 10− 4 m 2 )  = (28.8 g mol) (8.3145 J mol ⋅ K)(293.15 K)       −3 = 1.23 × 10 g. m , where ρ is the density of the V ambient air and m is the load. The density is inversely proportional to the temperature, so 18.58: The density ρ' of the hot air must be ρ′ = ρ −  ρ ρ m   T′ = T = = T 1 −  ρ′ ρ − (m V ) ρV   
−1

  (290 kg) = (288.15 K)1 −  (1.23 kg m 3 )(500 m 3 )     which is 272°C.

−1

= 545 K,

18.59:

 (0.0150 m 3 )(318.15 K )  V T  p 2 = p1  1 2  = (2.72 atm) V T   (0.0159 m 3 )(278.15 K )  = 2.94 atm,   2 1  

so the gauge pressure is 1.92 atm.

18.60: (Neglect the thermal expansion of the flask.) a) (1.013 × 10 Pa)(300 380) = 8.00 × 10 Pa.
5 4

p2 = p1 (T2 T1 ) =

b)

pV mtot = nM =  2  RT  2

 M  

 (8.00 × 104 Pa)(1.50 L)  =  (8.3145 J mol ⋅ K)(300 K ) (30.1 g mol) = 1.45 g.   

18.61: a) The absolute pressure of the gas in a cylinder is (1.20 × 106 + 1.013 × 105 ) Pa = 1.30 × 106 Pa. At atmospheric pressure, the volume of hydrogen will increase by a factor of 1.30 × 10 6 , so the number of cylinders is 1.01 × 10 5 750 m 3 = 31. (1.90 m 3 )((1.30 × 10 6 ) (1.01 × 10 5 )) b) The difference between the weight of the air displaced and the weight of the hydrogen is pM H 2   Vg ( ρair − ρH 2 )Vg =  ρair −  RT     (1.01 × 105 Pa)(2.02 × 10−3 kg mol)  1.23 kg m 3 −  = (8.3145 J mol ⋅ K)(288.15 K)    × (9.80 m s 2 )(750 m 3 ) = 8.42 × 10 3 N. c) Repeating the above calculation with M = 4.00 × 10 −3 kg mol gives a weight of 7.80 × 10 3 N.

18.62: If the original height is h and the piston descends a distance y, the final pressure  h  of the air will be p atm   h − y . This must be the same as the pressure at the bottom of the    mercury column, patm + ( ρg ) y. Equating these two, performing some minor algebra and solving for y gives patm (1.013 × 105 Pa) y =h− = (0.900 m) − = 0.140 m. ρg (13.6 × 103 kg m 3 )(9.80 m s 2 )

18.63: a) The tank is given as being “large,” so the speed of the water at the top of the surface in the tank may be neglected. The efflux speed is then obtained from 1 2 ρv = ρg∆h + ∆ p, or 2   ∆p  (3.20 × 105 Pa)  2   = 2 (9.80 m s ) (2.50 m) + v = 2 g∆h +   ρ  (1000 kg m3 )      = 26.2 m s. b) Let h0 = 3.50 m and p 0 = 4.20 × 10 5 Pa. In the above expression for  4.00 m − h0  v, ∆h = h − 1.00 m and ∆p = p 0   4.00 m − h  − p a . Repeating the calculation for    h = 3.00 m gives v = 16.1 m s and with h = 2.00 m, v = 5.44 m s. c) Setting v 2 = 0 in the above expression gives a quadratic equation in h which may be re-expressed as (h − 1.00 m) = pa p0 0.50 m − . ρg ρg 4.00 m − h

Denoting

pa p (0.50 m) = y = 10.204 m and 0 = z 2 = 21.43 m 2 , this quadratic becomes ρg ρg h 2 − (5.00 m + y )h + ((4.00 m) y + ( 4.00 m 2 ) − z 2 ) = 0,

which has as its solutions h = 1.737 m and h = 13.47 m. The larger solution is unphysical (the height is greater than the height of the tank), and so the flow stops when h = 1.74 m. Although use of the quadratic formula is correct, for this problem it is more efficient for those with programmable calculators to find the solution to the quadratic by iteration. Using h = 2.00 m (the lower height in part (b)) gives convergence to three figures after four iterations. (The larger root is not obtained by a convergent iteration.)

18.64: a)

N nN A pVN A (1.00 atm)(14.5 L)(6.023 × 1023 molecules mol) = = = ∆t ∆t RT∆t (0.08206 L ⋅ atm mol ⋅ K) (293.15 K) (3600 s) = 1.01 × 1020 molecule [ − 36 pt ]

b)

(14.5 L) 60 min) = 10 min . (0.5 L)(0.210 − 0.163)

c) The density of the air has decreased by a factor of (0.72 atm 1.00 atm) × 1 (293 K 273 K) = 0.773, and so the respiration rate must increase by a factor of 0.733 , to 13 breaths min . If the breathing rate is not increased, one would experience “shortness of breath.” (6.023 × 1023 molecules mol)(50 kg ) (18.0 × 10− 3 kg mol) = 5.0 × 10 27 atoms.

18.65: 3N = 3nN A = 3(m M ) N A = 3

18.66: The volume of gas per molecule (see Problem 18.28) is

RT N Ap

, and the volume of a

4 molecule is about V0 = π (2.0 × 10 −10 m)3 = 3.4 × 10− 29 m3 . Denoting the ratio of these 3 volumes as f, p= f RT (8.3145 J mol ⋅ K)(300 K) = f = (1.2 × 108 Pa) f . 23 − 29 3 N AV0 (6.023 × 10 molecules mol)(3.4 × 10 m )

“Noticeable deviations” is a subjective term, but f on the order of unity gives a pressure of 10 8 Pa. Deviations from ideality are likely to be seen at values of f substantially lower than this.

18.67: a) Dividing both sides of Eq. (18.7) by the product RTV gives the result. b) The algorithm described is best implemented on a programmable calculator or computer; for a calculator, the numerical procedure is an interation of  (9.8 × 105 )  (0.448) x= + x 2  1 − (4.29 × 10 − 5 ) x .  (8.3145)(400.15) (8.3145)(400.15)  Starting at x = 0 gives a fixed point at x = 3.03 × 10 2 after four iterations. The number density is 3.03 × 10 2 mol m 2 . c) The ideal-gas equation is the result after the first iteration, 295 mol m 3 . The vander Waals density is larger. The term corresponding to a represents the attraction of the molecules, and hence more molecules will be in a given volume for a given pressure. 18.68: a) U = mgh =   M 28.0 × 10−3 kg mol 2 − 22 gh =   6.023 × 1023 molecules mol (9.80 m s )(400 m) = 1.82 × 10 J.  NA  

[

]

3 2 1.82 × 10−22 J b) Setting U = kT , T = = 8.80 K. c) It is possible, but not at 2 3 1.38 × 10− 23 J K all likely for a molecule to rise to that altitude. This altitude is much larger than the mean free path.

18.69: a), b) (See figure.) The solid curve is U (r ), in units of U 0 , and with x = r R0 . The dashed curve is F (r ) in units of U 0 R0 . Note that r1 < r2 . R  R  c) When U = 0,  0  = 2 0  , or r1 = R0 21 6. Setting F = 0 in Eq. (18.26) r   r   1   1  r1 gives r2 = R0 and = 2 −1 6. d) U (r2 ) = U ( R0 ) = −U 0 , so the work required is U 0 . r2
12 6

18.70: gas is 1 2

a)
MpV RT MpV RT 3 2

3 2

nRT = 3 pV = 2

3 2

(1.01× 10 Pa )(5.00 × 10
5

−3

m 3 = 758 J. b) The mass of the

)

, and so the ratio of the energies is

v 2 1 Mv 2 1 2.016 × 10 −3 kg mol ( 30.0 m s ) 2 = = = 2.42 × 10 − 4 = 0.0242%. ( 8.3145 J mol ⋅ K )( 300 K ) pV 3 RT 3 a) From Eq. (18.19 ) , v s = 3 (8.3145 J mol ⋅ K ) (300.15 K) (28.0 × 10−3 kg mol) = 517 m s.

(

)

18.71:

b) v s

3 = 299 m s.

18.72:

a)

3kT 3(1.38 × 10−23 J K ) (5800 K) = = 1.20 × 104 m s. − 27 m (1.67 × 10 kg) 2(6.673 × 10 −11 N ⋅ m 2 kg 2 ) (1.99 × 1030 kg) = 6.18 × 105 m s. 8 (6.96 × 10 m)

b)

2GM = R

c) The escape speed is about 50 times the rms speed, and any of Fig. (18.20 ) , Eq. (18.32) or Table (18.2) will indicate that there is a negligibly small fraction of molecules with the escape speed. 18.73: a) To escape, the total energy must be positive, K + U > 0. At the surface of the earth, U = − GmM R = −mgR, so to escape K > mgR. b) Setting the average kinetic energy equal to the expression found in part (a), ( 3 2) kT = mgR, or T = ( 2 3)( mgR k ). For nitrogen, this is T= 2 ( 28.0 × 10−3 kg mol)(9.80 m s 2 ) (6.38 × 106 m) 3 (6.023 × 10 23 molecules mol)(1.381 × 10- 23 J K)

= 1.40 × 10 5 K and for hydrogen the escape temperature is ( 2.02 ) times this, or 1.01 × 10 4 K. 28.0 c) For

nitrogen, T = 6.36 × 10 3 K and for hydrogen, T = 459 K. d) The escape temperature for hydrogen on the moon is comparable to the temperature of the moon, and so hydrogen would tend to escape until there would be none left. Although the escape temperature for nitrogen is higher than the moon’s temperature, nitrogen would escape, and continue to escape, until there would be none left.

18.74: (See Example 12.5 for calculation of the escape speeds) a) Jupiter: v s = 3(8.3145 J mol ⋅ K)(140 K) (2.02 × 10−3 kg mol) = 1.31 × 103 m s = (0.0221)ve . Earth: v s = 3(8.3145 J mol ⋅ K)(220 K) ((2.02 × 10−3 kg mol) = 1.65 × 103 m s = (0.146)ve . b) Escape from Jupiter is not likely for any molecule, while escape from earth is possible for some and hence possible for all. c) v s = 3(8.3145 J mol ⋅ K)(200 K) (32.0 × 10−3 kg mol) = 395 m s. The radius of the asteroid is R = (3M 4πρ)
1 3

= 4.68 × 105 m, and the escape speed is

2GM R = 542 m s, so there can be no such atmosphere.

18.75:

a) From Eq. (18.19), m= 3kT 3(1.381 × 10−23 J K)(300 K) = = 1.24 × 10 −14 kg. v s2 (0.001 m s) 2

b) mN A M = (1.24 × 10 −14 kg)(6.023 × 10 23 molecules mol) (18.0 × 10 −3 kg mol) = 4.16 × 1011 molecules.  3V  3  3m ρ  c) D = 2r = 2  = 2   4π   4π 
1 1 3

 3(1.24 × 10 −14 kg)  −6 = 2  4π (920 kg m3 )  = 2.95 × 10 m,    which is too small to see.
3

1

18.76: From x = A cos ωt , v = −ωA sin ωt , U ave = 1 2 kA (cos2 ωt ) ave , 2
1 2

K ave =

1 mω2 A2 (sin 2 ωt ) ave . 2

Using (sin 2 θ ) ave = (cos2 θ )ave =

and mω2 = k shows that K ace = U ave .

18.77: a) In the same manner that Eq. (18.27) was obtained, the heat capacity of the two-dimensional solid would be 2R = 16.6 J mol ⋅ K. b) The heat capcity would behave qualitatively like those in Fig. (18.18), and heat capacity would decrease with decreasing temperature. 18.78: a) The two degrees of freedom associated with the rotation for a diatomic molecule account for two-fifths of the total kinetic energy, so K rot = nRT = (1.00) (8.3145 J mol ⋅ K)(300 K ) = 2.49 × 103 J. b)   16.0 × 10 −3 kg mol -11 2 I = 2m(L 2) 2 = 2  23  6.023 × 10 molecules mol  (6.05 × 10 m)    = 1.94 ×10 −46 kg ⋅ m 2 . c) Using the results of parts (a) and (b),

ωs =

2 K rot N A 2(2.49 × 103 J) = I (1.94 × 10− 46 kg ⋅ m 2 )(6.023 × 1023 molecules mol)
= 6.52 × 10 12 rad s,

much larger than that of machinery. 18.79: For CO 2 , the contribution to CV other than vibration is 5 5 R = 20.79 J mol ⋅ K, and CV − R = 0.270 CV 2 2 For both SO2 and H2S, the contribution to CV other than vibration is 6 R = 24.94 J mol ⋅ K, 2 and the respective fractions of CV are 0.25 and 0.039.

∞

18.80: a)

∫
0

 m  f (v) dv = 4π    2πkT   m  = 4π    2πkT 

3 2∞

∫v
0

2

e −mv

2

/ 2 kT

dv

32

  1 π   4(m 2 KT  m 2 KT = 1   

where the tabulated integral (given in Problem18.81) has been used. b) f (v)dv is the probability that a particle has speed between v and v + dv; the probability that the particle has some speed is unity, so the sum (integral) of f (v) dv must be 1. With n = 2 and a = m/2kT , the integral is  m  4π    2πkT  which is Eq. (18.16).
3 2∞ 32

18.81:

  3 π 3kT  3  2 (m 2kT ) 2  (m 2kT ) = m ,   

∞

18.82:

∫
0

 m  f (v)dv = 4π    2πkT 

∫v e
0

3 − mv 2 2 kT

dv.

Making the suggested change of variable, v 2 = x, 2v dv = dx, v 3dv = (1 2) x dx, the integral becomes
∞

 m  ∫ vf (v)dv = 2π  2πkT    0  m  = 2π    2πkT  2 = π

3

2

∞

∫
0

xe

− mx

2 kT

dx

3

 2kT     m  2 KT 8KT = , m πm
2

2

which is Eq. (18.35).

18.83: a) See Problem 18.80. Because f (v)dv is the probability that a particle has a speed between v and v + dv, f (v)dv is the fraction of the particles that have speed in that range. The number of particles with speeds between v and v + dv is therefore dN = Nf (v) dv and ∆N = N ∫ v + ∆v f (v )dv. v b) vmp =
2 kT m

, and
3

2 4  m   2kT  −1 f (vmp ) = 4π  .   e = e π vmp  2πkT   m 

For oxygen gas at 300 K, vmp = 3.95 × 10 2 m s, and f (v)∆v = 0.0421, keeping an extra figure. c) Increasing v by a factor of 7 changes f by a factor of 7 2 e −48 , and f (v)∆v = 2.94 × 10−21. d) Multiplying the 2, and the

temperature by a factor of 2 increases the most probable speed by a factor of

answers are decreased by 2 ; 0.0297 and 2.08 × 10-21. e) Similarly, when the temperature is one-half what it was parts (b) and (c), the fractions increase by 2 to 0.0595 and 4.15 × 10 −21. f) At lower temperatures, the distribution is more sharply peaked about the maximum (the most probable speed), as is shown in Fig. (18.20). 18.84: a) (0.60)(2.34 × 10 3 Pa) = 1.40 × 10 3 Pa. b) m = MpV (18.0 × 10 −3 kg mol)(1.40 × 10 3 Pa)(1.00 m 3 ) = = 10 g . RT (8.3145 J mol ⋅ K)(293.15 K)

18.85: The partial pressure of water in the room is the vapor pressure at which condensation occurs. The relative humidity is 1..81 = 42.6%. 4 25

18.86: a) The partial pressure is (0.35)(3.78 × 10 3 Pa) = 1.323 × 10 3 Pa. This is close to the vapor pressure at 12°C, which would be at an altitude (30°C − 12°C) (0.6° C 100 m) = 3 km above the ground (more precise interpolation is not warranted for this estimate). b) The vapor pressure will be the same as the water pressure at around 24°C, corresponding to an altitude of about 1 km.

18.87:

a) From Eq. (18.21),

λ = (4π 2r 2 ( N V )) −1 = (4π 2 (5.0 × 10 −11 m) 2 (50 × 10 6 m −3 )) −1 = 4.5 × 1011 m. b) 3(8.3145 J mol ⋅ K)(20 K) (1.008 × 10 −3 kg mol) = 703 m s, and the time between

collisions is then (4.5 × 1011 m) (703 m s) = 6.4 × 10 8 s, about 20 yr. Collisions are not very important. c) p = ( N V )kT = (50 × 10 6 m −3 )(1.381 × 10 −23 J K)(20 K) = 1.4 × 10 −14 Pa. d) ve =
2GM = R 2G ( Nm V )( 4πR 3 3) = (8π 3)G ( N V )mR R

= (8π 3)(6.673 × 10 −11 N ⋅ m 2 kg 2 )(50 × 10 6 m −3 )(1.67 × 10 −27 kg) × (10 × 9.46 × 1015 m) = 650 m s. This is lower than v s, and the cloud would tend to evaporate. e) In equilibrium (clearly not thermal equilibrium), the pressures will be the same; from pV = NkT , kTISM ( N V ) ISM = kTnebula ( N V ) nebula and the result follows. f) With the result of part (e),  (V N ) nebula   50 × 10 6 m 3  5  = (20 K)  TISM = Tnebula   (200 × 10 −6 m 3 ) −1  = 2 × 10 K,   (V N )  ISM     more than three times the temperature of the sun. This indicates a high average kinetic energy, but the thinness of the ISM means that a ship would not burn up.

18.88: a) Following Example 18.4,

dP dy

= − pM , which in this case becomes RT

dp Mg dy =− , p R T0 − αy which integrates to  p  Mg  αy   αy  Rα  = 1 − , or p = p0 1 −  . ln   ln    T  0   p0  Rα  T0   b) Using the first equation above, for sufficiently small α, ln(1 − αy ) ≈ − αy , and this T0 T0 gives the expression derived in Example 18.4.  (0.6 × 10 −2 C° m)(8863 m)  1 −  = 0.8154, c)   ( 288 K)   −3 2 Mg (28.8 × 10 )(9.80 m s ) = = 5.6576 Rα (8.3145 J mol ⋅ K)(0.6 × 10 − 2 C° m) (the extra significant figures are needed in exponents to reduce roundoff error), and p 0 (0.8154) 5.6576 = 0.315 atm, which is 0.95 of the result found in Example 18.4. Note: for calculators without the x y function, the pressure in part (c) must be found from p = p 0 exp((5.6576) ln( 0.8154)).
Mg

∂P 18.89: a) A positive slope ∂V would mean that an increase in pressure causes an increase in volume, or that decreasing volume results in a decrease in pressure, which cannot be the case for any real gas. b) See Fig. (18.5). From part (a), p cannot have a positive slope along an isotherm, and so can have no extremes (maxima or minima) along an ∂p isotherm. When ∂V vanishes along an isotherm, the point on the curve in a p - V diagram

must be an inflection point, and c) p= nRT an 2 − 2 V − nb V

∂2 p ∂V 2

=0

∂p nRT 2an 2 =− + 3 ∂V (V − nb) 2 V ∂2 p 2nRT 6an 2 = − 4 . ∂V 2 (V − nb)3 V Setting the last two of these equal to zero gives V 3 nRT = 2an 2 (V − nb) 2 , c) V 4 nRT = 3an 2 (V − nb) 3 .

Following the hint, V = (3 2)(V − nb), which is solved for (V n) c = 3b. Substituting this into either of the last two expressions in part (c) gives Tc = 8a 27 Rb. pc = R( 8 a ) RT a a a − = 27 Rb − 2 = . (V n) c − b (V n) c 2b 9b 27b 2
8 a 27 b a 27 b 2

d)

e)

RTc = p c (V n) c

8 = . 3b 3

g) H 2 : 3.28. N 2 : 3.44. H 2O : 4.35. h) While all are close to 8 3, the agreement is not good enough to be useful in predicting critical point data. The van der Waals equation models certain gases, and is not accurate for substances near critical points.

18.90:

a) vav = 1 (v1 + v2 ) and vrms = 2

1 2

2 v12 + v2 , and

1 1 2 2 2 2 vrms − vav = (v12 + v2 ) − (v12 + v2 + 2v1v2 ) 2 4 1 2 2 (v1 + v 2 − 2v1v 2 ) 4 1 = (v1 − v 2 ) 2 . 4 = This shows that v rms ≥ v av , with equality holding if and only if the particles have the same speeds. b) v′ 2 = rms
1 N +1 2 ′ ( Nvrms + u 2 ), vav = 1 N +1

( Nvav + u ), and the given forms follow immediately.

c) The algebra is similar to that in part (a); it helps somewhat to express ′2 vav = 1 2 ( N (( N + 1) − 1)vav + 2Nvav u + (( N + 1) − N )u 2 ) 2 ( N + 1) N 2 N 1 2 = vav + (−vav + 2vav u − u 2 ) + u2. 2 N +1 ( N + 1) N +1

Then, ′2 ′2 vrms − vav = N N 2 2 2 (vrms − vav ) + (vav − 2vav u + u 2 ) ( N + 1) ( N + 1) 2 N N 2 2 = (vrms − vav ) + (vav − u ) 2 . 2 N +1 ( N + 1)

′ If v rms > v av , then this difference is necessarily positive, and v′ > vav . rms d) The result has been shown for N = 1, and it has been shown that validity for N implies validity for N + 1; by induction, the result is true for all N.

Chapter 19

19.1: a)

b) p∆V = nR∆T = (2.00 mol)(8.3145 J mol ⋅ K )(80 C °) = 1.33 × 10 3 J. 19.2: a)

b) If the pressure is reduced to 40.0% of its original value, the final volume is (5 2) of its original value. From Eq. (19.4), V  5 W = nRT ln 2 = (3)(8.3145 J mol ⋅ K )(400.15 K ) ln   = 9.15 × 103 J. V1  2 19.3:

pV = nRT T constant, so when p increases, V decrease

19.4: At constant pressure, W = p∆V = nR∆T , so ∆T = W 1.75 × 10 3 J = = 35.1 K nR (6 mol) (8.3145 J mol ⋅ K)

and ∆TK = ∆TC , so T2 = 27.0°C + 35.1°C = 62.1°C. 19.5: a)

b) At constant volume, dV = 0 and so W = 0. 19.6:

b) p∆V = (1.50 × 10 5 Pa) (0.0600 m 3 − 0.0900 m 3 ) = −4.50 × 10 3 J.

19.7: a)

b) In the first process, W1 = p∆V = 0 . In the second process, W2 = p∆V = (5.00 × 10 5 Pa) (−0.080 m 3 ) = −4.00 × 10 4 J. 19.8: a) W13 = p1 (V2 − V1 ), W32 = 0, W24 = p 2 (V1 − V2 ) and W41 = 0. The total work done by the system is W13 + W32 + W24 + W41 = ( p1 − p 2 )(V2 − V1 ), which is the area in the p- V plane enclosed by the loop. b) For the process in reverse, the pressures are the same, but the volume changes are all the negatives of those found in part (a), so the total work is negative of the work found in part (a). 19.9: Q = 254 J, W = −73 J (work is done on the system), and so ∆U = Q − W = 327 J.

19.10: a) p∆V = (1.80 × 10 5 Pa)(0.210 m 2 ) = 3.78 × 10 4 J. b) ∆U = Q − W = 1.15 × 10 5 J − 3.78 × 10 4 J = 7.72 × 10 4 J. c) The relations W = p∆V and ∆U = Q − W hold for any system.

19.11: The type of process is not specified. We can use ∆U = Q − W because this applies to all processes. Q is positive since heat goes into the gas; Q = +1200 J W positive since gas expands; W = +2100 J ∆U = 1200 J - 2100 J = −900 J We can also use ∆U = n( 3 R ) ∆T since this is true for any process for an ideal gas. 2 ∆T = 2( −900 J) 2∆U = = −14.4 C° 3nR 3(5.00 mol)(8.3145 J mol ⋅ K)

T2 = T1 + ∆T = 127°C − 14.4 C° = 113°C 19.12: At constant volume, the work done by the system is zero, so ∆U = Q − W = Q. Because heat flows into the system, Q is positive, so the internal energy of the system increases. 19.13: a) p∆V = (2.30 × 10 5 Pa)(-0.50 m 3 ) = −1.15 × 10 5 J . (b Q = ∆U + W = −1.40 × 10 5 J + (−1.15 × 10 5 J) = −2.55 × 10 5 J (heat flows out of the gas). c) No; the first law of thermodynamics is valid for any system. 19.14: a) The greatest work is done along the path that bounds the largest area above the V-axis in the p- V plane (see Fig. (19.8)), which is path 1. The least work is done along path 3. b) W > 0 in all three cases; Q = ∆U + W , so Q > 0 for all three, with the greatest Q for the greatest work, that along path 1. When Q > 0, heat is absorbed. 19.15: a) The energy is (2.0 g )(4.0 kcal g ) + (17.0 g)(4.0 kcal g) + (7.0 g)(9.0 kcal g) = 139 kcal, and the time required is (139 kcal) (510 kcal h) = 0.273 h = 16.4 min. b) v = 2 K m = 2(139 × 10 3 cal) (4.186 J cal) (60 kg) = 139 m s = 501 km h.

19.16: a) The container is said to be well-insulated, so there is no heat transfer. b) Stirring requires work. The stirring needs to be irregular so that the stirring mechanism moves against the water, not with the water. c) The work mentioned in part (b) is work done on the system, so W < 0, and since no heat has been transferred, ∆U = −W > 0. 19.17: The work done is positive from a to b and negative from b to a; the net work is the area enclosed and is positive around the clockwise path. For the closed path ∆U = 0, so Q = W > 0. A positive value for Q means heat is absorbed. b) Q = 7200 J, and from part (a), Q > 0 and so Q = W = 7200 J. c) For the counterclockwise path, Q = W < 0. W= − 7200 J, so Q = −7200 J and heat is liberated, with |Q|=7200 J. 19.18: a), b) The clockwise loop (I) encloses a larger area in the p-V plane than the counterclockwise loop (II). Clockwise loops represent positive work and counterclockwise loops negative work, so WI > 0 and WII < 0. Over one complete cycle, the net work WI + WII > 0, and the net work done by the system is positive. c) For the complete cycle, ∆U = 0 and so W = Q. From part (a), W > 0 so Q > 0, and heat flows into the system. d) Consider each loop as beginning and ending at the intersection point of the loops. Around each loop, ∆U = 0, so Q = W ; then, QI = WI > 0 and QII = WII < 0. Heat flows into the system for loop I and out of the system for loop II. 19.19: a) Yes; heat has been transferred form the gasses to the water (and very likely the can), as indicated by the temperature rise of the water. For the system of the gasses, Q < 0. b) The can is given as being constant-volume, so the gasses do no work. Neglecting the thermal expansion of the water, no work is done. c) ∆U = Q − W = Q < 0. 19.20: a) p∆V = (2.026 × 10 5 Pa)(0.824 m 3 − 1.00 × 10 −3 m 3 ) = 1.67 × 10 5 J. b) ∆U = Q − W = mLv − W = (1.00 kg)(2.20 × 10 6 J kg ) − 1.67 × 10 5 J = 2.03 × 10 6 J.

19.21: a) Using Equation (19.12), dT =

dQ nCV

=

645 J (0.185 mol)(20.76 J mol⋅K)

= 167.9 K, or T = 948 K.

b) Using Equation (19.14), dT =

dQ nC p

=

645 J (0.185 mol)(29.07 J mol⋅.K)

= 119.9 K, or T = 900 K.

19.22: a) nCV ∆T = (0.0100 mol)(12.47 J mol ⋅ K)(40.0 C°) = 4.99 J.

19.23: n = 5.00 mol. ∆T = +30.0 C° a) For constant p, Q = nC p ∆T = (5.00 mol)(20.78 J mol ⋅ K)(30.0 C°) = +3120 J

Q > 0 so heat goes into gas.
b) For constant V, Q = nC v ∆T = (5.00 mol)(12.47 J mol ⋅ K)(30.0 C°) = +1870 J Q > 0 so heat goes into gas. c) For constant p, Q = nC p ∆T = (5.00 mol)(36.94 J mol ⋅ K)(30.0 C°) = +5540 J Q > 0 so heat goes into gas. 19.24: For an ideal gas, ∆U = CV ∆T , and at constant pressure, p∆V = nR∆T . Using CV = 3 R for a monatomic gas, 2 3 3 3  ∆U = n R  ∆T = p∆V = (4.00 × 10 4 Pa)(8.00 × 10 −3 m 3 − 2.00 × 10 −3 m 3 ) = 360 J. 2 2 2  19.25: For constant p, Q = nC p ∆T Since the gas is ideal, pV = nRT and for constant p, p∆V = nR∆T .  p∆V   C p   p∆V Q = nC p  =  nR   R    Since the gas expands, ∆V > 0 and therefore Q > 0. Q > 0 means heat goes into gas. 19.26: For an ideal gas, ∆U = CV ∆T , and at constant pressure, W = p∆V = nR∆T . Using CV = 3 R for a monatomic gas, ∆U = n( 3 R )∆T = 2 2 5 2 Q = ∆U + W = 2 W , so W Q = 5 .
3 2

p∆V = 3 W . Then 2

19.27: a) For an isothermal process, W = nRT ln (V2 V1 ) = (0.150 mol)(8.3145 J mol ⋅ K)(350.15 K)ln(1 4) = −605 J. b) For an isothermal process for an ideal gas, ∆T = 0 and ∆U = 0. c) For a process with ∆U = 0, Q = W = −605 J ; 605 J are liberated. 19.28: For an isothermal process, ∆U = 0, so W = Q = −335 J. 19.29: For an ideal gas γ = C p C V = 1 + R C V , and so C V = R (γ − 1) = (8.3145 J mol ⋅ K) (0.127) = 65.5 J mol ⋅ K and C p = CV + R = 73.8 J mol ⋅ K.

19.30: a)

b) pV2 − pV1 = nR(T2 − T1 ) = (0.250 mol)(8.3145 J mol ⋅ K)(100.0 K) = 208 J. c) The work is done on the piston. d) Since Eq. (19.13) holds for any process, ∆U = nCV ∆T = (0.250 mol)(28.46 J mol ⋅ K)(100.0 K) = 712 J. e) Either Q = nC P ∆T or Q = ∆U + W gives Q = 924 × 10 3 J to three significant figures. f) The lower pressure would mean a correspondingly larger volume, and the net result would be that the work done would be the same as that found in part (b).

19.31: a) C p = R (1 − (1 γ ) ) , and so Q = nC p ∆T =

( 2.40 mol)( 8.3145 J mol ⋅ K )( 5.0 C°) = 553 J.
1 − 1 1.220

b) nCV ∆T = nC P ∆T γ = ( 553 J ) (1.220 ) = 454 J. (An extra figure was kept for these calculations.) 19.32: a) See also Exercise 19.36; V   0.0800 m 3  3 5 p 2 = p1  1  = 1.50 × 10 5 Pa   0.0400 m 3  = 4.76 × 10 Pa.  V     2
γ

(

)

5

b) This result may be substituted into Eq. (19.26), or, substituting the above form for p 2 ,   V  γ −1  1 W = p1V1 1 −  1     V2   γ −1       0.0800  3  3 4 = 1.50 × 10 5 Pa 0.0800 m 3 1 −    = −1.60 × 10 J.   0.0400   2  

(

)(

)

2

c) From Eq. (19.22), ( T2 T1 ) = (V2 V1 ) = ( 0.0800 0.0400 ) = 1.59, and since the final temperature is higher than the initial temperature, the gas is heated (see the note in Section 19.8 regarding “heating” and “cooling.”)
γ −1 23

19.33: a)

b) ( Use γ = 1.400, as in Example 19.6) From Eq. (19.22),
T2 = T1 (V1 V2 )
γ −1

= ( 293.15 K )(11.1)
γ

0.400

= 768 K = 495°C
1.400

and from Eq. (19.24), p 2 = p1 (V1 V2 ) = (1.00 atm )(11.1)

= 29.1 atm.

19.34: γ = 1.4 for ideal diatomic gas Q = ∆U + W = 0 for adiabatic process ∆U = −W = − ∫ PdV PV γ = const = PiVi γ
iV ∆U = − ∫ 10L PV γi dV = −Pi Viγ 30 L γ

= −(1.2 atm) (30 L)1.4

[

(

V − γ +1 − γ +1

)

10L1 30 L

(10 L)1-1.4 −( 30 L)1-1.4 1−1.4

]

= 50 L ⋅ atm = 5.1 × 10 3 J. The internal energy increases because work is done on the gas (∆U > 0). The temperature increases because the internal energy has increased. 19.35: For an ideal gas ∆U = nCV ∆T . The sign of ∆U is the same as the sign of ∆T . T1V1γ −1 = T2V2γ −1 and V = nRT p so,
1 T1γ p1 −γ = T2γ p1−γ and T2γ = T1γ ( p2 p1 ) γ −1 2

p2 < p1 and γ − 1 is positive so T2 < T1 . ∆T is negative so ∆U is negative; the energy of the gas decreases.

19.36: Equations (19.22) and (19.24) may be re-expressed as T2  V1  =  T1  V2   
5

γ −1

V  p , 2 = 1 . p1  V2   
2

γ

a) γ = 5 , p2 = (4.00 atm)(2 3) 3 = 2.04 atm, T2 = (350 K)( 2 3) 3 = 267 K. 3 b) γ = 7 , p2 = (4.00 atm) (2 3) 5 = 2.27 atm, T2 = (350 K) ( 2 3) 5 = 298 K. 5
7 2

19.37: a)

b) From Eq. (19.25), W = nCV ∆T = (0.450 mol) (12.47 J mol ⋅ K) (40.0 C°)

= 224 J. For an adiabatic process, Q = 0 and there is no heat flow. ∆U = Q − W = −W = −224 J.

19.38: a) T =

pV nR

=

(1.00×105 Pa) (2.50×10 −3 m3 ) ( 0.1 mol) (8.3145 J mol⋅K )

= 301 K.

b) i) Isothermal: If the expansion is isothermal, the process occurs at constant temperature and the final temperature is the same as the initial temperature, namely 301 K. ii) Isobaric: pV (1.00 × 10 5 Pa) (5.00 × 10 −3 m 3 ) = nR (0.100 mol) (8.3145 J mol ⋅ K) T = 601 K. γ −1 (301 K )(V1.67 ) 1 iii) Adiabatic: Using Equation (19.22), T2 = TVVγ1−1 = = (301 K )( 1 ).67 = 189 K . 2 .67 2 (2V1 ) T= 19.39: See Exercise 19.32. a) p2 = p1 (V1 V2 ) γ = (1.10 × 10 5 Pa) ((5.00 × 10 −3 m 3 1.100 × 10 −2 m 3 ))1.29 = 4.50 × 10 4 Pa. b) Using Equation (19.26), ( p V − p2V2 ) W= 1 1 γ −1 [(1.1 × 10 5 N m 3 )(5.0 × 10 −3 m 3 ) − (4.5 × 10 4 N m 3 )(1.0 × 10 −2 m 3 )] = , (1.29 − 1)
and thus W = 345 J

c) (T2 T1 ) = (V2 V1 ) γ −1 = ((5.00 × 10 −3 m 3 ) (1.00 × 10 −2 m 3 )) 0.29 = 0.818. The final temperature is lower than the initial temperature, and the gas is cooled. 19.40: a) The product pV increases, and even for a non-ideal gas, this indicates a temperature increase. b) The work is the area in the p − V plane bounded by the blue line representing the process and the vericals at Va and Vb. The area of this trapeziod is 1 1 ( pb + pa )(Vb − Va ) = (2.40 × 10 5 Pa) (0.0400 m 3 ) = 4800 J. 2 2

19.41:

W

is the area under the path from A to B in the pV -graph. The volume

decreases, so W < 0. W = − 1 (500 × 10 3 Pa + 150 × 10 3 Pa)(0.60 m 3 ) = −1.95 × 10 5 J 2 ∆U = nCV ∆T pV PV p V − p1V1 T1 = 1 1 , T2 = 2 2 so ∆T = T2 − T1 = 2 2 nR nR nR ∆U = (CV R )( p 2V2 − p1V1 ) ∆U = (20.85 8.315)[(500 × 10 3 Pa)(0.20 m 3 ) − (150 × 10 3 Pa)(0.80 m 3 )] = −5.015 × 10 4 J Then ∆U = Q − W gives Q = ∆U + W = −5.015 × 10 4 J − 1.95 × 10 5 J = −2.45 × 105 J Q is negative, so heat flows out of the gas.

19.42: (a) Qabc = ∆U ac + Wabc = nCv ∆Tac + Wabc get ∆Tac : PV = nRT → T = PV nR PcVc PaVa PcVc − PaVa − = nR nR nR 5 3 (1.0 × 10 Pa)(0.010 m ) − (1.0 × 10 5 Pa)(0.0020 m 3 ) ∆Tac = = 289 K ( 1 mole)(8.31 J mole K) 3 1 Wabc = Area under PV graph = (0.010 − 0.002) m 3 (2.5 × 10 5 Pa) 2 3 + (0.010 − 0.002) m (1.0 × 10 5 Pa) ∆Tac = Tc − Ta = Wabc = 1.80 × 103 J 3  ∆U ac = nCv ∆Tac = n  R  ∆Tac 2  J  1  3  3 =  mole     8.31  ( 289 K ) = 1.20 × 10 J mole K  3  2  3 Qabc = 1.20 × 10 J + 1.8 × 103 J = 3000 J into the gas

(b) ∆U ac in the same = 1200 J Wac = area = ( 0.010 − 0.002) m 3 1.0 × 10 5 Pa = 800 J Qac = ∆U ac + Wac = 1200 J + 800 J = 2000 J into the gas (c) More heat is transfered in abc than in ac because more work is done in abc. 19.43: a) ∆U = Q − W = ( 90.0 J ) − ( 60.0 J ) = 30.0 J for any path between a and b. If W=15.0 J along path abd, then Q = ∆U + W = 30.0 J + 15.0 J = 45.0 J. b) Along the return path, ∆U = −30.0 J, and Q = ∆U + W = ( − 30.0 J ) + ( − 35.0 J ) = −65.0 J; the negative sign indicates that the system liberates heat. c) In the process db, dV = 0 and so the work done in the process ad is 15.0 J; Qad = (U d − U a ) + Wad = ( 8.00 J ) + (15.0 J ) = 23.0 J. In the process db, W = 0 and so Qdb = U b − U d = 30.0 J − 8.0 J = 22.0 J.

(

)

19.44: For each process, Q = ∆U + W . No work is done in the processes ab and dc, and so Wbc = Wabc and Wad = Wadc , and the heat flow for each process is: for ab, Q = 90 J : for bc, Q = 440 J + 450 J = 890 J : for ad , Q = 180 J + 120 J = 300 J : for dc, Q = 350 J. for Q = 350 each process, heat is absorbed in each process. Note that the arrows representing the processes all point the direction of increasing temperature (increasing U). 19.45: We will need to use Equations (19.3), W = p(V2 − V1 ) and (17 - 4), ∆U = Q − W . a) The work done by the system during the process: Along ab or cd, W=0. Along bc, Wbc = p c (Vc − Va ). Along ad , Wad = p a (Vc − Va ). b) The heat flow into the system during the process: Q = ∆U + W . ∆U ab = U b − U a , so Qab = U b − U a + 0. ∆U bc = U c − U b , so Qbc = (U c − U b ) + pc (Vc − Va ). ∆U dc = U c − U d , so Qdc = (U c − U d ) + 0.

∆U ad = U d − U a , so Qad = (U d − U a ) + pa (Vc − Va ). c) From state a to state c along path abc : Wabc = pc (Vc − Va ). Qabc = U b − U a + (U c − U b ) + pc (Vc − Va ) = (U c − U a ) + pc (Vc − Va ) From state a to state c along path adc : Wadc = pa (Vc − Va ). Qadc = (U c − U a ) + pa (Vc − Va ) Assuming pc > pa , Qabc > Qadc , and Wabc > Wadc . d) To understand this difference, start from the relationship Q = W + ∆U . The internal energy change ∆U is path independent and so it is the same for path abc and path adc. The work done by the system is the area under the path in the pV-plane and is not the same for the two paths. Indeed, it is larger for path abc. Since ∆U is the same and W is different, Q must be different for the two paths. The heat flow Q is path dependent.

19.46: a)

b)

W = Wab + Wbc + Wcd + Wda = 0 + area(bc) + 0 + area(da ) = (7.00 m 3 )(2000 Pa) + (−7.00 m 3 )(6000 Pa) = −28,000 J; on the gas since W < 0

c) Q = ∆U + W = 0 + (−28,000 J ) = −28,000 J Heat comes out of the gas since Q < 0. 19.47: a) We aren’t told whether the pressure increases or decreases in process bc. The cycle could be

In cycle I, the total work is negative and in cycle II the total work is positive. For a cycle, ∆U = 0, so Qtot = Wtot The net heat flow for the cycle is out of the gas, so heat Qtot < 0 and Wtot < 0. Sketch I is correct. b) Wtot = Qtot = −800 J Wtot = Wab + Wbc + Wca Wbc = 0 since ∆V = 0. Wab = p∆V since p is constant. But since it is an ideal gas, p∆V = nR∆T Wab = nR (Tb − Ta ) = 1660 J Wca = Wtot − Wab = −800 J − 1660 J = −2460 J

19.48: Path ac has constant pressure, so Wac = p∆V = nR∆T , and Wac = nR(Tc − Ta ) = (3 mol)(8.3145 J mol ⋅ K)(492 K − 300 K) = 4.789 × 103 J. Path cb is adiabatic (Q = 0), so Wcb = Q − ∆U = −∆U = −nCV ∆T , and using CV = C p − R , Wcb = −n(C p − R)(Tb − Tc ) = −(3 mol)(29.1 J mol ⋅ K − 8.3145 J mol ⋅ K)(600 K − 492 K) = −6.735 × 10 3 J. Path ba has constant volume, so Wba = 0. So the total work done is W = Wac + Wcb + Wba = 4.789 × 103 J − 6.735 × 103 J + 0 = −1.95 × 10 3 J. 19.49: a)

T a = Tc 19.50: a) n = C pQ = ∆T

( +2.5 ×10 4 J) ( 29.07 J mol⋅K)(40.0 K)

= 21.5 mol.

C 20.76 b) ∆U = nCV ∆T = Q CV = (−2.5 × 10 4 J) 29.07 = −1.79 × 10 4 J. P

c) W = Q − ∆U = −7.15 × 10 3 J. d) ∆U is the same for both processes, and if dV = 0, W = 0 and Q = ∆U = −1.79 × 10 4 J.

19.51: ∆U = 0, and so Q = W = p∆V and ∆V = W (−2.15 × 105 J) = = −0.226 m 3 , p (9.50 × 105 Pa)

with the negative sign indicating a decrease in volume. 19.52: a)

b) At constant temperature, the product pV is constant, so V2 = V1 ( p1 p 2 ) = (1.5 L)

(

1.00×10 5 Pa 2.50×10 4 Pa

) = 6.00 L. The final pressure is given as being the same

as p3 = p 2 = 2.5 × 10 4 Pa. The final volume is the same as the initial volume, so T3 = T1 ( p3 p1 ) = 75.0 K. c) Treating the gas as ideal, the work done in the first process is nRT ln(V2 V1 ) = p1V1 ln( p1 p2 )  1.00 × 105 Pa  = (1.00 × 105 Pa)(1.5 × 10 −3 m 3 ) ln   2.50 × 10 4 Pa     = 208 J, keeping an extra figure. For the second process, p 2 (V3 − V2 ) = P2 (V1 − V2 ) = p 2V1 (1 − ( p1 p 2 ))  1.00 × 10 5 Pa  = (2.50 × 10 4 Pa)(1.5 × 10 −3 m 3 )1 −  2.50 × 10 4 Pa  = −113 J.    The total work done is 208 J − 113 J = 95 J. d) Heat at constant volume.

19.53: a) The fractional change in volume is ∆V = V0 β∆T = (1.20 × 10 −2 m 3 )(1.20 × 10 −3 K −1 )(30.0 K) = 4.32 × 10 −4 m 3 . b) p∆V = ( F A)∆V = ((3.00 × 10 4 N) 0.0200 m 2 ))(4.32 × 10 −4 m 3 ) = 648 J.

c) Q = mC p ∆T = V0 ρC p ∆T = (1.20 × 10 −2 m 3 )(791 kg m 3 )(2.51 × 10 3 J kg ⋅ K)(30.0 K) = 7.15 × 105 J. d) ∆U = Q − W = 7.15 × 10 5 J to three figures. e) Under these conditions, there is no substantial difference between cV and c p .

19.54: a) β∆TV0 = (5.1 × 10 −5 (C°) −1 )(70.0 C°)(2.00 × 10 −2 ) 3 = 2.86 × 10 −8 m 3 . b) p∆V = 2.88 × 10−3 J. c) Q = mC∆T = ρV0C∆T = (8.9 × 10 3 kg m 3 )(8.00 × 10 −6 m 3 )(390 J kg ⋅ K)(70.0 C°) = 1944 J. a) To three figures, ∆U = Q = 1940 J. e) Under these conditions, the difference is not substantial. 19.55: For a mass m of ejected spray, the heat of reaction L is related to the temperature rise and the kinetic energy of the spray by mL = mC∆T − (1 2)mv 2 , or 1 1 L = C∆T − v 2 = (4190 J kg ⋅ K ) (80 C°) − (19 m s) 2 = 3.4 × 105 J kg. 2 2 19.56: Solving Equations (19.22) and (19.24) to eliminate the volumes, p  γ γ p1γ−1T1γ = p2−1T2γ , or T1 = T2  1  . p   2 Using γ = 7 for air, T1 = (273.15 K)( 1.60×10 5 ) 7 = 449 K, which is 176°C. 5 2.80×10
6 2

1− 1

19.57: a) As the air moves to lower altitude its density increases; under an adiabatic compression, the temperature rises. If the wind is fast-moving, Q is not as likely to be signigicant, and modeling the process as adiabatic (no heat loss to the surroundings) is more accurate. b) See Problems 19.59 and 19.56: The temperature at the higher pressure is T2 = (258.15 K)((8.12 ×10 4 Pa)/(5.60 × 10 4 Pa)) 2 7 = 287.1 K, which is 13.9°C and so the temperature would rise by 11.9 C°. 19.58: a)

b) The work done is W = p0 (2V0 − V0 ) + p3 = p0 (2V0 4V0 ) γ and so  C  W = p0V0 1 + V (2 − 2 2−γ ) R   Note that p0 is the absolute pressure. c) The most direct way to find the temperature is to find the ratio of the final pressure and volume to the original and treat the air as an ideal gas;  V3  1   = T0   4 = T0 ( 2 ) 2−γ V   2  1 pV pV C  d) Since n = 0 0 , Q = 0 0 ( CV + R )( 2T0 − T0 ) = p 0V0  V + 1. This amount of RT0 RT0  R  heat flows into the gas. V  pV T3 = T0 3 3 = T0  2  V  p1V1  3
γ γ

CV ( p0 (2V0 ) − p3 (4V0 )). R

19.59: a) From constant cross-section area, the volume is proportional to the length, and 1/ γ Eq. (19.24) becomes L2 = L1 ( p1 p 2 ) and the distance the piston has moved is
1 / 1.400 5   p 1/ γ    1 −  1   = ( 0.250 m ) 1 −  1.01 × 10 Pa     L1 − L2 = L1   p2     5.21 × 105 Pa           = 0.173 m. b) Raising both sides of Eq. (19.22) to the power γ and both sides of Eq. (19.24) to

the power γ − 1 , dividing to eliminate the terms V1γ ( γ −1) and V2γ ( γ −1) and solving for the ratio of the temperatures, p  T2 = T1  2  p   1
1− ( 1 / γ )

 5.21 × 10 5 Pa  = ( 300.15 K )   1.01 × 10 5 Pa    

1− ( 1 1.400 )

= 480 K = 206°C.

Using the result of part (a) to find L2 and then using Eq. (19.22) gives the same result. c) Of the many possible ways to find the work done, the most straightforward is to use the result of part (b) in Eq. (19.25), W = nCV ∆T = ( 20.0 mol )( 20.8 J mol ⋅ K )(179.0 C°) = 7.45 × 10 4 J, where an extra figure was kept for the temperature difference.

19.60: a)

b) The final temperature is the same as the initial temperature, and the density is proportional to the absolute pressure. The mass needed to fill the cylinder is then m = ρ0V p 1.45 × 105 Pa = (1.23 kg m 3 ) (575 × 10 −6 m 3 ) = 1.02 × 10 −3 kg. 5 pa 1.01 × 10 Pa

The increase in power is proportional to the increase in pressure; the percentage increases is 1..45 − 1 = 0.44 = 44%. c) The temperature of the compressed air is not the same as the original 1 01 temperature; the density is proportional to the pressure, and for the process, and modeled as abiabatic, the volumes are related to the pressure by Eq. (19.24), and the mass of air needed to fill the cylinder is  p m = ρ0V   p   a
1γ

 1.45 × 105 Pa  = (1.23 kg m ) (575 × 10 m )  1.01 × 105 Pa     −4 = 9.16 × 10 kg,
3 −6 3

1 1.40

an increase of (1.45 1.01)1 1.04 − 1 = 0.29 = 29% 19.61: a) For as isothermal process for an ideal gas, ∆T = 0 and ∆U = 0, so Q = W = 300 J. b) For an adiabatic process, Q = 0, and W ∆U = −W = −300 J. c) For isobaric, W = pdV = nRdT , or dT = nR . Then, Q = nCp dT and substituting for dT gives W Q = nC p nR = C p W , or Q = 5 R W = 5 (300 J). Thus, Q = 750 J. To find ∆U , use ∆U = nCV dT . R 2 R 2 Substituting for dT and CV , ∆U = n( 3 R) 2 W 3 = W = 450 J. nR 2

19.62: a)

b) The isobaric process doubles the temperature to 710 K, and this must be the temperature of the isothermal process. c) After the isothermal process, the oxygen is at its original volume but twice the original temperature, so the pressure is twice the original pressure, 4.80 × 10 5 Pa. d) Break the process into three steps. W1 = −nRTo = −(0.25 mol)(8.3145 J mol ⋅ K)(335 K) = −738 J; W2 = nRT ln (p1 p2 ) = nR (2To ) ln(1 2) = (0.250 mol)(8.3145 J mol ⋅ K )(710 K )(.693) = 1023 J; W3 = 0 (because dV = 0). Thus, W = 285 J. 19.63: a) During the expansion, the Kelvin temperature doubles and ∆T = 300 K. W = p∆V = nR∆T = (0.250 mol)(8.3145 J mol ⋅ K)(355 K) = 738 J, Q = nC p ∆T = (0.250 mol)(29.17 J mol ⋅ K)(355 K) = 2590 J and ∆U = nCV ∆T = Q − W = 1850 J. b) The final cooling is isochoric; dV = 0 and so W = 0. The temperature change is ∆T = −355 K, and Q = ∆U = nCV ∆T = −1850 J. c) for the isothermal compression, ∆T = 0 and so ∆U = 0.

19.64: a)

b) At constant pressure, halving the volume halves the Kelvin temperature, and the temperature at the beginning of the adiabatic expansion is 150 K. The volume doubles during the adiabatic expansion, and from Eq. (19.22), the temperature at the end of the expansion is (150 K)(1 2) 0.40 = 114 K. c) The minimum pressure occurs at the end of the adiabatic expansion. During the heating the volume is held constant, so the minimum pressure is proportional to the Kelvin temperature, p min = (1.80 × 10 5 Pa)(113.7 K 300 K) = 6.82 × 10 4 Pa. 19.65: a) W = p∆V = nR∆T = (0.150 mol)(8.3145 J mol ⋅ K)( −150 K) = −187 J, Q = nC p ∆T = (0.150 mol)(29.07 J mol ⋅ K)( −150 K) = −654 J, ∆U = Q − W = −467 J. b) From Eq. (19.24), using the expression for the temperature found in Problem 19.64, W = 1 0.40 (0.150 mol)(8.3145 J mol ⋅ K)(150 K)(1 − (1 2 ) = 113 J, 0.40

Q = 0 for an adiabatic process, and ∆U = Q − W = −W = −113 J. c) dV = 0, so W = 0. Using the temperature change as found in Problem 19.64 and part (b), Q = nCV ∆T = (0.150 mol)(20.76 J mol ⋅ K)(300 K - 113.7 K) = 580 J, and ∆U = Q − W = Q = 580 J.

19.66: a) W = nRT ln

( ) = nRT ln(3) = 3.29 × 10
V2 V1
2

3

J.

b) See Problem 19.32(b); nCV T1 (1 − (1 3) 3 ) = 2.33 × 10 3 J. c) V2 = 3V1 , so p∆V = 2 pV1 = 2nRT1 = 6.00 × 10 3 J. d)

The most work done is in the isobaric process, as the pressure is maintained at its original value. The least work is done in the abiabatic process. e) The isobaric process involves the most work and the largest temperature increase, and so requires the most heat. Adiabatic processes involve no heat transfer, and so the magnitude is zero. f) The isobaric process doubles the Kelvin temperature, and so has the largest change in internal energy. The isothermal process necessarily involves no change in internal energy.

19.67:

a)

b) No heat is supplied during the adiabatic expansion; during the isobaric expansion, the heat added is nC p ∆T . The Kelvin temperature doubles, so ∆T = 300.15 K and Q = (0.350 mol)(34.60 J mol ⋅ K)(300.15 K) = 3.63 × 10 3 J. c) For the entire process, ∆T = 0 and so ∆U = 0. d) If ∆U = 0, W = Q = 3.63 × 10 3 J. e) During the isobaric expansion, the volume doubles. During the adiabatic expansion, the temperature decreases by a factor of two, and from Eq. (19.22) the volume changes by a factor of 21 ( γ −1) = 21 0.33 , and the final volume is (14 × 10 −3 m 3 )21 0.33 = 0.114 m 3 .

19.68: a) The difference between the pressure, multiplied by the area of the piston, mg must be the weight of the piston. The pressure in the trapped gas is p 0 + mg = p 0 + πr 2 . A b) When the piston is a distance h + y above the cylinder, the pressure in the trapped gas is mg  h     p 0 + 2  πr  h + y     h y −1 y = 1 + h ~ 1 − h . The net force, taking the and for values of y small compared to h, h+ y positive direction to be upward, is then

(

)

  mg  y F =  p 0 + 2 1 −  − p 0  πr 2 − mg πr  h     y = −  p 0 πr 2 + mg . h This form shows that for positive h, the net force is down; the trapped gas is at a lower pressure than the equilibrium pressure, and so the net force tends to restore the piston to equilibrium. c) The angular frequency of small oscillations would be

( )

(

)

ω

2

( p πr =
0

2

+ mg h g  p πr 2  1 + 0 . =  m h mg  

)

If the displacements are not small, the motion is not simple harmonic. This can be seen be considering what happens if y ~ −h; the gas is compressed to a very small volume, and the force due to the pressure of the gas would become unboundedly large for a finite displacement, which is not characteristic of simple harmonic motion. If y >> h (but not so large that the piston leaves the cylinder), the force due to the pressure of the gas becomes small, and the restoring force due to the atmosphere and the weight would tend toward a constant, and this is not characteristic of simple harmonic motion.

19.69: a) Solving for p as a function of V and T and integrating with respect to V, nRT an 2 p= − V − nb V 2 V2 V − nb  1 2 1 W = ∫ pdV = nRT ln  2  + an  − . V1  V1 − nb  V2 V1  When a = b = 0, W = nRT ln (V2 V1 ) , as expected. b) Using the expression found in part (a), i) W = (1.80 mol )( 8.3145 J mol ⋅ K )( 300 K )

 4.00 × 10 −3 m 3 − (1.80 mol ) 6.38 × 10 −5 m 2 / mol  × ln   −3 3 −5 2  2.00 × 10 m − (1.80 mol ) 6.38 × 10 m / mol  1 1  2 + 0.554 J ⋅ m 3 mol 2 (1.80 mol )  − −3 3 −3 3 2.00 × 10 m   4.00 × 10 m = 2.80 × 10 3 J.

( (

(

) ) )

( (

) )

ii) nRT ln(2) = 3.11 × 10 3 J. c) 300 J to two figures, larger for the ideal gas. For this case, the difference due to nonzero a is more than that due to nonzero b. The presence of a nonzero a indicates that the molecules are attracted to each other and so do not do as much work in the expansion.

Chapter 20

20.1: a) 2200 J + 4300 J = 6500 J. 20.2: a) 9000 J − 6400 J = 2600 J. 20.3: a) c)
3700 16 ,100

b)

2200 6500

= 0.338 = 33.8%.

2600 b) 9000 J = 0.289 = 28.9%. J

= 0.230 = 23.0%. = 0.350 g.

b) 16,100 J − 3700 J = 12,400 J.
16 ,100 J 4.60 ×10 4 J kg

d) (3700 J)(60.0 s) = 222 kW = 298 hp.

20.4: a) Q = 1 Pt = e

(180×10 3 W)(1.00 s) ( 0.280 )

= 6.43 × 10 5 J.

b) Q − Pt = 6.43 × 10 5 J − (180 × 10 3 W)(1.00 s) = 4.63 × 10 5 J.

330 20.5: a) e = 1300 MW = 0.25 = 25%. b) 1300 MW − 330 MW = 970 MW. MW

20.6: Solving Eq. (20.6) for r , (1 − γ) ln r = ln(1 − e) or

r = (1 − e) 1−γ = (0.350) − 2.5 = 13.8. If the first equation is used (for instance, using a calculator without the x y function), note that the symbol “e” is the ideal efficiency , not the base of natural logarithms.
20.7: a) Tb = Ta r γ −1 = (295.15 K)(9.5)0.40 = 726 K = 453°C. b) pb = pa r γ = (8.50 × 104 Pa)(9.50)γ = 1.99 × 106 Pa. 20.8: a) From Eq. (20.6), e = 1 − r1−γ = 1 − (8.8) −0.40 = 0.58 = 58%. b) 1 − (9.6) −0.40 = 60%, an increase of 2%. If more figures are kept for the efficiencies, the difference is 1.4%.

1

20.9: a) W =

QC K

=

3.40×10 4 J 2.10

= 1.62 × 104 J.

1 b) QH = QC + W = QC (1 + K ) = 5.02 × 10 4 J.

20.10: P =

Q W 1  ∆m  = C =  ( Lf + c p ∆T ) ∆t K∆t K  ∆t  1  8.0 kg  5   3600 s  (1.60 × 10 J kg) + (485 J kg ⋅ K)(2.5 K) = 128 W.  2.8  

=

(

)

20.11: a) EER =

1.44 × 10 5 J − 9.80 × 10 4 J = 767 W. b) EER = H P, or 60.0 s

(9.8 × 104 J) (60 s) 1633 W (3.413) = (3.413) = 7.27. 5 4 [(1.44 × 10 J) (60 s) − (9.8 × 10 J) (60 s)] 767 W

20.12: a) QC = m( L f + cice ∆Tice + cwater ∆Twater ) = (1.80 kg) 334 × 103 J kg + (2100 J kg ⋅ K)(5.0 K) + (4190 J kg ⋅ K)(25.0 K) = 8.90 × 10 J. |Q | 10 5 b) W = KC = 8.08×40 J = 3.37 × 105 J. 2.
5

(

)

c) | QH |= W + | QC |= 3.37 × 105 J + 8.08 × 105 J = 1.14 × 106 J (note that | QH |= 1 | QC | (1 + K ).) 20.13: a) | QH | − | QC |= 550 J − 335 J = 215 J. b) TC = TH (| QC | | QH |) = (620 K)(335 J 550 J) = 378 K. c) 1 − (| QC | | QH |) = 1 − (335 J 550 J) = 39 %. 20.14: a) From Eq. (20.13), the rejected heat is ( 300 K )(6450 J) = 3.72 × 10 3 J. 520 K b) 6450 J − 3.72 × 103 J = 2.73 × 103 J. c) From either Eq. (20.4) or Eq. (20.14), e=0.423=42.3%. TH T = mLf H TC TC

20.15: a) | QH |=| QC |

(287.15 K) = 3.088 × 107 J, (273.15 K) 7 or 3.09 × 10 J to two figures. b) | W |=| QH | − | QC |=| QH | (1 − (TC TH )) = = (85.0 kg)(334 × 103 J kg) (3.09 × 107 J) × (1 − (273.15 297.15)) = 2.49 × 106 J.

20.16: a) From Eq. (20.13), ( 320 K )(415 J) = 492 J. b) The work per cycle is 270 K 77 492 J − 415 J = 77 J, and P = (2.75) × 1.00Js = 212 W, keeping an extra figure. c) TC (TH − TC ) = (270 K) (50 K) = 5.4. 20.17: For all cases, | W |=| QH | − | QC | . a) The heat is discarded at a higher temperature, and a refrigerator is required; | W |=| QC | ((TH TC ) − 1) = (5.00 × 10 3 J) × ((298.15 263.15) − 1) = 665 J. b) Again, the device is a refrigerator, and | W |=| QC | ((273.15 / 263.15) − 1) = 190 J. c) The device is an engine; the heat is taken form the hot reservoir, and the work done by the engine is | W |= (5.00 × 10 3 J) × ((248.15 263.15) − 1) = 285 J. 20.18: For the smallest amount of electrical energy, use a Carnot cycle. Qin = QCool water to 0° C + Qfreeze water = mc∆T + mLF = (5.00 kg) 4190 kgJ⋅K (20 K) + (5.00 kg)(334 × 103 J K) Carnot cycle: Qin Tcold = 2.09 × 106 J Q 2.09 × 106 J Q = out → = out Thot 268 K 293 K

(

)

Qout = 2.28 × 10 6 J(into the room) W = Qout − Qin = 2.28 × 10 6 J − 2.09 × 10 6 J W = 1.95 × 10 5 J(electrical energy) 20.19: The total work that must be done is Wtot = mgy = (500 kg)(9.80 m s 2 )(100 m) = 4.90 × 105 J QH = 250 J Find QC so can calculate work W done each cycle: QC T =− C QH TH QC = −(TC TH )QH = −(250 J)[ (373.15 K) (773.15 K) ] = −120.7 J W = QC + QH = 129.3 J The number of cycles required is Wtot 4.09 × 10 5 J = = 3790 cycles. W 129.3 J

20.20: For a heat engine, QH = −QC / (1 − e ) = − (−3000 J) (1 − 0.600) = 7500 J, and then W = eQH = (0.600)(7500 J ) = 4500 J. This does not make use of the given value of TH . If TH is used, then for a Carnot engine, TC = TH (1 − e ) = ( 800 K )(1 − 0.600) = 320 K and QH = −QCTH / TC , which gives the same result. 20.21: QC = −mLf = −( 0.0400 kg ) 334 × 103 J/kg = −1.336 × 10 4 J QC T =− C QH TH

(

)

QH = −( TH TC ) QC = − − 1.336 × 104 J [ ( 373.15 K ) ( 273.15 K ) ] = +1.825 × 10 4 J

(

)

W = QC + QH = 4.89 × 103 J
1.51×10 8 J 2.60×10 8 J

20.22: The claimed efficiency of the engine is

= 58%. While the most efficient

engine that can operate between those temperatures has efficiency eCarnot = 1 − 250 K = 38%. 400 K The proposed engine would violate the second law of thermodynamics, and is not likely to find a market among the prudent. 20.23: a) Combining Eq. (20.14) and Eq. (20.15), K= TC / TH 1− e 1− e = = . 1 − ( TC / TH ) (1 − (1 − e ) ) e

b) As e → 1, K → 0; a perfect (e = 1) engine exhausts no heat (QC = 0), and this is useless as a refrigerator. As e → 0, K → ∞; a useless (e = 0) engine does no work (W = 0), and a refrigerator that requires no energy input is very good indeed. Q mLf ( 0.350 kg ) 334 × 103 J kg = = = 428 J K . TC TC ( 273.15 k )

20.24: a) b)

(

)

− 1.17 × 105 J = −392 J K . 298.15 K c) ∆S = 428 J K + (−392 J K ) = 36 J K. (If more figures are kept in the intermediate calculations, or if ∆S = Q((1 273.15 K) − (1 298.15 K)) is used, ∆S = 35.6 J K.

20.25: a) Heat flows out of the 80.0° C water into the ocean water and the 80.0° C water cools to 20.0° C (the ocean warms, very, very slightly). Heat flow for an isolated system is always in this direction, from warmer objects into cooler objects, so this process is irreversible. b) 0.100 kg of water goes form 80.0°C to 20.0° C and the heat flow is Q = mc∆T = (0.100 kg)(4190 J kg ⋅ K)( −60.0C°) = −2.154 × 10 4 J This Q comes out of the 0.100 kg of water and goes into the ocean. For the 0.100 kg of water, ∆S = mc ln(T2 T1 ) = (0.100 kg)(4190 J kg ⋅ K) ln( 293.15 353.15) = −78.02 J K For the ocean the heat flow is Q = +2.154 × 104 J and occurs at constant T: ∆S = ∆S net Q 2.154 × 104 J = = +85.76 J K T 293.15 K = ∆S water + ∆S ocean = −78.02 J K + 85.76 J K = +7.7 J K

20.26: (a) Irreversible because heat will not spontaneously flow out of 15 kg of water into a warm room to freeze the water. (b) ∆S = ∆S ice + ∆S room = mLF mLF + Tice Troom (15.0 kg)(334 × 103 J kg ) − (15.0 kg)(334 × 103 J kg) + 273 K 293 K

=

= + 1,250 J K This result is consistent with the answer in (a) because ∆S > 0 for irreversible processes. 20.27: The final temperature will be
(1.00 kg)(20.0°C) + (2.00 kg)(80.0°C) = 60°C, (3.00 kg)

and so the entropy change is   333.15 K   333.15 K    + (2.00 kg) ln  (4190 J kg ⋅ K) (1.00 kg) ln   293.15 K   353.15 K   = 47.4 J K.      

20.28: For an isothermal expansion, ∆T = 0, ∆U = 0 and Q = W . The change of entropy is Q = T
1850 J 293.15 K

= 6.31 J K.

20.29: The entropy change is ∆S =

∆Q , and ∆Q = mLv . Thus, T

∆S =

− mLv − (0.13 kg)(2.09 × 104 J kg ) = = −644 J K. T (4.216 K)

20.30: a) ∆S =

Q T

=

mL v T

=

(1.00 kg)(2256 × 10 3 J kg ) ( 373 .15 K)

= 6.05 × 103 J K. Note that this is the change

of entropy of the water as it changes to steam. b) The magnitude of the entropy change is roughly five times the value found in Example 20.5. Water is less ordered (more random) than ice, but water is far less random than steam; a consideration of the density changes indicates why this should be so.

20.31: b)

a)

∆S =

Q mLv (18.0 × 10−3 kg)(2256 × 103 J kg ) = = = 109 J K. T T (373.15 K) (28.0 × 10 −3 kg)(201× 103 J kg) = 72.8 J K (77.34 K) (107.9 × 10−3 kg)(2336 × 103 J kg) = 102.2 J K (2466 K) (200.6 × 10−3 kg)(272 × 103 J kg) = 86.6 J K (630 K)

N2 :

Ag :

Hg :

c) The results are the same order or magnitude, all around 100 J K .The entropy change is a measure of the increase in randomness when a certain number (one mole) goes from the liquid to the vapor state. The entropy per particle for any substance in a vapor state is expected to be roughly the same, and since the randomness is much higher in the vapor state (see Exercise 20.30), the entropy change per molecule is roughly the same for these substances.

20.32: a) The final temperature, found using the methods of Chapter 17, is T= (3.50 kg)(390 J kg ⋅ K)(100 C°) = 28.94°C, (3.50 kg)(390 J kg ⋅ K) + (0.800 kg)(4190 J kg ⋅ K)

or 28.9°C to three figures. b) Using the result of Example 20.10, the total change in entropy is (making the conversion to Kelvin temperature)  302.09 K  ∆S = (3.50 kg)(390 J kg ⋅ K) ln    373.15 K   302.09 K  + (0.800 kg)(4190 J kg ⋅ K) ln   273.15 K     = 49.2 J K. (This result was obtained by keeping even more figures in the intermediate calculation. Rounding the Kelvin temperature to the nearest 0.01 K gives the same result.

20.33: As in Example 20.8, V   0.0420 m3  ∆S = nR ln  2  = (2.00 mol)(8.3145 J mol ⋅ K) ln  V   0.0280 m3  = 6.74 J K.     1 20.34: a) On the average, each half of the box will contain half of each type of molecule, 250 of nitrogen and 50 of oxygen. b ) See Example 20.11. The total change in entropy is ∆S = kN1 ln(2) + kN 2 ln(2) = ( N1 + N 2 )k ln(2) = (600)(1.381 × 10−23 J K) ln(2) = 5.74 × 10−21 J K. c) See also Exercise 20.36. The probability is (1 2) × (1 2 ) = (1 2 ) = 2.4 × 10 −181 , and is not likely to happen. The numerical result for part (c) above may not be obtained directly on some standard calculators. For such calculators, the result may be found by taking the log base ten of 0.5 and multiplying by 600, then adding 181 and then finding 10 to the power of the sum. The result is then 10 −181 × 10 0.87 = 2.4 × 10 −181.
500 100 600

20.35: a) No; the velocity distribution is a function of the mass of the particles, the number of particles and the temperature, none of which change during the isothermal N expansion. b) As in Example 20.11, w1 = 1 w2 (the volume has increased, and w2 < w1 ); 3 ln( w2 w1 ) = ln (3N ) = N ln(3), and ∆S = kN ln(3) = knN A ln( 3) = nR ln(3) = 18.3 J K. c) As in Example 20.8, ∆S = nR ln (V2 V1 ) = nR ln(3), the same as the expression used in part (b), and ∆S = 18.3 J K. 20.36: For those with a knowledge of elementary probability, all of the results for this exercise are obtained from P(k ) =

( ) p (1 − p)
n k k

n−k

=

4! 1   , k!(4 − k )!  2 
1 2

4

where P(k) is the probability of obtaining k heads, n = 4 and p = 1 − p = This is of course consistent with Fig. (20.18). a)
4! 4!0! 4 1 (1 2) 4 = 04!4!! (1 2) 4 = 16

for a fair coin.

for all heads or all tails. b)

4! 3 3 1 c) 2!2! (1 2) = 8 . d) 2 × 16 + 2 × 1 + 8 = 1. The number of heads must be one of 0, 1, 2, 3 or 4 4, and there must be unit probability of one and only one of these possibilities.

4! 3!1!

(1 2) 4 = 143!! = 1 . ! 4

20.37: a) QH = +400 J, W = +300 J W = QC + QH , so QC = W − QH = −100 J Q T Since it is a Carnot cycle, C = − C QH TH TC = −TH (QC QH ) = −(800.15 K)[ (−100 J) (400 J) ] = +200 K = −73°C b) Total QC required is − mLf = −(10.0 kg ) 334 × 103 J kg = −3.34 × 106 J QC for one cycle is − 100 J, so the number of cycles required is − 3.34 × 106 J = 3.34 × 104 cycles − 100 J cycle

(

)

20.38: a) Solving Eq. (20.14) for TH , TH = TC

1 so the temperature change 1 − e, 1  1   1  1 ′ TH − TH = TC  − −  = (183.15 K )   = 27.8 K.  1 − e′ 1 − e   0.55 0.600 

′ b) Similarly, TC = TH (1 − e ) , and if TH = TH, ′ TC − TC = TC e′ − e  0.050  = (183.15 K )   = 15.3 K. 1− e  0.600 
nRT1 p1

20.39: The initial volume is V1 =

= 8.62 × 10−3 m 3 . a) At point 1, the pressure
T2 T1

is given as atmospheric, and p1 = 1.01 × 10 5 Pa, with the volume found above, V1 = 8.62 × 10−3 m 3 . V2 = V1 = 8.62 × 10−3 m3 , and p2 = p1 = 2 p1 = 2.03 × 105 Pa (using pa = 1.013 × 10 5 Pa). p3 = p1 = 1.01 × 10 5 Pa and V3 = V1 T13 = 1.41 × 10 −2 m 3 . T

( 8.3145 J/mol ⋅ K )( 300K ) = 2.18 × 103 J. The process 2 - 3 is adiabatic, Q = 0,

b) Process 1 - 2 is isochoric, ∆V = 0 so W = 0. ∆U = Q = nCV ∆T = ( 0.350 mol )( 5 2 ) × and ∆U = − W = nCV ∆T = (0.350 mol)(5 2)(8.3145 J mol ⋅ K )(−108 K ) = −786 J (W > 0). The process 3 - 1 is isobaric; W = p∆V = nR∆T = (0.350 mol)(8.3145 J mol ⋅ K )(−192 K ) = − 559 J, ∆U = nCV ∆T = n(5 2)(8.3145 J mol ⋅ K )(−192 K ) = −1397 J and Q = nC p ∆T = (0.350 mol)(7 2)(8.3145 J mol ⋅ K )(−192 K ) = −1956 J = ∆U + W . c) The net work done is 786 J − 559 J = 227 J. d) Keeping extra figures in the calculatio ns for the process 1 - 2, the heat flow into the engine for one cycle is 2183 J − 1956 J = 227 J.

e) e =

227 J 2183 J

= 0.104 = 10.4%. For a Carnot - cycle engine operating between 300 K

and 600 K, the thermal efficiency is 1 − 300 = 0.500 = 50%. 600

20.40: (a) The temperature at point c is Tc = 1000 K since from pV = nRT , the maximum temperature occurs when the pressure and volume are both maximum. So pcVc 6.00 × 105 Pa 0.0300m 3 = = 2.16 mol. RTc ( 8.3145 J mol ⋅ K ) (1000 K ) (b) Heat enters the gas along paths ab and bc, so the heat input per cycle is QH = Qac = Wac + ∆U ac . Path ab has constant volume and path bc has constant pressure, so n= Wac = Wab + Wbc = 0 + p c (Vc − Vb ) = (6.00 × 10 5 Pa )(0.0300 m 3 − 0.0100 m 3 ) = 1.20 × 10 4 J. For an ideal gas, ∆U ac = nCV (Tc − Ta ) = CV ( pcVc − paVa ) R, using nT = pV R. For CO 2 , CV = 28.46 J mol.K, so 28.46 J mol ⋅ K ∆U ac = ((6.00 × 105 Pa)(0.0300 m3 ) − (2.00 × 105 Pa)(0.0100 m 3 )) = 5.48 × 104 J. 8.3145 J mol ⋅ K Then QH = 1.20 × 104 J + 5.48 × 104 J = 6.68 × 10 4 J. (c) Heat is removed from the gas along paths cd and da, so the waste heat per cycle is QC = Qca = Wca + ∆U ca . Path cd has constant volume and path da has constant pressure, so Wca = Wcd + Wda = 0 + pd (Va − Vd ) = (2.00 × 105 Pa)(0.0100 m3 − 0.0300 m 3 ) = −0.400 × 104 J. From (b), ∆U ca = −∆U ac = −5.48 × 104 J, so QC = −0.400 × 104 J − 5.48 × 10 4 J = −5.88 × 104 J. (d) The work is the area enclosed by the rectangular path abcd, W = ( pc − pa )(Vc − Va ), or W = QH + QC = 6.68 × 104 J − 5.86 × 104 J = 8000 J. (e) e = W QH = (8000 J) (6.68 × 104 J) = 0.120.

(

)(

)

20.41: a) W = 1.00 J, TC = 268.15 K, TH = 290.15 K For the heat pump QC > 0 and Q H < 0 Q T W = QC + QH ; combining this with C = − C gives QH TH W 1.00 J QH = = = 13.2 J 1 − TC TH 1 − (268.15 290.15) b) Electrical energy is converted directly into heat, so an electrical energy input of 13.2 J would be required. c) From part (a), QH = W ⋅ QH decrease as TC decreases. 1 − TC TH

The heat pump is less efficient as the temperature difference through which the heat has to be “pumped” increases. In an engine, heat flows from TH to TC and work is extracted. The engine is more efficient the larger the temperature difference through which the heat flows.

20.42: (a) Qin = Qab + Qbc Qout =Qca Tmax = Tb = Tc = 327°C = 600 K PaVa PbVb P 1 = → Ta = a Tb = (600 K) = 200 K Ta Tb Pb 3 PbVb = nRTb → Vb =
J nRTb (2 moles)(8.31 mole K )(600 K) = = 0.0332 m 3 5 Pb 3.0 × 10 Pa

PbVb PcVc P  3 = → Vc = Vb b = (0.0332 m3 )  = 0.0997 m3 = Va Tb Tc Pc 1 Monatomic gas : CV = 3 R and CP = 5 R 2 2 J   3  3 Qab = nCV ∆Tab = (2 moles)   8.31 (400 K) = 9.97 × 10 J mole K   2  c c nRT V b Qbc = Wbc = ∫ PdV = ∫ dV = nRTb ln c = nRTb ln 3 b b V Vb J   4 = (2.00 moles)  8.31 (600 K) ln 3 = 1.10 × 10 J mole K   Qin = Qab + Qbc = 2.10 × 104 J J  5 4 Qout = Qca = nC p ∆Tca = (2.00 moles)   8.31 (400 K) = 1.66 × 10 J mole K   2 (b) Q = ∆U + w = 0 + W → W = Qin − Qout = 2.10 × 104 J − 1.66 × 104 J = 4.4 × 103 J 4.4 × 103 J e = W Qin = = 0.21 = 21% 2.10 × 104 J 200 K (c) emax = ecannot = 1 − TC = 1 − 600 K = 0.67 = 67% Th

20.43: a)

b) QH = +500 J W = mgy = (15.0 kg)(9.80 m s 2 )(2.00 m) = 294 J W = QC + QH , QC = W − QH = 294 J − 500 J = −206 J QC T =− C QH TH TC = −TH (QC QH ) = −(773 K)[(−206 J) (500 J)] = +318 K = 45°C c) e = W QH = (294 J) (500 J) = 58.8% d) QC = −206 J; wastes 206 J of heat each cycle e) From part (a), state a has the maximum pressure and minimum volume. nRT (2.00 mol)(8.3145 J mol ⋅ K)(773 K) pV = nRT , p = = = 2.57 × 106 Pa −3 3 V 5.00 × 10 m
279.15 K 20.44: a) e = 1 − 300.15 K = 7.0%. b) ( 1 − 1)(210 kW) = 2.8 MW. e pout e

=

210 kW 0.070

= 3.0 MW, 3.0 MW − 210 kW =

c)

dm d QC dt (2.8 × 106 W) (3600 s hr) = = = 6 × 105 kg hr = 6 × 105 L hr. dt c∆T (4190 J kg ⋅ K) (4 K)

20.45: There are many equivalent ways of finding the efficiency; the method presented here saves some steps. The temperature at point 3 is T3 = 4T0 , and so 5 19 QH = ∆U13 + W13 = nCV (T3 − T0 ) + (2 p0 )(2V0 − V0 ) = nRT0 (3) + 2 p0V0 = p0V0 , 2 2 where nRT0 = p 0V0 has been used for an ideal gas. The work done by the gas during one cycle is the area enclosed by the blue square in Fig. (20.22), W = p 0V0 , and so the W 2 efficiency is e = QH = 19 = 10.5%.

20.46: a) p 2 = p1 = 2.00 atm, V2 = V1 T12 = (4.00 L)(3/2) = 6.00 L. V3 = V2 = 6.00 L, p3 = p 2 T
V p 4 = p3 V3 = p3 (3 / 2) = 1.67 atm. As a check, p1 = p 4 4 T1 T4 T3 T2

= p 2 (5 / 9) = 1.111 atm,

= p 4 (6 / 5) = 2.00 atm. To

summarize, ( p1 , V1 ) = (2.00 atm, 4.00 L) ( p3 , V3 ) = (1.111 atm, 6.00 L) ( p 2 , V2 ) = (2.00 atm, 6.00 L) ( p 4 , V4 ) = (1.67 atm, 4.00 L).

b) The number of moles of oxygen is n = CP R

p1V1 RT1

, and the heat capacities are those in

Table (19.1). The product p1V1 has the value x = 810.4 J; using this and the ideal gas law, i: T  x 2 − 1 = (3.508)(810.4 J)(1 2) = 1422 J, T   1  T  W = p1∆V = x 2 − 1 = (810.4 J)(1 2) = 405 J. T   1  Q = nCP ∆T = T −T  x 3 2  = (2.508)(810.4 J)(− 2 3) = −1355 J, W = 0.  T   1  V  T V  W = nRT3 ln  4  = x 3 ln  4  = (810.4 J)(5 6) ln ( 2 3) = −274 J, Q = W V  T1  V3   3   Q = nCV ∆T = CV R Q = nCV ∆T = CV R  T  x1 − 4  = (2.508)(810.4 J)(1 6) = 339 J, W = 0.  T  1  

ii : iii : iv :

In the above, the terms are given to nearest integer number of joules to reduce roundoff error. c) The net work done in the cycle is 405 J - 274 J = 131 J. d) Heat is added in steps i and iv, and the added heat is 1422 J + 339 J = 1761 J and the efficiency is
131 J 1761 J

= 0.075, or 7.5%. The efficiency of a Carnot-cycle engine operating

between 250 K and 450 K is 1 − 250 = 0.44 = 44%. 450

20.47: a) ∆U = 1657 kJ − 1005 kJ = 6.52 × 105 J, W = p∆V = (363 × 103 Pa) × (0.4513 m 3 − 0.2202 m 3 ) = 8.39 × 10 4 J, and so Q = ∆U + W = 7.36 × 10 5 J. b) Similarly, QH = ∆U − p∆V = (1171 kJ − 1969 kJ) + (2305 × 103 Pa)(0.00946 m 3 − 0.0682 m 3 ) = −9.33 × 105 J. c) The work done during the adiabatic processes must be found indirectly (the coolant is not ideal, and is not always a gas). For the entire cycle, ∆U = 0, and so the net work done by the coolant is the sum of the results of parts (a) and (b), − 1.97 × 10 5 J. The work done by the motor is the negative of this, 1.97 × 105 J. d) K =
Qc W

=

7.36 ×10 5 J 1.97 ×10 5 J

= 3.74.

20.48: For a monatomic ideal gas, CP = 5 R and CV = 3 R. 2 2 a) ab: The temperature changes by the same factor as the volume, and so C Q = nCP ∆T = P pa (Va − Vb ) = (2.5)(3.00 × 105 Pa)(0.300 m 3 ) = 2.25 × 105 J. R The work p∆V is the same except for the factor of 5 , so W = 0.90 × 105 J. 2 ∆U = Q − W = 1.35 × 105 J. bc: The temperature now changes in proportion to the pressure change, and Q = 32 ( pc − pb )Vb = (1.5)(−2.00 × 105 Pa)(0.800 m 3 ) = −2.40 × 105 J, and the work is zero (∆V = 0). ∆U = Q − W = −2.40 × 105 J. ca: The easiest way to do this is to find the work done first; W will be the negative of area in the p-V plane bounded by the line representing the process ca and the verticals from points a and c. The area of this trapezoid is 1 (3.00 × 10 5 Pa + 1.00 × 105 Pa) × 2 (0.800 m 3 − 0.500 m 3 ) = 6.00 × 104 J, and so the work is − 0.60 × 105 J. ∆U must be 1.05 × 105 J (since ∆U = 0 for the cycle, anticipating part (b)), and so Q must be ∆U + W = 0.45 × 10 5 J. b) See above; Q = W = 0.30 × 105 J, ∆U = 0. c) The heat added, during process ab and ca, is 2.25 × 105 J + 0.45 × 105 J 5 W = 2.70 × 105 J and the efficiency is QH = 0..30×105 = 0.111 = 11.1%. 2 70×10

20.49: a) ab: For the isothermal process, ∆T = 0 and ∆U = 0. W = nRT1 ln(Vb Va ) = nRT1ln(1/ r ) = − nRT1 ln( r ), and Q = W = −nRT1 ln( r ). bc: For the isochoric process, ∆V = 0 and W = 0; Q = ∆U = nCV ∆T = nCV (T2 − T1 ). cd: As in the process ab, ∆U = 0 and W = Q = nRT2ln( r ). da: As in process bc, ∆V = 0 and W = 0; ∆U = Q = nCV (T1 − T2 ). b) The values of Q for the processes are the negatives of each other. c) The net work for one cycle is Wnet = nR (T2 − T1 )ln( r ), and the heat added (neglecting the heat exchanged during the isochoric expansion and compression, as W mentioned in part (b)) is Qcd = nRT2 ln( r ), and the efficiency is Qnet = 1 − (T1 T2 ). This is cd the same as the efficiency of a Carnot-cycle engine operating between the two temperatures.
T −T − 20.50: The efficiency of the first engine is e1 = HTH and that of the second is e2 = T T TC , ′ ′ ′

and the overall efficiency is  T − T ′   T ′ − TC  e = e1e2 =  H  .  TH   T ′ 

The first term in the product is necessarily less than the original efficiency since T ′ > TC , and the second term is less than 1, and so the overall efficiency has been reduced. 20.51: a) The cylinder described contains a mass of air m = ρ(πd 2 4 )L, and so the total kinetic energy is K = ρ(π 8 )d 2 Lv 2. This mass of air will pass by the turbine in a time t = L v, and so the maximum power is K P = = ρ(π 8)d 2v 3 . t Numerically, the product ρair (π 8 ) ≈ 0.5 kg m 3 = 0.5 W ⋅ s 4 m 5 .  (3.2 × 106 W) (0.25)   P e b) v =  2  =   (0.5 W ⋅ s 4 m 5 )(97 m) 2  = 14 m s = 50 km h.   kd    c) Wind speeds tend to be higher in mountain passes.
1/ 3 1/ 3

 1 gal  1 mi  3.788 L  20.52: a) (105 km h )   = 9.89 L h.      25 mi  1.609 km  1 gal  b) From Eq. (20.6), e = 1 − r1−γ = 1 − (8.5) −0.40 = 0.575 = 57.5%.  9.89 L h  7 4 c)   3600 s hr ( 0.740 kg L ) 4.60 × 10 J kg ( 0.575) = 5.38 × 10 W = 72.1 hp.   

(

)

d) Repeating the calculation gives 1.4 × 10 4 W = 19 hp, about 8% of the maximum power.

20.53: (Extra figures are given in the numerical answers for clarity.) a) The efficiency is e = 1 − r −0.40 = 0.611 , so the work done is QH e = 122 J and | QC |= 78 J. b) Denote the length of the cylinder when the piston is at point a by L0 and the stroke as s. Then,
L0 L0 − s

= r , L0 = L0 A =

r r −1

s and volume is

r 10.6 sA = (86.4 × 10 −3 m)π (41.25 × 10 −3 m) 2 = 51.0 × 10 −4 m 3 . r −1 9.6

c) The calculations are presented symbolically, with numerical values substituted at the end. At point a, the pressure is p a = 8.50 × 10 4 Pa, the volume is Va = 5.10 × 10 −4 m 3 as found in part (b) and the temperature is Ta = 300 K. At point b, the volume is Vb = Va r , the pressure after the adiabatic compression is pb = pa r γ and the temperature is Tb = Ta r γ −1 . During the burning of the fuel, from b to c, the volume remains constant and so Vc = Vb = Va r . The temperature has changed by an amount Q QH RQH ∆T = H = = Ta nCV ( paVa RTa ) CV paVaCV Ta = f Ta , mol ⋅ K ) where f is a dimensionless constant equal to 1.871 to four figures. The temperature at c is then Tc = Tb + f Ta = Ta r γ −1 + f . The pressure is found from the volume and =

(8.50 × 10 ( (

4

( 8.3145 J mol ⋅ K ) ( 200 J ) Pa ) ( 5.10 × 10− 4 m 3 ) ( 20.5 J

temperature, pc = pa r r γ −1 + f . Similarly, the temperature at point d is found by considering the temperature change in going from d to a, QC Q = (1 − e) H = (1 − e) f Ta , so Td = Ta (1 + (1 − e) f ). The process from d to a is nCV nCV isochoric, so Vd = Va , and pd = pa (1 + (1 − e) f ). As a check, note that pd = pc r − γ . To summarize, p pa pa r γ p a (1 + ( 1 − e ) f ) pa r (r γ−1 + f ) V Va Va r Va r Va T Ta Ta r γ −1 Ta (r γ−1 + f ) Ta (1 + (1 − e) f )

)

)

a b c d

Using numerical values (and keeping all figures in the intermediate calculations),

20.54: (a)

∆Q ∆T =k A for furnace and water ∆t L

∆S ∆Sfurnace ∆S water = + ∆t ∆t ∆t kA∆T L kA∆T L =− + Tf Tw kA∆T  1 1  − +   T T  L  f w  2 (79.5 W m ⋅ K )  1 1     2 1m = +   ( 210 K )  −  15 cm  0.65m  100 cm    523 K 313 K     = +0.0494 J K ⋅ s = (b) ∆S > 0 means that this process is irreversible. Heat will not flow spontaneously from the cool water into the hot furnace.

20.55: a) Consider an infinitesimal heat flow dQH that occurs when the temperature of the hot reservoir is T ′ : dQC = −(TC / T ′)dQH dQH T′ dQ | QC |= TC ∫ H = TC | ∆S H | T′

∫ dQ

C

= −TC ∫

b) The 1.00 kg of water (the high-temperature reservoir) goes from 373 K to 273 K. QH = mc∆T = (1.00 kg)(4190 J kg ⋅ K )(100 K ) = 4.19 × 105 J ∆Sh = mcln (T2 T1 ) = (1.00 kg )(4190 J kg ⋅ K )ln( 273 373) = −1308 J/K The result of part (a) gives | QC |= (273 K )(1308 J K ) = 3.57 × 105 J QC comes out of the engine, so QC = −3.57 × 105 J Then W = QC + QH = −3.57 × 105 J + 4.19 × 105 J = 6.2 × 10 4 J. c) 2.00 kg of water goes from 323 K to 273 K QH = mc∆T = (2.00 kg )(4190 J kg ⋅ K )(50 K ) = 4.19 × 10 5 J ∆S h = mc ln (T2 T1 ) = (2.00 kg )(4190 J kg ⋅ K )ln (273 / 323) = −1.41 × 10 3 J K QC = −TC | ∆S h |= −3.85 × 10 5 J W = QC + QH = 3.4 × 10 4 J d) More work can be extracted from 1.00 kg of water at 373 K than from 2.00 kg of water at 323 K even though the energy that comes out of the water as it cools to 273 K is the same in both cases. The energy in the 323 K water is less available for conversion into mechanical work. 20.56: See Figure (20.15(c)), and Example 20.8. a) For the isobaric expansion followed by the isochoric process, follow a path from T to 2T to T . Use dQ = nCV dT or dQ = nC p dT to get ∆S = nC p ln 2 + nCV ln 1 = 2 n(C p − CV ) ln2 = nR ln2. b) For the isochoric cooling followed by the isobaric expansion, follow a path from T to T / 2 to T . Then ∆S = nCV ln 1 + nC p ln 2 = n(C p − CV ) ln = nR ln 2. 2

20.57: The much larger mass of water suggests that the final state of the system will be water at a temperature between 0°C and 60.0°C. This temperature would be  ( 0.600 kg )( 4190 J kg ⋅ K )( 45.0C°)     − ( 0.0500 kg ) ( ( 2100 J kg ⋅ K )(15.0C°)    3  + 334 × 10 J kg )  = 34.83°C, T= (0.650 kg)(4190 J kg ⋅ K ) keeping an extra figure. The entropy change of the system is then  307.98  ∆S = (0.600 kg)(4190 J kg ⋅ K)ln    318.15    273.15   (2100 J kg ⋅ K) ln  258.15       3   334 × 10 J kg + (0.0500 kg)  +  = 10.5 J K. 273.15 K     307.98  + (4190 J kg ⋅ K) ln  273.15    

(Some precision is lost in taking the logarithms of numbers close to unity.) 20.58: a) For constant-volume processes for an ideal gas, the result of Example 20.10 may be used; the entropy changes are nCV ln(Tc Tb ) and nCV ln(Ta Td ). b) The total entropy change for one cycle is the sum of the entropy changes found in part (a); the other processes in the cycle are adiabatic, with Q = 0 and ∆S = 0. The total is then TT  T T ∆S = nCV ln c + nCV ln a = nCV ln  c a . T T  Tb Td  a d From the derivation of Eq. (20.6), Tb = r γ −1Ta and Tc = r γ −1Td , and so the argument of the logarithm in the expression for the net entropy change is 1 identically, and the net entropy change is zero. c) The system is not isolated, and a zero change of entropy for an irreversible system is certainly possible.

20.59: a)

b) From Eq. (20.17), dS =

dQ T

, and so dQ = T dS , and

which is the area under the curve in the TS plane. c) QH is the area under the rectangle bounded by the horizontal part of the rectangle at TH and the verticals. | QC | is the area bounded by the horizontal part of the rectangle at TC and the verticals. The net work is then QH − | QC |, the area bounded by the rectangle that represents the process. The ratio of the areas is the ratio of the lengths of the vertical sides of the respective rectangles, and W the efficiency is e = QH = THT−TC . d) As explained in problem 20.49, the substance that H mediates the heat exchange during the isochoric expansion and compression does not leave the system, and the diagram is the same as in part (a). As found in that problem, the ideal efficiency is the same as for a Carnot-cycle engine. = − mLf = − ( 0.160 kg) (334×10 T ( 373 .15 K ) =
mLf T
3

Q = ∫ dQ = ∫ T dS

20.60: a) ∆S = b) ΔS =

Q T Q T

J kg)

= −143 J K.

=

( 0.160 kg)(334 × 10 J kg ) ( 273 .15 K)

3

= 196 J K.

c) From the time equilbrium has been reached, there is no heat exchange between the rod and its surroundings (as much heat leaves the end of the rod in the ice as enters at the end of the rod in the boiling water), so the entropy change of the copper rod is zero. d) 196 J K − 143 J K = 53 J K.

20.61: a)

∆S = mcln(T2 T1 ) = (250 × 10−3 kg)(4190 J kg ⋅ K)ln(338.1 5 K 293.15 K) = 150 J K.

b) ∆S =

− mc∆T Telement

=

− ( 250 ×10 −3 kg)(4190 J kg ⋅ K)(338.15 K − 293.15 K) 393 .15 K

= −120 J K. c) The sum of the result of

parts (a) and (b) is ∆S system = 30 J K. d) Heating a liquid is not reversible. Whatever the energy source for the heating element, heat is being delivered at a higher temperature than that of the water, and the entropy loss of the source will be less in magnitude than the entropy gain of the water. The net entropy change is positive.

20.62: a) As in Example 20.10, the entropy change of the first object is m1c1ln(T T 1 ) and that of the second is m2c2ln(T ′ T 2 ) , and so the net entropy change is as given. Neglecting heat transfer to the surroundings, Q1 + Q2 = 0, m1c1 (T − T1 ) + m2c2 (T ′ − T2 ) = 0, which is the given expression. b) Solving the energy-conservation relation for T ′ and substituting into the expression for ∆S gives  T  m c  T T  ∆S = m1c1ln   + m2 c2 1n 1 − 1 1  − 1  . T   m c  T T  2 2  2 2   1  Differentiating with respect to T and setting the derivative equal to 0 gives mc (m2 c2 )(m1c1 m2 c2 )(− 1 T2 ) . 0= 1 1 + T   T T1   1 − (m1c1 m2 c2 )  −    T T   2   2  This may be solved for m1c1T1 + m2 c 2T2 , m1c1 + m2 c 2 which is the same as T ′ when substituted into the expression representing conservation of energy. T= Those familiar with Lagrange multipliers can use that technique to obtain the relations ∂ ∂Q ∂ ∂Q ∆S = λ , ∆S = λ ∂T ∂T ∂T ′ ∂T ′ and so conclude that T = T ′ immediately; this is equivalent to treating the differentiation as a related rate problem, as d m c m c dT ′ ∆S = 1 1 + 2 2 =0 dT ′ T T ′ dT and using
dT ′ dT m = − m21c12 gives T = T ′ with a great savings of algebra. c

c) The final state of the system will be that for which no further entropy change is possible. If T < T ′, it is possible for the temperatures to approach each other while increasing the total entropy, but when T = T ′, no further spontaneous heat exchange is possible.

20.63: a) For an ideal gas, CP = CV + R, and taking air to be diatomic, C P = 7 R, CV = 5 R and γ = 7 . Referring to Fig. (20.6), 2 2 5 7 7 QH = n 2 R (Tc − Tb ) = 2 ( p cVc − pbVb ). Similarly, QC = n 5 R ( p aVa − p d Vd ). What needs 2 to be done is to find the relations between the product of the pressure and the volume at the four points. pcV pbV Tc For an ideal gas, Tc c = Tb b , so p cVc = p aVa Ta . For a compression ratio r, and

( )

given that for the Diesel cycle the process ab is adiabatic, V  pbVb = paVa  a  V   b
γ −1 γ −1

= paVa r γ −1.

V  Similarly, pdVd = pcVc  c  . Note that the last result uses the fact that process da is V   a Tc isochoric, and Vd = Va ; also, pc = pb (process bc is isobaric), and so Vc = Vb Ta . Then,

( )

Vc Tc Vb Tb Ta Va = ⋅ = ⋅ ⋅ Va Tb Va Ta Tb Vb T = c Ta =  T V γ −1  V  ⋅  a aγ −1  a   T V  V   b b  b 
−γ

Tc γ r Ta
γ

Combining the above results, T  2 pdVd = paVa  c  r γ − γ T   a Subsitution of the above results into Eq. (20.4) gives   5 e = 1− 7    =1− where
Tc Ta

( )

  r − 1  Tc  γ −1   −r  T    a 
Tc γ γ −γ 2 Ta

1  (5.002)r −0.56 − 1 , 1.4  (3.167) − r 0.40   

= 3.167, γ = 1.4 have been used. Substitution of r = 21.0 yields

e = 0.708 = 70.8%.

Chapter 21

21.1:

mlead = 8.00 g and charge = −3.20 × 10 −9 C − 3.20 × 10 −9 C = 2.0 × 1010 . −19 − 1.6 × 10 C n 8.00 g = NA × = 2.33 × 10 22 and e = 8.58 × 10 −13. 207 nlead

a) ne = b) nlead

21.2:

current = 20,000 C s and t = 100 µs = 10 −4 s Q = It = 2.00 C Q ne = = 1.25 × 1019 . −19 1.60 × 10 C

21.3: The mass is primarily protons and neutrons of m = 1.67 × 10 −27 kg, so: 70.0 kg np and n = = 4.19 × 1028 − 27 1.67 × 10 kg About one-half are protons, so n p = 2.10 × 10 28 = ne and the charge on the electrons is given by: Q = (1.60 × 10 −19 C) × (2.10 × 10 28 ) = 3.35 × 10 9 C.

21.4: Mass of gold = 17.7 g and the atomic weight of gold is 197 g mol. So the number of atoms N A × mol = (6.02 × 10 23 ) × a) np = 79 × 5.41 × 10 22 = 4.27 × 1024 q = np × 1.60 × 10−19 C = 6.83 × 105 C b) ne = n p = 4.27 × 10 24.

(

17.7 g 197 g mol

) = 5.41 × 10

22

.

21.5: 1.80 mol = 1.80 × 6.02 × 1023 H atoms = 1.08 × 1024 electrons. charge = −1.08 × 1024 × 1.60 × 10−19 C = −1.73 × 105 C. 21.6: First find the total charge on the spheres: 1 q2 F= ⇒ q = 4πε0 Fr 2 = 4πε0 ( 4.57 × 10 − 21 )(0.2) 2 = 1.43 × 10−16 C 4πε0 r 2 And therefore, the total number of electrons required is n = q e = 1.43 × 10 −16 C 1.60 × 10−19 C = 890.

21.7: a) Using Coulomb’s Law for equal charges, we find: 1 q2 F = 0.220 N = ⇒ q = 5.5 × 10 −13 C 2 = 7.42 × 10− 7 C. 2 4πε0 (0.150 m) b) When one charge is four times the other, we have: 1 4q 2 F = 0.220 N = ⇒ q = 1.375 × 10−13 C 2 = 3.71 × 10− 7 C 4πε0 (0.150 m) 2 So one charge is 3.71 × 10 −7 C, and the other is 1.484 × 10 −6 C. 21.8: a) The total number of electrons on each sphere equals the number of protons. ne = np = 13 × N A × 0.0250 kg = 7.25 × 10 24. 0.026982 kg mol

b) For a force of 1.00 × 10 4 N to act between the spheres, F = 104 N = 1 q2 ⇒ q = 4πε0 (104 N) (0.08 m) 2 = 8.43 × 10 − 4 C. 2 4πε0 r ′ ⇒ ne = q e = 5.27 × 1015 ′ c) ne is 7.27 × 10 −10 of the total number. 21.9: The force of gravity must equal the electric force. 1 q2 1 (1.60 × 10−19 C) 2 mg = ⇒ r2 = = 25.8 m 2 ⇒ r = 5.08 m. 2 − 31 4πε0 r 4πε0 (9.11 × 10 kg)(9.8 m s) 21.10: a) Rubbing the glass rod removes electrons from it, since it becomes positive. 7.50 nC = (7.50 × 10 −9 C) (6.25 × 1018 electrons C) = 4.69 × 1010 electrons (4.69 × 1010 electrons) (9.11 × 10 −31 kg electron) = 4.27 × 10 −20 kg. The rods mass decreases by 4.27 × 10 −20 kg. b) The number of electrons transferred is the same, but they are added to the mass of the plastic rod, which increases by 4.27 × 10 −20 kg.

  21.11: F2 is in the + x - direction, so F1 must be in the − x - direction and q1 is positive. q 2 q3 qq F1 = F2 , k 12 3 = k 2 r13 r23 q1 = ( 0.0200 0.0400 ) q 2 = 0.750 nC
2

21.12: a)

F=

1 q1q2 1 (0.550 × 10−6 C) q2 ⇒ 0.200 N = 4πε0 r 2 4πε0 (0.30 m) 2

⇒ q 2 = + 3.64 × 10 −6 C. b) F = 0.200 N, and is attractive. 21.13: Since the charges are equal in sign the force is repulsive and of magnitude: kq 2 (3.50 × 10−6 C) 2 F= 2 = = 0.172 N r 4πε0 (0.800 m) 2 21.14: We only need the y-components, and each charge contributes equally. 1 (2.0 × 10 −6 C) (4 × 10−6 C) F= sin α = 0.173 N (since sin α = 0.6). 4πε0 (0.500 m) 2 Therefore, the total force is 2 F = 0.35 N , downward.

  21.15: F2 and F3 are both in the + x-direction.
F2 = k q1q2 qq = 6.749 × 10 − 5 N, F3 = k 1 2 3 = 1.124 × 10− 4 N 2 r12 r13

F = F2 + F3 = 1.8 × 10 −4 N, in the + x-direction.

21.16: F21 =

(9 × 10 9 N ⋅ m 2 C 2 ) (20. × 10 −6 C) (2.0 × 10 −6 C)

( 0.60m ) 2

= 0.100 N

FQ1 is equal and opposite to F1Q (Ex. 21.4), so

(F ) (F )

Q1 x Q1 y

= −0.23 N = 0.17 N

Overall: Fx = −0.23 N Fy = 0.100 N + 0.17 N = 0.27 N The magnitude of the total force is ( 0.23 N ) + ( 0.27 N ) = 0.35 N. The direction of the force, as measured from the +y axis is 0.23 θ= tan −1 = 40 0.27
2 2

 21.17: F2 is in the + x − direction. F2 = k q1 q 2
2 r12

= 3.37 N, so F2 x = +3.37 N

Fx = F2 x + F3 x and Fx = −7.00 N F3 x = Fx − F2 x = −7.00 N − 3.37 N = −10.37 N For F3 x to be negative, q3 must be on the –x-axis. F3 = k q1q3 , so x = x2 k q1q3 = 0.144 m, so x = −0.144 m F3

21.18: The charge q3 must be to the right of the origin; otherwise both q 2 and q 3 would exert forces in the + x direction. Calculating the magnitude of the two forces: 1 q1q2 (9 × 109 N ⋅ m 2 C 2 )(3.00 × 10 −6 C)(5.00 × 10 −6 C) F21 = = 2 4πε0 r12 (0.200 m) 2 = 3.375 N in the + x direction. (9 × 10 9 N ⋅ m 3 C 2 ) (3.00 × 10 −6 C) (8.00 × 10 −6 C) F31 = 2 r13 0.216 N ⋅ m 2 in the − x direction 2 r13 We need F21 − F31 = −7.00 N : = 0.216 N ⋅ m 2 3.375 N − = −7.00 N 2 r13 0.216 N ⋅ m 2 r = = 0.0208 m 2 3.375 N + 7.00 N r13 = 0.144 m to the right of the origin
2 13

   21.19: F = F1 + F2 and F = F2 + F1 since they are acting in the same direction at y = − 0.400 m so, F=  1.50 × 10−9 C 3.20 × 10 −9 C  1 −6 (5.00 × 10− 9 C)   (0.200 m) 2 + (0.400 m) 2  = 2.59 × 10 N downward.  4πε0  

   21.20: F = F1 + F2 and F = F1 − F2 since they are acting in opposite directions at x = 0 so,  4.00 × 10−9 C 5.00 × 10 −9 C  1 −9 −6 F= (6.00 × 10 C)   (0.200 m) 2 + (0.300 m) 2  = 2.4 × 10 N to the right.  4πε0  

21.21: a)

1 qQ 1 2qQa sin θ 2 2 2 4πε 0 (a + x ) 4πε 0 (a + x 2 ) 3 2 1 2qQ c) At x = 0, Fy = in the + y direction. 4πε0 a 2 b) Fx = 0, Fy = 2

d)

21.22: a)

b) Fx = −2

1 qQ 1 − 2qQx cos θ = , Fy = 0 2 2 2 4πε 0 (a + x ) 4πε 0 (a + x 2 ) 3 / 2

c) At x = 0, F = 0. d)

21.23:

1 q2 1 q2 1 q2 + 2 = 1+ 2 2 at an angle of 45° below the 4πε0 2 L2 4πε0 L2 4πε0 2 L2 positive x-axis b) F =

(

)

21.24: a) E =

1 q 1 (3.00 × 10−9 C) = 432 N C , down toward the particle. 4πε0 r 2 4πε0 (0.250 m) 2 1 q ⇒ r= 4πε0 r 2 1 (3.00 × 10−9 C) = 1.50 m. 4πε0 (12.0 N C)

b) E = 12.00 N C =

21.25: Let +x-direction be to the right. Find a x : v0 x = +1.50 × 103 m s , vx = −1.50 × 103 m s , t = 2.65 × 10−6 s, ax = ? vx = v0 x + axt gives a x = −1.132 × 109 m s 2 Fx = max = −7.516 × 10−18 N   F is to the left ( − x - direction ) , charge is positive, so E is to the left. E = F q = (7.516 × 10 −18 N)

[( 2) (1.602 × 10

−19

C) = 23.5 N C

]

21.26: (a) x = 1 at 2 2 2(4.50 m) 2x 2 a= 2 = = 1.00 × 1012 m s -6 2 t (3.00 × 10 s) E= F ma (9.11 × 10 −31 kg) (1.00 × 1012 m s ) = = q q 1.6 × 10 −19 C
2

= 5.69 N C The force is up, so the electric field must be downward since the electron is negative. (b) The electron’s acceleration is ~ 1011 g, so gravity must be negligibly small compared to the electrical force. (0.00145 kg) (9.8 m s 2 ) 21.27: a) q E = mg ⇒ q = = 2.19 × 10 − 5 C, sign is negative. 650 N C (1.67 × 10−27 kg) (9.8 m s 2 ) b) qE = mg ⇒ E = = 1.02 × 10− 7 N / C, upward. −19 1.60 × 10 C

21.28: a) b)

1 q 1 (26 × 1.60 × 10−19 C) E= = = 1.04 × 1011 N C . 2 −10 2 4πε0 r 4πε0 (6.00 × 10 m) Eproton 1 q 1 (1.60 × 10−19 C) = = = 5.15 × 1011 N C . 2 −11 2 4πε0 r 4πε0 (5.29 × 10 m)

21.29: a) q = −55.0 × 10 −6 C, and F is downward with magnitude 6.20 × 10 −9 N. Therefore, E = F q = 1.13 × 10 −4 N C, upward. b) If a copper nucleus is placed at that point, it feels an upward force of magnitude F = qE = ( 29 ) ⋅ 1.6 × 10 −19 C ⋅ 1.13 × 10 −4 N C = 5.24 × 10 −22 N.

21.30: a) The electric field of the Earth points toward the ground, so a NEGATIVE charge will hover above the surface. 2 (60.0 kg) (9.8 m s ) mg = qE ⇒ q = − = −3.92 C. 150 N C 1 q2 1 (3.92 C) 2 = = 1.38 × 107 N. The magnitude of the charge is 4πε0 r 2 4πε0 (100.00 m) 2 too great for practical use. b) F = 21.31: a) Passing between the charged plates the electron feels a force upward, and just misses the top plate. The distance it travels in the y-direction is 0.005 m. Time of flight 0200 = t = 1.600.× 10 6 m s = 1.25 × 10 −8 s and initial y-velocity is zero. Now, m

y = v0 yt + 1 at 2 so 0.005 m = 1 a(1.25 × 10 −8 s) 2 ⇒ a = 6.40 × 1013 m s . But also 2 2
a=
F m

2

=

eE me

⇒E=

( 9.11 × 10 −31 kg)( 6.40 × 1013 m s 2 ) 1.60 × 10 −19 C

= 364 N C .

b) Since the proton is more massive, it will accelerate less, and NOT hit the plates. To find the vertical displacement when it exits the plates, we use the kinematic equations again: 1 1 eE y = at 2 = (1.25 × 10 −8 s) 2 = 2.73 × 10 −6 m. 2 2 mp c) As mention in b), the proton will not hit one of the plates because although the electric force felt by the proton is the same as the electron felt, a smaller acceleration results for the more massive proton. d) The acceleration produced by the electric force is much greater than g; it is reasonable to ignore gravity.

21.32: a)  q1 ˆ (9 × 10 9 N ⋅ m 2 C 2 ) (−5.00 × 10 −9 C) E1 = j= = (−2.813 × 10 4 N C) ˆ j 2 2 ( 0.0400 m ) 4πε 0 r1  q 2 (9 × 10 9 N ⋅ m 2 C 2 ) (3.00 × 10 −9 C) E2 = 2 = = 1.08 × 10 4 N C 2 2 r2 ( 0.0300m ) + (0.0400 m)  The angle of E2 , measured from the x - axis, is 180 − tan −1

(

4.00 cm 3.00 cm

) = 126.9° Thus

 ˆ E 2 = (1.080 × 104 N C) ( i cos 126.9° + ˆ sin 126.9°) j ˆ = ( − 6.485 × 103 N C) i + (8.64 × 103 N C) ˆ j b) The resultant field is   ˆ E1 + E 2 = ( − 6.485 × 103 N C) i + ( − 2.813 × 104 N / C + 8.64 × 103 N C) ˆ j ˆ = ( − 6.485 × 103 N / C) i − (1.95 × 104 N C) ˆ j 21.33: Let + x be to the right and + y be downward. Use the horizontal motion to find the time when the electron emerges from the field: x − x0 = 0.0200 m, a x = 0, v 0 x = 1.60 × 10 6 m s , t = ? x − x0 = v0 x t + 1 a x t 2 gives t = 1.25 × 10 −8 s 2 v x = 1.60 × 10 6 m s y − y 0 = 0.0050 m, v 0y = 0, t = 1.25 × 10 −8 s, v y = ?  v0 y + v y y − y0 =   2    t gives v y = 8.00 × 10 5 m s  

2 v = v x + v 2 = 1.79 × 10 6 m s y

 ˆ 21.34: a) E = −11 N Ci + 14 N Cˆ, so E = (−11) 2 + (14) 2 = 17.8 N C. j θ = tan −1 ( − 14 11) = − 51.8°, so θ = 128° counterclockwise from the x-axis   b) F = E q so F = (17.8 N C) (2.5 × 10−9 C) = 4.45 × 10−8 N, i) at − 52° (repulsive ) ii ) at + 128° (repulsive ).

21.35: a) Fg = me g = (9.11 × 10 −31 kg) (9.8 m s ) = 8.93 × 10 −30 N. Fe = eE = (1.60 × 10−19 C) (1.00 × 10 4 N C) = 1.60 × 10−15 N. Yes, ok to neglect Fg because Fe >> Fg . b) E = 10 4 N C ⇒ Fe = 1.6 × 10 −15 N = mg ⇒ m = 1.63 × 10 −16 kg ⇒ m = 1.79 × 1014 me . c) No. The field is uniform. 2(0.0160 m) (1.67 × 10 −27 kg ) 1 2 1 eE 2 21.36: a) x = at = t ⇒E= = 148 N C . 2 2 mp (1.60 × 10 −19 C) (1.50 × 10 −6 s) 2 eE b) v = v 0 + at = t = 2.13 × 10 4 m s . mp π  π 2ˆ 2 ˆ  − 1.35  −1  12  ˆ j i+ j 21.37: a) tan −1   = − , = r − ˆ b) tan   = , r = 2 2 2  .2  4  0   2.6  ˆ ˆ ˆ c) tan −1   + 1.10  = 1.97 radians = 112.9°, r = − 0.39i + 0.92 j (Second quadrant).    21.38: a) E = 614 N C , F = qE = 9.82 × 10 −17 N. b) F = e 2 4πε0 (1.0 × 10−10 ) 2 = 2.3 × 10−8 N. c) Part (b) >> Part (a), so the electron hardly notices the electric field. A person in the electric field should notice nothing if physiological effects are based solely on magnitude.

2

21.39: a) Let + x be east.   E is west and q is negative, so F is east and the electron speeds up. Fx =| q | E = (1.602 × 10 −19 C) (1.50 V m) = 2.403 × 10 −19 N ax = Fx m = (2.403 × 10−19 N) (9.109 × 10− 31kg ) = + 2.638 × 1011 m s 2 v0 x = + 4.50 × 105 m s , ax = + 2.638 × 1011 m s 2 , x − x0 = 0.375 m, vx = ?
2 2 vx = v0 x + 2ax ( x − x0 ) gives vx = 6.33 × 105 m s  b) q > 0 so F is west and the proton slows down.

Fx = − | q | E = − (1.602 × 10 −19 C) (1.50 V m) = − 2.403 × 10 −19 N ax = Fx m = ( − 2.403 × 10−19 N) (1.673 × 10− 27 kg ) = − 1.436 × 108 m s
2 2

v0 x = + 1.90 × 104 m s , ax = − 1.436 × 108 m s , x − x0 = 0.375 m, vx = ? v 2 x = v 2 0 x + 2ax ( x − x0 ) gives vx = 1.59 × 10 4 m s 21.40: Point charges q1 (0.500 nC) and q 2 (8.00 nC) are separated by x = 1.20 m. The electric field is zero when E1 = E2 ⇒
kq1 r12

=

kq 2 (1.20 − r1 ) 2

⇒ q2 r12 = q1 (1.2 − r1 ) 2 =

q1r12 − 2q1 (1.2)r1 + 1.22 q1 ⇒ (q2 − q1 )r12 + 2(1.2) q1r1 − (1.2) 2 q1 = 0 or 7.5r12 + 1.2r1 − 0.72 = 0 r1 = + 0.24, − 0.4 r1 = 0.24 is the point between.

21.41: Two positive charges, q , are on the x-axis a distance a from the origin. a) Halfway between them, E = 0.  1   q q    (a + x ) 2 − (a − x) 2  , | x | < a    4πε0   1   q q   , x > a b) At any position x, E =  + 2 2  4πε0  (a + x ) (a − x)     −1   q q    (a + x) 2 + (a − x) 2  , x < − a   4πε0    For graph, see below.

21.42: The point where the two fields cancel each other will have to be closer to the negative charge, because it is smaller. Also, it cant’t be between the two, since the two fields would then act in the same direction. We could use Coulomb’s law to calculate the actual values, but a simpler way is to note that the 8.00 nC charge is twice as large as the –4.00 nC Charge. The zero point will therefore have to be a factor of 2 farther from the 8.00 nC charge for the two fields to have equal magnitude. Calling x the distance from the –4.00 nC charge: 1.20 + x = 2 x x = 2.90 m

21.43: a) Point charge q1 (2.00 nC) is at the origin and q 2 ( − 5.00 nC ) is at x = 0.800 m. k | q1 | k | q2 | i) At x = 0.200 m, E = + = 575 N C right. 2 (0.200 m) (0.600 m) 2 k | q2 | k | q1 | ii) At x = 1.20 m, E = + = 269 N C left. 2 (0.400 m) (1.20 m) 2 k | q1 | k | q2 | iii) At x = − 0.200 m, E = + = 405 N C left. 2 (0.200 m) (1.00 m) 2 b) F = − eE i) F = 1.6 × 10−19 C ⋅ 575 N C = 9.2 × 10−17 N left, ii ) F = 1.6 × 10−19 C ⋅ 269 N C = 4.3 × 10 −17 N right, iii) F = 1.6 × 10−19 ⋅ 405 = 6.48 × 10−17 N right.

21.44: A positive and negative charge, of equal magnitude q , are on the x-axis, a distance a from the origin. 1 2q a) Halfway between them, E = , to the left. 4πε0 a 2  1   4πε0  1  b) At any position x, E =   4πε0  1   4πε0  with “+” to the right. This is graphed below.  −q  q   (a + x) 2 − (a − x) 2  , | x | < a     −q  q   (a + x) 2 + (a − x) 2  , x > a     −q  q   (a + x) 2 − (a − x) 2  , x < − a   

21.45: a) At the origin, E = 0. b) At x = 0.3 m, y = 0 :   ˆ 1 1 1 ˆ E= (6.00 × 10 −9 C)   (0.15 m) 2 + (0.45 m) 2 i = 2667i N C .  4πε0   c) At x = 0.15 m, y = −0.4 m :   −1 ˆ 1 1 0.3 ˆ 1 0.4 ˆ  E= (6.00 × 10− 9 C)   (0.4 m) 2 j + (0.5 m) 2 0.5 i − (0.5 m) 2 0.5 j   4πε0    ˆ − 510.3 ˆ) N C ⇒ E = 526.5 N C and θ = 75.7 down from the x-axis. ⇒ E = (129.6i j  0.2  2(6.00 × 10− 9 C) ⋅    1  0.25  = 1382 ˆ N C d) x = 0, y = 0.2 m : E = j 4πε0 (0.25 m) 2 21.46: Calculate in vector form the electric field for each charge, and add them.  − 1 (6.00 × 10−9 C) ˆ ˆ E− = i = −150i N C 4πε0 (0.6 m) 2   −1 1 1 ˆ ˆ ˆ ˆ E+ = (4.00 × 10− 9 C)  (1.00 m) 2 (0.6)i + (1.00 m) 2 (0.8) j  = 21.6i + 28.8 j N C  4πε0    28.8   ⇒ E = (128.4) 2 + (28.8) 2 = 131.6 N C , at θ = tan −1   = 12.6 up from 128.4   − x axis.  1 2(6.0 × 10−9 C) ˆ ˆ 21.47: a) At the origin, E = − i = −4800i N C . 4πε0 (0.15 m) 2 b) At x = 0.3 m, y = 0 :   ˆ 1 1 1 ˆ E= (6.0 × 10− 9C)   (0.15 m) 2 − (0.45 m) 2  i = 2133i N C .  4πε0   c) At x = 0.15 m, y = −0.4 m :   −1 ˆ 1 1 0.3 ˆ 1 0.4 ˆ  E= (6.0 × 10− 9 C) j− i+ j 2 2 2  (0.4 m) 4πε0 (0.5 m) 0.5 (0.5 m) 0.5     ˆ ⇒ E = (−129.6i − 164.5 ˆ) N C ⇒ E = 209 N C and θ = 232° clockwise from j + x - axis.  1 2(6.00 × 10−9 C) ( 0.15 ) 0.25 ˆ d) x = 0, y = 0.2 m : E y = 0, E = − = − 1037 i N C 4πε0 (0.25 m) 2

21.48: For a long straight wire, E =

λ 1.5 × 10−10 C m ⇒r= = 1.08 m. 2πε0 r 2πε0 (2.5 N C)

21.49: a) For a wire of length 2a centered at the origin and lying along the y-axis, the electric field is given by Eq. (21.10).  1 λ ˆ E= i 2 2 2πε0 x x a + 1 b) For an infinite line of charge:  λ ˆ E= i 2πε0 x Graphs of electric field versus position for both are shown below.

21.50: For a ring of charge, the electric field is given by Eq. (21.8).  1 Qx ˆ a) E = i so with 2 4πε0 ( x + a 2 )3 2  ˆ Q = 0.125 × 10−9 C, a = 0.025 m and x = 0.4 m ⇒ E = 7.0i N C .

   ˆ ˆ b) Fon ring = − Fon q = − q E = − ( − 2.50 × 10−6 C) (7.0i N C) = 1.75 × 10-5 i N.

21.51: For a uniformly charged disk, the electric field is given by Eq. (21.11):   σ  1 1 − i ˆ E= 2 2  2 ε0  R x +1  The x -component of the electric field is shown below.

21.52: The earth’s electric field is 150 N C , directly downward. So, σ 2 E = 150 = ⇒ σ = 300ε0 = 2.66 × 10− 9 C m , and is negative. 2ε0 21.53: For an infinite plane sheet, E is constant and is given by E = perpendicular to the surface. e−  C   100 cm  −9 C σ = 2.5 × 10 − 1.6 × 10−19 −  ⋅  2   1 m  = − 4 × 10 m 2  cm  e   
6 2

σ directed 2ε0

so E =

4 × 10−9 2 ε0

C m2

= 226 N C directed toward the surface.

21.54: By superposition we can add the electric fields from two parallel sheets of charge. a) E = 0. b) E = 0. σ σ c) E = 2 = , directed downward. 2 ε0 ε0

21.55:

21.56: The field appears like that of a point charge a long way from the disk and an infinite plane close to the disk’s center. The field is symmetrical on the right and left (not shown).

21.57: An infinite line of charge has a radial field in the plane through the wire, and constant in the plane of the wire, mirror-imaged about the wire: Cross section through the wire: Plane of the wire:

Length of vector does not depend on angle.

Length of vector gets shorter at points further away from wire.

21.58: a) Since field lines pass from positive charges and toward negative charges, we can deduce that the top charge is positive, middle is negative, and bottom is positive. b) The electric field is the smallest on the horizontal line through the middle charge, at two positions on either side where the field lines are least dense. Here the ycomponents of the field are cancelled between the positive charges and the negative charge cancels the x-component of the field from the two positive charges. 21.59: a) p = qd ⇒ (4.5 × 10−9 C)(0.0031 m) = 1.4 × 10−11 C ⋅ m , in the direction from and towards q 2 .  b) If E is at 36.9°, and the torque τ = pE sin φ , then: τ 7.2 × 10 −9 N ⋅ m E= = = 856.5 N C . p sin φ (1.4 × 10−11 C ⋅ m) sin 36.9°

21.60: a) d = p q = (8.9 × 10−30 C ⋅ m) (1.6 × 10−19 C) = 5.56 × 10−11 m. b) τ max = pE = (8.9 × 10−30 C ⋅ m)(6.0 × 105 N C) = 5.34 × 10−24 N ⋅ m. Maximum torque:

21.61: a) Changing the orientation of a dipole from parallel to perpendicular yields: ∆U = U f − U i = − ( pE cos 90° − pE cos 0°) = + (5.0 × 10−30 C ⋅ m)(1.6 × 106 N C) = + 8 × 10−24 J. b) 3 2(8 × 10 −24 J) kT = 8 × 10 − 24 J ⇒ T = = 0.384 K. 2 3(1.38 × 10 − 23 J K )

p 6.17 × 10−30 C ⋅ m −9 21.62: Edipole ( x) = ⇒ Edipole (3.00 × 10 m) = = 4.11 2πε0 x 3 2πε0 (3.0 × 10 − 9 m) 3 × 106 N C . The electric force F = qE = (1.60 × 10−19 C)(4.11 × 106 N C) = 6.58 × 10 −13 N and is toward the water molecule (negative x-direction). ( y + d 2) 2 − ( y − d 2) 2 1 1 2 yd − = = 2 2 2 2 2 2 ( y − d 2) ( y + d 2) ( y − d 4) ( y − d 2 4) 2 q 2 yd qd y p ⇒ Ey = = ≈ 2 2 2 2 2 2 4πε0 ( y − d 4 ) 2πε0 ( y − d 4) 2πε0 y 3⋅ b) This also gives the correct expression for E y since y appears in the full expression’s denominator squared, so the signs carry through correctly. 21.63: a)

21.64: a) The magnitude of the field the due to each charge is  1 q q  1  , E= = 2 2 2  4πε0 r 4πε0  (d 2) + x )   where d is the distance between the two charges. The x-components of the forces due to the two charges are equal and oppositely directed and so cancel each other. The two fields have equal y-components, so:  2q  1   sin θ E = 2Ey = 2 2  4πε0  (d 2) + x   where θ is the angle below the x-axis for both fields. d 2 sin θ = ; (d 2) 2 + x 2 thus   2q   1 d 2 qd =   Edipole =  2 2 3 2  4πε  (d 2) 2 + x 2   (d 2) 2 + x 2  4πε0 ((d 2) + x ) 0    The field is the − y directions. b) At large x, x 2 >> (d 2) 2 , so the relationship reduces to the approximations qd Edipole ≈ 4πε0 x 3 21.65:

The dipoles attract. Fx = F1x + F2 x = 0, b) Opposite charges are closest so the dipoles attract. Fy = F1 y + F2 y = 2 F1 y

21.66: a)

  The torque is zero when p is aligned either in the same direction as E or in the opposite directions   b) The stable orientation is when p is aligned in the same direction as E c)

21.67:

sin θ = 1.50 2.00 so θ = 48.6° Opposite charges attract and like charges repel. Fx = F1x + F2 x = 0 qq′ (5.00 × 10−6 C)(10.0 × 10 −6 C) F1 = k 2 = k = 1.124 × 103 N r (0.0200 m) 2 F1 y = − F1 sin θ = −842.6 N F2 y = −842.6 N so Fy = F1 y + F2 y = −1680 N (in the direction from the + 5.00 - μC charge toward the − 5.00 - μC charge). b)

The y − components have zero moment arm and therefore zero torque. F1x and F2 x both produce clockwise torques. F1x = F1 cos θ = 743.1 N τ = 2( F1x )(0.0150 m) = 22.3 N ⋅ m, clockwise

 1 q1q3 1 q1q3 ˆ 21.68: a) F13 = + cos θi + sin θˆ j 2 2 4πε0 r13 4πε0 r13  1 (5.00 nC)(6.00 nC) 4 ˆ 1 (5.00 nC)(6.00 nC) 3 ˆ ⇒ F13 = + i+ j −4 4πε0 ((9.00 + 16.0) × 10 m) 5 4πε0 ((9.00 + 16.0) × 10− 4 m) 5  ˆ ⇒ F = +(8.64 × 10 −5 N)i + (6.48 × 10−5 N) ˆ. j
13

Similarly for the force from the other charge:  − 1 q2 q3 ˆ − 1 ( 2.00 nC)(6.00 nC) ˆ F23 = j= j = −(1.20 × 10− 4 N ) ˆ j 2 2 4πε0 r23 4πε0 (0.0300 m) Therefore the two force components are: Fy = 6.48 × 10−5 − 12.0 × 10 −5 = −5.52 × 10 −5 N Fx = 8.64 × 10 −5 N b) Thus, F = Fx2 + Fy2 = (8.64 × 10−5 N) 2 + (−5.52 × 10−5 N) 2 = 1.03 × 10−4 N, and the angle is θ = arctan( Fy Fx ) = 32.6, below the x axis  1 qQ 1 qQ 1 qQ  1 1   − = − 2 2 2  2 2  4πε0 ( a + x) 4πε0 (a − x) 4πε0 a  (1 + x a) (1 − x a)   qQ  1 qQ x x 1 qQ  x ⇒ Fq ≈ (1 − 2 . . . − (1 + 2 . . .)) = − 4  = − 2 2   πε a 3  x. But this is  4πε0 a a a 4πε0 a  a  0  the equation of a simple harmonic oscillator, so: qQ 1 qQ kqQ ω = 2πf = ⇒ f = = . 3 3 mπ ε0 a 2π mπ ε0 a mπ 2 a 3 b) If the charge was placed on the y-axis there would be no restoring force if q and Q had the same sign. It would move straight out from the origin along the y-axis, since the x-components of force would cancel. 21.69: a) Fq = 21.70: Examining the forces: ∑ Fx = T sin θ − Fe = 0 and ∑ Fy = T cos θ − mg = 0. So
mg sin θ cos θ

= Fe =

kq 2 d2

But tan θ ≈

d 2L

⇒ d3 =

2 kq 2 L mg

⇒d =

(

q2L 2 πε 0 mg

)

1 3

.

21.71: a)

b) Using the same force analysis as in problem 21.70, we find: q = 4πε0 d 2 mg tan θ and
2

d = 2 ⋅ (1.2)sin25 ⇒ q = 4πε0 ( 2 ⋅ (1.2) ⋅ sin 25°) 2 tan 25°(0.015 kg)(9.80 m s ) ⇒ q = 2.79 × 10 −6 C. c) From Problem 21.70, mg tan θ = sin θ =
kq 2 d2

2

.

d kq 2 (8.99 × 109 Nm 2 C)(2.79 × 10-6 C) 2 ⇒ tan θ = = 2L mg ( 2 L) 2 sin 2 θ 4(0.6m) 2 (0.015 kg)(9.8 m s 2 ) sin − 2 θ 0. Therefore tan θ = sin331 . Numerical solution of this transcendental equation leads to 2 θ
θ = 39.5°.

21.72: a) Free body diagram as in 21.71. Each charge still feels equal and opposite electric forces. kq q b) T = mg cos 20° = 0.0834 N so Fe = T sin 20° = 0.0285 N = 1 2 . (Note: r 21 r1 = 2(0.500 m)sin20 ° = 0.342 m.) c) From part (b), q1 q 2 = 3.71 × 10 −13 C 2 . d) The charges on the spheres are made equal by connecting them with a wire, but 2 1 we still have Fe = mg tan θ = 0.0453 N = 4 πε0 Q2 where Q = q1 + q 2 . But the separation r2 is 2 r
2

known: r2 = 2(0.500 m) sin 30° = 0.500 m. Hence: Q =

q1 + q 2 2

= 4πε0 Fe r 2 2

= 1.12 × 10−6 C. This equation, along with that from part (b), gives us two equations in q1 and q 2 . q1 + q 2 = 2.24 × 10 −6 C and q1 q 2 = 3.70 × 10 −13 C 2 . By elimination, substitution and after solving the resulting quadratic equation, we find: q1 = 2.06 × 10 −6 C and q 2 = 1.80 × 10 −7 C .

21.73: a) 0.100 mol NaCl ⇒ m Na = (0.100 mol)(22.99 g mol) = 2.30 g ⇒ m Cl = (0.100 mol)(35.45 g mol) = 3.55 g Also the number of ions is (0.100 mol) N A = 6.02 × 10 22 so the charge is: q = (6.02 × 10 22 )(1.60 × 10 −19 C) = 9630 C. The force between two such charges is: F= 1 q2 1 (9630) 2 = = 2.09 × 1021 N. 2 2 4πε0 r 4πε0 (0.0200 m)
2

b) a = F / m = (2.09 × 10 21 N) (3.55 × 10−3 kg ) = 5.89 × 1023 m s . c) With such a large force between them, it does not seem reasonable to think the sodium and chlorine ions could be separated in this way.  (2.5 × 10 −9 C) 4.5 × 10−9 C  kq1q3 kq2 q3 N= 2 + = kq3  2  (−0.3 m) 2 + (+0.2 m) 2  ⇒  r13 r23  

21.74: a) F3 = 4.0 × 10 q3 =

−6

4.0 × 10−6 N = 3.2 nC. (1262 N C) b) The force acts on the middle charge to the right. c) The force equals zero if the two forces from the other charges cancel. Because of the magnitude and size of the charges, this can only occur to the left of the negative kq1 kq 2 charge q 2 . Then: F13 = F23 ⇒ where x is the distance = 2 (0.300 − x) ( −0.200 − x) 2 from the origin. Solving for x we find: x = −1.76 m. The other value of x was between the two charges and is not allowed.

1 q(3q) 1 6q 2 = , toward the lower the left charge. The other 4πε0 ( L 2 ) 2 4πε0 L2 two forces are equal and opposite. 21.75: a) F = +

b) The upper left charge and lower right charge have equal magnitude forces at right angles to each other, resulting in a total force of twice the force of one, directed toward the lower left charge. So, all the forces sum to: 1  q (3q ) 2 q(3q )  q2  3  = F= + 3 2 +  N. 2 2  2   4πε0  L 2 ( 2 L)  4πε0 L  1  q q 2q    ( y − a)2 + ( y + a)2 − y 2  .  4πε0   1 q b) E ( p) = ((1 − a y ) − 2 + (1 + a y ) − 2 − 2). Using the binomial expansion: 2 4πε0 y

21.76: a) E ( p) =

 1 q  2a 3a 2 2a 3a 2 1 6qa 2 1 + + 2 + ⋅ ⋅ ⋅ + 1 − = ⇒ E ( p) ≈ + 2 + ⋅ ⋅ ⋅ − 2 . 4 4πε0 y 2  y y y y   4πε0 y 1 1 Note that a point charge drops off like 2 and a dipole like 3 . y y

21.77: a) The field is all in the x -direction (the rest cancels). From the + q charges: 1 q 1 q x 1 qx E= ⇒ Ex = = . 2 2 2 2 2 2 3 2 4πε0 a + x 4πε0 a + x a 2 + x 2 4πε0 (a + x ) (Each + q contributes this). From the − 2q : Ex = − 1 2q 1  2qx  2 ⇒ Etotal = 2  ( a + x 2 )3 4πε0 x 4πε0 
2

−

2q  1 2q = ((a 2 x 2 + 1) − 3 2  x  4πε0 x 2

2

− 1).

b) Etotal ≈

 1 2q  3a 2 1 3qa 2 1 − 2 + ⋅ ⋅ ⋅ − 1 =  4πε x 4 , for x >> a.. 4πε0 x 2  2 x   0 1 1 Note that a point charge drops off like 2 and a dipole like 3 . x x 20.0 g = 1.67 mol carbon ⇒ 6(1.67) = 10.0 mol 12.0 g mol

21.78: a) 20.0 g carbon ⇒

electrons ⇒ q = (10.0)N A (1.60 × 10−19 C) = 0.963 × 106 C. This much charge is placed at the earth’s poles (negative at north, positive at south), leading to a force: 1 q2 1 (0.963 ×106C) 2 F= = = 5.13 × 107 N. 2 7 2 4πε0 (2 Rearth ) 4πε0 (1.276 ×10 m) b) A positive charge at the equator of the same magnitude as above will feel a force in the south-to-north direction, perpendicular to the earth’s surface: 1 q2 F =2 sin 45 4πε0 (2 Rearth ) 2 ⇒F =2 1 4 (0.963 × 106C) 2 = 1.44 × 108 N. 7 2 4πε0 2 (1.276 × 10 m)

21.79: a) With the mass of the book about 1.0 kg, most of which is protons and neutrons, we find: #protons = 1 (1.0 kg) (1.67 × 10 −27 kg) = 3.0 × 10 26. Thus the charge 2 difference present if the electron’s charge was 99.999% of the proton’s is ∆q = (3.0 × 10 26 )(0.00001)(1.6 × 10 −19 C) = 480 C. b) F = k (∆q) 2 r 2 = k (480 C) 2 (5.0 m) 2 = 8.3 × 1013 N − repulsive. The acceleration a = F m = (8.3 × 1013 N) (1 kg ) = 8.3 × 1013 m s . c) Thus even the slightest charge imbalance in matter would lead to explosive repulsion! 21.80: (a)
2

Fnet (on central charge) = =

1 q2 1 q2 − 4πε0 (b − x) 2 4πε0 (b + x) 2

q2  1 1   (b − x) 2 − (b + x ) 2  4πε0   2 2 2 q (b + x) − (b − x) q2 4bx = = 2 2 4πε0 (b − x ) (b + x) 4πε0 (b − x) 2 (b + x) 2 For x << b, this expression becomes q 2 bx q2 Fnet ≈ = x Direction is opposite to x. πε0 b 2b 2 πε0b3 (b) ∑ F = ma : −
−q2 πε 0b 3

x = m d 2x dt
2

 q2  d 2x =−  mπ ε b3  x  dt 2 0   ω= q2 1 = 2π f → f = 3 mπ ε0b 2π q2 mπ ε0b3

(c) q = e, b = 4.0 × 10 −10 m, m = 12 amu = 12(1.66 × 10−27 kg) f = 1 (1.6 × 10 −19 C) 2 = 4.28 × 1012 Hz 2π 12(1.66 × 10 − 27 kg )π (8.85 × 10−12 C 2 Nm 2 )(4.0 × 10−10 m)3

21.81: a) m = ρV = ρ( 4 πr 3 ) = (8.9 × 103 kg m 3 )( 4 π )(1.00 × 10−3 m)3 = 3 3 3.728 × 10 −5 kg n = m M = (3.728 × 10 −5 kg )(63.546 × 10 −3 kg mol) = 5.867 × 10 −4 mol N = nN A = 3.5 × 1020 atoms (b) N e = (29)(3.5 × 10 20 ) = 1.015 × 10 22 electrons and protons qnet = eN e − (0.99900)eN e = (0.100 × 10 −2 )(1.602 × 10−19 C)(1.015 × 10 22 ) = 1.6 C (1.6 C) 2 q2 F =k 2 =k = 2.3 × 1010 N 2 r (1.00 m) 21.82: First, the mass of the drop: −6 m )3  3  4π (15.0 × 10  = 1.41 × 10−11 kg. m = ρV = (1000 kg m )   3   Next, the time of flight: t = D v = 0.02 20 = 0.00100 s and the acceleration : d= So: (1.41 × 10−11 kg )(600 m s ) a = F m = qE m ⇒ q = ma E = = 1.06 × 10 −13 C. 8.00 × 104 N C
2

1 2 2d 2(3.00 × 10 −4 m) 2 at ⇒ a = 2 = = 600 m s . 2 2 t (0.001 s)

21.83: Fy = eE ay = a) Fy mp =

Fx = 0 eE ax = 0 mp 2eE ∆y mp ∆ y = hmax when v y = 0

2 2 v 2 = v0 y + 2a y ∆y = v0 sin 2 α + y 2 vo m p sin 2 α

2eE 1 2 b) ∆y = v0 y t + a y t 2 t = torig when ∆y = 0 1   ⇒ 0 =  − v0 sin α + a y torig  torig 2   2v sin α = 0, 0 ay

⇒ hmax =

so torig or torig =

2v0 mp sin α eE 2 2v0 mp d = v0 xtorig = cos α sin α eE

c)

d) hmax

(4 × 105 m s) 2 (1.67 × 10−27 kg) sin 2 30° = = 0.42 m 2(1.6 × 10 −19 C)(500 N C)

2(4 × 105m s) 2 (1.67 × 1027 kg ) cos 30° sin 30° d= = 2.89 m (1.6 × 10−19 C)(500 N C)

21.84: a) E = 50 N C =

 1 q1 1 q2 1  q1 2  2 + q2  ⇒ q2 = + = 4πε0 r12 4πε0 r22 4πε0  r1 r2     (4.00 × 10−9 C)   4πε0 50.0 N C −  = − 8 × 10− 9 C. 2   (0.6 m)  

 q  r22  4πε0 E − 1  ⇒ q2 = (1.2 m) 2  r12    b) E = −50 N C =

 1 q1 1 q2 1  q1 2  2 + q2  ⇒ q2 = + = 4πε0 r12 4πε0 r22 4πε0  r1 r2     (4.00 × 10−9 C)   4πε0 (−50.0) −  = − 24.0 × 10− 9 C. 2   (0.6 m)  

 − 50 q1  2  r22   k − r 2  ⇒ q2 = (1.2 m) 1  

21.85: E = 12.0 N C =

− k (16.0 nC) k (12.0 nC) kq + + 2 2 (3.00 m) (8.00 m) (5.00 m) 2

 12 1.60 × 10 −8 C 1.20 × 10 −8 C   = +7.31 × 10−8 C = +73.1 nC. ⇒ q = 25.0 m 2  + − 2 2 k 9.0 m 64.00 m    21.86: a) On the x-axis: dE x = 1 dq 1 Qdx ⇒ Ex = 2 ∫ a(a + r − x) 2 = 4πε0 (a + r ) 4πε0 0
a

  1 Q 1 1 1 qQ  1 1 ˆ   x − a − x  ⇒ F = qE = 4πε a  x − a − x  i .    4πε0 a     0 kqQ kqQ kqQ c) For x >> a, F = ((1 − a x) −1 − 1) = (1 + a x + ⋅ ⋅ ⋅ − 1) ≈ 2 ≈ ax ax x 1 qQ . ( Note that for x >> a, r = x − a ≈ x.) Charge distributi on looks like a point 4πε0 r 2 from far away. b) If a + r = x, then E =

1 Q1 1   −  r a + r . And E y = 0 .  4πε0 a  

21.87: a) dE =

k dq kQ dy kQx dy = with dE x = 2 2 2 ( x + y ) a( x + y ) a( x 2 + y 2 ) 3
2

2

and dE y =

− KQy dy . Thus: a( x 2 + y 2 ) 3 2 Ex = 1 Qx dy ∫ ( x 2 + y 2 )3 4πε0 a 0
a a 2

= =

1 Qx 1 2 4πε0 a ( x + a 2 )1

2

a 1 Q = 2 2 x 4πε0 x( x + a 2 )1

2

−1 Q ydy Ey = ∫ ( x 2 + y 2 )3 4πε0 a 0

 −1 Q  1 1  − 2  x ( x + a 2 )1 2   4πε0 a   b) Fx = −qE x and Fy = −qE y where E x and E y are given in (a).
2

1 qQ 1 qQ a 2 1 qQa 2 2 −1 2 c) For x >> a, Fy = (1 − (1 + a / x ) ) ≈ = . 2 4πε0 ax 4πε0 ax 2 x 4πε0 2 x 3 1 qQ  a2  1 + 2  Looks dipole-like in y-direction Fx = − 4πε0 x 2  x    Looks point-like along x-direction  1 Q ˆ 21.88: a) From Eq. (22.9), E = i 4πε0 x x 2 + a 2  1 (−9.00 × 10 −9 C ) ˆ ˆ ⇒E= i = ( −1.29 × 106 N C)i . −3 −3 2 2 4πε0 (2.5 × 10 m) (2.5 × 10 m) + (0.025 m) (b) The electric field is less than that at the same distance from an infinite line of charge ( Ea → ∞ = 1 2λ − 1 2Q = = − 1.30 × 106 N C). This is because in the 4πε0 x 4πε0 x 2a 1 λ x2 2πε0 x 1 + a 2
−1 2

≈

qQ . 4πε0 x 2

approximation, the terms left off were negative. E ∞ − (Higher order terms).
Line

(

)

12

≈

 λ  x2 1 − 2 + ⋅ ⋅ ⋅ =  2a  2πε0 x  

c) For a 1% difference, we need the next highest term in the expansion that was left off to be less than 0.01: x2 < 0.01 ⇒ x < a 2(0.01) = 0.025m 2(0.01) ⇒ x < 0.35 cm. 2a 2

 1 Q ˆ 21.89: (a) From Eq. (22.9), E = i 2 2 4πε0 x x + a  1 (−9.0 × 10 −9 C) ˆ ⇒E= = (−7858 N C) i . 4πε0 (0.100 m) (0.100 m) 2 + (0.025 m) 2 b) The electric field is less than that at the same distance from a point charge (8100 N C). Since E x → ∞ =  1 Q a2 1 − 2 + ⋅ ⋅ ⋅  = Epoint –(Higher order terms). 2   4πε0 x  2 x 

c) For a 1% difference, we need the next highest term in the expansion that was left off to be less than 0.01: a2 ≈ 0.01 ⇒ x ≈ a 1 (2 (0.01)) = 0.025 1 0.02 ⇒ x ≈ 0.177 m. 2x2

21.90: (a) On the axis,
−1 2   4.00 p C π (0.025 m) 2 σ   R2 E= 1 −  + 1  =  2 ε0   x 2 2 ε0      −1 2   (0.025 m) 2   1 −  + 1  2     (0.0020 m)    

⇒ E = 106 N C , in the + x-direction. b) The electric field is less than that of an infinite sheet E∞ = σ = 115 N C . 2 ε0

c) Finite disk electric field can be expanded using the binomial theorem since the expansion terms are small: ⇒ E ≈ σ 2 ε0  x  x3 1 − + 3 − ⋅ ⋅ ⋅ So the difference between the  R 2R   x . Thus: R

infinite sheet and finite disk goes like

∆E ( x = 0.20 cm) ≈ 0.2 2.5 = 0.08 = 8% and ∆E ( x = 0.40 cm) ≈ 0.4 2.5 = 0.16 = 16%.

−1  σ   R2 21.91: (a) As in 22.72: E = 1 −  + 1  2 ε0   x 2   

2

   
2

−1  4.00 pC π (0.025 m) 2   (0.025 m) 2 = 1 −  + 1 2   2 ε0   (0.200 m)  

 ⇒E  

= 0.89 N C in the + x-direction. b) x >> R, E = σ [1 − (1 − R 2 2 x 2 + 3R 4 8 x 4 − ⋅ ⋅ ⋅)] 2 ε0 σ R2 σπR 2 Q = = . 2 2 2ε0 2 x 4πε0 x 4πε0 x 2

≈

c) The electric field of (a) is less than that of the point charge (0.90 N C) since the correction term that was omitted was negative. d) From above x = 0.2 m (0.9 − 0.89) = 0.01 = 1%. For x = 0.1 m 0.89 Edisk = 3.43 N C Epoint = 3.6 N C so (3.6 − 3.43) = 0.047 ≈ 5%. 3.6
a 0

21.92: a) f ( x ) = f (− x) : ∫ f ( x )dx = ∫
−a

−a a

f ( x)dx + ∫ f ( x)dx = ∫
0 a 0

a

−a

0

f (− x ) d ( − x ) +
a 0

∫

a

0

f ( x)dx. Now replace − x with y : ⇒ ∫ f ( x )dx = ∫ f ( y )d ( y ) + ∫ f ( x )dx =
−a a 0

2 ∫ f ( x)dx. b) g ( x ) = − g (− x ) : ∫ g ( x)dx = ∫ g ( x )dx + ∫ g ( x)dx = − ∫ − g (− x)(− d (− x )) +
−a −a 0 0 a 0 a −a

∫

a

0

g ( x)dx. Now replace − x with y : ⇒ ∫ g ( x)dx = − ∫ g ( y )d ( y ) + ∫ g ( x )dx = 0.
−a 0 0

a

a

a

c) The integrand in E y for Example 21.11 is odd, so E y =0.

21.93: a) The y-components of the electric field cancel, and the x-components from both charges, as given in problem 21.87 is: Ex = 1 − 2Q 4πε0 a 1 1  − 2  y ( y + a 2 )1    1 − 2Qq  1 1 ⇒F =  − 2 2   y ( y + a 2 )1 4πε0 a  
2

ˆ i .  

 1 − 2Qq 1 Qqa ˆ If y >> a, F ≈ (1 − (1 − a 2 2 y 2 + ⋅ ⋅ ⋅))i = − . 4πε0 ay 4πε0 y 3 b) If the point charge is now on the x-axis the two charged parts of the rods provide different forces, though still along the x-axis (see problem 21.86).     1 Qq  1 1 ˆ 1 Qq  1 1 ˆ F+ = qE + = − i and F− = q E − = −   − i 4πε0 a  x − a x  4πε0 a  x x + a  So,    1 Qq  1 2 1 ˆ F = F+ + F− = − +  i 4πε0 a  x − a x x + a 
2    ˆ 1 Qq   a a 2 ˆ  1 + + 2 + . . .  − 2 + 1 − a + a 2 − . . . i = 1 2Qqa i . For x >> a, F ≈   x x   4πε0 ax   x x 4πε0 x 3   

21.94: The electric field in the x-direction cancels the left and right halves of the semicircle. The remaining y-component points in the negative y-direction. The charge per unit length of the semicircle is: λ= Q kλ dl kλ dθ kλ sin θ dθ and dE = 2 = but dE y = dE sin θ = . πa a a a 2kλ π a ∫0
2

So, E y =

sin θ dθ =

2kλ [−cosθ]π 0 a

2

=

2kλ 2kQ = , downward. a πa 2

21.95: By symmetry, E x = E y . For E y , compared to problem 21.94, the integral over the angle is halved but the charge density doubles─giving the same result. Thus, Ex = E y = 2kλ 2kQ = . a πa 2

21.96:

∑F

mg cos α qσ qσ ∑ Fy = 0 ⇒ T sin α = 2ε ⇒ T = 2ε sin α 0 0
x

= 0 ⇒ T cos α = mg ⇒ T =

⇒

mg qσ qσ = ⇒ tan α = cos α 2ε0 sin α 2ε0 mg

 qσ  ⇒ α = arctan  2ε mg    0 

21.97: a) b) 1429

qE = 10 mg ⇒

1.4 × 10 5 N C m E = = = 1429 kg C. q 10g 10(9.8 m s 2 )

kg 1 mol 6.02 × 1023 carbons 1.6 × 10−19 C carbons . . . = 1.15 × 1010 . −3 − C 12 × 10 kg mol excess e excess e − =2 1 Q , where 4πε0 x x 2 + ( a ) 2 2

21.98: a) E x = E y , and E x = 2 Elength of wire a , charge x=

Q

a 2Q 2Q ⇒ Ex = − , Ey = − . 2 2 πε0 a πε0 a 2 b) If all edges of the square had equal charge, the electric fields would cancel by symmetry at the center of the square.

21.99: a) E ( P) = − σ1 σ σ 0.0200 C m 2 0.0100 C m 2 0.0200 C m 2 − 2 + 3 =− − + 2 ε0 2 ε0 2 ε0 2 ε0 2 ε0 2 ε0 0.0100 C m 2 = 5.65 × 108 N C, in the − x-direction. 2 ε0 σ1 σ σ 0.0200 C m 2 0.0100 C m 2 0.0200 C m 2 − 2 + 3 =+ − + 2 ε0 2 ε0 2ε 0 2ε0 2 ε0 2 ε0

⇒ E ( P) =

b) E ( R ) = + ⇒ E ( R) = c) ⇒ E (S ) =

0.0300 C m 2 = 1.69 × 109 N C, in the + x-direction. 2 ε0 E (S ) = + σ1 σ σ 0.0200 C m 2 0.0100 C m 2 0.0200 C m 2 + 2 + 3 =+ + + 2 ε0 2 ε0 2 ε0 2 ε0 2ε0 2 ε0

0.0500 C m 2 = 2.82 × 109 N C, in the + x-direction. 2 ε0 σ1 σ σ 0.0200 C m 2 0.0100 C m 2 0.0200 C m 2 + 2 − 3 =+ + − 2 ε0 2 ε0 2 ε0 2 ε0 2ε0 2 ε0

d) E (T ) = + ⇒ E (S ) = 21.100:

0.0100 C m 2 = 5.65 × 108 N C, in the + x-direction. 2 ε0

 − σ 2 + σ 3  2.00 × 10− 4 C m 2 = = σ1  = +1.13 × 107 N m.   2 ε0 2 ε0    + σ1 + σ 3  4.00 × 10− 4 C m 2 Fon II qEat II = = = σ2  = +2.26 × 107 N m   A A 2 ε0 2 ε0    + σ1 + σ 2  − 6.00 × 10− 4 C m 2 Fon III qEat III = = = σ3  = −3.39 × 107 N m   A A 2 ε0 2 ε0   (Note that “+” means toward the right, and “–” is toward the left.) Fon I qEat = A A
I

21.101:

By inspection the fields in the different regions are as shown below:  σ  ˆ ˆ  σ  ˆ ˆ EI =   2ε (−i + k ), EII =  2ε (+ i + k )     0  0  σ  ˆ ˆ  σ  ˆ ˆ EIII =   2ε (+ i − k ), EIV =  2ε (− i − k )     0  0   σ  x ˆ z ˆ ( − ∴E =  i+ k ).  2ε  x z  0

21.102:

2 2 a) Q = Aσ = π ( R2 − R2 )σ

b) Recall the electric field of a disk, Eq. (21.11): E =  σ E ( x) = 1 −1 2ε0

([

( R2 x) 2 + 1 − 1 − 1 ( R1 x) 2 + 1 ( R1 x ) 2 + 1 =

] [

( R1 x) 2 + 1

] ) x iˆ ⇒ E ( x) = −εσ × x 2
0

σ 1−1 2 ε0

[

( R x) 2 + 1 . So,

]

(1

( R2 x) 2 + 1 − 1 c) Note that 1

) x iˆ. x

x x  ( x R1 ) 2  (1 + ( x R1 ) 2 ) −1 2 ≈ 1 − + ...   R1 R1  2   σ  x x x ˆ σ 1 1  x ˆ  −  i=  −  ⇒ E ( x) = R R  x  R R  2 i , and sufficiently close means that 2 ε0  1 2 ε0  1 2  2  x ( x R1 ) 2 << 1. d) F = qE ( x) = − qσ  1 1  ω 1  −  x = m ⇒ f = x = R R  2 ε0  1 2π 2π 2  qσ  1 1   − . R R  2 ε0 m  1 2 

21.103:

a) The four possible force diagrams are:

Only the last picture can result in an electric field in the –x-direction. b) q1 = −2.00 μ C, q3 = +4.00, μC, and q2 > 0. 1 q1 1 q2 c) E y = 0 = sin θ1 − sin θ2 2 4πε0 (0.0400 m) 4πε0 (0.0300 m) 2 9 sin θ1 9 3 5 27 ⇒ q2 = q1 = q1 = q1 = 0.843 μC. 16 sin θ2 16 4 5 64 d) F3 = q3 E x = q3 1  q1 4 q2 3  +   = 56.2 N 4πε0  0.0016 5 0.0009 5 

21.104:

(a) The four possible diagrams are:

The first diagram is the only one in which the electric field must point in the negative ydirection. b) q1 = −3.00 μC, and q2 < 0. kq1 kq 2 kq 2 kq1 5 12 5 c) E x = 0 = − ⇒ = 2 2 2 2 (0.050 m) 13 (0.120 m) 13 (0.120 m) (0.050 m) 12 E = Ey = kq1 kq 2 kq1 12 5 + = 2 2 (0.050 m) 13 (0.120 m) 13 (0.05 m) 2  12  5  5    +      13 12 13      

⇒ E = E y = 1.17 × 10 7 N C.

21.105: a) For a rod in general of length L, E = So, E left rod =

kQ  1 1  a  −  and here r = x + . L r L+r 2

 2kQ  1 kQ  1 1 1    x + a 2 − L + x + a 2  = L  2 x + a − 2 L + 2 x + a .  L     L + a 2 EQ 2kQ 2 L + a 2 b) dF = dq E ⇒ F = ∫ E dq = ∫ dx = 2 ∫ × a 2 L L a 2 1  1  −   dx  2x + a 2L + 2 x + a  2kQ 2 1 L a L a ⇒F= 2 [ln (a + 2 x)]a +2 2 − [ln( 2 L + 2 x + a )]a +2 2 L 2  ( a + L) 2    a + 2 L + a   2 L + 2a   kQ 2 kQ 2 ⇒ F = 2 1n      = 2 1n     a( a + 2 L) .  2a L L   4 L + 2a     

(

)

kQ 2  a 2 (1 + L a) 2  kQ 2 c) For a >> L : F = 2 1n  2  a (1 + 2 L a)  = L2 (2 1n (1 + 2 L a) − ln(1 + 2 L a))  L   ⇒F≈   2 L 2 L2  kQ 2   L L2 kQ 2  2 − 2 + ⋅ ⋅ ⋅  −  − 2 + ⋅ ⋅ ⋅  ⇒ F ≈ 2 .   a  L2   a 2a a a     

Chapter 22

  22.1: a) Φ = E ⋅ A = (14 N/C) (0.250 m 2 ) cos 60° = 1.75 Nm 2 C. b) As long as the sheet is flat, its shape does not matter. ci) The maximum flux occurs at an angle φ = 0° between the normal and field. cii) The minimum flux occurs at an angle φ = 90° between the normal and field. In part i), the paper is oriented to “capture” the most field lines whereas in ii) the area is oriented so that it “captures” no field lines.    ˆ 22.2: a) Φ = E ⋅ A = EA cos θ where A = An ˆ nS1 = − ˆ (left) Φ S1 = −(4 × 103 N C) (0.1 m) 2 cos (90 − 36.9°) = −24 N ⋅ m 2 C j ˆ ˆ nS2 = + k ( top)Φ S2 = −(4 × 10 3 N C) (0.1 m) 2 cos 90° = 0 ˆ nS3 = + ˆ (right) Φ S3 = +(4 × 10 3 N C) (0.1 m) 2 cos (90° − 36.9°) = +24 N ⋅ m 2 C j ˆ ˆ nS4 = −k (bottom)Φ S4 = (4 × 10 3 N C) (0.1 m) 2 cos 90° = 0 ˆ ˆ nS5 = + i (front)Φ S5 = +(4 × 10 3 N C) (0.1 m) 2 cos 36.9° = 32 N ⋅ m 2 C ˆ ˆ nS6 = −i (back)Φ S6 = −(4 × 10 3 N C) (0.1 m) 2 cos 36.9° = −32 N ⋅ m 2 C b) The total flux through the cube must be zero; any flux entering the cube must also leave it.    ˆ ˆ 22.3: a) Given that E = −Bi + Cˆ − Dk , Φ = E ⋅ A, edge length L, and j  ˆ ˆ nS1 = − ˆ ⇒ Φ1 = E ⋅ AnS1 = −CL2 . j  ˆ ˆ ˆ nS2 = + k ⇒ Φ 2 = E ⋅ AnS2 = −DL2 .  ˆ ˆ nS3 = + ˆ ⇒ Φ 3 = E ⋅ AnS3 = + CL2 . j  ˆ ˆ ˆ nS4 = − k ⇒ Φ 4 = E ⋅ AnS4 = + DL2 .  ˆ ˆ ˆ nS5 = + i ⇒ Φ 5 = E ⋅ AnS5 = − BL2 .  ˆ ˆ ˆ nS6 = − i ⇒ Φ 6 = E ⋅ AnS6 = + BL2 . b) Total flux = ∑i =1 Φ i = 0
6

22.4:

  Φ = E ⋅ A = (75.0 N C) (0.240 m 2 ) cos 70° = 6.16 Nm 2 C.

  22.5: a) Φ = E ⋅ A =

λ 2 πε0 r

(2πrl ) =

λl ε0

=

( 6.00×10 −6 C/m) (0.400 m) ε0

= 2.71 × 10 5 Nm 2 C.

b) We would get the same flux as in (a) if the cylinder’s radius was made larger—the field lines must still pass through the surface. c) If the length was increased to l = 0.800 m, the flux would increase by a factor of two: Φ = 5.42 × 105 Nm 2 C. 22.6: a) Φ S1 = q1 ε0 = (4.00 × 10 −9 C) ε 0 = 452 Nm 2 C. b) Φ S2 = q2 ε0 = (−7.80 × 10 −9 C) ε0 = −881 Nm 2 C. c) Φ S3 = (q1 + q2 ) ε0 = ((4.00 − 7.80) × 10 −9 C) ε0 = −429 Nm 2 C. d) Φ S4 = (q1 + q2 ) ε0 = ((4.00 + 2.40) × 10 −9 C) ε0 = 723 Nm 2 C. e) Φ S5 = (q1 + q2 + q3 ) ε 0 = ((4.00 − 7.80 + 2.40) × 10 −9 C) ε0 = −158 Nm 2 C. f) All that matters for Gauss’s law is the total amount of charge enclosed by the surface, not its distribution within the surface. 22.7: a) Φ = q ε0 = (−3.60 × 10 −6 C) ε0 = −4.07 × 105 Nm 2 C. b) Φ = q ε0 ⇒ q = ε0 Φ = ε0 (780 Nm 2 C) = 6.90 × 10 −9 C. c) No. All that matters is the total charge enclosed by the cube, not the details of where the charge is located. 22.8: a) No charge enclosed so Φ = 0 q − 6.00 × 10 −9 C b) Φ= 2 = = − 678 Nm 2 C. −12 2 2 ε0 8.85 × 10 C Nm c) Φ= q1 + q2 (4.00 − 6.00) × 10 −9 C = = −226 Nm 2 C. −12 2 2 ε0 8.85 × 10 C Nm

 22.9: a) Since E is uniform, the flux through a closed surface must be zero. That is:   Φ = ∫ E ⋅ dA = εq0 = ε10 ∫ ρdV = 0 ⇒ ∫ ρdV = 0. But because we can choose any volume we

want, ρ must be zero if the integral equals zero. b) If there is no charge in a region of space, that does NOT mean that the electric field is uniform. Consider a closed volume close to, but not including, a point charge. The field diverges there, but there is no charge in that region.

22.10: a) If ρ > 0 and uniform, then q inside any closed surface is greater than zero.   ⇒ Φ > 0 ⇒ ∫ E ⋅ dA > 0 and so the electric field cannot be uniform, i.e., since an

arbitrary surface of our choice encloses a non-zero amount of charge, E must depend on position. b) However, inside a small bubble of zero density within the material with density ρ , the field CAN be uniform. All that is important is that there be zero flux through the surface of the bubble (since it encloses no charge). (See Exercise 22.61.) 22.11: Φ 6sides = q ε0 = (9.60 × 10 −6 C) ε0 = 1.08 × 10 6 Nm 2 C. But the box is symmetrical, so for one side, the flux is: Φ1 side = 1.80 × 10 5 Nm 2 C . b) No change. Charge enclosed is the same. 22.12: Since the cube is empty, there is no net charge enclosed in it. The net flux, according to Gauss’s law, must be zero. 22.13: Φ E = Qencl ε0 The flux through the sphere depends only on the charge within the sphere. Qencl = ε0 Φ E = ε0 (360 N ⋅ m 2 C) = 3.19 nC 22.14: a) E (r = 0.450 m + 0.1 m) = 1 q 1 (2.50 × 10 −10 C) = = 7.44 N C . 4πε0 r 2 4πε0 (0.550 m) 2

 b) E = 0 inside of a conductor or else free charges would move under the influence of forces, violating our electrostatic assumptions (i.e., that charges aren’t moving).

1 |q| 1 q 1 (0.180 × 10 −6 C) ⇒r= = = 1.62 m. 4πε0 r 2 4πε0 E 4πε0 614 N C b) As long as we are outside the sphere, the charge enclosed is constant and the sphere acts like a point charge. 22.15: a) | E |= 22.16: a) Φ = EA = q / ε0 ⇒ q = ε 0 EA = ε0 (1.40 × 105 N C) (0.0610 m 2 ) = 7.56 × 10 −8 C. b) Double the surface area: q = ε0 (1.40 × 105 N C) (0.122 m 2 ) = 1.51 × 10 −7 C.

22.17: E =

q 1 4 πε 0 r 2

⇒ q = 4πε0 Er 2 = 4πε0 (1150 N C) (0.160 m) 2 = 3.27 × 10−9 C. So the
−9

27 10 number of electrons is: ne = 13..60 ××10 −19 C = 2.04 × 1010. C

22.18: Draw a cylindrical Gaussian surface with the line of charge as its axis. The cylinder has radius 0.400 m and is 0.0200 m long. The electric field is then 840 N/C at every point on the cylindrical surface and directed perpendicular to the surface. Thus   E ⋅ d s = ( E )( Acylinder ) = ( E )(2πrL ) ∫ = (840 N/C) (2π ) (0.400 m) (0.0200 m) = 42.2 N ⋅ m 2 /C   The field is parallel to the end caps of the cylinder, so for them ∫ E ⋅ d s = 0 . From Gauss’s law: q = ε0 Φ E = (8.854 × 10 −12 = 3.74 × 10 −10 C 22.19: C2 N ⋅ m2 ) (42.2 ) N ⋅ m2 C

E=

1 λ 2πε 0 r

22.20: a) For points outside a uniform spherical charge distribution, all the charge can be considered to be concentrated at the center of the sphere. The field outside the sphere is thus inversely proportional to the square of the distance from the center. In this case:  0.200 cm  E = (480 N C)  0.600 cm  = 53 N C    b) For points outside a long cylindrically symmetrical charge distribution, the field is identical to that of a long line of charge: λ E= , 2πε0 r that is, inversely proportional to the distance from the axis of the cylinder. In this case  0.200 cm  E = (480 N C)  0.600 cm  = 160 N C    c) The field of an infinite sheet of charge is E = σ / 2ε0 ; i.e., it is independent of the distance from the sheet. Thus in this case E = 480 N C . 22.21: Outside each sphere the electric field is the same as if all the charge of the sphere  were at its center, and the point where we are to calculate E is outside both spheres.   E1 and E 2 are both toward the sphere with negative charge. E1 = k | q1 | 1.80 × 10−6 C =k = 2.591 × 105 N C 2 2 (0.250 m) r1
2

| q2 | 3.80 × 10− 6 C E2 = k 2 = k = 5.471 × 105 N C 2 (0.250 m) r2 E = E1 + E2 = 8.06 × 105 N C , toward the negatively charged sphere.

22.22: For points outside the sphere, the field is identical to that of a point charge of the same total magnitude located at the center of the sphere. The total charge is given by charge density × volume: 4 q = (7.50 n C m 3 )( π )(0.150 m) 3 = 1.60 × 10 −10 C 3 a) The field just outside the sphere is E= q (9 × 10 9 N ⋅ m 2 /C 2 ) (1.06 × 10 −10 C) = = 42.4 N C 4πε0 r 2 (0.150 m) 2

b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will be 1/4 as strong: 10.6 N C c) Inside the sphere, only the charge inside the radius in question affects the field. In this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to the field: (9 × 10 9 N ⋅ m 2 / C 2 ) (1 / 8) (1.06 × 10 −10 C) E= = 21.2 N C (0.075 m) 2

22.23: The point is inside the sphere, so E = kQr / R 3 (Example 22.9) Q= ER 3 (950 N C) (0.220 m) 3 = = 10.2 nC kr k (0.100 m)

22.24: a) Positive charge is attracted to the inner surface of the conductor by the charge in the cavity. Its magnitude is the same as the cavity charge: qinner = + 6.00 nC, since E = 0 inside a conductor. b) On the outer surface the charge is a combination of the net charge on the conductor and the charge “left behind” when the + 6.00 nC moved to the inner surface: qtot = qinner + qouter ⇒ qouter = qtot − qinner = 5.00 nC − 6.00 nC = − 1.00 nC.

22.25: S 2 and S3 enclose no charge, so the flux is zero, and electric field outside the plates is zero. For between the plates, S1 shows that: EA = q ε0 = σ A ε0 ⇒ E = σ ε0 .

22.26: a) At a distance of 0.1 mm from the center, the sheet appears “infinite,” so:   q q 7.50 × 10 −9 C E ⋅ dA = E 2 A = ⇒ E = = = 662 N C . ∫ ε0 2ε0 A 2ε0 (0.800 m) 2 b) At a distance of 100 m from the center, the sheet looks like a point, so: 1 q 1 (7.50 × 10 −9 C) E≈ = = 6.75 × 10 −3 N C . 2 2 4πε0 r 4πε 0 (100 m) c) There would be no difference if the sheet was a conductor. The charge would automatically spread out evenly over both faces, giving it half the charge density on any as the insulator (σ :). Ec = εσ0 = 2σε:0 near one face. Unlike a conductor, the insulator is the charge density in some sense. Thus one shouldn’t think of the charge as “spreading over each face” for an insulator. Far away, they both look like points with the same charge. Q Q Q = ⇒ = σ 2πR = λ. A 2πRL L Q σ 2πRL σR b) ∫ E ⋅ d A = E (2πrL) = ε0 = ε0 ⇒ E = rε0 . λ c) But from (a), λ = σ 2π R, so E = 2 πε0r , same as an infinite line of charge. σ=

22.27: a)

22.28: All the σ ' s are absolute values. σ σ σ σ (a) at A : E A = 2 ε20 + 2 ε30 + 2 ε40 − 2 ε10 EA = = 1 (σ 2 + σ 3 + σ 4 − σ 1 ) 2ε 0 1 (5 μ C m 2 + 2 μ C m 2 + 4 μ C m 2 − 6 μ C m 2 ) 2ε 0

= 2.82 × 10 5 N C to the left. (b) EB = = σ σ1 σ σ 1 + 3 + 4 − 2 = (σ 1 + σ 3 + σ 4 − σ 2 ) 2ε 0 2ε 0 2ε 0 2ε0 2ε 0 1 (6 μ C m 2 + 2 μ C m 2 + 4 μ C m 2 − 5 μ C m 2 ) 2ε 0

= 3.95 × 10 5 N C to the left. (c) EC = = σ2 σ σ σ 1 + 3 − 4 − 1 = (σ 2 + σ 3 − σ 4 − σ1 ) 2ε0 2ε0 2ε0 2ε0 2ε0 1 (5 μ C m 2 + 2 μ C m 2 − 4 μ C m 2 − 6 μ C m 2 ) 2 ε0

= 1.69 × 105 N C to the left 22.29: a) Gauss’s law says +Q on inner surface, so E = 0 inside metal. b) The outside surface of the sphere is grounded, so no excess charge. c) Consider a Gaussian sphere with the –Q charge at its center and radius less than the inner radius of the metal. This sphere encloses net charge –Q so there is an electric field flux through it; there is electric field in the cavity. d) In an electrostatic situation E = 0 inside a conductor. A Gaussian sphere with the − Q charge at its center and radius greater than the outer radius of the metal encloses zero net charge (the − Q charge and the + Q on the inner surface of the metal) so there is no flux through it and E = 0 outside the metal. e) No, E = 0 there. Yes, the charge has been shielded by the grounded conductor. There is nothing like positive and negative mass (the gravity force is always attractive), so this cannot be done for gravity.

 ˆ ˆ 22.30: Given E = (−5.00 ( N C) ⋅ m) x i + (3.00 ( N C) ⋅ m) z k , edge length  ˆ ˆ j L = 0.300 m, L = 0.300 m, and ns1 = − ˆ ⇒ Φ1 = E ⋅ nS1 A = 0.  ˆ ˆ ˆ nS1 = + k ⇒ Φ 2 = E ⋅ nS2 A = (3.00 ( N C) ⋅ m)(0.300 m) 2 z = (0.27 ( N C)m) z = (0.27 ( N C)m)(0.300 m) = 0.081 ( N C) m 2 .  ˆ ˆ nS3 = + ˆ ⇒ Φ 3 = E ⋅ nS3 A = 0. j  ˆ ˆ ˆ nS4 = −k ⇒ Φ 4 = E ⋅ nS4 A = −(0.27 ( N C) ⋅ m) z = 0 ( z = 0).  ˆ ˆ ˆ n = + i ⇒ Φ = E ⋅ n A = (−5.00 ( N C) ⋅ m)(0.300 m) 2 x = −(0.45( N/C) ⋅ m) x
S5 5 S5

= − (0.45 ( N/C) ⋅ m)(0.300 m) = − (0.135 ( N C) ⋅ m 2 ).  ˆ ˆ ˆ nS6 = −i ⇒ Φ 6 = E ⋅ nS6 A = + (0.45 ( N C) ⋅ m) x = 0 ( x = 0). b) Total flux: Φ = Φ 2 + Φ 5 = (0.081 − 0.135) ( N C) ⋅ m 2 = −0.054 Nm 2 C q = ε0 Φ = −4.78 × 10 −13 C 22.31: a)

b) Imagine a charge q at the center of a cube of edge length 2L. Then: Φ = q / ε0 . Here the square is one 24th of the surface area of the imaginary cube, so it intercepts 1/24 of the flux. That is, Φ = q 24ε 0 . 22.32: a) Φ = EA = (125 N C)(6.0 m 2 ) = 750 N ⋅ m 2 C . b) Since the field is parallel to the surface, Φ = 0. c) Choose the Gaussian surface to equal the volume’s surface. Then: 750 – EA= q ε0 ⇒ E = 6.01m 2 (2.40 × 10−8 C ε0 + 750) = 577 N C , in the positive x -direction. Since q < 0 we must have some net flux flowing in so EA → − E A on second face. d) q < 0 but we have E pointing away from face I. This is due to an external field that does not affect the flux but affects the value of E.

22.33: To find the charge enclosed, we need the flux through the parallelepiped: Φ1 = AE1 cos 60° = (0.0500 m)(0.0600 m)(2.50 × 10 4 N C) cos 60° = 37.5 N ⋅ m 2 C Φ 2 = AE2 cos120° = (0.0500 m)(0.0600 m)(7.00 × 10 4 N C) cos 60° = − 105 N ⋅ m 2 C So the total flux is Φ = Φ1 + Φ 2 = (37.5 − 105) N ⋅ m 2 C = −67.5 N ⋅ m 2 C , and q = Φε 0 = (−67.5 N ⋅ m 2 C)ε0 = −5.97 × 10 −10 C. b) There must be a net charge (negative) in the parallelepiped since there is a net flux flowing into the surface. Also, there must be an external field or all lines would point toward the slab. 22.34:

The α particle feels no force where the net electric field is zero. The fields can cancel only in regions A and B. Eline = E sheet λ σ = 2πε0 r 2ε0 50 µC/m r = λ πσ = = 0.16m = 16cm π (100 µC/m 2 ) The fields cancel 16 cm from the line in regions A and B.

22.35:

 field E 2 of the sphere must be away from the sheet. This is true above the center of the sphere. Let r be the distance above the center of the sphere for the point where the electric field is zero. σ 1 Q2 r E1 = E2 so 1 = 2ε0 4πε0 R 3 r= 2πσ1R 3 2π (8.00 × 10−9 C/m 2 )(0.120 m)3 = = 0.097 m Q2 0.900 × 10− 9 C

 The electric field E1 of the sheet of charge is toward the sheet, so the electric

22.36: a) For r < a, E = 0, since no charge is enclosed. For a < r < b, E = For r > c, E = b)
q 1 4 πε 0 r 2

, since there is +q inside a radius r.

For b < r < c, E = 0, since now the –q cancels the inner +q.
q 1 4 πε 0 r 2

, since again the total charge enclosed is +q.

c) Charge on inner shell surface is –q. d) Charge on outer shell surface is +q. e)

22.37: a) r < R, E = 0, since no charge is enclosed. b) R < r < 2 R, E = charge enclosed is 2Q.
1 Q 4 πε 0 r 2

, since charge enclosed is Q. r > 2 R, E =

1 2Q 4 πε 0 r 2

, since

22.38: a) r < a, E =

1 Q 4 πε 0 r 2

, since the charge enclosed is Q.

a < r < b, E = 0, since the –Q on the inner surface of the shell cancels the +Q at the center of the sphere. 1 r > b, E = − 4 πε0 2Q , since the total enclosed charge is –2Q. r2
Q b) The surface charge density on inner surface: σ = − 4πa 2 .

c) The surface charge density on the outer surface: σ = − 42Q2 . πb d)

e)

22.39: a)(i) r < a, E = 0, since Q = 0 (ii) a < r < b, E = 0, since Q = 0. (iii) b < r < c, E = (v) r > d , E =
1 2q 4 πε 0 r 2

, since Q = + 2q.

(iv) c < r < d , E = 0, since Q = 0.
1 6q 4 πε 0 r 2

, since Q = + 6q.

b)(i) small shell inner: (ii) small shell outer: (iii) large shell inner: (iv) large shell outer:

Q=0 Q = + 2q Q = − 2q Q = + 6q

22.40: a)(i) r < a, E = 0, since the charge enclosed is zero. (ii) a < r < b, E = 0, since the charge enclosed is zero. (iii) b < r < c, E =
1 2q 4 πε 0 r 2

, since charge enclosed is + 2q.

(iv) c < r < d , E = 0, since the net charge enclosed is zero. (v) r > d , E = 0, since the net charge enclosed is zero.

b)(i) small shell inner: (ii) small shell outer: (iii) large shell inner: (iv) large shell outer:

Q=0 Q = + 2q Q = − 2q Q=0

22.41: a)(i) r < a, E = 0, since charge enclosed is zero. (ii) a < r < b, E = 0, since charge enclosed is zero. (iii) b < r < c, E =
1 2q 4 πε 0 r 2

, since charge enclosed is + 2q.

(iv) c < r < d , E = 0, since charge enclosed is zero.
1 (v) r > d , E = − 4πε0 2q r2

, since charge enclosed is − 2q.

b)(i) small shell inner: (ii) small shell outer: (iii) large shell inner: (iv) large shell outer: 22.42: a) We need:

Q=0 Q = + 2q Q = − 2q Q = − 2q

4π ρ − 28π ρR 3 3Q ((2 R )3 − R 3 ) ⇒ Q = ⇒ ρ=− . 3 3 28πR 3 b) r < R, E = 0 and r > 2 R, E = 0, since the net charges are zero. Q 4π ρ 3 Q ρ R < r < 2 R, Φ = E (4πr 2 ) = + (r − R 3 ) ⇒ E = + (r 3 − R 3 ). 2 2 ε 0 3ε0 4πε 0 r 3ε 0 r −Q = Substituting ρ from (a) E =
2 Q 7 πε 0 r 2 Qr − 28 πε R 3 .
0

c) We see a discontinuity in going from the conducting sphere to the insulator due to the thin surface charge of the conducting sphere—but we see a smooth transition from the uniform insulator to the outside.

22.43: a) The sphere acts as a point charge on an external charge, so: 1 F = qE = 4 πε0 qQ , radially inward. r2 (b) If the point charge was inside the sphere (where there is no electric field) it would feel zero force. +q q 3q  1  = 4 3 4 3 =   Vb − Va 3 πb − 3 πa 4π  b3 − a 3  −q −q − 3q  1  = 4 3 4 3 =   Vd − Vc 3 πd − 3 πc 4π  d 3 − c 3 

22.44: a) ρinner =

ρouter =

  b) (i) r < a ∫ E ⋅ dA = 0 ⇒ E = 0.   1 (ii) a < r < b ∫ E ⋅ dA = ε0

∫ρ

inner

dV ⇒ E 4π r 2 =

4 π (r 3 − a 3 ) ρ inner 3ε0

1 (r 3 − q 3 ) q (r 3 − a 3 ) E= ρ inner = 3ε0 r2 4πε0 (b 3 − a 3 )   q q q (iii) b < r < c ∫ E ⋅ dA = ⇒ E 4πr 2 = ⇒ E = ε0 ε0 4πε0 r 2   q 1 (iv) c < r < d ∫ E ⋅ dA = + ∫ ρouter dV ⇒ ε0 ε0 q 4π 3 q q(r 3 − c 3 ) 3 E 4π r = + (r − c ) ρouter, so E = − ε0 3ε0 4πε0 r 2 4πε0 r 2 (d 3 − c 3 )  q q  (v) r > d ∫ E ⋅ d A = − = 0 ⇒ E = 0 ε0 ε0
2

22.45: a) a < r < b, E = b) r > c, E = same as in part (a). c)
1 4 πε0

1 2λ , radially outward, as in 22.48 (b). 4πε 0 r 2λ r , radially outward, since again the charge enclosed is the

d) The inner and outer surfaces of the outer cylinder must have the same amount of charge on them: λl = −λ inner l ⇒ λ inner = −λ, and λ outer = λ. q αl α = ⇒E= . ε0 ε0 2πε0 r (ii) a < r < b, there is no net charge enclosed, so the electric field is zero. (iii) r > b, E ( 2πrl ) = q 2αl α = ⇒E= . ε0 ε0 πε0 r

22.46: a) (i) r < a, E (2πrl ) =

b) (i) Inner charge per unit length is − α. (ii) Outer charge per length is + 2α .

22.47: a) (i) r < a, E (2πrl ) =

q ε0

=

αl ε0

⇒E=

α 2 πε 0 r

(ii) a < r < b, there is not net charge enclosed, so the electric field is zero. (iii) r > b, there is no net charge enclosed, so the electric field is zero.

b) (i) Inner charge per unit length is − α . (ii) Outer charge per length is ZERO.
ρπr 2 l ε0

22.48: a) r < R, E (2πrl ) =

q ε0

=

⇒E=

ρr 2ε 0

ρπR 2l ε0

b) r > R, and λ = ρπR 2 , E (2πrl ) = d)

q ε0

=

⇒E=

ρR 2 2 ε0 r ρR 2ε 0

=

λ 2 πε0 r

=

2 kλ r

.

c) r = R. the electric field for BOTH regions is E =

, so they are consistent.

22.49: a) The conductor has the surface charge density on BOTH sides, so it has twice the enclosed charge and twice the electric field. b) We have a conductor with surface charge density σ on both sides. Thus the electric field outside the plate is Φ = E (2 A) = (2σA) ε0 ⇒ E = σ ε0 . To find the field inside the conductor use a Gaussian surface that has one face inside the conductor, and one outside. Then: Φ = Eout A + Ein A = (σA) ε0 but Eout = σ ε 0 ⇒ Ein A = 0 ⇒ Ein = 0.

22.50: a) If the nucleus is a uniform positively charged sphere, it is only at its very center where forces on a charge would balance or cancel   q e b) Φ = ∫ E ⋅ dA = ⇒ E 4πr 2 = ε0 ε0  r3  er  3⇒ E = R  4πε0 R 3  

1 e2r ⇒ F = qE = − . 4πε0 R 3 So from the simple harmonic motion equation: F = −mω2 r = − 1 e2r ⇒ω= 4πε0 R 3 1 e2 1 ⇒ f = 3 4πε0 mR 2π 1 e2 . 4πε0 mR 3

c) If f = 4.57 × 1014 Hz =

1 1 e2 2π 4πε0 mR 3

⇒R=3

1 (1.60 × 10 −19 C) 2 = 3.13 × 10 −10 m. 2 −31 14 2 4πε 0 4π (9.11 × 10 kg )(4.57 × 10 Hz)

ractual r Thompson ≈ 1 d) If r > R then the electron would still oscillate but not undergo simple harmonic motion, because for r > R, F ∝ 1 r 2 , and is not linear. 22.51: The electrons are separated by a distance 2d , and the amount of the positive nucleus’s charge that is within radius d is all that exerts a force on the electron. So: ke 2 Fe = = Fnucleus = 2ke 2 Rd3 ⇒ d 3 = R 3 / 8 ⇒ d = R / 2. 2 ( 2d )

22.52: a) Q(r ) = Q − ∫ ρdV = Q − ⇒ Q(r ) = Q −

Q 3 πa0

∫∫∫e

− 2 r / a0

r 2 dr sin θ dθ dφ = Q −

4Q r 2 −2 x / a0 x e dx 3 a0 ∫0

4Qe −α r (2eα r − α 2 r 2 − 2α r − 2) = Qe −2 r / a0 [2(r / a0 ) 2 + 2(r / a0 ) + 1]. 3 3 a0 α Note if r → ∞, Q(r ) → 0. b) The electric field is radially outward, and has magnitude: ⇒E= kQe −2 r / a0 (2(r a0 ) 2 + 2(r a0 ) + 1). r2

22.53: a) At r = 2 R, F = qe E =

1 qe qFe 4 πε0 4 R 2

=

−19 2 1 ( 82 )(1.6 × 10 C) 4 πε0 4 ( 7.1 × 10 −15 m) 2

= 94 N.

So: a = F m = 94 N 9 .11 × 10−31 kg = 1.0 × 1032 m/s 2 . b) At r = R, a = 4a(a) = 4.1 × 1032 m/s 2 . c) At r = R 2 , Q = 1 (82e) ( 1 because the charge enclosed goes like r 3 ) so with the 8 8 radius decreasing by 2, the acceleration from the change in radius goes up by (2) 2 = 4, 4 but the charge decreased by 8, so a = 8 a( b ) = 2.1 × 1032 m/s 2 . d) At r = 0, Q = 0, so F = 0.

22.54: a) The electric field of the slab must be zero by symmetry. There is no preferred direction in the y -z plane, so the electric field can only point in the x -direction. But at the origin in the x -direction, neither the positive nor negative directions should be singled out as special, and so the field must be zero. b) Use a Gaussian surface that has one face of area A on in the y -z plane at x = 0, and the other face at a general value x. Then: Q ρAx ρx x ≤ d : Φ = EA = encl = ⇒E= , ε0 ε0 ε0 ˆ with direction given by x i .
| x|

Note that E is zero at x = 0. Now outside the slab, the enclosed charge is constant with x : Q ρAd ρd x ≥ d : Φ = EA = encl = ⇒E= , ε0 ε0 ε0 ˆ again with direction given by x i .
| x|

22.55: a) Again, E is zero at x = 0 , by symmetry arguments. b) x ≤ d : Φ = EA = x ≥ d : Φ = EA = Qencl ρ A = 0 2 ε0 ε0 d
d 2 ∫ x' dx' = 0 2 x

ρ0 Ax 3 ρ x3 x ˆ ⇒ E = 0 2 , in i direction. 2 3ε0 d 3ε0 d |x|

Qencl ρ A = 0 2 ε0 ε0 d

∫ x'
0

dx ' =

ρ0 Ad ρd x ˆ ⇒ E = o , in i direction. 3ε0 3ε0 |x|

22.56: a) We could place two charges + Q on either side of the charge + q :

b) In order for the charge to be stable, the electric field in a neighborhood around it must always point back to the equilibrium position. c) If q is moved to infinity and we require there to be an electric field always pointing in to the region where q had been, we could draw a small Gaussian surface there. We would find that we need a negative flux into the surface. That is, there has to be a negative charge in that region. However, there is none, and so we cannot get such a stable equilibrium. d) For a negative charge to be in stable equilibrium, we need the electric field to always point away from the charge position. The argument in (c) carries through again, this time inferring that a positive charge must be in the space where the negative charge was if stable equilibrium is to be attained.

22.57: a) The total charge: q = 4π

∫
0 3

R

ρ 0 (1 − r / R )r 2 dr = 4π [ ∫ r 2 dr − ∫ r 3 / Rdr ]
0 0

R

R

4πR 3 ρ0 4πR 3 3Q ⇒ q = 4πρ0 [ R 3 − R 4] = = = Q. 12 12 πR 3
3

b) r ≥ R, all the charge Q is enclosed, and: Φ = E (4πr 2 ) = Q / ε 0 ⇒ E = the same as a point charge. c) r ≤ R, then Q(r) = q( r 3 R 3 ). Also, Q(r ) = 4π ∫ ρ0 (1 − r R )r 2 dr =4πρ0 ⇒ E (r ) = d)

1 Q 4 πε0 r 2

,

(

r3 3

−

r4 4R

).

12kQ  1 r 3 1 r 4  kQ  4r 3 3r 4  r  r   = 2  3 − 4  = kQ 3  4 − 3  − 2  3 4  R  r 3 R 4R  r  R  R  R

e)

∂E 4kQ 6kQ 2 2 kQ 4kQ = 0 (r ≤ R ) ⇒ 3 − 4 r = 0 ⇒ rmax = R. So Emax = ( 4 − 2) = 2 ∂r R R 3 3R 3R 2

22.58: a)
∞ R R 2 4r  2 4 R 3   Q = 4π ∫ ρ(r )r 2 dr = 4πρ0 ∫ 1 − r dr   r dr = 4πρ0  ∫ r dr − 3R  3R ∫0 0 0  0   R3 4 R 4  = 4πρ0  − ⋅ ⇒Q=0  3 3R 4    q b) r ≥ R, ∫ E ⋅ dA = encl = 0 ⇒ E = 0 ε0

  4π c) r ≤ R, ∫ E ⋅ dA = ε0

∫

r

0

ρ(r ′)r ′ 2 dr ′ ⇒ E 4πr 2 =

4πρ0  r 2 4  ∫0 r ′ dr ′ − 3R ε0  r  1 − R   

∫

r

0

 r ′3 dr ′ 

ρ0 1  r 3 r 4  ρ0 ⇒E= − = r ε0 r 2  3 3R  3ε0   d)

e)

ρ 2ρ r ∂E R = 0 ⇒ 0 − 0 max = 0 rmax = ∂r 3ε0 3ε 0 R 2 R ρ R   E r −  = 0 1− 2  3ε0 2    1  ρ0 R = 2  12ε0 

  r 2sin θ dr dθ dφ 22.59: a) Φ g = ∫ g ⋅ dA = −Gm ∫ = −4πGm. r2 b) For any closed surface, mass OUTSIDE the surface contributes zero to the flux passing through the surface. Thus the formula above holds for any situation where m is the mass enclosed by the Gaussian surface.   That is: Φ g = ∫ g ⋅ dA = − 4πGM encl .

GM , which is the same as for a point mass. r2 b) Inside a hollow shell, the M encl = 0, so g = 0. c) Inside a uniform spherical mass:  r3  GMr Φ g = g 4πr 2 = − 4πGM encl = − 4πG  M 3  ⇒ g = − 3 ,  R  R   which is linear in r. 22.60: a) Φ g = g 4πr 2 = −4πGM ⇒ g = − 22.61: a) For a sphere NOT at the coordinate origin:    Q ρ 4πr ′3 ρr ′ r ′ = r − b ⇒ Φ = 4πr ′ 2 E = encl = ⇒E= , ε0 ε0 3 3ε0 ˆ in the r ′ - direction.    ρ(r − b ) ⇒E= . 3ε0

b) The electric field inside a hole in a charged insulating sphere is:        ρr ρ ( r − b ) ρb E hole = Esphere − E(a) = − = . 3ε0 3ε0 3ε0  Note that E is uniform. 22.62: Using the technique of 22.61, we first find the field of a cylinder off-axis, then the electric field in a hole in a cylinder is the difference between two electric fields—that of a solid cylinder on-axis, and one off-axis.    ρ(r − b )    Qencl ρ ρr ′ 2 r ′ = r − b ⇒ Φ = 2πr ′l E = = lπ r ′ ⇒ E = ⇒E= . ε0 ε0 2ε 0 2ε 0        ρ r ρ(r − b ) ρ b E hole = E cylinder − Eabove = − = . Note that E is uniform. 2 ε0 2 ε0 2 ε0

22.63: a) x = 0 : no field contribution from the sphere centered at the origin, and the other sphere produces a point-like field:  1 Q ˆ 1 Q ˆ E ( x = 0) = − i =− i. 2 4πε0 (2 R ) 4πε0 4 R 2 b) x = R 2 : the sphere at the origin provides the field of a point charge of charge q = Q 8 since only one-eighth of the charge’s volume is included. So:  ˆ 1  (Q 8) Q 1 Q 1 Q ˆ ˆ  E ( x = R 2) =  ( R 2) 2 − (3 R 2) 2  i = 4πε R 2 (1 / 2 − 4 / 9)i = 4πε 18 R 2 i .  4πε0   0 0  c) x = R : the two electric fields cancel, so E = 0. d) x = 3R : now both spheres contribute fields pointing to the right:  1  Q Qˆ 1 10Q ˆ  E ( x = 3R ) =  (3R) 2 + R 2  i = 4πε 9 R 2 i .  4πε0   0 22.64: (See Problem 22.63 with Q → − Q for terms associated with right sphere)  1 Q ˆ a) E ( x = 0) = + i 4πε0 4 R 2  ˆ R 1  (Q 8) Q 1  Q 4Q  ˆ 1 17Q ˆ b) E  x =  =  ( R 2) 2 + (3 R 2) 2  i = 4πε  2 R 2 + 9 R 2  i = 4πε 18 R 2 i 2  4πε0     0  0  1 Q Qˆ Q ˆ c) E ( x = R) =  R 2 + R 2  i = 2πε R 2 i 4πε0   0  1  Q Qˆ 1  Q Q  ˆ − 1 8Q ˆ d) E ( x = 3R ) =  (3R ) 2 − R 2  i = 4πε  9 R 2 − R 2  i = 4πε 9 R 2 i 4πε0    0  0

22.65: a) The charge enclosed: R 4π ( R 2)3 απR 3 Q = Qi + Q0 , where Qi = α = , and Q0 = 4π (2α ) ∫ (r 2 − r 3 / R )dr R/2 3 6 3 3 4 4 3  ( R − R 8) ( R − R 16)  11απR = = 8απ  −   3 4R 24   ⇒Q= 15απR 3 8Q ⇒α= . 24 5πR 3 α 4πr 3 αr 8Qr b) r ≤ R 2 : Φ = E 4πr 2 = ⇒E= = . 3ε0 3ε0 15πε0 R 3
3 4 4  3  Qi 1   8απ  (r − R 8) − (r − R 16)   R 2 ≤ r ≤ R : Φ = E 4πr = +    ε0 ε0  3 4R   3 απR kQ ⇒E= (64(r R ) 3 − 48(r R ) 4 − 1) = (64(r R ) 3 − 48(r R) 4 − 1). 2 2 24ε0 (4πr ) 15r 2

r ≥ R:E = c)

Q , since all charge is enclosed. 4πε0 r 2 Qi (4Q / 15) 4 = = = 0.267. Q Q 15
0

d) r ≤ R / 2 : F = − eE = − 158eQR 3 r , so the restoring force depends upon displacement to πε the first power, and we have simple harmonic motion. 8eQ k 8eQ 2π 15πε0 R 3me ,ω= = ,T = = 2π . 15πε0 R 3 me 15πε0 R 3me ω 8eQ f) If the amplitude of oscillation is greater than R / 2, the force is no longer linear in r , and is thus no longer simple harmonic. e) F = − kr , k =

22.66: a) Charge enclosed: Q = Qi + Q0 where Qi = 4π and Q0 = 4πα ∫
R R/2

3αr 3 6πα 1 R 4 3 dr = = παR 3 . ∫0 2R R 4 16 32 31  47  7 (1 − (r R ) 2 )r 2 dr = 4παR 3  − παR 3 . =  24 160  120
R/2

47  233 480Q  3 3 Therefore, Q =  + παR 3 ⇒ α = .  παR = 480 233πR 3  32 120  4π r 3αr ′3 3παr 4 6αr 2 180Qr 2 b) r ≤ R 2 : Φ = E 4πr 2 = dr ′ = ⇒E= = . ε0 ∫0 2 R 2ε 0 R 16ε0 R 233πε0 R 4 Q 4πα r 2 2 R 2 ≤ r ≤ R : Φ = E 4πr 2 = i + ∫R / 2 (1 − (r ′ / R) )r ′ dr ′ ε0 ε0  r 3 R3 r5 R3  3 4παR 3  − = − +  3 24 5 R 2 160  128 ε   0 3 5  4παR 3  1  r     − 1  r  − 17  +   ε0  3  R  5 R 480    3 5  480Q  1  r     − 1  r  − 23  . ⇒E=   233πε 0 r 2  3  R  5  R  1920    Q r ≥ R:E = , since all charge is enclosed. 4πε0 r 2 c) The fraction of Q between R 2 ≤ r ≤ R : Qo 47 480 = = 0.807. Q 120 233 = Qi 4πα + ε0 ε0 d) E (r = R / 2) = 180 233
Q 4 πε 0 R 2

, using either of the electric field expressions above,

evaluated at r = R / 2. e) The force an electron would feel never is proportional to − r which is necessary for simple harmonic oscillations. It is oscillatory since the force is always attractive, but it has the wrong power of r to be simple harmonic.

Chapter 23

23.1:

 1 1 1   1 ∆U = kq1q2  −  = k (2.40 μC)( − 4.30 μC) −  = 0.357J r r   0.354m 0.150m   2 1 ⇒ W = − ∆U = − 0.357 J.

23.2: W = − 1.9 × 10 −8 J = − ∆U = U i − U f ⇒ U f = 1.9 × 10 −8 J + 5.4 × 10 −8 J =

7.3 × 10−8 J
23.3: a) 1 k (2.80 × 10 −6 C)(7.50 × 10 −6 C) (0.0015 kg )(22.0 m s) 2 + = 0.608 J 2 0.800 m

Ei = K i + U i = Ei = E f =

1 2 kq1q2 2(0.608 J − 0.491 J) mv f + ⇒ vf = = 12.5 m s . 2 rf 0.0015 kg b) At the closest point, the velocity is zero: kq q k (2.80 × 10 −6 C)(7.80 × 10 −6 C) ⇒ 0.608 J = 1 2 ⇒ r = = 0.323 m. r 0.608 J

23.4: U = − 0.400 J =

kq1 q 2 − k (2.30 × 10 −6 C)(7.20 × 10 −6 C) ⇒r= = 0.373 m. r − 0.400 J

23.5: a) U =

kQq k (4.60 × 10 −6 C) (1.20 × 10 −6 C) = = 0.199 J. r 0.250 m b) (i) K f = K i + U i − U f  1 1  = 0 J + k (4.60 × 10 −6 C) (1.20 × 10 −6 C) −  0.25 m 0.5 m  = 0.0994 J    ⇒ K f = 0.0994 J = 1 2 mv f ⇒ v f = 2 2(0.0994 J) = 26.6 m s. 2.80 × 10 −4 kg

(ii ) K f = 0.189 J, v f = 36.7 m s. (iii ) K f = 0.198 J, v f = 37.6 m s.

23.6: U =

kq 2 2kq 2 + = 6kq 2 = 6k (1.2 × 10 −6 C) 2 = 0.078 J. 0.500m 0.500 m

23.7: a) qq qq U = k 1 2 + 1 2  r r13  12  (4.00 nC)(−3.00 nC) (4.00 nC)(2.00 nC) +  (0.200 m) (0.100m) q1q2   =k +  (−3.00 nC)(2.00 nC) r23   +  (0.100 m)  = − 3.60 × 10− 7 J.      

q q qq q q  b) If U = 0, 0 = k  1 2 + 1 3 + 2 3 . So solving for x we find:  r x r12 − x   12  8 6 0 = − 60 + − ⇒ 60 x 2 − 26 x + 1.6 = 0 ⇒ x = 0.074 m, 0.360 m. Therefore x 0.2 − x x = 0.074 m since it is the only value between the two charges.

23.8: From Example 23.1, the initial energy Ei can be calculated: 1 Ei = K i + U i = (9.11 × 10−31 kg )(3.00 × 106 m s) 2 2 k ( − 1.60 × 10−19 C)(3.20 × 10−19 C) + 10−10 m ⇒ Ei = − 5.09 × 10−19 J. When velocity equals zero, all energy is electric potential energy, so: k 2e 2 − 5.09 × 10 −19 J = − ⇒ r = 9.06 × 10 −10 m. r 23.9: Since the work done is zero, the sum of the work to bring in the two equal charges q must equal the work done in bringing in charge Q. kq 2 2kqQ q Wqq = WqQ ⇒ − = ⇒ Q=− . d d 2

23.10: The work is the potential energy of the combination. U = U p ∝ + U pe + U e ∝ = = = ke( 2e) ke( − e) k ( − e)(2e) + + −10 −10 −10 5 2 × 10 m 5 × 10 m 5 × 10 m ke 2  2  − 1 − 2  −10 5 × 10 m  2 

(9.0 × 109 Nm 2 C 2 ) (1.6 × 10−19 C) 2  2  − 3  −10 5 × 10 m  2  −19 = − 7.31 × 10 J Since U is negative, we want do + 7.31 × 10 −19 J to separate the particles

23.11: K1 + U1 = K 2 + U 2 ; K1 = U 2 = 0 so K 2 = U1 U1 = e2 4πε0 1 5e 2 1 2 2 , with r = 8.00 × 10−10 m  + + =  r r r  4πε0 r

U1 = 1.44 × 10 −18 J = 9.00 eV 1 1 ke 2 23.12: Get closest distance γ. Energy conservation: mv 2 + mv 2 = 2 2 γ (9 × 10 9 Nm 2 C 2 )(1.6 × 10 −19 C) 2 ke 2 = = 1.38 × 10 −13 m mv 2 (1.67 × 10 −27 kg )(10 6 m s) Maximum force: ke 2 F= 2 γ γ= = (9 × 109 Nm 2 C 2 ) (1.6 × 10 −19 C) 2 (1.38 × 10−13 m) 2

= 0.012 N 23.13: K A + U A = K B + U B U = qV , so K A + qV A = K B + qVB K B = K A + q(V A − VB ) = 0.00250 J + ( − 5.00 × 10 −6 C) (200 V − 800 V) = 0.00550 J v B = 2 K B m = 7.42 m s It is faster at B; a negative charge gains speed when it moves to higher potential.

23.14: Taking the origin at the center of the square, the symmetry means that the potential is the same at the two corners not occupied by the + 5.00 μC charges (The work done in moving to either corner from infinity is the same). But this also means that no net work is done is moving from one corner to the other.

 23.15: E points from high potential to low potential, so VB > V A and VC < V A . The force on a positive test charge is east, so no work is done on it by the electric force when it moves due south (the force and displacement are perpendicular); V D = V A .
23.16: a) W = − ∆U = qEd = ∆K = 1.50 × 10 −6 J. b) The initial point was at a higher potential than the latter since any positive charge, when free to move, will move from greater to lesser potential. ∆V = ∆ U q = (1.50 × 10−6 J) (4.20 nC) = 357 V. c) qEd = 1.50 × 10
−6

1.50 × 10−6 J J⇒E= = 5.95 × 103 N C. (4.20 nC)(0.06 m)

23.17: a) Work done is zero since the motion is along an equipotential, perpendicular to the electric field. V  b) W = qEd = (28.0 nC)  4.00 × 104 (0.670 m) = 7.5 × 10− 4 J m  V  c) W = qEd = (28.0 nC)  4.00 × 104 ( − 2.60 cos 45°) = − 2.06 × 10− 3 J m 

23.18: Initial energy equals final energy: Ei = E f ⇒ − keq1 keq2 keq1 keq2 1 2 − =− − + me v f r1i r2i r1 f r2 f 2

 (3.00 × 10 −9 C) ( 2.00 × 10 −9 C)   = − 2.88 × 10 −17 J Ei = k ( − 1.60 × 10 −19 C)  +   0.25 m 0.25 m   −9 −9  (3.00 × 10 C) ( 2.00 × 10 C  1  + me v 2 E f = k ( − 1.60 × 10 −19 C)  + f   2 0.10 m 0.40 m   1 2 = − 5.04 × 10 −17 J + me v f 2 2 ⇒ vf = (5.04 × 10 −17 J − 2.88 × 10 −17 J) −31 9.11 × 10 kg = 6.89 × 10 6 m s. kq kq k (2.50 × 10 −11 C) ⇒r= = = 2.5 × 10 −3 m. r V 90.0 V

23.19: a) V = b) V =

kq kq k (2.50 × 10 −11 C) ⇒r= = = 7.5 × 10 −3 m. r V 30.0 V

23.20: a) V =

kq rV (0.250 m) (48.0 V) ⇒q= = = 1.33 × 10 −9 C. r k k −9 k (1.33 × 10 C) b) V = = 16 V (0.750 m)

 2.40 × 10 −9 C − 6.50 × 10 −9 C  q q   = − 738 V. 23.21: a) At A : V A = k  1 + 2  = k  + r   r2  0.05 m 0.05 m  1    q  2.40 × 10 −9 C − 6.50 × 10 −9 C  q   = − 705 V. b) At B : V B = k  1 + 2  = k  + r   r2  0.08 m 0.06 m  1    c) W = q∆V = (2.50 × 10 −9 C)( − 33 V) = − 8.25 × 10 −8 J. The negative sign indicates that the work is done on the charge. So the work done by the field is 8.25 × 10 −8 J.

23.22: a)

b) V = 2

1 q . 4πε0 a 1 q 1 =2 4πε0 r 4πε0 q a2 + x2

c) Looking at the diagram in (a): V ( x ) = 2 d)

e) When x >> a, V = 23.23: a)

1 2q , just like a point charge of charge + 2q. 4πε0 x

kq k ( − q) + = 0. r r c) The potential along the x-axis is always zero, so a graph would be flat. d) If the two charges are interchanged, then the results of (b) and (c) still hold. The potential is zero b) V x =

23.24: a) | y | < a : V =

kq kq 2kqy − = 2 . (a + y ) (a − y ) y − a 2 kq kq − 2kqa y > a :V = − = 2 . (a + y ) y − a y − a 2 − kq kq 2kqa y < − a :V = − = 2 . (a + y ) (− y + a ) y − a 2

 −q q  Note: This can also be written as V = k  | y − a | + | y + a |    b)

kq kq − 2kqa − = . (a + y ) ( y − a ) y2 d) If the charges are interchanged, then the potential is of the opposite sign. c) y >> a : V =

23.25: a)

b)

− kq ( x + a ) kq 2kq − = . x x−a x( x − a ) kq (3 x − a) kq 2kq 0 < x < a :V = − = . x a−x x( x − a ) − kq kq( x + a) 2kq x < 0 :V = + = . x x−a x( x − a ) x > a :V = Note: This can be also be written as V = k ( |q| − x
2q | x − a|

)

c) The potential is zero at x = − a and a / 3. d)

− kqx − kq = , which is the same as the potential of a point charge x x2 –q. (Note: The two charges must be added with the correct sign.) e) For x >> a : V ≈

23.26:a) V =

 1  kq 2kq 2 . − = kq  − 2 2  | y | | y| r a +y   2 2 a +y a ⇒ 3y2 = a2 ⇒ y = ± . b) V = 0, when y 2 = 4 3 c)

1 2 kq d) y >> a : V ≈ kq −  = − , which is the potential of a point charge − q .  y y y   23.27: W = − ∆U = − Vq = (295 V) (1.60 × 10−19 C) = 4.72 × 10−17 J. But also: 1 W = ∆K = mv 2 ⇒ v = 2 2(4.72 × 10 −17 J) = 1.01 × 10 7 m s. −31 9.11 × 10 kg

23.28:

V V 4.98 V ⇒d = = = 0.415 m. d E 12.0 N C kq Vd (4.98 V) (0.415 m) b) V = ⇒q= = = 2.30 × 10−10 C. d k k c) The electric field is directed away from q since it is a positive charge. a) E =

23.29: a) Point b has a higher potential since it is “upstream” from where the positive charge moves. Va − Vb = E (b − a ) = − | E | (b − a ) ⇒ Vb − Va = | E | (b − a ) > 0 V 240 V b) E = = = 800 N C. d 0.3 m c) W = − ∆U = − q∆V = − (−0.20 × 10−6 C)( − 240 V) = − 4.8 × 10−5 J.

23.30:(a) V = VQ + V2Q > 0, so V is zero nowhere except for infinitely far from the charges.

The fields can cancel only between the charges kQ k (2Q) EQ = E 2Q → 2 = → (d − x) 2 = 2 x 2 2 x (d − x) d d x = 1+ 2 . The other root, x = 1− 2 , does not lie between the charges. (b) V can be zero in 2 places, A and B. at A : at B : k ( − Q) k (2Q) + =0→ x=d 3 x d−x

k ( − Q) k (2Q) + =0→ y=d y d+y EQ = E2Q to the left of − Q. kQ k (2Q) = →x= 2 x (d + x) 2 d 2 −1

(c) Note that E and V are not zero at the same places. 23.31: a) K1 + qV1 = K 2 + qV2 q(V1 − V2 ) = K 2 − K1; q = − 1.602 × 10 −19 C 2 K1 = 1 me v12 = 4.099 × 10−18 J; K 2 = 1 mev2 = 2.915 × 10−17 J 2 2 K 2 − K1 = − 156 V q The electron gains kinetic energy when it moves to higher potential. b) Now K 1 = 2.915 × 10 −17 J, K 2 = 0 K − K1 V1 − V2 = 2 = + 182 V q The electron loses kinetic energy when it moves to lower potential V1 − V2 =

kq k (3.50 × 10 −9 C) 23.32: a) V = = = 65.6 V. r 0.48 m k (3.50 × 10−9 C) = 131.3 V 0.240 m c) Since the sphere is metal, its interior is an equipotential, and so the potential inside is 131.3 V. b) V = 23.33: a) The electron will exhibit simple harmonic motion for x << a, but will otherwise oscillate between ± 30.0 cm. b) From Example 23.11, 1  kQ 1  V = ⇒ ∆V = kQ − 2 2 2 2  a x +a x +a    1 1 ⇒ ΔV = k (24.0 × 10 −9 C)  −  0.150 m (0.300 m) 2 + (0.150 m) 2  = 796 V −19 1 (796 But W = − q∆V = mv 2 ⇒ v = 2(1.60 × 10 10 −C) kg V) = 1.67 × 107 m s. 31 9.11 × 2 23.34: Energy is conserved: 1 2 (1.67 × 10−27 kg) (1500 m s) 2 mv = q∆V ⇒ ∆V = = 0.0117 V. 2 2(1.60 × 10−19 C) But: λ  2πε0 ∆V   2πε0 ∆V  ∆V = ln( r0 r ) ⇒ r0 = r exp   ⇒ r = r0 exp −  2πε0 λ  λ     2πε0 (0.0117 V)  ⇒ r = (0.180 m)exp  −  5.00 × 10 −12 C / m  = 0.158 m.    V 360 V = = 8000 N C. d 0.0450 m    

23.35: a) E =

b) F = Eq = (8000 N C) (2.40 × 10−9 C) = 1.92 × 10−5 N. c) W = Fd = (1.92 × 10 −5 N) (0.0450 m) = 8.64 × 10 −7 J. d) ∆U = ∆Vq = ( − 360 V) (2.40 × 10−9 C) = − 8.64 × 10−7 J.

23.36: a) V = Ed = (480 N C) (3.8 × 10 −2 m) = 18.2 V. b) The higher potential is at the positive sheet. σ c) E = ⇒ σ = ε0 (480 N C) = 4.25 × 10 − 9 C m 2 . ε0 V V 4750 V ⇒d = = = 1.58 × 10 −3 m. 6 d E 3.00 × 10 V m σ b) E = ⇒ σ = ε0 (3.00 × 106 V m) = 2.66 × 10− 5 C m 2 . ε0

23.37:a) E =

σ 47.0 × 10−9 C m 2 23.38: a) E = = = 5311 N C. ε0 ε0 b) V = Ed = (5311 N / C) (0.0220 m) = 117 V. c) The electric field stays the same if the separation of the plates doubles, while the potential between the plates doubles. 23.39: a) The electric field outside the shell is the same as for a point charge at the center of the shell, so the potential outside the shell is the same as for a point charge: q V= for r > R. 4πε0 r The electric field is zero inside the shell, so no work is done on a test charge as it moves inside the shell and all points inside the shell are at the same potential as the q surface of the shell: V = for r ≤ R. 4πε0 R kq RV (0.15 m) ( − 1200 V) b) V = so q = = = − 20 nC R k k c) No, the amount of charge on the sphere is very small.

23.40: For points outside this spherical charge distribution the field is the same as if all the charge were concentrated at the center. Therefore q E= 4πε0 r 2 and (3800 N C) (0.200 m) 2 q = 4πε0 Er 2 = = 1.69 × 10−8 C 9 × 109 N.m 2 / C 2 Since the field is directed inward, the charge must be negative. The potential of a point charge, taking ∞ as zero, is q (9 × 109 N.m 2 / C 2 ) ( − 1.69 × 10 −8 C) V= = = − 760 V 4πε0 r 0.200 m at the surface of the sphere. Since the charge all resides on the surface of a conductor, the field inside the sphere due to this symmetrical distribution is zero. No work is therefore done in moving a test charge from just inside the surface to the center, and the potential at the center must also be − 760 V. a) E = − ∇V . ∂V ∂ Ex = − =− ( Axy − Bx 2 + Cy ) = − Ay + 2 Bx. ∂x ∂x ∂V ∂ Ey = − =− ( Axy − Bx 2 + Cy ) = − Ax − C. ∂y ∂y ∂V ∂ Ez = − = − ( Axy − Bx 2 + Cy ) = 0. ∂z ∂z

23.41:

2B C 2B  − C  x, − Ax − C = 0 ⇒ x = − so y = . = A A A  A  − 2 BC  − C − 2 BC  , E = 0 at  , , z 2 A A2  A  b) − Ay + 2 Bx = 0 ⇒ y =

23.42: a) E = − ∇V Ex = −  ∂V ∂  kQ kQx kQx  = =− = 3 . 2 2 2 32 2 2 2  ∂x ∂x  x + y + z (x + y + z ) r  

Similarly, E y =

kQy kQz and Ez = 3 . 3 r r ˆ ˆ kQ  xi yˆ zk  kQ j = ˆ b) So from (a), E = 2  + + r , which agrees with Equation (21.7). r  r r r  r2  

23.43: a) There is no dependence of the potential on x or y, and so it has no components in those directions. However, there is z dependence: ∂V ˆ = − C ⇒ E = − Ck , for 0 < z < d . ∂z  and E = 0, for z > d , since the potential is constant there. E = −∇V ⇒ E z = − (b) Infinite parallel plates of opposite charge could create this electric field, where the surface charge is σ = ± Cε0 .

23.44: a) 1 1 kq kq − = kq  − . r r  ra rb b   a 1 1  kq kq (ii) ra < r < rb : V = − = kq  − . r r  r rb b   (iii) r > rb : V = 0, since outside a sphere the potential is the same as for point charge. Therefore we have the identical potential to two oppositely charged point charges at the same location. These potentials cancel. (i) r < ra : V = b) Va = 1 4πε0 q q 1 1 1  −  and Vb = 0 ⇒ Vab = q −  r  4πε0  ra rb   a rb   

Vab ∂V q ∂ 1 1  1 q 1  − =+ =− = . 2 r r  ∂r 4πε0 ∂r  4πε 0 r  1 1  r2 b   −  r r  b   a d) From Equation (24.23): E = 0, since V is zero outside the spheres. e) If the outer charge is different, then outside the outer sphere the potential is no 1 q 1 Q 1 (q − Q) longer zero but is V = − = . All potentials inside the outer 4πε0 r 4πε0 r 4πε0 r 1 Q shell are just shifted by an amount V = − . Therefore relative potentials within the 4πε0 rb shells are not affected. Thus (b) and (c) do not change. However, now that the potential does vary outside the spheres, there is an electric field there: ∂V ∂  kq − kQ  kq  Q E=− =−  +  = 2 1 − .  ∂r ∂r  r r  r  q  c) ra < r < rb : E = −

23.45: a)

1 1 Vab = kq  −  = 500 V r   a rb  500 V ⇒q= = 7.62 × 10 −10 C.   1 1 k  0.012 m − 0.096 m    

b)

c) The equipotentials are closest when the electric field is largest.

23.46: a) E x = −

 kQ  a 2 + x 2 + a     ln   2a  a 2 + x 2 − a      kQ  ∂ ∂  ⇒ Ex = − ln( a 2 + x 2 + a ) − ln( a 2 + x 2 − a )  ∂x 2a  ∂x  ∂V ∂ =− ∂x ∂x kQ  x(a 2 + x 2 ) −1 2 x(a 2 + x 2 ) −1 / 2  kQ −  2 = 2 2 2 2a  a + x + a a + x − a  x a2 + x2 (2aλ ) 1 λ ⇒ Ex = = . 4πε0 xa 1 + x 2 a 2 2πε 0 x 1 + x 2 a 2 =−

b) The potential was evaluated at y and z equal to zero, and thus shows no dependence on them. However, the electric field depends upon the derivative of the potential and the potential could still have a functional dependence on the variables y and z, and hence E y and E z may be non-zero.

23.47:

a) Equipotentials and electric field lines of two large parallel plates are shown above. b) The electric field lines and the equipotential lines are mutually perpendicular.

23.48:

(a) ΣF = m1a = F12 + F13 kq1q2 kq1q 3 + 2 = m1a 2 r12 r13 q1 = q2 = q3 = q  1 1 ma =kq 2  2 + 2  r   12 r13    1 1 (0.02 kg )a = (9 × 109 Nm 2 C 2 ) ( 2.0 × 10 −6 C) 2  + 2 2 (0.16 m)   (0.08 m) 2 a = 352 m s (b) Maximum speed occurs at “infinity”. The center charge does not move since the forces on it balance. Energy conservation gives U i = K f . kq1q2 kq1q3 kq2 q3 1 1 2 + + = m1v12 + m3v3 . r12 r13 r23 2 2 v1 = v3 , m1 = m3 , and q1 = q2 = q3 = q v1 = = kq 2 m1  1 1 1  + + r  12 r13 r23      1 1 1    0.08 m + 0.16 m + 0.08 m  = 7.5 m s   

(9 × 10 9 Nm 2 C 2 ) (2 × 10 −6 C) 2 0.020 kg

23.49: a) WE = ∆K − WF = 4.35 × 10−5 J − 6.50 × 10−5 J = − 2.15 × 10 −5 J. − WE 2.15 × 10 −5 J = = + 2829 V. So the initial point is q 7.60 × 10 −9 C –2829 V with respect to the final point. b) WE = − q∆V ⇒ ∆V = c) E = V 2829 V V = = 3.54 × 10 4 . d 0.08 m m

23.50: a)

mv 2 ke 2 = 2 ⇒v= r r

ke 2 . mr

1 2 1 ke 2 1 mv = = − U. 2 2 r 2 1 1 ke 2 1 k (1.60 × 10−19 C) 2 c) E = K + U = U = − =− = − 2.17 × 10 −18 J = − 13.6 eV. −11 2 2 r 2 5.29 × 10 m b) K =

23.51: a) V = Cx 4 / 3 ⇒ C = (240 V) (0.0130 m) −4/3 = 7.85 × 10 4 V m . ∂V 4 4 V   4/3 b) E = − = − Cx1 / 3 = − (7.85 × 10 4 V m ) x1 / 3 =  − 1.05 × 10 5 4/3 x1 / 3  V m , ∂x 3 3 m   toward cathode. c) F = − eE = ((1.05 × 10 5 ) (0.00650)1 / 3 V m) (1.60 × 10 −19 C) = 3.14 × 10 −15 N, toward anode.

4/3

23.52: From Problem 22.51, the electric field of a sphere with radius R and q distributed qr q uniformly over its volume is E = for r ≤ R and E = for r ≥ R 3 4πε0 R 4πε0 r 2 the center of the sphere. R ∞ qr q q Vr = ∫ dr + ∫ dr = 3 2 r 4πε R R 4πε r 8πε0 R 0 0 Va − Vb = ∫ Edr. Take b at infinity and V∞ = 0. Let point a be a distance r < R from
a b

 r2  3 − 2   R    Set q = + 2e to get Vr for the sphere. The work done by the attractive force of the sphere when one electron is removed from r = d to ∞ is 2e 2  d2  3 − 2  8πε0 R  R    The total work done by the attractive force of the sphere when both electrons are removed is twice this, 2Wsphere . The work done by the repulsive force of the two electrons Wsphere = − eVr = − e2 The total work done by the electrical forces is 2Wsphere + Wee . The 4πε0 (2d ) energy required to remove the two electrons is the negative of this, e2  R d2  3 −  − 2πε 0 R  4d R 2    We can check this result in the special case of d = R, when the electrons initially sit on the surface of the sphere. The potential due to the sphere is the same as for a point charge + 2e at the center of the sphere. is Wee = Wq → b = U a − U b  − 2e 2  e2 e2 + U b = 0. U a = 2  =  4πε R  4πε (2 R) 4πε R 0  0 0  1  − 7e 2  −2+ =  4  8πε0 R 

The work done by the electric forces when the electrons are removed is − 7e 2 8πε0 R and the energy required to remove them is 7e 2 8πε0 R . Setting d =R in our general expression yields this same result.

23.53: a) 3 1  3 1  2  3  2  2 U = kq 2  − + − − −  +kq 2  − +  + kq 2  − + 2d 3d  2d 3d  2d  d  d  d 2 1  1  1   1  2  1 + kq 2  − + −  + kq 2  − +  + kq 2  − +  + kq 2 2d 3d  2d  2d   d  d  d 2 12 4  12kq  1 1   12 ⇒ U = kq 2  − + − + =− 1 −  = − 1.46 q 2 πε 0 d d  2d 3d  2 3 3  d b) The fact that the electric potential energy is less than zero means that it is energetically favourable for the crystal ions to be together. 1 1 2kq 2  1  23.54: a) U = 2kq 2  − + − + ⋅ ⋅ ⋅ = − d  d 2d 3d  2kq 2 ln( 2) b) U = − d c) The potential energy is the same for the negative ions—the equations are identical if we examine (a). 2k (1.60 × 10 −19 C) 2 ln (2) d) If d = 2.82 × 10 −10 m, then U = − = −1.13 × 10 −18 J. −10 2.82 × 10 m e) The real energy (−0.80 × 10 −18 J ) is about 70% of that calculated above. − 2ke 2 − 2k (1.60 × 10 −19 C) 2 = = − 8.61 × 10−18 J. r 0.535 × 10−10 m b) If all the kinetic energy goes into potential energy: 2ke 2 d +x
2 2

1   3d   1 −   d

(−1) i −1 ∑ i . i =1
∞

23.55: a) U e =

U t = U e + K = − 8.61 × 10−18 J + 1.02 × 10−18 J = ⇒ x2 =
2 4

= − 7.59 × 10−18 J

4k e − d 2 = 8.24 × 10− 22 m 2 (d = 5.35 × 10−11 m) 2 Ut

(Note that we must be careful to keep all digits along the way.) ⇒ x = 2.87 × 10 −11 m.

23.56: Fe = mg tan θ = (1.50 × 10 −3 kg) (9.80 m s ) tan (30°) = 0.0085 N . (Balance forces in x and y directions.) But also: Vq Fd (0.0085 N) (0.0500 m) Fe = Eq = ⇒V = = = 47.8 V. d q 8.90 × 10 − 6 C 23.57: a) (i) V = λ λ (ln(b a ) − ln (b b)) = ln( b a ). 2πε0 2πε0 λ λ (ii) V = (ln(b r ) − ln( b b)) = ln( b r ) . 2πε 0 2πε0 (iii) V = 0. λ b) Vab = V (a ) − V ( b ) = ln( b a). 2πε0 c) Between the cylinders: Vab λ V = ln( b r ) = ln( b r ) 2πε0 ln (b a) ∴E = − Vab ∂ Vab 1 ∂V =− (ln(b r )) = . ∂r ln( b a ) ∂r ln( b a) r

2

d) The potential difference between the two cylinders is identical to that in part (b) even if the outer cylinder has no charge. 23.58: Using the results of Problem 23.57, we can calculate the potential difference: Vab 1 E= ⇒ Vab = E ln (b/a)r ln( b a ) r ⇒ Vab = (2.00 × 10 4 N C) (ln (0.018 m 145 × 10 −6 m)) 0.012 m = 1157 V.

23.59: a) F = Eq = (1.10 × 103 V m) (1.60 × 10−19 C) = 1.76 × 10−16 N, downward. b) a = F me = (1.76 × 10 −16 N) (9.11 × 10 −31 kg) = 1.93 × 1014 m s , downward. 0.060 m 1 1 2 c) t = = 9.23 × 10− 9 s, y − y0 = at 2 = (1.93 × 1014 m s ) (9.23 × 10 − 9 s) 2 6 6.50 × 10 m s 2 2 = 8.22 × 10−3 m. d) Angle θ = arctan(v y vx ) = arctan(at vx ) = arctan(1.78 6.50) = 15.3°. e) The distance below center of the screen is: 0.120 m D = d y + v y t = 8.22 × 10 −3 m + (1.78 × 106 m s) = 0.0411 m. 6.50 × 106 m s
2

23.60:

(a) Use ∆Vab to get λ : ∆V = ∫ E ⋅ dl = ∫
a

b

b

a

λ λ dr = ln 2πε0 2πε0

b a

λ= E=

2πε 0 ∆V ln b a 2πε0 ∆V ln b a λ ∆V = = 2πε 0 r 2πε 0 r r ln b a

at outer surface of the wire, r = a = 0.127 mm 2 850 V E= = 2.65 × 10 6 V m 0.000127 m  1.00 cm  ( 2 ) ln ( 0.0127 cm )  2    (b) at the inner surface of the cylinder, r = 1.00 cm , which gives E = 1.68 × 10 4 V m

23.61: a) From Problem 23.57, Vab 1 50,000 V 1 E= = −5 ln (b a) r ln (0.140 9.00 × 10 ) 0.070 m ⇒ E = 9.72 × 10 4 V m. b) F = Eq = 10 mg ⇒ q = 10 (3.00 × 10−8 kg)(9.80 m s ) = 3.02 × 10 −11 C. 9.72 × 104 V/ m
2

23.62: Recall from Example 23.12 for a line of charge of length a : kQ  a 2 4 + x 2 + a 2  V= ln  2  a  a 4 + x2 − a 2    a) For a square with two sets of oppositely charged sides, the potentials cancel and V = 0. b) If all sides have the same charge we have: 2 2 4kQ  a 4 + x + a 2  V= ln  , but here x = a 2 , so: a  a2 4 + x2 − a 2    ⇒V = 4kQ  a 2 + 4 x 2 + a  4kQ  ( 2 + 1)  ln  ln  = . a a  a2 + 4x2 − a   ( 2 − 1)   

23.63: a) dV =
R

 2πr dr  2kQ = 2 2   x 2 + r 2  πR  R kQ 2kQ R2

r dr x2 + r 2
z=x2 +R2 z=x2

V = ∫ dV =
0

∫
0

R

r dr x +r
2 2

=

2kQ 1 / 2 (z ) R2

=

2kQ R2

[

x2 + R2 − x =

]

σ 2 ε0 b)

[

x2 + R2 − x .   ∂V 2kQ  x σ  1 =− 2  − 1 = 1 − . ∂x R  x2 + R2  2ε 0  1 + R2 x2     

]

Ex = −

23.64: a) From Example 23.12:
2 2 2 2 kQ  a + x + a  kQ  1 + a x + a x  V ( x) = ln  = ln   2a  a 2 + x 2 − a  2a  1 + a 2 x 2 − a x     

If a << x, 1 + a x
2

2

1 a a a 1 ± a x ≈ 1 +   ± ≈ 1 + , and ln (1 + α ) ≈ α + α 2 + ⋅ ⋅ ⋅ 2 x x x 2

2

That is, the finite rod acts like a point charge when you are a long way from it. b) From Example 23.12: kQ  a 2 + x 2 + a  kQ  1 + x 2 a 2 + 1  V ( x) = ln  ln  = . 2 a  a 2 + x 2 − a  2a  1 + x 2 a 2 − 1      1 x 1 If x << a, 1 + x a ± 1 ≈ 1 ± 1 +   , and ln (1 + α ) ≈ α + α 2 + ⋅ ⋅ ⋅ 2a 2
2 2 2

2 2    kQ  a 1  a   +   + ⋅ ⋅ ⋅  −  − a + 1  a  + ⋅ ⋅ ⋅  = kQ  2a  = kQ . ⇒ V ( x) ≈      x 2 x   2a  x  2a  x 2  x  x      

 kQ kQ   (2 + x 2 2a 2 )  kQ   4a 2  = In(2a x) = ln  ln  2 + 1 ≈ 2 2  2a   ( x 2a )   2a   x a     Q λ ln( 2a x ) = ln (2a / x). 4πε0 a 2πε 0 Q Thus λ ≡ , and R = 2a, which is the only natural length in the problem. 2a ⇒ V ( x) ≈

23.65: a) Recall: r ≤ R : E =
2

r   ρr ρ ⇒V = − ∫ E ⋅d r = − 2ε 0 2ε 0 R 2 2

∫
R

r

r dr = −

ρ 2 (r − R 2 ) 4ε 0

So with λ = πR ρ, V = − kλ (r / R − 1). r   ρR 2 ρR 2 For r ≥ R : E = ⇒V = − ∫ E ⋅d r = − 2ε 0 r 2ε 0 R b)

∫

dr λ r r =− ln [ R ] = − 2kλ ln [ R ]. r 2πε0 R

r

k (5.00 × 10 −9 C) k ( − 2.00 × 10−9 C) 23.66: a) V (0.03, 0) = + = − 300 V. 0.0300 m 0.01 m

V (0.03, 0.05) =

k (5.00 × 10 −9 C) (0.0300 2 + 0.0500 2 ) m

+

k ( − 2.00 × 10 −9 C) 0.0100 2 + 0.0500 2

= 419 V.

b) W = − q∆V = − ( + 6.00 × 10−9 C) (718 V) = − 4.31 × 10−6 J. Note that the work done by the field is negative, since the charge is moved AGAINST the electric field. 23.67: From Example 21.10, we have: Ex = Q ⇒V = − 4π ε 0 (23.16). 1 Qx 2 4πε0 ( x + a 2 )3 / 2
u = x2 + a2 u =∞

∞

∫ ( x′

x

2

x′ Q −1 / 2 dx′ = u 2 3/ 2 +a ) 4πε0

=

1 4πε0

Q x2 + a2

= Equation

23.68: dV = 1 dq 1 λ dl 1 Q dl 1 Q dθ 1 = = = ⇒V = 4πε0 r 4πε 0 a 4πε0 πa a 4πε0 πa 4πε0

∫

π

Q dθ 1 Q = . πa 4πε0 a 0

ˆ ˆ j 23.69: a) S1 and S3 : V13 = − ∫ ( − 5 xi + 3zk ) ⋅ ˆ dy = 0; S1 and S3 are at equal potentials.
0

0.3

− 3 2 0.3 − 3 ˆ ˆ ˆ b) S 2 and S 4 : V24 = − ∫ ( − 5 xi + 3 zk ) ⋅ kdz = − 3 ∫ z dz = z |0 = (0.3) 2 = 2 2 0 0 − 0.135 V . S 4 is higher. ˆ ˆ ˆ c) S 5 and S 6 : V56 = − ∫ (−5 xi + 3zk ) ⋅ i dx = 5
0 0.3 0.3

0.3

0.3

∫ x dx = 2 x
0

5

2 0.3 0

| =

5 (0.3) 2 = 0.225 V. 2

S 5 is higher. 23.70: From Example 22.9, we have: r kQ dr ′ kQ r > R : E = 2 ⇒ V = − kQ ∫ 2 = ′ r r ∞ r R  r    kQ kQ r kQr ′ − ∫ E ⋅ d r′ = r < R : E = 3 ⇒ V = − ∫ E ⋅d r − 3 ∫ r ′ dr ′ R R R R ∞ R kQ kQ 1 2 kQ kQ kQr 2 ⇒V = − 3 r′ = + − R R 2 R R 2R 2R3 ∴V = b) kQ 2R  r2  3− 2  R  
r

23.71: a) Problem 23.70 shows that Q Q Vr = (3 − r 2 R 2 ) for r ≤ R and Vr = for r ≥ R 8πε0 R 4πε0 r 3Q Q Q V0 = , VR = , and V0 − VR = 8πε0 R 4πε 0 R 8πε 0 R b) If Q > 0, V is higher at the center. If Q < 0, V is higher at the surface. 23.72: (a) Points a, b, and c are all at the same potential because E = 0 inside the spherical shell of charge on the outer surface. So ∆Vab = ∆Vbc = ∆Vac = 0. kq (9 × 109 Nm 2 C ) (150 × 10 −6 C) ∆Vc∞ = = R 0.60 m = 2.25 × 10 6 V. (b) They are all at the same potential (c) Only ∆Vc∞ would change; it would be − 2.25 × 10 6 V. 23.73: a) The electrical potential energy for a spherical shell with uniform surface charge density and a point charge q outside the shell is the same as if the shell is replaced by a point charge at its center. Since Fr = − dU dr , this means the force the shell exerts on the point charge is the same as if the shell were replaced by a point charge at its center. But by Newton’s 3rd law, the force q exerts on the shell is the same as if the shell were a point charge. But q can be replaced by a spherical shell with uniform surface charge and the force is the same, so the force between the shells is the same as if they were both replaced by point charges at their centers. And since the force is the same as for point charges, the electrical potential energy for the pair of spheres is the same as for a pair of point charges. b) The potential for solid insulating spheres with uniform charge density is the same outside of the sphere as for a spherical shell, so the same result holds. c) The result doesn’t hold for conducting spheres or shells because when two charged conductors are brought close together, the forces between them causes the charges to redistribute and the charges are no longer distributed uniformly over the surfaces.
2

23.74: Maximum speed occurs at “infinity” Energy conservation gives kq1 q 2 1 1 2 2 = m50 v50 + m150 v150 r 2 2 Momentum conservation: m50 v50 = m150 v150 and v50 = 3v150 Solve for v50 and v150 , where r = 0.50 m v50 = 12.7 m s, v150 = 4.24 m s Maximum acceleration occurs just after spheres are released. ∑ F = ma gives kq1q2 = m150 a150 r2 (9 × 10 9 Nm 2 C 2 ) (10 −5 C) (3 × 10 −5 C) = (0.15 kg )a150 (0.50 m) 2 a150 = 72.0 m s
2 2

a50 = 3a150 = 216 m s

23.75: Using the electric field from Problem 22.37, the potential difference between the conducting sphere and insulating shell is: R  R  kQ 1  kQ 1 V = − ∫ E ⋅ d r = − ∫ 2 dr = kQ  −  ⇒ V = 2R . r  R 2R  2R 2R 23.76: a) At r = c : V = − ∫ kq kq dr = . 2 c ∞ r c   b   kq kq b) At r = b : V = − ∫ E ⋅ d r − ∫ E ⋅ d r = −0= . c c ∞ c
c c

a   b   a   kq dr 1 1 1  c) At r = a : V = − ∫ E ⋅ d r − ∫ E ⋅ d r − ∫ E ⋅ d r = − kq ∫ 2 = kq  − + . c r c b a  ∞ c b b 1 1 1  d) At r = 0 : V = kq  − +  since it is inside a metal sphere, and thus at the c b a  same potential as its surface.

23.77: Using the electric field from Problem 22.54, the potential difference between the two faces of the uniformly charged slab is: V =−   ∫E ⋅d r = −
d

−d

ρx ρ  x2    dx = ∫ 2 ε0 2ε0  2    −d
d

d −d

⇒ V = 0.

kQ k ( − 1.20 × 10 −12 C) 23.78: a) V = = = − 16.6 V. r 6.50 × 10 − 4 m b) The volume doubles, so the radius increases by the cube root of two: Rnew = 3 2 R = 8.19 × 10−4 m and the new charge is Qnew = 2Q = − 2.40 × 10 −12 C. So the new potential is: kQnew k ( − 2.40 × 10−12 C) Vnew = = = − 26.4 V. Rnew 8.19 × 10− 4 m 23.79: a) dV p = b) kQ dz kQ dV R = = a r a c) x >> a : V p ≈ kQ ⇒ VR = a z2 + y2 dz kdq kQ dz kQ = ⇒V = z+x a z+x a

∫z+x=
0

a

dz

kQ  x + a  kQ  a 1n  ln 1 + . = a x  x  a 
2 2 kQ  a + y + a   . = 1n 2 2   a y z +y  

∫
0

a

dz

kQ a kQ = , Since ln (1 + α ) ≈ α. a x x  a2 + y2 + a  kQ a kQ  ≈ ln  y + a  = 1n 1 + a  ≈ a .    y >> a : VR ≈ = , Since ln   y      a y y y y y      

23.80: Set the alpha particle’s kinetic energy equal to its potential energy: K = U ⇒ 11.0 Me V = k (2e) (82e) k (164) (1.60 × 10−19 C) 2 ⇒r= r (11.0 × 106 eV)(1.60 × 10 −19 J eV) = 2.15 × 10 −14 m. 23.81: a) V = kQB kQ A kQB Q 1 = = ⇒ Q A = 3QB ⇒ B = . RB RA RA 3 QA 3

b) EB = −

∂V ∂r

=
r = RB

kQB k (QA 3) 3kQA E = = = 3E A ⇒ B = 3. 2 2 2 ( RA 3) EA RB RA

23.82: a) From Problem 22.57 we have the electric field: r kQ kQ kQ r ≥ R : E = 2 ⇒ V = − ∫ 2 dr ′ = , r r r′ ∞ which is the potential of a point charge. R r kQ  r 3 r4  b) r ≤ R : E = 2 4 3 − 3 4  ⇒ V = − ∫ E dr ′ − ∫ E dr ′ r  R R  ∞ R kQ ⇒V = R   r2 R 2 r 3 R 3  kQ  r 3 r2 1 − 2 R 2 + 2 R 2 + R 3 − R 3  = R  R 3 − 2 R 2 + 2 .     kQ1 , so V = RE R1

23.83: a) E =

kQ1
2

R1 b) After electrostatic equilibrium is reached, with charge Q1′ now on the original sphere we have: Q′ Q R Q1 = Q1′ + Q2 and V1 = V2 ⇒ 1 = 2 ⇒ Q1′ = Q2 1 R1 R2 R2 R Q1 R2 Q1 ( R R )Q R1Q1 Q1 = Q2 1 + Q2 ⇒ Q2 = = and Q1′ = 1 2R1 1 = R1 R2 ( R2 + R1 ) (1 + R2 ) ( R2 + R1 ) 1 + R2

, V=

(

)

c) The new potential is the same at each sphere’s surface: kQ ′ kQ1 kQ1 V1 = 1 = = = V2 R1 R1 ( R2 + R1 ) R2 1 + R2

(

)

d) The new electric field is not the same at each sphere’s surface: kQ ′ kQ1 kQ1 E1 = 21 = = R1 R1 R2 (1 + R2 ) R1 ( R2 + R1 ) R1 E2 = kQ2 R2
2

=

kQ1 R2 (1 +
2 R1 R2

)

=

kQ1 R2 ( R2 + R1 )

23.84: a) We have V ( x, y, z ) = A( x 2 − 3 y 2 + z 2 ). So : ∂V ˆ ∂V ˆ ∂V ˆ ˆ ˆ E=− i− j− k = − 2 Axi + 6 Ayˆ − 2 Azk j ∂x ∂y ∂z b) A charge is moved in along the z -axis. So the work done is given by: 0 0  ˆ dz = q ( − 2 Az ) dz = + ( Aq) z 2 ⇒ A = W W = q ∫E ⋅k 0 2 ∫ qz 0 z0 z0 A= 6.00 × 10 −5 J = 640 V m 2 . −6 2 (1.5 × 10 C)(0.250 m)

ˆ ˆ c) E (0, 0, 0.250) = − 2(640 V m 2 ) (0.250 m) k = - 320 V mk . d) In every plane parallel to the x - z plane , y is constant, so: V +C V ( x, y, z ) = Ax 2 + Az 2 − C ⇒ x 2 + z 2 = ≡ R2, A which is the equation for a circle since R is constant as long as we have constant potential on those planes. 2 1280 V + 3(640 V m )(2.00 m) 2 e) V = 1280 V, and y = 2m: x 2 + z 2 = = 14 m 2 . 2 640 V m Thus the radius of the circle is 3.74 m.

23.85: a)

2 1  ke E i = E f ⇒ 2  mp v 2  = ⇒v= 2  2rp

k (1.60 × 10 −19 C) 2 2(1.2 × 10 −15 m)(1.67 × 10 −27 kg)

⇒ v = 7.58 × 106 m s . b) For a helium-helium collision, the charges and masses change from (a): v= k (2(1.60 × 10−19 C)) 2 = 7.26 × 106 m s . (3.5 × 10−15 m)(2.99)(1.67 × 10 − 27 kg)

c) mp v 2 (1.67 × 10 −27 kg)(7.58 × 10 6 m s) 2 3kT mv 2 K= = ⇒ Tp = = = 2.3 × 10 9 K 2 2 3k 3(1.38 × 10 −23 J K) ⇒ THe = mHe v 2 (2.99)(1.67 × 10 −27 kg)(7.26 × 10 6 m s) 2 = = 6.4 × 10 9 K. 3k 3(1.38 × 10 −23 J K )

d) These calculations were based on the particles’ average speed. The distribution of speeds ensures that there are always a certain percentage with a speed greater than the average speed, and these particles can undergo the necessary reactions in the sun’s core.

23.86: a) The two daughter nuclei have half the volume of the original uranium nucleus, so their radii are smaller by a factor of the cube root of 2: 7.4 × 10 −15 m r= = 5.9 × 10 −15 m. 3 2 2 k (46) 2 (1.60 × 10 −19 C) 2 k (46e) b) U = = = 4.14 × 10 −11 J −14 2r 1.17 × 10 m Each daughter has half of the potential energy turn into its kinetic energy when far from each other, so: K = U 2 = (4.15 × 10 −11 J) 2 = 2.07 × 10 −11 J. c) If we have 10.0 kg of uranium, then the number of nuclei is: 10.0 kg n= = 2.55 × 10 25 nuclei. − 27 236 u (1.66 × 10 kg u ) And each releases energy U : E = nU = (2.55 × 1025 )(4.15 × 10 −11 J) = 1.06 × 1015 J = 253 kilotons of TNT. d) We could call an atomic bomb an “electric” bomb since the electric potential energy provides the kinetic energy of the particles. 23.87: Angular momentum and energy must be conserved, so: 1 kq q 2 mv1b = mv2 r2 and E1 = E2 ⇒ E1 = mv2 + 1 2 and E1 = 11 MeV = 1.76 × 10 −12 J. 2 r2 Substituting in for v 2 we find: E1 = E1 b 2 kq1q2 2 + ⇒ ( E1 )r2 − (kq1q2 )r2 − E1b 2 = 0, and note q1 = 2e and q2 = 82e. 2 r2 r2

(i ) b = 10−12 m ⇒ r2 = 1.01 × 10 −12 m (ii) b = 10−13 m ⇒ r2 = 1.11 × 10−13 m. (iii) b = 10−14 m ⇒ r2 = 2.54 × 10−14 m.

23.88: a) r ≤ a : V =

 r2 r3  ∂V 1 − 3 2 + 2 3  and E = −  a a  ∂r  ρ0 a 2  r r 2  ρ0 a  r r 2  ⇒E=− −6 2 + 6 3 = − . 18ε0  a a  3ε0  a a 2     ∂V r ≥ a : V = 0 and E = − = 0. ∂r Q ρ a  r r2  b) r ≤ a : Er 4πr 2 = r = 0  − 2  4πr 2 ε0 3ε0  a a  ρ0 a 2 18ε0 ρ0 a  r + dr ε0 3ε0  a  2 Qr + dr − Qr ρ(r )4πr dr ρ0 a 4πr 2 dr ⇒ = ≈ ε0 ε0 3ε0 Er + dr 4π (r 2 + 2rdr ) = Qr + dr = ⇒ ρ(r ) = ρ0 3 4r    4r  3 − a  = ρ0 1 − 3a .     − (r 2 + 2r dr )  2  4π (r + 2r dr ) 2 a 

 2r 2 2 r 1   − a2 + a − a2 + a   

c) r ≥ a : ρ(r ) = 0, so the total charge enclosed will be given by:  2 4r 3  1 3 r 4  Q = 4π ∫ ρ(r )r dr = 4πρ0 ∫ r − dr = 4πρ0  r −  = 0. 0 3a  3a  0   3 0 Therefore, by Gauss’s Law, the electric field must equal zero for any position r ≥ a.
a 2 a a

23.89: a) Fg = mg = b) Fg = mg =

4πr 3 4π ρr 3 gd ρg = qVab d = qE = Fe ⇒ q = . 3 3 V ab

4πr 3 9ηvt ρg = 6πηrvt = Fv ⇒ r = 3 2 ρg
3

4π ρgd  9ηvt  d ⇒q=   = 18π 3 Vab  2 ρg  Vab c) q = 18π r= 10−3 m 9.16 V

η3vt . 2 ρg

3

(1.81 × 10 −5 Ns m 2 )3 (10 −3 m 39.3 s) 3 = 4.80 × 10 −19 C = 3e. 2(824 kg m3 )(9.80 m s 2 )

9(1.81 × 10 −5 Ns m 2 )(10−3 m 39.3 s) = 5.07 × 10− 7 m 3 2 2(824 kg m )(9.80 m s )

23.90: For an infinitesimal slice of a finite cylinder, we have the potential: dV = k dQ ( x − z)2 + R 2 kQ L
L 2

=

kQ dz L ( x − z)2 + R 2 dz
2 2

⇒V = ⇒V =

−L 2

∫

( x − z) + R

=

kQ L

L 2−x

−L 2 − x

∫

du u + R2
2

where u = x − z.

kQ  ( L 2− x) 2 + R 2 + ( L 2 − x )  ln   on the cylinder' s axis. L  ( L 2 + x)2 + R 2 − L 2 − x   

b) For L << R : V ≈
2 2 2 2 kQ  (L 2 − x) + R + L 2 − x  kQ  x − xL + R + L 2 − x  ≈  ln  ln  L  ( L 2 + x) 2 + R 2 − L 2 − x  L  x 2 + xL + R 2 − L 2 − x      2 2 2 2 kQ  1 − xL (R + x ) + ( L 2 − x) R + x  ⇒V ≈ ln   L  1 + xL ( R 2 + x 2 ) + (− L 2 − x) R 2 + x 2    2 2 2 2 kQ  1 − xL 2( R + x ) + ( L 2 − x) R + x  ⇒V ≈ ln   L 1 + xL 2( R 2 + x 2 ) + (− L 2 − x) R 2 + x 2    2 2    kQ 1 + L 2 R + x  kQ   L L  ln 1 + =  − ln 1 −  ⇒V ≈ ln  2 2 2 2    2 R2 + x2  L 1 − L 2 R + x  L    2 R + x       kQ 2L kQ ⇒V ≈ = , which is the same as for a ring. L 2 x2 + R2 x2 + R2

c)

2 2 2 2 ∂V 2kQ ( L − 2 x ) + 4 R − ( L + 2 x) + 4 R E=− = . ∂x ( L − 2 x) 2 + 4 R 2 ⋅ ( L + 2 x ) 2 + 4 R 2

(

)

23.91: a) m1v1 + m2v2 (6 × 10−5 kg )(400 m s) + (3 × 10 −5 kg)(1300 m s) vcm = = = 700 m s m1 + m2 6.0 × 10− 5 kg + 3.0 × 10− 5 kg kq q 1 1 1 2 2 2 b) E rel = m1v1 + m2 v2 + 1 2 − (m1 + m2 )vcm . 2 2 r 2 After expanding the center of mass velocity and collecting like terms: 1 m1m2 kq q 1 kq q 2 2 ⇒ Erel = [v1 + v2 − 2v1v2 ] + 1 2 = μ(v1 − v2 ) 2 + 1 2 . 2 m1 + m2 r 2 r 1 k (2.0 × 10−6 C)(−5.0 × 10−6 C) c) Erel = (2.0 × 10− 5 kg)(900 m s) 2 + = − 1.9 J. 2 0.0090 m d) Since the energy is less than zero, the system is “bound.” e) The maximum separation is when the velocity is zero: kq q k (2.0 × 10−6 C)( − 5.0 × 10−6 C) − 1.9 J = 1 2 ⇒ r = = 0.047 m. r − 1.9 J f) Now using v1 = 400 m s and v2 = 1800 m s , we find : Erel = + 9.6 J. so the particles do escape, and the final relative velocity is : v1 − v2 = 2 Erel = μ 2(9.6 J ) = 980 m s. 2.0 × 10− 5 kg

Chapter 24

24.1:

Q = CV = (25.0 V)(7.28 μF) = 1.82 × 10 −4 C.

A 0.00122 m 2 24.2: a) C = ε 0 = ε0 = 3.29 pF. d 0.00328 m Q 4.35 × 10 −8 C b) V = = = 13.2 kV. C 3.29 × 10 −12 F V 13.2 × 10 3 V c) E = = = 4.02 × 10 6 V m . d 0.00328 m 24.3: a) V = b) A = Q 0.148 × 10 −6 C = = 604 V. C 2.45 × 10 −10 F

Cd ( 2.45 × 10−10 F)(0.328 × 10−3 m) = = 0.0091 m 2 . ε0 ε0 604 V V c) E = = = 1.84 × 10 6 V m . −3 d 0.328 × 10 m σ d) E = ⇒ σ = ε0 E = ε0 (1.84 ×106 V m) = 1.63 × 10− 5 C/m 2 . ε0 24.4: ∆V = Ed = = σ d ε0
2

(5.60 × 10 −12 C m )(0.00180 m) 8.85 × 10 −12 C 2 Nm 2 =1.14 mV

24.5: a) Q = CV = 120 μC b) C = ε0 A d d → d 2 means C → C 2 and Q → Q 2 = 60 μC c) r → 2r means A → 4 A, C → 4C , and Q → 4Q = 480 μC

24.6: (a) 12.0 V since the plates remain charged. (b) (i) V = Q C Q does not change since the plates are disconnected from the battery. ε⋅A C= d 1 If d is doubled, C → 2 C , so V → 2V = 24.0 V (ii) A = πr 2 , so if r → 2r , then A → 4 A, and C → 4C which means that 1 V → V = 3.00 V 4 24.7: Estimate r = 1.0 cm ε0 A ε πr 2 ε0 π (0.010 m) 2 so d = 0 = = 2.8 mm d C 1.00 × 10 −12 F The separation between the pennies is nearly a factor of 10 smaller than the diameter of a penny, so it is a reasonable approximation to treat them as infinite sheets. C= 24.8: (a) ∆V = Ed 100 V = (10 4 N C)d d = 10 −2 m = 1.00 cm ε A ε π R2 C= 0 = 0 d d Cd 4Cd R= = πε0 4πε0 Nm 2 R = 4(5.00 × 10 F)(10 m)(9 × 10 ) C2 R = 4.24 × 10 −2 m = 4.24 cm (b) Q = CV = (5pF)(100 V) = 500 pC
−12 −2 9

24.9: a)

C 2πε0 = L ln (rb ra ) (0.180 m)2πε0 C= = 4.35 × 10−12 F ln (5.00 0.50)

b) V = Q / C = (10.0 × 10 −12 C) /(4.35 × 10 −12 F) = 2.30 V

24.10: a)

C 2πε0 2πε0 2πε0 r = ⇒ ln (rb ra ) = = = 1.77 ⇒ b = 5.84. −12 L ln (rb ra ) C L 31.5 × 10 F m ra Q C b) = V = (2.60 V)(31.5 × 10 −12 F m) = 8.19 × 10 −11 C m . L L

2πε0 2πε0 = = 6.56 × 10 −11 F/m. ln ( rb ra ) ln (3.5 mm / 1.5 mm ) b) The charge on each conductor is equal but opposite. Since the inner conductor is at a higher potential it is positively charged, and the magnitude is: 2πε0 LV 2πε0 (2.8 m)(0.35 V) Q = CV = = = 6.43 × 10 −11 C. ln (rb ra ) ln (3.5 mm 1.5 mm ) 24.11: a) C L = 24.12: a) For two concentric spherical shells, the capacitance is: 1 r r  kCra C =  a b  ⇒ kCrb − kCra = ra rb ⇒ rb = r −r  k b a kC − ra k (116 × 10−12 F)(0.150 m) ⇒ rb = = 0.175 m. k (116 × 10−12 F) − 0.150 m b) V = 220 V, and Q = CV = (116 × 10 −12 F)(220 V) = 2.55 × 10−8 C.

1  rb ra  1  (0.148 m)(0.125 m)  −11     r − r  = k  0.148 m − 0.125 m  = 8.94 × 10 F.  k b a   b) The electric field at a distance of 12.6 cm: kQ kCV k (8.94 × 10−11 F)(120 V) E= 2 = 2 = = 6082 N/C. r r (0.126 m) 2 c) The electric field at a distance of 14.7 cm: kQ kCV k (8.94 × 10−11 F)(120 V) E= 2 = 2 = = 4468 N/C. r r (0.147 m) 2 d) For a spherical capacitor, the electric field is not constant between the surfaces. 24.13: a) C =

24.14: a)

1 1 1 1 1 = + = + −6 Ceq C1 + C2 C3 ((3.0 + 5.0) × 10 F) (6.0 × 10− 6 F)

⇒ Ceq = 3.42 × 10 −6 F. The magnitude of the charge for capacitors in series is equal, while the charge is distributed for capacitors in parallel. Therefore, Q3 = Q1 + Q2 = VCeq = ( 24.0 V)(3.42 × 10−6 F) = 8.21 × 10−5 C. Since C1 and C 2 are at the same potential, Q1 Q2 C 5 = ⇒ Q2 = 2 Q1 = Q, C1 C 2 C1 3

Q3 = 8 Q1 = 8.21 × 10−5 C ⇒ Q1 = 3.08 × 10 −5 C, and Q2 = 5.13 × 10−5 C. 3 b) V2 = V1 = Q1 C1 = (3.08 × 10−5 C) /(3.00 × 10−6 F) = 10.3 V. And V3 = 24.0 V − 10.3 V = 13.7 V. c) The potential difference between a and d: Vad = V1 = V2 = 10.3 V. 1 1 1 1 1 = 1 + = + 1 Ceq ( C1 + C 2 ) + C3 C4 (2.00 μF + 4.0 μF) (4.0 μF) ⇒ Ceq = 2.40 μF. Then, Q12 + Q3 = Q4 = Qtotal = CeqV = (2.40 × 10 −6 F)(28.0 V) = 6.72 × 10 −5 C and Qtotal 6.72 × 10 −5 C = = 2.24 × 10 −5 C, and Q3 = 4.48 × 10 −5 C. But 3 3 −5 also, Q1 = Q2 = Q12 = 2.24 × 10 C. b) V1 = Q1 C1 = (2.24 × 10 −5 C) (4.00 × 10 −6 F) = 5.60 V = V2 V3 = Q3 C3 = (4.48 × 10 −5 C) (4.00 × 10−6 F) = 11.2 V. 2Q12 = Q3 ⇒ Q12 = V4 = Q4 C4 = (6.72 × 10−5 C) (4.00 × 10−6 F) = 16.8 V. c) Vad = Vab − V4 = 28.0 V − 16.8 V = 11.2 V. 24.16: a) 1 1 1 1 1 = + = + −6 Ceq C1 C 2 (3.0 × 10 F) (5.0 × 10 −6 F) = 5.33 × 10 5 F −1 ⇒ Ceq = 1.88 × 10 −6 F ⇒ Q = VC eq = (52.0 V)(1.88 × 10 −6 F) = 9.75 × 10 −5 C

24.15: a)

b) V1 = Q / C1 = 9.75 × 10−5 C 3.0 × 10−6 F = 32.5 V. V2 = Q / C2 = 9.75 × 10 −5 C 5.0 × 10 −6 F = 19.5 V.

24.17: a) Q1 = VC1 = (52.0 V)(3.0 × 10 −6 F) = 1.56 × 10 −4 C. Q2 = VC2 = (52.0 V)(5.0 × 10 −6 F) = 2.6 × 10 −4 C. b) For parallel capacitors, the voltage over each is the same, and equals the voltage source: 52.0 V.

24.18: Ceq =

(

1 C1

+

1 −1 C2

)

=

(

d1 ε0 A

+

d 2 −1 ε0 A

)

=

ε0 A d1 + d 2

. So the combined capacitance for two

capacitors in series is the same as that for a capacitor of area A and separation (d1 + d 2 ) . 24.19: Ceq = C1 + C 2 =

ε0 A1 d

+

ε0 A2 d

=

ε0 ( A1 + A2 ) d

. So the combined capacitance for two

capacitors in parallel is that of a single capacitor of their combined area ( A1 + A2 ) and common plate separation d. 24.20: a) and b) The equivalent resistance of the combination is 6.0 µF, therefore the

total charge on the network is: Q = CeqVeq (6.0 μF)(36 V ) = 2.16 × 10 −4 C. This is also the charge on the 9.0 μF capacitor because it is connected in series with the point b. So: Q9 2.16 × 10 −4 C = = 24 V. C9 9.0 × 10 −6 F Then V3 = V11 = V12 + V6 = V − V9 = 36 V − 24 V = 12 V. V9 = ⇒ Q3 = C3V3 = (3.0 µF)(12 V) = 3.6 × 10 −5 C. ⇒ Q11 = C11V11 = (11 μF)(12 V) = 1.32 × 10−4 C. ⇒ Q6 = Q12 = Q − Q3 − Q11 = 2.16 × 10 −4 C − 3.6 × 10 −5 C − 1.32 × 10 −4 C. = 4.8 × 10 −5 C. So now the final voltages can be calculated: Q6 4.8 × 10−5 C V6 = = = 8 V. C6 6.0 × 10− 6 F Q12 4.8 × 10− 5 C V12 = = = 4 V. C12 12 × 10− 6 F c) Since the 3 μF, 11 μF and 6 μF capacitors are connected in parallel and are in series with the 9 μF capacitor, their charges must add up to that of the 9 μF capacitor. Similarly, the charge on the 3 μF, 11 μF and 12 μF capacitors must add up to the same as that of the 9 μF capacitor, which is the same as the whole network. In short, charge is conserved for the whole system. It gets redistributed for capacitors in parallel and it is equal for capacitors in series.

24.21: Capacitances in parallel simply add, so:  1 1 1 1   ⇒ (15 + x) μF = 72 μF ⇒ x = 57 μF. = = + Ceq 8.0 μF  (11 + 4.0 + x) μF 9.0 μF    24.22: a) C1 and C 2 are in parallel and so have the same potential across them: V= Q2 40.0 × 10 −6 C = = 13.33 V C2 3.00 × 10− 6 F

Thus Q1 = VC1 = (13.33 V)(3.00 × 10 −6 F) = 80.0 × 10 −6 C. Since Q3 is in series with the parallel combination of C1 and C2 , its charge must be equal to their combined charge: 40.0 × 10−6 C + 80.0 × 10−6 C = 120.0 × 10−6 C b) The total capacitance is found from: 1 1 1 1 1 = + = + −6 Ctot C|| C3 9.00 × 10 F 5.00 × 10− 6 F Ctot = 3.21 μF and Vab = Qtot 120.0 × 10−6 C = = 37.4 V Ctot 3.21 × 10 − 6 F

24.23: V1 = Q1 C1 = (150 μC) (3.00 μF) = 50 V C1 and C2 are in parallel, so V2 = 50 V V3 = 120 V − V1 = 70 V 24.24: a) V = Q / C = (2.55 μC) (920 × 10−12 F) = 2772 V. b) Since the charge is kept constant while the separation doubles, that means that the capacitance halves and the voltage doubles to 5544 V. c) U = 1 CV 2 = 1 (920 × 10 −12 F)(2772 V) 2 = 3.53 × 10−3 J. Now if the separation 2 2 is doubled, the capacitance halves, and the energy stored doubles. So the amount of work done to move the plates equals the difference in energy stored in the capacitor, which is 3.53 × 10 −3 J. 24.25: E = V d = (400 V) (0.005 m) = 8.00 × 104 V m. And u = 1 ε0 E 2 = 1 ε0 (8.00 × 10 4 V m) 2 = 0.0283 J m 3 . 2 2

24.26: a) C = Q V = (0.0180 μC) (200 V) = 9.00 × 10 −11 F. b) C = ε0 A Cd (9.00 × 10 −11 F)(0.0015 m) ⇒ A= = = 0.0152 m 2 . d ε0 ε0 Q 2 (1.80 × 10 −8 C) 2 = = 1.80 × 10 −6 J. −11 2C 2(9.00 × 10 F)

c) E max = Vmax d ⇒ Vmax = E max d = (3.00 × 10 6 V m)(0.0015 m) = 4500 V. d) U =

24.27: U = 1 CV 2 = 1 (4.50 × 10 −4 F)(295 V) 2 = 19.6 J. 2 2 24.28: a) Q = CV0 . b) They must have equal potential difference, and their combined charge must add up to the original charge. Therefore: Q Q V = 1 = 2 and also Q1 + Q2 = Q = CV0 C1 C 2 C Q Q2 Q C1 = C and C2 = so 1 = ⇒ Q2 = 1 2 C (C 2) 2 3 2 Q 2Q 2 ⇒ Q = Q1 ⇒ Q1 = Q so V = 1 = = V0 2 3 C 3C 3 2 2 1Q Q  1  ( 2 Q) 2 2( 1 Q) 2  1 Q 2 1 2 c) U =  1 + 2  =  3 + 3 = CV0 = C  2 2 1 C2  C  3 C 3  C d) The original U was U = 1 CV0 ⇒ ∆U = −61 CV0 . 2 e) Thermal energy of capacitor, wires, etc., and electromagnetic radiation. Q2 xQ 2 = . 2C 2ε0 A
( x + dx ) Q 2 2ε0 A 2 2

24.29: a) U 0 =

b) Increase the separation by dx ⇒ U = then
Q2 2ε0 A

= U 0 (1 + dx x). The change is

dx .

c) The work done in increasing the separation is given by: dxQ 2 Q2 dW = U − U 0 = = Fdx ⇒ F = . 2 ε0 A 2ε0 A d) The reason for the difference is that E is the field due to both plates. The force is QE if E is the field due to one plate is Q is the charge on the other plate.

24.30: a) If the separation distance is halved while the charge is kept fixed, then the capacitance increases and the stored energy, which was 8.38 J, decreases since U = Q 2 2C. Therefore the new energy is 4.19 J. b) If the voltage is kept fixed while the separation is decreased by one half, then the doubling of the capacitance leads to a doubling of the stored energy to 16.76 J, using U = CV 2 2, when V is held constant throughout. 24.31: a) U = Q 2 2C Q = 2UC = 2(25.0 J)(5.00 × 10−9 F) = 5.00 × 10−4 C The number of electrons N that must be removed from one plate and added to the other is N = Q e = (5.00 × 10 −4 C) /(1.602 × 10−19 C) = 3.12 × 1015 electrons. b) To double U while keeping Q constant, decrease C by a factor of 2. C = ε0 A / d ; halve the plate area or double the plate separation. Q 8.20 × 10 −12 C = = 3.417 × 10 −12 farad V 2.40 V Since C = Kε0 A d for a parallel plate capacitor d= Kε0 A (1.00)(8.85 × 10 −12 C 2 / N ⋅ m 2 )(2.60 × 10 −3 m 2 ) = C 3.417 × 10 −12 farad

24.32: C =

= 6.734 × 10 −3 m The energy density is thus
1 CV 2 (3.42 × 10−12 farad )(2.40 V) 2 J 2 u= = = 5.63 × 10 − 7 3 −3 2 −3 Ad (2.60 × 10 m )(6.734 × 10 m) m 1 2

1 2U 2(3.20 × 10 −9 J ) 24.33: a) U = QV ⇒ Q = = = 1.60 × 10 −9 C. 2 V 4.00 V 2πε0 r C b) = ⇒ a = exp( 2πε0 L C ) = exp( 2πε0 LV Q) L ln (ra rb ) rb r ⇒ a = exp( 2πε0 (15.0 m) (4.00 V) (1.60 × 10 − 9 C)) = 8.05. rb

24.34: a) For a spherical capacitor: 1 ra rb 1 (0.100 m)(0.115 m) C= = = 8.53 × 10−11 F k rb − ra k (0.115 m − 0.100 m) ⇒ V = Q C = (3.30 × 10− 9 C) (8.53 × 10−11 F) = 38.7 V. (8.53 × 10 −11 F)(38.7 V) 2 1 2 b) U = CV = = 6.38 × 10 −8 J. 2 2 1 ε  kq  ε  kVC  ε k 2 (120 V) 2 (8.94 × 10−11 F) 2 24.35: a) u = ε0 E 2 = 0  2  = 0  2  = 0 2 2r  2 r  2 (0.126 m) 4 ⇒ u = 1.64 × 10−4 J m 3 . b) The same calculation for r = 14.7 cm ⇒ u = 8.83 × 10 −5 J m 3 . c) No, the electric energy density is NOT constant within the spheres.
2 2 2

1 1  1 q 1 (8.00 × 10 −9 C) 2  = 24.36: a) u = ε0 E 2 = ε0  = 1.11 × 10 −4 J m 3 . 2 2  4πε0 r 2  32π 2 ε0 (0.120 m) 4   b) If the charge was –8.00 nC, the electric field energy would remain the same since U only depends on the square of E. 24.37: Let the applied voltage be V. Let each capacitor have capacitance C. U = 1 CV 2 2 for a single capacitor with voltage V. a) series Voltage across each capacitor is V 2. The total energy stored is 1  U s = 2 C [V 2] 2  = 1 CV 2 4 2  parallel Voltage across each capacitor is V. The total energy stored is U p = 2 1 CV 2 = CV 2 2

(

)

U p = 4U s b) Q = CV for a single capacitor with voltage V . Qs = 2( C [V 2] ) = CV ; Qp = 2(CV ) = 2CV ; Qp = 2Qs c) E = V d for a capacitor with voltage V Es = V 2d ; Ep = V d ; Ep = 2 Es

24.38: a) C = Kε0 A d gives us the area of the plates: Cd (5.00 × 10 −12 farad )(1.50 × 10 −3 m) = 8.475 × 10 − 4 m 2 −12 2 2 Kε0 (1.00)(8.85 × 10 C / N ⋅ m ) We also have C = Kε0 A d = Q V , so Q = Kε 0 A(V d ). V d is the electric field A= between the plates, which is not to exceed 3.00 × 10 4 N C. Thus Q = (1.00)(8.85 × 10−12 C 2 N ⋅ m 2 )(8.475 × 10−4 m 2 )(3.00 × 104 N C) = 2.25 × 10 −10 C b) Again, Q = Kε0 A(V d ) = 2.70ε0 A(V d ). If we continue to think of V d as the electric field, only K has changed from part (a); thus Q in this case is (2.70)(2.25 × 10 −10 C) = 6.08 × 10 −10 C. 24.39: a) σ i = ε0 ((3.20 − 2.50) × 105 V m) = 6.20 × 10−7 C m 2 . The field induced in the dielectric creates the bound charges on its surface. E 3.20 × 105 V m b) K = 0 = = 1.28. E 2.50 × 105 V m

24.40: a) E0 = KE = (3.60)(1.20 × 106 V m) = 4.32 × 106 V m ⇒ σ = ε0 E0 = 3.82 × 10 −5 C m 2 . 1  b) σ i = σ 1 −  = (3.82 × 10 −5 C m 2 )(1 − 1 3.60) = 2.76 × 10− 5 C m 2 .  K 1 c) U = 2 CV 2 = uAd = 1 Kε0 E 2 Ad 2 ⇒ U = 1 (3.60)ε0 (1.20 × 106 V m) 2 (0.0018 m)(2.5 × 10−4 m 2 ) = 1.03 × 10−5 J. 2

24.41: C =

Kε0 A Kε0 AE CV (1.25 × 10 −9 F)(5500 V) = ⇒ A= = = 0.0135 m 2 . d V Kε0 E (3.60)ε0 (1.60 × 10 7 V m)

24.42: Placing a dielectric between the plates just results in the replacement of ε for ε 0 in the derivation of Equation (24.20). One can follow exactly the procedure as shown for Equation (24.11).

24.43: a) ε = Kε0 = ( 2.6) ε0 = 2.3 × 10 −11 C 2 Nm 2 . b) Vmax = Emax d = (2.0 × 107 V m)(2.0 × 10−3 m) = 4.0 × 10 4 V. σ c) E = ⇒ σ = εE = (2.3 × 10 −11 C 2 Nm 2 )(2.0 × 10 7 V m) = 0.46 × 10 −3 C m 2 . Kε0 1  And σ i = σ 1 −  = (0.46 × 10 −3 C m 2 )(1 − 1 2.6) = 2.8 × 10 −4 C m 2 .  K

24.44: a) ∆Q = Q − Q0 = ( K − 1)Q0 = ( K − 1)C0V0 = (2.1)(2.5 × 10−7 F)(12 V) = 6.3 × 10−6 C. 1 b) Qi = Q(1 − K ) = (9.3 × 10−6 C)(1 − 1 3.1) = 6.3 × 10−6 C. c) The addition of the mylar doesn’t affect the electric field since the induced charge cancels the additional charge drawn to the plates.

24.45: a) U 0 =

1 C0V 2 ⇒ V = 2

2U 0 = C0

2(1.85 × 10 −5 J) = 10.1 V. (3.60 × 10 −7 F)

2(2.32 × 10 −5 + 1.85 × 10 −5 J ) 1 U 2 b) U = KC 0V ⇒ K = = = 2.27. 2 C 0V 2 (3.60 × 10 −7 F)(10.1 V) 2

24.46: a) The capacitance changes by a factor of K when the dielectric is inserted. Since V is unchanged (The battery is still connected), Cafter Q 45.0 pC = after = = K = 1.80 C before Qbefore 25.0 pC b) The area of the plates is πr 2 = π (0.0300 m) 2 = 2.827 × 10 −3 m 2 , and the separation between them is thus Kε A (1.00)(8.85 × 10 −12 C 2 N ⋅ m 2 )(2.827 × 10 −3 m 2 ) d= 0 = C 12.5 × 10 −12 farad = 2.002 × 10− 3 m Before the dielectric is inserted, Kε A Q C= 0 = d V Qd (25.0 × 10−12 C)(2.00 × 10 −3 m) V= = Kε0 A (1.00)(8.85 × 10 −12 C2 N ⋅ m 2 )(2.827 × 10 −3 m 2 ) = 2.000 V The battery remains connected, so the potential difference is unchanged after the dielectric is inserted. c) Before the dielectric is inserted, Q 25.0 × 10−12 C E= = ε0 KA (8.85 × 10−12 C 2 N ⋅ m 2 )(1.00)(2.827 × 10− 3 m 2 ) = 999 N C Again, since the voltage is unchanged after the dielectric is inserted, the electric field is also unchanged.

24.47: a) before: V0 = Q0 C 0 = (9.00 × 10 −6 C) (3.00 × 10 −6 F) = 3.00 V after: C = KC 0 = 15.0 F; Q = Q0 V = Q C = 0.600 V; V decreases by a factor of K b) E = V d , the same at all points between the plates (as long as far from the edges of the plates) before: E = (3.00 V) (2.00 × 10 −3 m) = 1500 V m after: E = (0.600 V) (2.00 × 10 −3 m) = 300 V m

24.48: a)

  Q free q q KE ⋅ A = ⇒ KE 4πd 2 = ⇒ E = . ∫ ε0 ε0 4πεd 2   q q + qb q + qb q + qb b) ∫ E ⋅ dA = total = f ⇒ E 4πd 2 = ⇒E= ε0 ε0 ε0 4πε0 d 2 ⇒ qtotal = q + qb = q / K .
1 c) The total bound change is qb = q( K − 1).

  24.49: a) Equation (25.22): ∫ KE ⋅ dA = Qd Qd = . Kε0 A εA ε A Q εA c) C = = = K 0 = KC 0 . V d d b) V = Ed =

Q free ε0

⇒ KEA =

Q ε0

⇒E=

Q Kε 0 A

=

Q εA

.

24.50: a) C =

ε0 A ε0 (0.16 m) 2 = = 4.8 × 10−11 F. d 4.7 × 10− 3 m

b) Q = CV = (4.8 × 10 −11 F) (12 V) = 0.58 × 10 −9 C. c) E= V d =(12 V)/(4.7 × 10 −3 m) =2553 V m . d) U = 1 CV 2 = 1 (4.8 × 10 −11 F)(12 V) 2 = 3.46 × 10−9 J. 2 2 e) If the battery is disconnected, so the charge remains constant, and the plates are pulled further apart to 0.0094 m, then the calculations above can be carried out just as before, and we find: a) C = 2.41 × 10 −11 F c) E = 2553 V m b) Q = 0.58 × 10 −9 C. d) U = Q 2 (0.58 × 10−9 C) 2 = = 6.91 × 10−9 J. 2C 2(2.41 × 10 −11 F)

24.51: If the plates are pulled out as in Problem 24.50 the battery is connected, ensuring that the voltage remains constant. This time we find: 12 V V V a) C = 2.4 × 10 −11 F b) Q = 2.9 × 10−10 C c) E = = = 1.3 × 10 3 d 0.0094 m d) U = CV 2 (2.4 × 10 −11 F) (12 V) 2 = = 1.73 × 10 −9 J. 2 2

24.52: a) System acts like two capacitors in series so Ceq =

(

1 C1

+

1 −1 C2

)

ε0 L2 ε0 L2 1 Q2 1 Q2 Q 2d C1 = C2 = so Ceq = ⋅ U= = ε L2 = . 0 d 2d 2 C 2 2d ε0 L2

( )

b) After rearranging, the E fields should be calculated. Use superposition recalling E = 2 εQ A for a single plate (not εQA since charge Q is only on one face). 0 0  Q   Q   Q   Q  Q between 1 and 3: E =   2ε L2  −  2ε L2  +  2ε L2  +  2ε L2  = ε L2        0  0 1  0 3  0  2  0  4  Q   Q   Q   Q  2Q between 3 and 2: E =   2ε L2  +  2ε L2  +  2ε L2  +  2ε L2  = ε L2        0  0 1  0 3  0  2  0  4  Q   Q   Q   Q  Q between 2 and 4: E =   2ε L2  +  2ε L2  −  2ε L2  +  2ε L2  = ε L2        0  0 1  0 3  0  2  0  4 1  Q2 4Q 2 Q2  3Q 2 d 1 2 2  2 4 + 2 4 + 2 4  L2 d = U new =  ε0 E  L d = ε0  2  ε0 L ε0 L2 ε0 L ε0 L  2   3Q 2 d Q 2 d 2Q 2 d ∆U = U new − U = − = ε0 L2 ε0 L2 ε0 L2 This is the work required to rearrange the plates. 24.53: a) The power output is 600 W, and 95% of the original energy is converted. ⇒ E = Pt = (2.70 × 105 W ) (1.48 × 10 −3 s) = 400 J ∴ E0 = 400 J = 421 J. 0.95 1 2U 2(421 J ) b) U = CV 2 ⇒ C = 2 = = 0.054 F 2 V (125 V) 2 24.54: C0 = Aε0 (4.20 × 10 −5 m 2 )ε0 = = 5.31 × 10 −13 F −4 d 7.00 × 10 m ⇒ C = C0 + 0.25 pF = 7.81 × 10 −13 F. Aε0 Aε (4.20 × 10−5 m 2 )ε0 ′= 0 = ⇒d = 4.76 × 10 − 4 m. −13 d′ C 7.81 × 10 F Therefore the key must be depressed by a distance of: 7.00 × 10 −4 m − 4.76 × 10 −4 m = 0.224 mm. But C =

24.55: a) d << ra : C =

2πε0 L 2πε0 L 2πε0 L 2πra Lε0 ε0 A = = ≈ = . ln( rb ra ) ln( (d + ra ) ra ) ln(1 + d ra ) d d b) At the scale of part (a) the cylinders appear to be flat, and so the capacitance should appear like that of flat plates.

24.56: Originally: Q1 = C1V1 = (9.0 μF) (28 V) = 2.52 × 10−4 C; Q2 = C2V2 = (4.0 μF) × (28 V) = 1.12 × 10 −4 C, and Ceq = C1 + C2 = 13.0 μF. So the original energy stored is U = 1 CeqV 2 = 1 (13.0 × 10−6 F) (28 V) 2 = 5.10 ×10 −3 J. Disconnect and flip the capacitors, 2 2 so now the total charge is Q = Q2 − Q1 = 1.4 × 10− 4 C, and the equivalent capacitance is still the same, Ceq = 13.0 μF. So the new energy stored is : U= Q2 (1.4 × 10 −4 C) 2 = = 7.54 × 10 −4 J −6 2Ceq 2(13.0 × 10 F)

⇒ ∆U = 7.45 × 10 −4 J − 5.10 × 10 −3 J = − 4.35 × 10 −3 J. 24.57: a) Ceq = 4.00 μF + 6.00 μF = 10.00 μF, and Qtotal = Ceq V = (10.00 μF) (660 V) = 6.6 × 10 −3 C. The voltage over each is 660 V since they are in parallel. So: Q1 = C1V1 = (4.00 μF) (660 V) = 2.64 × 10−3 C. Q2 = C2V2 = (6.00 μF) (660 V) = 3.96 × 10− 3 C. b) Qtotal = 3.96 × 10 −3 C − 2.64 × 10 −3 C = 1.32 × 10 −3 C, and still Ceq = 10.00 µ F, so the voltage is V = Q/C = (1.32 × 10 −3 C) (10.00 μF) = 132 V, and the new charges: Q1 = C1V1 = (4.00 μF)(132 V) = 5.28 × 10−4 C. Q2 = C2V2 = (6.00 μF)(132 V) = 7.92 × 10− 4 C.

24.58:

a)

C eq = C + C = C. So the total capacitance is the same as each individual capacitor, and 2 2 the voltage is spilt over each so that V = 480 V. Another solution is two capacitors in parallel that are in series with two others in parallel. b) If one capacitor is a moderately good conductor, then it can be treated as a “short” and thus removed from the circuit, and one capacitor will have greater than 600 V over it. 1 1 1 = + 1 Ceq C1 C2 + C3 + 1 ⇒ C1 = C5 = 2C 2 and C5

24.59: a)

(

1 −1 C4

)

+

C2 = C3 = C4 so b)

1 2 2 5 3 = + = C2 ⇒ Ceq = C 2 = 2.52 μF. Ceq C1 3C2 3 5

Q = CV = (2.52 μF)(220 V) = 5.54 × 10−4 C = Q1 = Q5 ⇒ V1 = V5 = (5.54 × 10−4 C) / (8.4 × 10−6 F) = 66 V.

So V2 = 220 − 2(66) = 88 V ⇒ Q2 = (88 V)(4.2 μF) = 3.70 × 10 −4 C. Also V3 = V4 =
1 2

(88 V) = 44 V ⇒ Q3 = Q4 = (44 V)(4.2 μF) = 1.85 × 10 −4 C.

24.60: a) With the switch open:

Ceq =

((

1 3 μF

+

1 −1 6 μF

)

+

(

1 3 μF

+

1 −1 6 μF

) ) = 4.00 μF

⇒ Qtotal = CeqV = (4.00 μF) (210 V) = 8.4 × 10 −4 C . By symmetry, each capacitor carries 4.20 × 10 −4 C. The voltages are then just calculated via V=Q/C. So: Vad = Q / C3 = 140 V, and Vac = Q / C6 = 70 V ⇒ Vcd = Vad − Vac = 70 V. b) When the switch is closed, the points c and d must be at the same potential, so the equivalent capacitance is:   1 1 Ceq =   (3 + 6) μF + (3 + 6) μF  = 4.5 μF.    ⇒ Qtotal = CeqV = (4.50 μF) (210 V) = 9.5 × 10−4 C, and each capacitor has the same potential difference of 105 V (again, by symmetry) c) The only way for the sum of the positive charge on one plate of C 2 and the negative charge on one plate of C1 to change is for charge to flow through the switch. That is, the quantity of charge that flows through the switch is equal to the charge in Q2 − Q1 = 0. With the switch open, Q1 = Q2 and Q2 − Q1 = 0. After the switch is closed, Q2 − Q1 = 315 μC; 315 μC of charge flowed through the switch.
−1 −1

 1 1 1  24.61: a) Ceq =   8.4 μF + 8.4 μF + 4.2 μF  = 2.1 μF    ⇒ Q = CeqV = (2.1 μ F) (36 V) = 7.50 × 10 −5 C. b) U = 1 CV 2 = 1 ( 2.1 μF) (36 V) 2 = 1.36 × 10 −3 J. 2 2 c) If the capacitors are all in parallel, then: Ceq = (8.4 μF + 8.4 μF + 4.2 μF) = 21 μF and Q = 3(7.56 × 10 −5 C) = 2.27 × 10 −4 C, and V = Q C = (2.27 × 10 −4 C) / (21 μF) = 10.8 V. d) U = 1 CV 2 = 1 (21 µF) (10.8 V) 2 = 1.22 × 10 −3 J. 2 2

 1 1  −6 24.62: a) Ceq =   4.0 μF + 6.0 μF  = 2.4 × 10 F    −6 ⇒ Q = CeqV = (2.4 × 10 F) (600 V) = 1.58 × 10−3 C and V2 = Q / C 2 = (1.58 × 10 −3 C) (4.0 μF) = 395 V ⇒ V3 = 660 V − 395 V = 265 V. b) Disconnecting them from the voltage source and reconnecting them to themselves we must have equal potential difference, and the sum of their charges must be the sum of the original charges: Q1 = C1V and Q2 = C2V ⇒ 2Q = Q1 + Q2 = (C1 + C2 ) V ⇒V = 2Q 2(1.58 × 10− 3 C) = = 316 V. C1 + C2 10.0 × 10 − 6 F

−1

⇒ Q1 = (4.00 × 10− 6 F)(316 V) = 1.26 × 10 − 3 C. ⇒ Q2 = (6.00 × 10 − 6 F)(316 V) = 1.90 × 10 − 3 C.

24.63: a) Reducing the furthest right leg yields C =

(

1 6.9 µ F

+

1 6.9 µ F

+

−1 1 6.9 µ F

)

=

2.3 μF = C1 / 3. It combines in parallel with a C 2 ⇒ C = 4.6 μF + 2.3 μF = 6.9 μF = C1. So the next reduction is the same as the first: C = 2.3 μF = C1 / 3. And the next is the same as the second, leaving 3 C1 ’s in series so Ceq = 2.3 μF = C1 / 3. b) For the three capacitors nearest points a and b: QC1 = C eqV = (2.3 × 10 −6 F)(420 V) = 9.7 × 10 −4 C and QC 2 = C2V2 = (4.6 × 10−6 F) (420 V) 3 = 6.44 × 10−4 C. c) Vcd = 1 ( 420 V ) = 46.7 V, since by symmetry the total voltage drop over the 3 3 equivalent capacitance of the part of the circuit from the junctions between a, c and d, b is 420 V, and the equivalent capacitance is that of three equal capacitors C1 in series. 3 Vcd is the voltage over just one of those capacitors, i.e., 1 3 of 420 V. 3 24.64: (a) Cequiv = C1 + C2 + C3 = 60 μF Q = CV = (60 μ F) (120 V) = 7200 μC 1 1 1 1 (b) = + + C equiv C1 C 2 C 3 Cequiv = 5.45 μF Q = CV = (5.45 μF)(120 V) = 654 μC

24.65: a) Q is constant. with the dielectric: V = Q C = Q ( KC0 ) without the dielectric: V0 = Q C0 V0 / V = K , so K = (45.0 V)/(11.5 V) = 3.91 b)

Let C0 = ε0 A d be the capacitance with only air between the plates. With the dielectric filling one-third of the space between the plates, the capacitor is equivalent to C1 and C 2 in parallel, where C1 has A1 = A / 3 and C 2 has A2 = 2 A / 3 C1 = K C0 3 , C2 = 2 C0 3; Ceq = C1 + C2 = (C0 3) ( K + 2) V=  3  Q Q 3   3   =  K + 2  = V0  K + 2  = (45.0 V)  5.91  = 22.8 V    Ceq C0      

24.66: a) This situation is analagous to having two capacitors C1 in series, each with separation
1 2

(d − a ). Therefore C =

(

1 C1

1 + C1

)

−1

= 1 C1 = 2

ε0 A 1 2 (d −a) 2

=

ε0 A d −a

.

ε0 A ε A d d = 0 = C0 . d −a d d −a d −a c) As a → 0, C → C0 . And as a → d , C → ∞. b) C = 24.67: a) One can think of “infinity” as a giant conductor with V = 0. Q b) C = V = ( Q / 4Q 0 R ) = 4πε0 R, where we’ve chosen V = 0 at infinity. πε c) Cearth = 4πε0 Rearth = 4πε0 (6.4 × 106 m) = 7.1 × 10 −4 F. Larger than, but comparable to the capacitance of a typical capacitor in a circuit.

24.68: a) r < R : u =

1 ε0 E 2 = 0. 2
1 2 2

1 b) r > R : u = ε0 E = ε0 2
∞ 2

 Q  Q2   =  4πε r 2  32π 2 ε r 4 . 0 0  
∞

2

Q 2 dr Q2 c) U = ∫ udV = 4π ∫ r udr = = 8πε0 ∫ r 2 8πε0 R ⋅ R R d) This energy is equal to
1 Q 2 4 πε 0 R
2

which is just the energy required to assemble all

the charge into a spherical distribution. (Note, being aware of double counting gives the factor of 1 2 in front of the familiar potential energy formula for a charge Q a distance R from another charge Q.) 2 Q2 e) From Equation (24.9): U = QC = 8πε0 R from part (c) ⇒ C = 4πε0 R, as in 2 Problem (24.67).
2

24.69:

a)

r < R:u =

1 1  kQr  kQ 2 r 2 ε0 E 2 = ε0  3  = 2 2  R  8πR 6 ⋅
2

1 1  kQ  kQ 2 2 b) r > R : u = ε0 E = ε0  2  = . 2 2  r  8πr 4 c) r < R : U = ∫ udV = 4π ∫ r 2udr =
0 R

kQ 2 4 kQ 2 r dr = . 2R6 ∫ 10 R 0 kQ 2 2
∞

R

r > R : U = ∫ udV = 4π ∫ r 2udr =
R

∞

dr kQ 2 3kQ 2 = ⇒U = . ∫ r 2 2R 5R R

 λ  λ2   = 2 2 24.70:  2πε r  8π ε0 r . 0   r Lλ 2 b dr U λ2 b) U = ∫ udV = 2πL ∫ urdr = ⇒ = ln (rb / ra ). ∫ 4πε0 ra r L 4πε0 1 1 a) u = ε0 E 2 = ε0 2 2 c) Using Equation (24.9): Q2 Q2 λ2L U= = ln (rb / ra ) = ln (rb / ra ) = U of part (b). 2C 4πε 0 L 4πε 0

2

  ε1 A  −1  ε2 A  −1   d   d   d  1 1        24.71: Ceq =    +   =    2ε A  +  2ε A   =  2ε A  K + K    d / 2  d /2  2   1   2   0  1   2ε A  K K  ⇒ Ceq = 0  1 2 . d  K1 + K 2   

−1

−1

−1

24.72: This situation is analagous to having two capacitors in parallel, each with an area A . So: 2 ε A 2 ε 2 A 2 ε0 A Ceq = C1 + C2 = 1 + = ( K1 + K 2 ). d d 2d σ 0.50 × 10−3 C/m 2 = = 1.0 × 107 V/m. Kε0 (5.4)ε0

24.73: a) E =

b) V = Ed = (1.0 × 107 V/m) (5.0 × 10 −9 m) = 0.052 V. The outside is at the higher potential. c) volume = 10 −16 m 3 ⇒ R ≈ 2.88 × 10 −6 m ⇒ shell volume = 4πR 2 d = 4π (2.88 × 10−6 m) 2 (5.0 × 10−9 m) = 5.2 × 10−19 m 3 ⇒ U = uV = ( 1 Kε0 E 2 )V = 1 (5.4)ε0 (1.0 × 107 V/m) 2 (5.2 × 10 −19 m 3 ) = 1.36 × 10 −15 J. 2 2

24.74: a) Q = CV =

Kε0 A ( 2.50)ε0 (0.200 m 2 ) (3000 V) V= = 1.33 × 10− 6 C. −2 d 1.00 × 10 m

b) Qi = Q(1 − 1 / K ) = (1.33 × 10−6 C) (1 − 1/2.50) = 7.98 × 10−7 C. σ Q 1.33 × 10−6 C = = = 3.01 × 105 V/m. ε Kε0 A (2.50) ε0 (0.200 m 2 ) 1 1 d) U = QV = (1.33 × 10− 6 C) (3000 V) = 2.00 × 10− 3 J. 2 2 2.00 × 10 −3 J U e) u = = = 1.00 J/m 3 ⇒ or Ad (0.200 m 2 ) (0.0100 m) c) E = u = 1 Kε0 E 2 = 1 (2.50)ε0 (3.01 × 105 V/m) 2 = 1.00 J/m 3 . 2 2 f) In this case, one does work by pushing the slab into the capacitor since the constant potential requires more charges to be brought onto the plates. When the charge is kept constant, the field pulls the dielectric into the gap, with the field (or charges) doing the work.

24.75: a) We are to show the transformation from one circuit to the other:

From Circuit 1: Vac = Vab =

q1 − q3 q + q3 and Vbc = 2 , where q3 is derived from Vab : Cy Cx

q C xC y C z  q1 q2  q3 q1 − q3 q2 − q3 q   = − ⇒ q3 = −  ≡ K 1 − 2  C  Cz Cy Cx Cx + C y + Cz  C y Cx     y Cx   1 q q + q2 1  1  + q2 From Circuit 2: Vac = 1 + 1 and = q1  + C  C1 C3 C3  1 C3   1 q 2 q1 + q 2 1 1  + = q1 + q2   C + C .  C2 C3 C3 3   2 Setting the coefficients of the charges equal to each other in matching potential equations from the two circuits results in three independent equations relating the two sets of capacitances. The set of equations are:    1 1 1  1 − 1 − 1  , 1 = 1 1 − 1 − 1  and 1 = = . C1 C y  KC y KC x  C2 C x  KC y KC x  C 3 KC y C x     From these, subbing in the expression for K , we get: C1 = (C xC y + C y C z + C z C x ) C x . Vbc = C2 = (C xC y + C yC z + C z C x ) C y . C3 = (C xC y + C y C z + C z C x ) C z .

24.76: a) The force between the two parallel plates is: 2 qσ q2 (CV ) 2 ε0 A2 V 2 ε AV 2 F = qE = = = = 2 = 0 2 . 2 ε0 2 ε0 A 2 ε0 A z 2ε0 A 2z b) When V = 0, the separation is just z 0 . So: ε0 AV 2 ε AV 2 ⇒ 2 z 3 − 2 z 2 z0 + 0 = 0. 2z2 4k c) For A = 0.300 m 2 , z0 = 1.2 × 10 −3 m, k = 25 N/m, and V = 120 V, F4 springs = 4k ( z0 − z ) = 2 z 3 − (2.4 × 10 −3 m) z 2 + 3.82 × 10−10 m 3 = 0 ⇒ z = 0.537 mm, 1.014 mm. d) Stable equilibrium occurs if a slight displacement from equilibrium yields a force back toward the equilibrium point. If one evaluates the forces at small displacements from the equilibrium positions above, the 1.014 mm separation is seen to be stable, but not the 0.537 mm separation. ε0 εL (( L − x) L + xKL ) = 0 ( L + ( K − 1) x). D D 1 εL b) ∆U = (∆C )V 2 where C = C0 + 0 (−dx + dxK ) D 2 2 1  ε0 L dx  2 ( K − 1)ε0V L ⇒ ∆U =  ( K − 1) V = dx. 2 D 2D  c) If the charge is kept constant on the plates, then: C ε LV 1 1 Q= 0 ( L + ( K − 1) x), and U = CV 2 = C0V 2   C  D 2 2  0  C0V 2  εL ( K − 1)ε0V 2 L 1 − 0 ( K − 1)dx  ⇒ ∆U = U − U 0 = − ⇒U ≈ dx.  2  DC0 2D   d) Since dU = − Fdx = − 2 D0 dx, then the force is in the opposite direction to the motion dx, meaning that the slab feels a force pushing it out.
( K − 1) ε V 2 L

24.77: a) C0 =

24.78: a) For a normal spherical capacitor: C0 = 4πε0 parallel capacitors, C L and CU .

(

ra rb rb − ra

). Here we have, in effect, two

 rr   rr  KC 0 C = 2πKε 0  a b  and CU = 0 = 2πε 0  a b . r −r  r − r  2 2 a   b a  b b) Using a hemispherical Gaussian surface for each respective half: 4πr 2 QL QL 4πr 2 QU QU and EU EL = ⇒ EL = = ⇒ EU = . 2 2 Kε0 2πKε0 r 2 ε0 2πε0 r 2 But QL = VC L and QU = VCU , QL + QU = Q. KQ VC0 K Q So: QL = and QL = . = KQU ⇒ QU (1 + K ) = Q ⇒ QU = 2 1+ K 1+ K⋅ KQ 1 2 Q Q 1 2 Q ⇒ EL = = and EU = = . 2 2 2 1 + K 2πKε0 r 1 + K 4πε0 r 1 + K 2πKε0 r 1 + K 4πKε0 r 2 c) The free charge density on upper and lower hemispheres are: Q Qu Q Q (σ f r )U = U 2 = and (σ f r )U = = . 2 2 2 a a 4πra 4πra (1 + K ) 4πrb 4πrb (1 + K ) QL QL KQ KQ and (σ f r ) L = (σ f rs ) L = = = . 2 2 2 2 a 4πra 4πra (1 + K ) 4πrb 4πrb (1 + K ) ( K − 1) Q K K−1 Q d) σ ir = σ f r (1 − 1 K ) = = . 2 a a K 4πra K + 1 K + 1 4πra 2 ( K − 1) Q K K−1 Q σ i rb = σ f rb (1 − 1 K ) = = . 2 K 4πra K + 1 K + 1 4πrb 2 e) There is zero bound charge on the flat surface of the dielectric-air interface, or else that would imply a circumferential electric field, or that the electric field changed as we went around the sphere. CL = 24.79: a)

2  εA  2(4.2)ε0 (0.120 m) b) C = 2  = ⇒ C = 2.38 × 10 −9 F. −4 4.5 × 10 m d 

24.80: a) The capacitors are in parallel so: εeff WL ε0W ( L − h) Kε0Wh ε0WL  Kh h  C= = + = −  ⇒ K eff 1 + d d d d  L L Kh h   = 1 + − . L L  b) For gasoline, with K = 1.95 : L 1 L 1   full: K eff  h =  = 1.24; full: K eff  h =  = 1.48; 4 2 2 4   3L  3  full: K eff  h =  = 1.71. 4  4  c) For methanol, with K = 33 : L 1 L 1   full: K eff  h =  = 9; full: K eff  h =  = 17; 4 2 2 4   3L  3  full: K eff  h =  = 25. 4  4  d) This kind of fuel tank sensor will work best for methanol since it has the greater range of K eff values.

Chapter 25

25.1:

Q = It = (3.6 A)(3)(3600 s) = 3.89 × 10 4 C.
Q t

25.2: a) Current is given by I = b) I = nqvd A ⇒ vd =

=

420 C 80 ( 60 s )

= 8.75 × 10 −2 A.

I 8.75 × 10−2 A = nqA (5.8 × 1028 )(1.6 × 10−19 C)(π(1.3 × 10−3 m) 2 )

=1.78 × 10−6 m s . 25.3: a) vd = I 4.85 A = 28 −19 nqA (8.5 × 10 )(1.6 × 10 C)(π 4)(2.05 × 10− 3 m) 2 )
d vd 71 = 1.08 0.10 −m m s = 6574 s = 110 min 4 ×

= 1.08 × 10−4 m s ⇒ travel time = b) If the diameter is now 4.12 mm, the time can be calculated using the formula above or comparing the ratio of the areas, and yields a time of 26542 s =442 min. c) The drift velocity depends on the diameter of the wire as an inverse square relationship. 25.4: The cross-sectional area of the wire is A = πr 2 = π (2.06 × 10−3 m) 2 = 1.333 × 10−5 m 2 . The current density is I 8.00A J= = = 6.00 × 10 5 A m 2 A 1.333 × 10 −5 m 2 We have vd = J ne ; Therefore J 6.00 × 105 A m 2 electrons n= = = 6.94 × 1028 −5 −19 vd e (5.40 × 10 m s)(1.60 × 10 C electron) m3
J = n q vd , so J vd is constant.

25.5:

J1 vd 1 = J 2 vd 2 , vd 2 = vd 1 ( J 2 J1 ) = vd 1 ( I 2 I1 ) = (1.20 × 10−4 m s)(6.00 1.20) = 6.00 × 10−4 m s

25.6: The atomic weight of copper is 63.55 g mole, and its density is 8.96 g cm 3 . The number of copper atoms in 1.00 m 3 is thus (8.96 g cm 3 )(1.00 × 106 cm3 m3 )(6.023 × 1023 atoms mole ) 63.55 g mole = 8.49 × 10 28 atoms m 3 Since there are the same number of free electrons m 3 as there are atoms of copper m3 (see Ex. 25.1), The number of free electrons per copper atom is one. 25.7: Consider 1 m 3 of silver. density = 10.5 × 10 3 kg m 3 , so m = 10.5 × 10 3 kg M = 107.868 × 10 −3 kg mol , so n = m M = 9.734 × 104 mol and N = nN A = 5.86 × 1028 atoms m 3 If there is one free electron per m 3 , there are 5.86 × 1028 free electrons m 3 . This agrees with the value given in Exercise 25.2. 25.8: a) Qtotal = (nCl + n Na )e = (3.92 × 1016 + 2.68 × 1016 )(1.60 × 10 −19 C) = 0.0106 C Qtotal 0.0106 C = = 0.0106 A = 10.6 mA. t 1.00 s b) Current flows, by convention, in the direction of positive charge. Thus, current flows with Na + toward the negative electrode. ⇒I= 25.9: a) Q = ∫ I dt = ∫ (55 − 0.65 t 2 ) dt = 55t | + 0.65 3 8 t | = 329 C. 0 0 3 0 0 b) The same charge would flow in 10 seconds if there was a constant current of: I = Q t = (329 C) (8 s) = 41.1 A.
8 8 8

25.10: a) J =

I A

=

3.6 A ( 2.3 × 10 − 3 m ) 2

= 6.81 × 10 5 A/m 2 .

b) E = ρJ = (1.72 × 10 −8 Ω ⋅ m)(6.81 × 105 A/m 2 ) = 0.012 V m . c) Time to travel the wire’s length: l l nqA (4.0 m)(8.5 × 1028 m3 )(1.6 × 10 −19 C)(2.3 × 10 −3 m) 2 t= = = = 8.0 × 104 s vd I 3.6 A = 1333 min ≈ 22 hrs! 25.11: R = ρL (1.72 × 10 −8 Ω ⋅ m)(24.0 m) = = 0.125 Ω. A (π 4)(2.05 × 10− 3 m) 2

25.12: R =

ρL RA (1.00 Ω)(π 4)(0.462 × 10−3 m) 2 ⇒L= = = 9.75 m. A ρ 1.72 × 10−8 Ω ⋅ m

25.13: a) tungsten: ρI (5.25 × 10−8 Ω m 3 )(0.820 A) E = ρJ = = = 5.16 × 10 −3 V m . −3 2 A (π 4)(3.26 × 10 m) b) aluminum: ρI (2.75 × 10 −8 Ω m 3 )(0.820 A) E = ρJ = = = 2.70 × 10− 3 V m . A (π 4)(3.26 × 10− 3 m) 2 25.14: R Al = RCu
2 πd ρ L ρ L πd ⇒ Al = Cu ⇒ A l = Cu ⇒ d Cu = d Al AAl ACu 4 ρ Al 4 ρCu 2

ρCu ρ Al

⇒ d Al = (3.26 mm )

1.72 × 10−8 Ω ⋅ m = 2.6 mm. 2.75 × 10 −8 Ω ⋅ m

25.15: Find the volume of one of the wires: ρL ρL R= so A = and A R ρL2 1.72 × 10 −8 Ohm ⋅ m)(3.50m) 2 volume = AL = = = 1.686 × 10 − 6 mcb R 0.125Ohm 3 m = (density )V = (8.9 × 10 kg m 3 )(1.686 × 10 −6 m 3 ) = 15 g

25.16:

3.5 cm = 1.75 cm 2 3.25 mm r2 = = 1.625 mm 2 r1 =

25.17: a) From Example 25.1, an 18-gauge wire has A = 8.17 × 10 −3 cm 2 I = JA = (1.0 × 105 A/cm 2 )(8.17 × 10 −3 cm 2 ) = 820 A b) A = I J = (1000 A) (1.0 × 106 A cm 2 ) = 1.0 × 10−3 cm 2 A = πr 2 so r = A π = (1.0 × 10−3 cm 2 π = 0.0178 cm d = 2r = 0.36 mm 25.18: Assuming linear variation of the resistivity with temperature: ρ = ρ0 [1 + α(T − T0 )] = ρ0 [1 + (4.5 × 10 −3 °C)(320 − 20)°C] = 2.35 ρ0 Since ρ = E J , the electric field required to maintain a given current density is proportional to the resistivity. Thus E = (2.35)(0.0560 V m) = 0.132 V m 25.19: R = ρL ρL ρ 2.75 × 10−8 Ω ⋅ m = 2 = = = 1.53 × 10−8 Ω A L L 1.80m

25.20: The ratio of the current at 20°C to that at the higher temperature is (0.860 A) (0.220 A) = 3.909. Since the current density for a given field is inversely proportional to ρ( ρ = E J ), The resistivity must be a factor of 3.909 higher at the higher temperature. ρ = 1 + α (T − T0 ) ρ0 T = T0 +
ρ ρ0

−1

α

= 20°C +

3.909 − 1 = 666°C 4.5 × 10 −3 °C

25.21:

R=

V ρL ρL = = ⇒r= I A πr 2

IρL = πV

(6.00 A)(2.75 × 10 −8 Ω ⋅ m)(1.20 m) π (1.50 V)

= 2.05 × 10 −4 m.

25.22: ρ =

RA VA (4.50 V)π (6.54 × 10 −4 m) 2 = = = 1.37 × 10 −7 Ω ⋅ m. L IL (17.6 A)(2.50 m)

25.23: a) I = JA = b) V = IR =

EA (0.49 V m)(π 4 (0.84 × 10 −3 m) 2 ) = = 11.1 A. ρ (2.44 × 10 −8 Ω ⋅ m)

IρL (11.1 A)(2.44 × 10 −8 Ω ⋅ m)(6.4 m) = = 3.13 V. A (π 4)(0.84 × 10 −3 m) 2 V 3.13 V c) R = = = 0.28 Ω. I 11.1A

25.24: Because the density does not change, volume stays the same, so LA = (2 L)( A 2) and the area is halved. So the resistance becomes: ρ( 2 L) ρL R= =4 = 4 R0 . A2 A That is, four times the original resistance. RAJ RI V 0.938 V = = = = 1.25 V m . L L L 0.75 m RA V 0.938 V b) ρ = = = = 2.84 × 10−8 Ω ⋅ m. 7 2 L JL (4.40 × 10 A m )(0.75 m) R − R0 = α (T f − Ti ) R0 R − R0 1.512 Ω − 1.484 Ω ⇒α = = = 1.35 × 10 −3 °C −1 .   (T f − Ti ) R0 (34.0 C − 20.0 C)(1.484 Ω)

25.25: a) E = ρJ =

25.26:

25.27:a) R f − Ri = Riα (T f − Ti ) ⇒ R f = 100 Ω − 100 Ω(0.0004°C −1 )(11.5°C) = 99.54 Ω. b) R f − Ri = Riα (T f − Ti ) ⇒ R f = 0.0160 Ω + 0.0160 Ω(−0.0005°C −1 )(25.8°C) = 0.0158 Ω. 25.28: T f − Ti = R f − Ri ; T f = Ti + R f − Ri

αRi

αRi 215.8 Ω − 217.3 Ω = + 4 o C = 17.8o C. o −1 (−0.0005 C )(217.3 Ω)

25.29: a) If 120 strands of wire are placed side by side, we are effectively increasing the area of the current carrier by 120. So the resistance is smaller by that factor: R = 5.60 × 10−6 Ω 120 = 4.67 × 10−8 Ω. b) If 120 strands of wire are placed end to end, we are effectively increasing the length of the wire by 120, and so R = (5.60 × 10 −6 Ω)120 = 6.72 × 10 −4 Ω. 25.30: With the 4.0 Ω load, where r = internal resistance 12.6 V = (r + 4.0 Ω) I Change in terminal voltage: ∆VT = rI = 12.6 V − 10.4 V = 2.2 V I= Substitute for I: Solve for r: 25.31: a) R = 2.2 V r

 2.2 V  12.6 V = ( r + 4.0 Ω)   r  r = 0.846 Ω

ρL 1.72 × 10 −8 Ωm)(100 × 10 3 m) = = 0.219Ω A π (0.050m) 2 V = IR = (125A)(0.219Ω) = 27.4V b) P = VI = (27.4 V)(125 A) = 3422 W = 3422 J/s Energy = Pt = (3422 J/s)(3600 s) = 1.23 × 10 7 J

25.32: a) Vr = ε − Vab = 24.0 V − 21.2 V = 2.8 V ⇒ r = 2.8 V 4.00 A = 0.700 Ω . b) VR = 21.2 V ⇒ R = 21.2 V 4.00 A = 5.30 Ω. 25.33: a) An ideal voltmeter has infinite
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