# A model train with a mass of 9 kg is moving along a track at 15 (cm)/s. If the curvature of the track changes from a radius of 46 cm to 35 cm, by how much must the centripetal force applied by the tracks change?

Feb 22, 2017

The variation in centripetal force is $= 0.14 N$

#### Explanation:

The centripetal force is

$F = \frac{m {v}^{2}}{r}$

$m = m a s s = 9 k g$

$v = 0.15 m {s}^{-} 1$

The variation in centripetal force is

$\Delta F = {F}_{2} - {F}_{1}$

${F}_{1} = 9 \cdot {0.15}^{2} / 0.46 = 0.44 N$

${F}_{2} = 9 \cdot {0.15}^{5} / 0.35 = 0.58 N$

$\Delta F = 0.58 - 0.44 = 0.14 N$
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