structural:expansion, information:classification
Derivation for a particle moving along a straight line

In the case the resultant force F is constant in both magnitude and direction, and parallel to the velocity of the particle, the particle is moving with constant acceleration a along a straight line. The relation between the net force and the acceleration is given by the equation F = ma (Newton's second law), and the particle displacement s can be expressed by the equation s = v 2 2 − v 1 2 2 a {\displaystyle s={\frac {v_{2}^{2}-v_{1}^{2}}{2a}}} which follows from v 2 2 = v 1 2 + 2 a s {\displaystyle v_{2}^{2}=v_{1}^{2}+2as} (see Equations of motion).

The work of the net force is calculated as the product of its magnitude and the particle displacement. Substituting the above equations, one obtains: W = F s = m a s = m a v 2 2 − v 1 2 2 a = m v 2 2 2 − m v 1 2 2 = Δ E k {\displaystyle W=Fs=mas=ma{\frac {v_{2}^{2}-v_{1}^{2}}{2a}}={\frac {mv_{2}^{2}}{2}}-{\frac {mv_{1}^{2}}{2}}=\Delta E_{\text{k}}}

Other derivation: W = F s = m a s = m v 2 2 − v 1 2 2 s s = 1 2 m v 2 2 − 1 2 m v 1 2 = Δ E k {\displaystyle W=Fs=mas=m{\frac {v_{2}^{2}-v_{1}^{2}}{2s}}s={\frac {1}{2}}mv_{2}^{2}-{\frac {1}{2}}mv_{1}^{2}=\Delta E_{\text{k}}}

In the general case of rectilinear motion, when the net force F is not constant in magnitude, but is constant in direction, and parallel to the velocity of the particle, the work must be integrated along the path of the particle: W = ∫ t 1 t 2 F ⋅ v d t = ∫ t 1 t 2 F v d t = ∫ t 1 t 2 m a v d t = m ∫ t 1 t 2 v d v d t d t = m ∫ v 1 v 2 v d v = 1 2 m (v 2 2 − v 1 2). {\displaystyle W=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot \mathbf {v} dt=\int _{t_{1}}^{t_{2}}F\,v\,dt=\int _{t_{1}}^{t_{2}}ma\,v\,dt=m\int _{t_{1}}^{t_{2}}v\,{\frac {dv}{dt}}\,dt=m\int _{v_{1}}^{v_{2}}v\,dv={\tfrac {1}{2}}m\left(v_{2}^{2}-v_{1}^{2}\right).}