The work–kinetic energy theorem
Work–kinetic energy theorem
The net work done on a system by the net force acting on the system equals the change in kinetic energy of the system:
\[\sum{W} = \Delta E_{\mathrm{K}} = \frac12 m v^2 - \frac12 m v_0^2,\]
where \(v_0\) is the initial speed and \(v\) is the final speed of the system.
Proof in one dimension
We have seen that the work–kinetic energy theorem always holds true for linear motion with constant acceleration. Can we prove it in a more general case when the acceleration \(a\) is not constant?
The total work done on an object is \(\sum W = \int_{x_0}^{x_1}\left(\sum\limits{F}\right)\mathrm{d}x.\)
Notice that according to Newton's 2nd law, \(\sum\limits{F} = m a\), so \(\sum W = \int_{x_0}^{x_1}{\left(m a\right)}\mathrm{d}x = m \int_{x_0}^{x_1}\left(\frac{\mathrm{d}v}{\mathrm{d}t}\right)\mathrm{d}x.\)
We can apply the chain rule: \(\frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}t} = v \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{v^2}{2}\right).\)
Thus the integral becomes \(\sum W = m\int_{x_0}^{x_1}{\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{v^2}{2}\right)}\mathrm{d}x = m\left[\frac{v^2}{2}\right]_{v_0}^{v_1} = \frac12 m v_1^2 - \frac12 m v_0^2 = \Delta E_{\mathrm{K}}.\) This is what we have set out to prove.