kinetic energy

\[E_{k}\] is given as \[\frac{1}{2}mv^{2}\]. Assuming \[a\] is constant:

\begin{align*} s&=vt-\frac{1}{2}at^{2}\,,v=at\\ &=(at)\cdot t-\frac{1}{2}at^{2}\\ &=\frac{at^{2}}{2}\\ \\ W&=F\cdot s\\ &=ma\cdot \frac{at^{2}}{2}\\ &=\frac{1}{2}mv^{2}\\ \end{align*}

A more generalised version proof is as such. Start with \[W=\int F\,ds\], \[F=ma\] and applying chain rule:

\begin{align*} W&=\int F\,ds\\ &=m\int a\,ds\\ &=m\int \frac{dv}{dt}\,ds\quad\text{substitite $a=\frac{dv}{dt}$}\\ &=m\int \frac{dv}{ds}\frac{ds}{dt}\,ds\\ &=m\int vv^{\prime}\,ds\quad\text{substitute $v=\frac{ds}{dt}$ and $v^{\prime}=\frac{dv}{ds}$}\\ \end{align*}

With this form, we can now include the definite limits for integration and apply integration by substitution. Let \[f(x)=x\],

\begin{align*} m\int^{s_{f}}_{s_{i}} v(s)v^{\prime}(s)\,ds&=m\int^{s_{f}}_{s_{i}}f(v(s))\cdot v^{\prime}(s)\,ds\\ &=m\int^{v(s_{f})}_{v(s_{i})}f(u)\,du\\ &=m\int^{v(s_{f})}_{v(s_{i})}u\,du\\ &=m\biggl[ \frac{u^{2}}{2} \biggr]^{v(s_{f})}_{v(s_{i})}\\ &=m \left( \frac{v(s_{f})^{2}}{2}-\frac{v(s_{i})^{2}}{2} \right)\\ \end{align*}

This result is known the work-energy theorem. Assuming initial kinetic energy is zero and let \[v=v(s_{f})\], we get \[W=\frac{1}{2}mv^{2}\].