Derive the work energy theorem for a variable force exerted on a body in one dimension.

If a body of mass $m$ and speed $v$ moving in $\mathrm{X}$-direction in one dimension then its kinetic energy, $\mathrm{K}=\frac{1}{2} m v^{2}$

Intergrating on both the side,

$\frac{d \mathrm{~K}}{d t}=\frac{d}{d t}\left(\frac{1}{2} m v^{2}\right)$

$=\frac{1}{2} m \times 2 v \cdot \frac{d v}{d t}$

$\therefore \frac{d \mathrm{~K}}{d t}=\operatorname{mav}\left[\because \frac{d v}{d t}=a\right]$

$\therefore \frac{d \mathrm{~K}}{d t}=\mathrm{F} \frac{d x}{d t}\left[\because m a=\mathrm{F}\right.$ and $\left.v=\frac{d x}{d t}\right]$

$\therefore d \mathrm{~K}=\mathrm{F} d x$

Intergrating on both side from initial position $x_{i}$ to final position $x_{f}$

$\int_{\mathrm{K}_{i}}^{\mathrm{K} f} d \mathrm{~K}=\int_{x_{i}}^{x_{f}} \mathrm{~F} d x$

$\mathrm{~K}_{f}-\mathrm{K}_{i}=\mathrm{F}\left(x_{f}-x_{i}\right)$

$\quad=\mathrm{F} \Delta x \quad \text { where } \Delta x=x_{f}-x_{i}$

which is a work energy theorem for a variable force.

Work energy theorem does not incorporate the complete dynamical information of Newton's second law.

It is an integral form of Newton's second law.

Wherk energy theorem involves an integral over an interval of time, the time information contained in the Newton's second law is integrated over and is not available explicitly. Newton's second law for two or three dimensions is in vector form whereas the work energy theorem is in scalar form.