Work energy theorem states that the change in kinetic energy of an object is equal to the net work done on it by the net force.
Let us suppose that a body is initially at rest and a force \(F^{1}\) is applied on the body to displace it through \(d \bar{S}\) along the direction of the force. Then, small amount of work done is given by
$$
d W=F^{\prime}-d s=F d S
$$
Also, according to Newton's second law of motion, we have
$$
F=m a
$$
where a is acceleration produced (in the direction of force) on applying the force. Therefore,
$$
\begin{aligned}
&d W=M a d S=M \frac{d v}{d t} d S \\
&O R d W=M \frac{d S}{d t} d v=M v d v
\end{aligned}
$$
Now, work done by the force in order to increase its velocity from u (initial velocity) to \(v\) (Final velocity) is given by
$$
\begin{aligned}
&\mathrm{W}=\int_{\mathrm{u}}^{v} M v d v=M \int_{\mathrm{u}}^{v} v d v=M\left|\frac{v^{2}}{2}\right|_{u}^{v} \\
&\mathrm{~W}=\frac{1}{2} M v^{2}-\frac{1}{2} M u^{2}
\end{aligned}
$$
Hence, work done on a body by a force is equal to the change in its kinetic energy.
