Consider that a body moves through a displacement $\vec s$ along positive x-direction under the action of a constant net force $\vec F$. Also note that ${v_{ix}}$ is the initial velocity and ${v_{fx}}$ is the final velocity of the body. Newton's second law gives,

\[{F} = m{a_x}\]

We have ${v_{fx}}^2 = {v_{ix}}^2 + 2{a_x}s{\rm{ }}$. Therefore, putting the value of $a_x$ from ${v_{fx}}^2 = {v_{ix}}^2 + 2{a_x}s{\rm{ }}$ in the above equation, we get,

\[\begin{align*} F &= m\left( {\frac{{{v_{fx}}^2 - {v_{ix}}^2}}{{2s}}} \right)\\ {\rm{or,}}\quad Fs &= \frac{1}{2}m{v_{fx}}^2 - \frac{1}{2}m{v_{ix}}^2\\ {\rm{or,}}\quad W &= \frac{1}{2}m{v_{fx}}^2 - \frac{1}{2}m{v_{ix}}^2 \tag{2} \label{2} \end{align*}\]

In Equation \eqref{2} the quantity $\frac{1}{2}m{v^2}$ which is the half of the product of mass and velocity squared is called kinetic energy of the body. It means the kinetic energy of a moving body with velocity $v$ at any instant of time is $\frac{1}{2}m{v}^2$. So, $\frac{1}{2}m{v_{ix}}^2$ is the initial kinetic energy and $\frac{1}{2}m{v_{fx}}^2$ is the final kinetic energy. Therefore, ${K_1} = \frac{1}{2}m{v_{ix}}^2$ and ${K_2} = \frac{1}{2}m{v_{fx}}^2$, therefore

\[W = {K_2} - {K_1} = \Delta K \tag{3} \label{3}\]

Equation \eqref{3} is also called work-energy theorem. You can conclude from Equation \eqref{3} that the work done by a net or resultant force on a body is equal to the change in kinetic energy of the body.

The above derivation of work done as the change in kinetic energy is based on a constant force applied but the work-energy theorem given by Equation \eqref{3} or Equation \eqref{2} is true even if the force is not constant or the motion is not always along a straight line.