Explain work energy theorem.

For straight line motion : If body is moving with initial velocity $u$ and constant acceleration $a$, it acquired velocity $v$ with displacement $s$ then from the third equation of motion, $v^{2}-u^{2}=2 a s$

multiplying $\frac{m}{2}$ on both sides of equation,

$\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}=m a s$

$\mathrm{~K}_{f}-\mathrm{K}_{i}=\mathrm{Fs}[\because m a=\mathrm{F}]$

For motion in three dimension :

$v^{2}-u^{2}=2 \vec{a} \cdot \vec{d}$

where $v=$ final velocity, $u=$ initial velocity, $\vec{a}=$ constant acceleration and $\vec{d}=$ displacement Multiplying by $\frac{m}{2}$ on both side of equation, $\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}=m \vec{a} \cdot \vec{d}$

$=\overrightarrow{\mathrm{F}} \cdot \vec{d} \quad[\because m \vec{a}=\overrightarrow{\mathrm{F}}]$

The left hand side is the change in kinetic energy of body, while right hand is the multiplication of force and the displacement. It is known as work and it is denoted by $\mathrm{W}$.

$\therefore \mathrm{K}_{f}-\mathrm{K}_{i}=\mathrm{W}$ and $\Delta \mathrm{K}=\mathrm{W}$

This relation is known as work energy theorem.