Explain the work-energy theorem.

(N/A) For straight-line motion: If a body is moving with initial velocity $u$ and constant acceleration $a$, it acquires velocity $v$ after displacement $s$. From the third equation of motion, $v^{2} - u^{2} = 2as$.
Multiplying by $\frac{m}{2}$ on both sides of the equation:
$\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2} = mas$
Since $F = ma$, we get:
$K_{f} - K_{i} = Fs = W$
For motion in three dimensions:
$v^{2} - u^{2} = 2\vec{a} \cdot \vec{d}$
where $v$ is final velocity, $u$ is initial velocity, $\vec{a}$ is constant acceleration, and $\vec{d}$ is displacement.
Multiplying by $\frac{m}{2}$ on both sides:
$\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2} = m\vec{a} \cdot \vec{d}$
$= \vec{F} \cdot \vec{d} = W$
The left-hand side represents the change in kinetic energy $(\Delta K)$ of the body, while the right-hand side represents the work done $(W)$ by the force.
Therefore, $\Delta K = W$. This relation is known as the work-energy theorem.