The velocity of a body moving in a straight line is increased by applying a constant force $F$, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

(N/A) According to the third equation of motion: $v^{2} - u^{2} = 2as$
From this, the displacement $s$ can be expressed as: $s = \frac{v^{2} - u^{2}}{2a}$
According to Newton's second law of motion, the force applied is: $F = ma$
Work done $(W)$ by the force $F$ over a distance $s$ is defined as: $W = F \times s$
Substituting the expressions for $F$ and $s$: $W = (ma) \times \left( \frac{v^{2} - u^{2}}{2a} \right)$
Simplifying the expression: $W = \frac{1}{2}m(v^{2} - u^{2}) = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$
Since kinetic energy $(K.E.)$ is given by $\frac{1}{2}mv^{2}$, we get: $W = (K.E.)_{f} - (K.E.)_{i}$
Thus, the work done is equal to the change in kinetic energy.