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) Let the initial velocity of the body be $u$ and the final velocity be $v$ after traveling a distance $S$ under the influence of a constant force $F$.
According to the third equation of motion, $v^{2} - u^{2} = 2aS$, where $a$ is the acceleration.
From this, the distance $S$ can be expressed as $S = \frac{v^{2} - u^{2}}{2a}$.
According to Newton's second law of motion, the force applied is $F = ma$.
The work done $W$ by the force is defined as $W = F \times S$.
Substituting the values of $F$ and $S$ into the work equation:
$W = (ma) \times \left( \frac{v^{2} - u^{2}}{2a} \right)$.
Simplifying the expression by canceling $a$:
$W = \frac{1}{2}m(v^{2} - u^{2}) = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$.
Since kinetic energy $K = \frac{1}{2}mv^{2}$, the expression becomes $W = K_{final} - K_{initial} = \Delta K$.
Thus, the work done by the force is equal to the change in the kinetic energy of the body.