$(a)$ $A$ person is holding a bucket of water by applying a vertical force of $100 \, N$. He first moves a horizontal distance of $5 \, m$ and then climbs up through stairs a vertical distance of $10 \, m$. Calculate the total work done by the person on the bucket.
$(b)$ Establish the relationship between kinetic energy and momentum of a body.

(A) Given: Force $F = 100 \, N$,horizontal displacement $S_H = 5 \, m$,vertical displacement $S_V = 10 \, m$.
Let $W_1$ be the work done during horizontal movement and $W_2$ be the work done during vertical movement.
For horizontal movement,the force is vertical and displacement is horizontal,so the angle $\theta = 90^{\circ}$.
$W_1 = F \cdot S_H \cdot \cos(90^{\circ}) = 100 \times 5 \times 0 = 0 \, J$.
For vertical movement,the force is vertical and displacement is vertical,so the angle $\theta = 0^{\circ}$.
$W_2 = F \cdot S_V \cdot \cos(0^{\circ}) = 100 \times 10 \times 1 = 1000 \, J$.
Total work done $W = W_1 + W_2 = 0 + 1000 = 1000 \, J$.
$(b)$ Kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$ and momentum $p$ is given by $p = mv$.
From $p = mv$,we get $v = \frac{p}{m}$.
Substituting $v$ in the kinetic energy formula:
$K = \frac{1}{2}m \left(\frac{p}{m}\right)^2 = \frac{1}{2}m \left(\frac{p^2}{m^2}\right) = \frac{p^2}{2m}$.
Thus,the relationship is $K = \frac{p^2}{2m}$.