$A$ rain drop of radius $2 \; mm$ falls from a height of $500 \; m$ above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height,it attains its maximum (terminal) speed,and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is $10 \; m s^{-1}$?

(N/A) Radius of the rain drop,$r = 2 \; mm = 2 \times 10^{-3} \; m$.
Volume of the rain drop,$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (2 \times 10^{-3})^3 \; m^3 = 3.35 \times 10^{-8} \; m^3$.
Mass of the rain drop,$m = \rho V = 10^3 \; kg/m^3 \times 3.35 \times 10^{-8} \; m^3 = 3.35 \times 10^{-5} \; kg$.
Gravitational force,$F_g = mg = 3.35 \times 10^{-5} \times 9.8 \; N = 3.283 \times 10^{-4} \; N$.
Work done by gravity in the first half $(h_1 = 250 \; m)$: $W_1 = F_g \times h_1 = 3.283 \times 10^{-4} \times 250 = 0.082 \; J$.
Work done by gravity in the second half $(h_2 = 250 \; m)$: $W_2 = F_g \times h_2 = 0.082 \; J$.
Total work done by gravity $W_g = W_1 + W_2 = 0.164 \; J$.
By Work-Energy Theorem,$W_g + W_r = \Delta K = \frac{1}{2}mv^2 - 0$.
$W_r = \frac{1}{2} \times (3.35 \times 10^{-5}) \times (10)^2 - 0.164 = 1.675 \times 10^{-3} - 0.164 = -0.1623 \; J$.