It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass $1.00 \; g$ falling from a height $1.00 \; km$. It hits the ground with a speed of $50.0 \; m s^{-1}$. $(a)$ What is the work done by the gravitational force? $(b)$ What is the work done by the unknown resistive force?

(N/A) The change in kinetic energy of the drop is $\Delta K = \frac{1}{2} m v^2 - 0 = \frac{1}{2} \times 10^{-3} \; kg \times (50.0 \; m s^{-1})^2 = 1.25 \; J$,assuming the drop starts from rest.
Taking $g = 10 \; m s^{-2}$,the work done by the gravitational force is $W_g = mgh = 10^{-3} \; kg \times 10 \; m s^{-2} \times 1000 \; m = 10.0 \; J$.
$(b)$ According to the work-energy theorem,$\Delta K = W_g + W_r$,where $W_r$ is the work done by the resistive force.
Thus,$W_r = \Delta K - W_g = 1.25 \; J - 10.0 \; J = -8.75 \; J$.