The rain drop of mass $1$ g, starts with zero velocity from a height of $1$ km. It hits the ground with a speed of $5\ \text{m/s}$. The work done by the unknown resistive force is ______ J. (Take $g = 10\ \text{m/s}^2$)
Answer: (D) $-9.98$
Concept. Two forces do work on the drop: gravity and the air resistance. The work–energy theorem, $W_g+W_r=\Delta K$, gives the unknown resistive work once the gravity work and the change in kinetic energy are known.
Step 1. Work by gravity. The drop falls $h=1$ km $=1000$ m, with $m=1$ g $=10^{-3}$ kg:
$$W_g=mgh=10^{-3}\times10\times1000=10\ \text{J}$$
Step 2. Change in kinetic energy. It starts at rest and lands at $5$ m/s:
$$\Delta K=\tfrac12\times10^{-3}\times5^2-0=0.0125\ \text{J}$$
Step 3. Work by the resistive force.
$$W_r=\Delta K-W_g=0.0125-10=-9.9875\ \text{J}$$
Answer. Option D, $-9.98$ J (to the given precision). The sign is negative because the resistance opposes the motion; almost all the potential energy is lost to the air.
Solution by Sreeraj P, M.Sc Physics
