Question

A rain drop of mass $1\text{g}$ falling from a height of $1\text{km}$ hits the ground with a speed of $50\text{m/s}$. If a resistive force acts on the drop, then the work done by the resistive force is (Take $g:10{\text{m/s}}^2)$

• A. $-10\text{J}$
• B. $10\text{J}$
• C. $8.75\text{J}$
• D. $-8.75\text{J}$

Answer 1

The correct option is D $-8.75\text{J}$

Given:
$m=1\text{g}={10}^{-3}\text{kg}$
$h=1\text{Km}=1000\text{m}={10}^3\text{m}$
The change in kinetic energy of the drop is
$\Delta K=\frac{1}{2}m{v}^2-0(\because u=0)$
= $\frac{1}{2}\times {10}^{-3}\times 50\times 50=1.25\text{J}$
The work done by the gravitational force is
$W_g=mgh={10}^{-3}\times 10\times {10}^3=10\text{J}$
According to work-energy theorem,
$\Delta K=W_g+W_r$
Where $W_r$ is the work done by the resistive force on the rain drop
$W_r=\Delta K-W_g=1.25\text{J}-10\text{J}=-8.75\text{J}$