Question

A raindrop of mass $1\, g$ falling from a height of $1 \,km$ hits the ground with a speed of $50\, ms$ . If the resistive force is proportional to the speed of the drop, then the work done by the resistive force is (Take $g= 10 \,ms^{-2}$)

• A. $10 \,J$
• B. $-10 \,J$
• C. $8.75 \,J$
• D. $-8.75 \,J$

Answer 1

Answer: (D)

$-8.75 \,J$

Answer 2

Here, $m = 1 \,g = 10^{-3}\, kg$ $h = 1\, km = 1000\, m = 10^3\, m$ The change in kinetic energy of the drop is $\Delta K = \frac{1}{2}mv^{2}-0$ $= \frac{1}{2}\times10^{-3}\times 50\times 50$ $= 1.25\,J \,\,\left(\because u = 0\right)$ The work done by the gravitational force is $W_{g} = mgh = 10^{-3} \times 10 \times 10^{3} = 10 \,J$ According to work-energy theorem $\Delta K = W_{g} + W_{r}$ where $W_{r}$ is the work done by the resistive force on the raindrop. Thus $W_{r} = \Delta K-W_{g} = 1.25\,J-10\,J$ $= -8.75\,J$