Question

Consider a rain drop of mass $1\text{g}$ falling from a height of $2\text{km}$. It hits the ground with a speed of $50\text{m/s}$. The work done by resistive force of air is
$(g=10{\text{m/s}}^2)$

• A. $-187.5\text{J}$
• B. $43.75\text{J}$
• C. $-18.75\text{J}$
• D. $14.375\text{J}$

Answer 1

The correct option is C $-18.75\text{J}$

We know that by the work energy theorem, net work done $=$ Change in kinetic energy

The sum of the workdone by gravity and work done by the resistive force in air will be equal to the change in the kinetic energy of a rain drop.

The initial velocity, $v_1=0$ as it is dropped and let the final velocity be $v_2=50\text{m/s}$ as given in the problem and the mass is given as $m=1\text{g}=\frac{1}{1000}\text{kg}$

We know, $W_{\text{gravity}=}mgh,$ Change in kinetic energy $=\frac{1}{2}m(v_2^2-v_1^2)$
$\Rightarrow W_{\text{gravity}}+W_{\text{resistive force}}=\frac{1}{2}m(v_2^2-v_1^2)$

$\Rightarrow mgh+W_{\text{resistive force}}=\frac{1}{2}m(v_2^2-0^2)$

$\Rightarrow \left(\frac{1}{1000}\right)\times 10\times 2000+W_{\text{resistive force}}=\frac{1}{2}\times \frac{1}{1000}\left({50}^2\right)$

$W_{\text{resistive force}}=1.25-20$

The work done by the resistive force is $=-18.75\text{J}$

Hence option C is the correct answer