Question

A rain drop of radius $2mm$ falls from a height of $500m$ above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed and moves with uniform speed there after. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on touching the ground is $10m{s}^{-1}$

Answer 1

Given:

Radius of the rain drop $r=2mm=2\times {10}^{-3}m$

Density of water, $\rho ={10}^{3}kg{m}^{-3}$

The mass contained in a rain drop,

$m=\rho V$

$\Rightarrow \frac{4}{3}\times 3.141\times (2\times {10}^{-3}{)}^3\times {10}^{3}kg$

Gravitational force experienced by the rain drop,

$F=mg$

$\Rightarrow \frac{4}{3}\times 3.141\times \left(\right(2\times {10}^{-3}{)}^3\times {10}^3\times 9.8)N$

The work done by the gravity on the drop is given by:

$W_1=Fs$

$=\frac{4}{3}\times 3.14\times (2\times {10}^3{)}^3\times {10}^3\times 9.8\times 250=0.082J$

The work done on the drop in the second half of the journey will be same as that in the first half.

$\therefore W_2=0.082J$

The total energy of the drop remains conserved during its motion.

Total energy at the top:

$E_T=mgh+0$

$=4/3\times 3.141\times ({10}^{-3}{)}^3\times {10}^3\times 500\times {10}^{-5}$

$=0.164J$

Due to the presence of a resistive force, the drop hits the ground with a velocity of $10m/s$.

∴Total energy at the ground:

$E_G=\frac{1}{2}m{v}^2+0=1/2\times 4/3\times 3.141\times (2\times {10}^{-3}{)}^3\times {10}^3\times 9.8\times {10}^2=1.675\times {10}^{-3}J$

So, the work done by the resistive force = $E_G-E_T=-0.162J$