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 thereafter. 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

Hint: Calculate the energy difference between the energy of rain drop on the top and energy of rain drop on the ground. To find energies, you need mass of the raindrop to calculate mass first from density and radius of the raindrop.
Formula used:
$density=\dfrac {mass}{volume}$
$F=mg$
$W= Force \times Displacement$

Complete answer:
Let’s first mention whatever information is given in the question.
Given: Radius of raindrop (r)= 2mm = $2\times{10}^{-3}$
             Height (h)=500m
             Velocity (v)= $10m{s}^{-1}$
To calculate mass (m) of the raindrop,
$Density=\dfrac {mass}{volume}$...(1)
We know, density of water ($\rho$) is ${10}^{3}kg{m}^{-3}$ …(2)
And volume (V) of the raindrop is given as,
$V=\dfrac {4}{3} \pi {r}^{3}$ …(3)
Now, rearranging equation.(1) and then substituting equation.(2) and equation.(3) in it.
$\Rightarrow m =\dfrac {4}{3} \pi {r}^{3} \times {10}^{3}$
$\Rightarrow m =\dfrac {4}{3} \pi { \left( 2\times {10}^{-3} \right) }^{3}\times {10}^{3}$
$\Rightarrow m =3.35 \times {10}^{-5}kg $
Force experienced by the raindrop is given as,
$F=mg$
$\Rightarrow F=3.35\times {10}^{-5}\times 9.8$
$\Rightarrow F=32.83\times {10}^{-5}N$
Now, work done by the gravity on the drop is given by.
 $W= Force \times Displacement$
 $\Rightarrow W= F \times s$ …(4)
 $s=\dfrac {500}{2} =250$
Then, substitute value of s in equation.(4)
$\Rightarrow W= 32.83 \times {10}^{-5} \times 250$
$\Rightarrow W= 82.07 \times {10}^{-3}J$
We know, work done by the resistive forces is the difference between the energy on ground and energy of the rain drop at the top.
$\Rightarrow W={E}_{1}-{E}_{2}$ …(5)
where, ${E}_{1}$: Energy of rain drop at top
             ${E}_{2}$: Actual energy
${E}_{1}=mgh$
$\Rightarrow {E}_{1}= 32.83 \times {10}^{-5}\times 500$
$\Rightarrow {E}_{1}= 0.164J$ …(6)
Now, ${E}_{2}= \dfrac {1}{2} m{v}^{2}$
$\Rightarrow {E}_{2}=\dfrac {1}{2} \times 3.35 \times {10}^{-5}\times {10}^{2}$
$\Rightarrow {E}_{2}= 1.675 \times {10}^{-3}J$ …(7)
Then, substitute equation.(6) and equation.(7) in equation.(5)
$\Rightarrow W= 0.164- \left( 1.675\times {10}^{-3} \right)$
$\therefore W= -0.162J$
Hence, work done by the resistive force in its entire journey is 0.162J.

Note:
Whether the drop moves with uniform velocity or decreasing acceleration, work done by the gravitational force on the rain drop remains the same. That is. The work done by the gravitational force on the rain drop in the first half of the journey is equal to the second half.