Question

A raindrop 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

Given the radius of raindrop is spherical in shape. Understand the question and find the work done in 2 cycles, the first cycle, where the droplet falls with viscous resistance of air for half of its original height, and the second cycle from half its height to ground.

Complete step by step answer:
Here it is given that the radius of the spherical droplet is $r = 2mm$ and the total height of the journey is 500m.
Now considering the density of water $\rho = {10^3}kg/{m^3}$, we calculate the mass of the raindrop. Now, mass is calculated as a product of density and volume.
Therefore,
$m = \rho \times V$,Where V is the volume of the sphere. We know the volume of sphere as $V = \dfrac{4}{3}\pi {r^3}$
$m = {10^3} \times \dfrac{4}{3}\pi {r^3}$
$m = {10^3} \times \dfrac{4}{3} \times \dfrac{{22}}{7} \times {(2 \times {10^{ - 3}})^3}$
$m = 3.35 \times {10^{ - 5}}Kg$
Now, we know that work done can be calculated using Force and distance.
$W = F \times s$
Where for the first cycle s is half the total. Hence , s is 250m.
$W = m \times s \times g$
$\Rightarrow W = 3.35 \times {10^{ - 5}} \times 250 \times 9.81$
Therefore the work done on the first cycle is given as
$W = 0.082J$.
Now work done for the second half of the cycle will remain the same as the work done on the first half of the cycle. Which means that W=W1=W2
Since the total work is the same in both half of the cycles, the total energy of the motion is conserved.
This means that the work done by gravitational force on the droplet remains constant.
Hence,
${E_1} = mgh$(Since, at the top, there won’t be any kinetic energy)
${E_1} = mgh = 3.35 \times {10^{ - 5}} \times 500 \times 9.81$
Therefore energy at top is equal to ${E_1} = 0.164J$
The resistive force to reach the ground is given as $10m{s^{ - 1}}$
Therefore, energy at the ground level
${E_2} = 0 + \dfrac{1}{2}m{v^2}$
${E_2} = \dfrac{1}{2} \times 3.35 \times {10^{ - 5}} \times {(10)^2}$
${E_2} = 1.675 \times {10^{ - 3}}J$
Now, work done on the drop is equal to the difference between energy of the particle before experiencing a drop and after experiencing a drop.
$W = {E_1} - {E_2}$
$W = 0.164 - 0.001675$
$W = 0.1623J$

Note: A resistive force is defined as the force whose direction of application is opposite to the velocity of the object. An example of a resistive force is friction.