Question

A rain drop of radius 2 mm falls from a height of 500 m 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 reaching the ground is 10 m s^–1 ?

Answer 1

Given

• Radius of raindrop: \( r = 2 \,\text{mm} = 2 \times 10^{-3} \,\text{m} \)
• Height from which it falls: \( H = 500 \,\text{m} \)
• At half the height (i.e. after falling 250 m), it attains terminal speed and then falls with uniform speed.
• Density of water: \( \rho = 10^{3} \,\text{kg m}^{-3} \)
• Speed on reaching the ground: \( v = 10 \,\text{m s}^{-1} \)
• Acceleration due to gravity: \( g \approx 9.8 \,\text{m s}^{-2} \)

1. Mass of the raindrop

Volume of the spherical drop:

\( V = \dfrac{4}{3} \pi r^{3} = \dfrac{4}{3} \pi (2 \times 10^{-3})^{3} = \dfrac{32\pi}{3} \times 10^{-9} \,\text{m}^{3} \)

Mass:

\( m = \rho V = 10^{3} \times \dfrac{32\pi}{3} \times 10^{-9} = \dfrac{32\pi}{3} \times 10^{-6} \,\text{kg} \approx 3.35 \times 10^{-5} \,\text{kg} \)

2. Work done by gravity in first and second half

Gravitational force is constant and conservative, so work done depends only on vertical displacement, not on the details of motion. Each half of the journey corresponds to a fall of \( 250 \,\text{m} \).

First half (from 500 m to 250 m)

\( W_{g1} = m g h_1 = (3.35 \times 10^{-5})(9.8)(250) \approx 0.082 \,\text{J} \)

Second half (from 250 m to 0 m)

\( W_{g2} = m g h_2 = (3.35 \times 10^{-5})(9.8)(250) \approx 0.082 \,\text{J} \)

Work by gravity in first half \( \approx 0.082 \,\text{J} \); Work by gravity in second half \( \approx 0.082 \,\text{J} \).

3. Work done by the resistive (drag) force

Total work done by all forces equals the change in kinetic energy:

\( W_{\text{gravity}} + W_{\text{resistive}} = \Delta K \)

Total work by gravity

\( W_{\text{gravity}} = W_{g1} + W_{g2} \approx 0.082 + 0.082 = 0.164 \,\text{J} \)

Change in kinetic energy

Raindrop starts from rest and reaches speed \( v = 10 \,\text{m s}^{-1} \) at the ground:

\( \Delta K = \dfrac{1}{2} m v^{2} - 0 = \dfrac{1}{2} (3.35 \times 10^{-5})(10^{2}) \approx 1.68 \times 10^{-3} \,\text{J} \)

Work by resistive force

\( W_{\text{resistive}} = \Delta K - W_{\text{gravity}} \approx 1.68 \times 10^{-3} - 0.164 \approx -0.162 \,\text{J} \)

Work done by the resistive force in the entire journey \( \approx -0.16 \,\text{J} \) (negative sign indicates energy loss due to drag).