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

Radius of the rain drop, r = 2 mm = 2 × 10 ^- 3 m

Volume of the rain drop, \(V = \frac{4 }{ 3} π r^3 \)

= \(\frac{4 }{ 3} \)× 3.14 × (2 × 10 ^-3)³m ^{- 3}
Density of water, ρ = 10³ kg m ^-3
Mass of the rain drop, m = ρV 

= \(\frac{4 }{ 3} \) × 3.14 × ( 2 × 10 ^{- 3})³ × 10 ³ kg
Gravitational force, F = mg 

= \(\frac{4 }{ 3} \)× 3.14 × ( 2 × 10^-3)³ × 10 ³ ×9.8 N
The work done by the gravitational force on the drop in the first half of its journey: 

W_I = Fs 

= \(\frac{4 }{ 3} \)× 3.14 × (2 × 10 ^{- 3})³ × 10 ³ × 9.8 × 250 

= 0.082 J
This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey, i.e., W_II, = 0.082 J
As per the law of conservation of energy, if no resistive force is present, then the total energy of the rain drop will remain the same. 

∴ Total energy at the top: ET = mgh + 0 = \(\frac{4 }{ 3} \) × 3.14 × (2 × 10 ^- 3)³ × 10 3 × 9.8 × 500 × 10 ^-5

= 0.164 J 

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

∴Total energy at the ground: E_G = \(\frac{1 }{ 2}\) mv² + 0 

= \(\frac{1 }{ 2}\) × \(\frac{4 }{ 3}\) × 3.14 × (2 × 10^-3)³ × 10³ × 9.8 × (10)² 

= 1.675 × 10 ^{- 3} J

∴Resistive force = E_G – E_T = –0.162 J.