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}?

#### Solution 1

Radius of the rain drop, r = 2 mm = 2 × 10^{–3} m

Volume of the rain drop, V = (4/3)πr^{3} = (4/3) × 3.14 × (2 × 10^{-3})^{3} m^{-3}

Density of water, ρ = 103 kg m^{–3 }

Mass of the rain drop, m = ρV = (4/3) × 3.14 × (2 × 10^{-3})^{3} × 10^{3} kg

Gravitational force, F = mg = (4/3) × 3.14 × (2 × 10^{-3})^{3} × 103 × 9.8 N

The work done by the gravitational force on the drop in the first half of its journey:

WI = Fs = (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

= (4/3) × 3.14 × (2 × 10-3)3 × 103 × 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:

EG = (1/2) mv^{2} + 0

= (1/2) × (4/3) × 3.14 × (2 × 10^{-3})^{3} × 10^{3} × 9.8 × (10)^{2}

= 1.675 × 10^{-3} J

∴Resistive force = E_{G} – E_{T} = –0.162 J

#### Solution 2

Here, r = 2 mm = 2 x 10^{-3 }m.

Distance moved in each half of the journey, S=500/2= 250 m.

Density of water, p = 10^{3} kg/ m^{3}

Mass of rain drop = volume of drop x density

m =4/3 π r^{2} x ρ =4/3 x 22/7 (2 x 10^{-3})^{3} x 10^{3} = 3.35 x 10^{-5} kg

∴ W = mg x s = 3.35 x 10^{-5} x 9.8 x 250 = 0.082 J

Note: Whether the drop moves with decreasing acceleration or with uniform speed, work done by the gravitational force on the drop remains the same.

If there was no resistive forces, energy of drop on reaching the ground.

E_{1}= mgh = 3.35 x 10^{-5} x 9.8 x 500 = 0.164 J

Actual energy, E_{2} = 1/2mv^{2} = 1/2 x 3.35 x 10^{-5} (10)^{2} = 1.675 x 10^{-3}J

Work done by the resistive forces, W =E_{1} – E_{2} = 0.164 – 1.675 x 10^{-3} W

= 0.1623 joule.