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

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 = 10³ kg/ m³

Mass of rain drop = volume of drop x density

m = `4/3 πr^3 xx ρ`

=`4/3 xx 22/7 (2 xx 10^-3)^3 xx 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

It is important to realize that the gravitational force continues to exert the same amount of force on the drop whether it moves at a constant speed or with decreasing acceleration.

If there were no resistive forces, energy would drop on reaching the ground.

E₁= mgh = 3.35 x 10^-5 x 9.8 x 500 = 0.164 J

Actual energy, E₂ = `1/2`mv² 

= `1/2 xx 3.35 xx 10^-5 (10)^2`

= 1.675 x 10^-3J

Work done by the resistive forces,

W =E₂ – E₁ = 1.675 x 10^-3 – 0.164

=−0.162 joule.