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?
Given,
Radius of the rain drop, r = 2 mm = 2×10-3 m
Total displacement of the rain drop, s = 500 m
Assuming the rain drop to be spherical,
Volume of the rain drop, V = (4/3)πr3
∴ V = (4/3)×π×(2×10-3)3
⇒ V = 3.35×10-8 m3
We know, Density of water, ρ = 103 kg m-3
So, Mass of rain drop, m = ρV
⇒ m = 103 kg m-3 × 3.35×10-8 m3
⇒ m = 3.35×10-5 kg
Gravitational force on the rain drop, F = mg
⇒ F = 3.35×10-5 kg × 9.8 m s-2
⇒ F = 3.28×10-4 N
Work done by the gravitational force in the first half of the motion, W1=F(s/2)
⇒ W1 = 3.28×10-4× (500/2)
⇒ W1 = 8.21×10-2 J
The work done by the gravitational force is equal in both the halves of the motion.
So, work done by the gravitational force during second half of motion, W2 = 8.21×10-2 J
If no resistive force is present, then the total energy of the rain drop will remain conserved according to law of conservation of energy.
Total energy at the starting point, E = mgh
⇒ E = 3.35×10-5 kg × 9.8 m s-2 × 500 m
⇒ E = 0.164 J
Due to the resistive force, some energy will be lost. The drop hits ground with velocity 10 m s-1.
Total energy of the drop when it hits the ground, E’ = (1/2)mv2
⇒ E’ = (1/2) × 3.35×10-5 kg × (10 m s-1)2
⇒ E’ = 1.675×10-3 J
So, loss of energy due to resistive force, ΔE = E’-E
⇒ ΔE = 1.675×10-3 J – 0.164 J
⇒ ΔE = -0.162 J
NOTE: The negative sign indicates that energy is lost due to the resistive forces.