A raindrop of radius 2 mm falls from a height of 500 m above the ground. It follows 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 resistive force in the entire journey if its speed on reaching the ground is 10 m s⁻¹.
Answers
Radius of the rain drop, r = 2 mm =0.002 m
rain drop falls from height, s = 500 m
Volume of the rain drop, V = (4/3)πr³
V = (4/3) × π × (0.002)³
⇒ V = 3.35×10^-8 m³
We know, Density of water, ρ = 10³ kg/m³
So, Mass of rain drop, m = ρV
m = 10³ kg/m³ × 3.35×10^-8 m³
m = 3.35×10^-5 kg
now, Gravitational force on the rain drop, F = mg
F = 3.35×10^-5 kg × 9.8 m/s²
F = 3.28×10^-4 N
Work done by the gravitational force in the first half of the motion,
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,
according to law of conservation of energy, If no resistive force is present, then the total energy of the rain drop will remain conserved.
Total energy at the starting point, E = mgh
E = 3.35×10^-5 kg × 9.8 m/s² × 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.
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)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.
Answer:
Radius of the rain drop, r = 2 mm = 2 × 10–3 m
Volume of the rain drop, V = (4/3)πr3
= (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 × 103 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 × 103 × 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., WII, = 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) mv2 + 0
= (1/2) × (4/3) × 3.14 × (2 × 10-3)3 × 103 × 9.8 × (10)2
= 1.675 × 10-3 J
∴Resistive force = EG – ET = –0.162 J