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Question
A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 ms–1. Calculate
- the loss of P.E. of the drop.
- the gain in K.E. of the drop.
- Is the gain in K.E. equal to a loss of P.E.? If not why.
Take g = 10 ms–2
Solution
Given, the mass of the raindrop (m) = 100 g
= 1 × 10–2 kg
Height of falling (h) = 1 km = 103 m
g = 10 m/s2
Speed of the raindrop (v) = 50 m/s
a. Loss of PE of the drop (v) = 50 m/s
= 1 × 10–3 × 10 × 103
= 10 J
b. Gain in KE of the drop = `1/2 mv^2`
= `1/2 xx 1 xx 10^-3 xx (50)^2`
= `1/2 xx 10^-3 xx 2500`
= 1.250 J
c. No, the gain in KE is not equal to the loss in its PE, because a part of PE is utilised in doing work against the viscous drag of air.
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