Work Energy Power

JEE Main level, single correct, medium

3 of 183

Question 3 of 183

The rain drop of mass 11 g, starts with zero velocity from a height of 11 km. It hits the ground with a speed of 55 m/s. The work done by the unknown resistive force is _______ J. (take g=10g = 10 m/s2^2)

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Answer: (D) −9.98-9.98

Using the work-energy theorem, the total work done by all forces equals the change in kinetic energy:

Wg+Wr=ΔKW_{g} + W_{r} = \Delta K

mgh+Wr=12mv2−0mgh + W_{r} = \dfrac{1}{2}mv^2 - 0

Substituting the given values (m=10−3m = 10^{-3} kg, g=10g = 10 m/s2^2, h=1000h = 1000 m, v=5v = 5 m/s):

(10−3)(10)(1000)+Wr=12(10−3)(5)2(10^{-3})(10)(1000) + W_{r} = \dfrac{1}{2}(10^{-3})(5)^2

10+Wr=0.012510 + W_{r} = 0.0125

Wr=0.0125−10=−9.9875W_{r} = 0.0125 - 10 = -9.9875 J

Wr≈−9.98W_{r} \approx -9.98 J

Answer: −9.98-9.98

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