Work Energy Power

JEE Main level, single correct, medium

38 of 183

Question 38 of 183

A spherical ball of mass 22 kg falls from a height of 1010 m and is brought to rest after penetrating 1010 cm into sand. The average force exerted by sand on the ball is _______ N. (Take g=10g = 10 m/s2^2)

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Answer: (B) 20202020

Using the work-energy theorem, the total work done on the ball is equal to its change in kinetic energy.

Since the ball starts from rest and finally comes to rest, the change in kinetic energy is zero.

Wgravity+Wsand=ΔK=0W_{\text{gravity}} + W_{\text{sand}} = \Delta K = 0

The work done by gravity is over the total distance h+dh + d, and the work done by the resistive force of the sand FF is over the distance dd.

mg(h+d)−Fd=0mg(h + d) - Fd = 0

F=mg(h+d)dF = \dfrac{mg(h + d)}{d}

Given m=2m = 2 kg, g=10g = 10 m/s2^2, h=10h = 10 m, and d=10d = 10 cm =0.1= 0.1 m:

F=2×10×(10+0.1)0.1F = \dfrac{2 \times 10 \times (10 + 0.1)}{0.1}

F=20×10.10.1F = \dfrac{20 \times 10.1}{0.1}

F=20×101=2020F = 20 \times 101 = 2020 N

Answer: 20202020

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