A ball of mass 100 g falls freely from a height of h = 20 m and hits t

A ball of mass 100 g falls freely from a height of h = 20 m and hits the ground at a speed of 1·4 √gh. Take g = 10 m s⁻². Which one among the following is the correct value of the work done on the ball by air friction ?

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0·4 J

0·5 J

0·3 J

1 J

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC Geoscientist – 2024

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The work done on the ball by air friction can be determined using the Work-Energy Theorem, which states that the total work done on an object equals the change in its kinetic energy. The total work done is the sum of work done by gravity and work done by air friction.

Work done by gravity (a conservative force) is W_g = mgh = (0.1 kg)(10 m/s²)(20 m) = 20 J. The initial kinetic energy is KE_i = 0. The final kinetic energy is KE_f = ½ m v_f², where v_f = 1.4√gh. With g=10 m/s² and h=20m, v_f = 1.4√(10*20) = 1.4√200. So, v_f² = (1.4)² * 200 = 1.96 * 200 = 392 m²/s². KE_f = ½ (0.1 kg)(392 m²/s²) = 0.05 * 392 = 19.6 J.

According to the Work-Energy Theorem, W_total = W_g + W_air = ΔKE = KE_f – KE_i. 20 J + W_air = 19.6 J – 0 J. W_air = 19.6 J – 20 J = -0.4 J. The question asks for the value, and options are positive, implying magnitude. The magnitude of the work done by air friction is |-0.4 J| = 0.4 J.

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