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 ?
0·4 J
0·5 J
0·3 J
1 J
Answer is Right!
Answer is Wrong!
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UPSC Geoscientist – 2024
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.