Block A has a mass of 5.0 kg and rests on a smooth table and is connected to block B, which has a mass of 3.0 kg, after passing over an ideal pulley, as shown. Block B is released from rest and it accelerates down. What is the velocity of the block A at the instant when block B falls through a distance of 10 m. Use g = 10 m/st I A. 10 m/s B. 25 m/s C. (75) m/s D. 5.0 m/s Solution Here, let the acceleration is a Using second law of motion a = net force/total mass a = (3 * 10)/(3 + 5) a = 3.75 m/s^2 Now , for the velocity after falling 10 m vf^2 = 2 * a * d vf^2 =2 * 3.75 * 10 vf = sqrt(75) the velocity of the block A at the instant is C) sqrt(75) m/s .