In the arrangement shown in the figure, friction exists only between the two blocks, A and B . The coefficient of static friction μ s = 0.6 and coefficient of kinetic friction μ k = 0.4 , the masses of the blocks A and B are m 1 = 20 kg and m 2 = 30 kg , respectively. Find the acceleration (in m s - 2 ) of m 1 , if a force F = 150 N is applied, as shown in the figure. [Assume that string and pulleys are massless]

Step-by-step solution

Let us assume that the two blocks move together, without slipping, relative to each other. The acceleration of the system in that case is a = F cos 60 ° m 1 + m 2 a = 150 × 1 2 20 + 30 = 1 . 5 m s - 2 In this case, if the frictional force acting between the two blocks is f , then writing the Newton's second law of motion, for the block of mass m 1, we get T - f = m 1 a 150 - f = 20 × 1 . 5 = 30 f = 120 N f max = 0 . 6 × 200 = 120 N Since f ≤ f max , our assumption about the two blocks moving together is correct and