Two blocks of masses 1   kg and 2   kg connected by a light rod and the system is slipping down a rough incline angle 45 ° with the horizontal. The frictional coefficient at both the contacts is 0 . 4 . If the acceleration of the system is α 2 , the value of α is (Use g = 10   m   s - 2 )

Step-by-step solution

Frictional force acting on both the block will be kinetic in nature. Therefore, f 1 = μ m 1 g cos 45 ° = 0 . 4 × 1 × 10 × 1 2 = 2 2   N Similarly, f 2 = μ m 2 g cos 45 ° = 0 . 4 × 1 × 10 × 1 2 = 4 2   N Now, net force along the incline in the downward direction will provide the required acceleration, m 1 g sin 45 ° + m 2 g sin 45 ° - f 1 - f 2 = m 1 + m 2 × α 2 ⇒ 5 2 + 10 2 - 2 2 - 4 2 = 3 α 2 ⇒ α = 3