In the arrangement shown in the figure, if the blocks of masses m and 2 m are released from the state of rest, tension in the string is ( = coefficient of friction, string is massless and inextensible, pulley is frictionless)

Step-by-step solution

Let N₁, N₂ are normal reaction force and f₁, f₂ are the friction force on two blocks. acceleration is a and tension is T . The respective FBD Friction force, f₁= N₁, f₂= N₂ As, N₁=m g 45^ , f₁= 2 3 m g 45^ = 2 3 m g and N₂=2 m g 45^ , f₂= 2 3 2 m g 45^ = 2 2 3 m g Now, by second law of motion for 2 m mass, 2 m g 45^ -T-f₂=2 m a 2 m g-T- 2 2 3 m g=2 m a For m₁, T-f₁-m g 45^ =m a T- 2 3 m g- m g 2 =m a Multiplying by 2 in Eq. (ii), we get 2 T- 2 2 3 m g- 2 m g=2 m a Subtracting Eq. (iii) from Eq. (i), we get 3 T-2 2