A block of mass 4 kg is placed on a smooth horizontal table and connected by a light inextensible string passing over a smooth pulley at the edge of the table to a hanging block of mass 3 kg. The system is released from rest. The distance traversed by the centre of mass of the system in 1.4 s is _______ m. (Take g = 10 m/s ^2 )

Step-by-step solution

Let the mass on the table be M = 4 kg and the hanging mass be m = 3 kg. The common acceleration of the blocks is given by: a = mg M + m = 3 10 4 + 3 = 30 7 m/s ^2 The block on the table accelerates horizontally, so its acceleration vector is a _M = 30 7 i . The hanging block accelerates vertically downwards, so its acceleration vector is a _m = - 30 7 j . The acceleration of the centre of mass is: a _ cm = M a _M + m a _m M + m a _ cm = 4 ( 30 7 i ) + 3 (- 30 7 j ) 7 = 120 49 i - 90 49 j The magnitude of the centre