A block of mass M is lying on a horizontal frictionless surface. One end of the uniform rope of half the mass of the block is fixed to the block which is pulled in the horizontal direction by applying a force F at the other end. The tension in the middle of the rope will be

Step-by-step solution

Mass of the block = M Mass of the rope = M 2 Total mass of the system = M + M 2 = 3M 2 Acceleration of the system a = F 3M 2 = 2F 3M The tension in the middle of the rope accelerates the block and the rear half of the rope. Mass behind the middle point m' = M + 1 2 ( M 2 ) = M + M 4 = 5M 4 Tension at the middle of the rope T = m' a = ( 5M 4 ) ( 2F 3M ) = 5F 6 Answer: 5F 6