A large slab of mass 5 kg lies on a smooth horizontal surface, with a block of mass 4 kg lying on the top of it, the coefficient of friction between the block and the slab is Q.25. If the block is pulled horizontally by a force of F=6 ~N , the work done by the force of friction on the slab between the instants t=2 ~s and t=3 ~s is

Step-by-step solution

Maximum frictional force between the slab and the block f_ = N= m g= 1 4 4 10=10 ~N Evidently, f f_ So, the two bodies will move together as a single unit. If a be their combined acceleration, then a= F m+M = 6 4+5 = 2 3 ~ms ⁻² Therefore, frictional force acting can be obtained as f=M a= 2 3 5= 10 3 N Using S= 1 2 a t^2S(2)= 1 2 2 3 (2)^2= 4 3 and S(3)= 1 2 2 3 (3)^2=3 Therefore, work done by friction =F[S(3)-S(2)]= 10 3 [3- 4 3 ]= 50 9 =5.55 ~J