question_answer

 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 0.25. If the block is pulled horizontally by a force of F = 6N, the work done by the force of friction on the slab between the instants t = 2 s and t = 3 s is

A)  2.4 J                     

B) 5.55 J

C) 4,44 J                    

D)  -10 J

Correct Answer: B

Solution :

 Maximum frictional force between the slab and the block \[20\Omega \] Evidently, \[\text{ }\!\!\alpha\!\!\text{ -}\] So, the two bodies will move together as a single unit. If a be their combined acceleration, then\[\text{ }\!\!\gamma\!\!\text{ -}\] Therefore, motional force acting can be obtained as\[\alpha \text{-}\] Using       \[\alpha \text{-}\] \[\text{ }\!\!\gamma\!\!\text{ -}\]and\[x\left( <<R \right)\] Therefore, work done by friction \[(x=5\lambda )\]\[{{v}_{1}}\]