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 $$=$$ $$6$$ N, the work done by the force of friction on the slab, between the instants t $$=$$ $$2$$ s and t $$=$$ $$3$$ s, is (g$$ $$=$$ $$10$$ $$ms-2)

Question

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 $$=$$ $$6$$ N, the work done by the force of friction on the slab, between the instants t $$=$$ $$2$$ s and t $$=$$ $$3$$ s, is (g$$ $$=$$ $$10$$ $$ms-2)

Infinity Learn · Accepted Answer

Maximum frictional force between the slab and the block $$fmax=$$μN $$=$$μmg $$=14$$× $$4$$ × $$10$$ $$=$$ $$10N$$. Evidently, f < fmax. So, the two bodies will move together as a single unit. If a be their combined acceleration, $$thena=Fm+M=64+5=23ms$$−$$1$$. Therefore, factional force acting can be obtained as $$f=Ma=23$$×$$5N=103N$$. Using $$s=12at2$$.$$s(2)=12$$×$$23(2)2=43$$ and $$s(3)=12$$×$$23(3)2=3$$. Therefore, work done by friction $$=$$ $$=1033$$−$$43=509=5.55$$ J

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