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Question

A large slab of mass 5 kg lies on a smooth horizontal surface, with a block of mass 4 kg lying on 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 ms2)


A

2.4 J

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B

5.55 J

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C

4.44 J

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D

10 J

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Solution

The correct option is B

5.55 J


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, then

a=Fm+M=64+5=23ms1

Therefore, frictional 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 = f[s(3) - s(2)]

= 103[343]

= 509=5.55J


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