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Question

A block of mass m, attached to a fixed position O on a smooth inclined wedge of mass M, oscillates with
amplitude A and linear frequency f . The wedge is located on a rough horizontal surface. If the angle of the
wedge is 600, then the force of friction acting on the wedge is given by (coefficient of static friction =μ


A
μ(M+m)g
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B
12mω2Asinωt
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C
μ(M+m)ω2Asinωt
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D
μ[(M+m)g+32mω2Asinωt]
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Solution

The correct option is B 12mω2Asinωt

The small block oscillates along the inclined plane with an amplitude A. As a result the centre of mass of the system undergoes SHM along the horizontal direction:
xcm=mAsinωtm+Mcos60=12mm+MAsinωt
The acceleration of the C.M is acm=ω2xcm,along the horizontal while the net horizontal force is =(M+m)acm,which is equal to the force of friction acting on it.


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