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Question

A block of mass m is lying on a wedge having inclination of α=tan1(15). Wedge is moving with a constant acceleration of 2 m/s2. Find the minimum value of coefficient of friction so that m remains stationary with respect to wedge. Take g=10 m/s2.

A
29
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B
512
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C
15
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D
25
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Solution

The correct option is B 512
In the FBD the pseudo force will be acting on the block towards left as the wedge is accelerating towards right,


Balancing the forces acting perpendicular to the plane:
N+masinα=mgcosα
N=mgcosαmasinα
Along the plane
mgsinα+macosα=f=μN
mgsinα+macosα=μ(mgcosαmasinα)
μ=gtanα+agatanα,
μ=g(15)+aga(15)
putting the values we get,
μ=512

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