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Question

A block of mass m is kept on an inclined plane inside a lift moving down with acceleration of 2 m/s2. What should be the coefficient of friction between block and inclined plane to let the block move down with constant velocity relative to lift?
(Take g=10 m/s2)


A
13
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B
15
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C
16
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D
12
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Solution

The correct option is A 13
F.B.D of block from lift frame.


Pseudo force,
Fp=ma=2m N
For constant velocity in accelerating frame.
Applying equilibrium condition perpendicular to inclined plane.
N+macosθ=mgcosθ
N=mcosθ(ga)
Kinetic friction will act on block.
f=μN=μmcosθ(ga)
Applying equilibrium condition along the inclined plane.
mgsinθ=f+masinθ
mgsinθ=μmcosθ(ga)+masinθ
μ=gsinθasinθcosθ(ga)
μ=10sin302sin30cos30(102)
μ=518×32
μ=13

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