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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 1√3F.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(g−a) Kinetic friction will act on block. f=μN=μmcosθ(g−a) Applying equilibrium condition along the inclined plane. mgsinθ=f+masinθ ⇒mgsinθ=μmcosθ(g−a)+masinθ ⇒μ=gsinθ−asinθcosθ(g−a) ⇒μ=10sin30∘−2sin30∘cos30∘(10−2) ⇒μ=5−18×√32 ⇒μ=1√3

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