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Question

A block of mass m slides down an inclined right angled trough. If the coefficient of kinetic friction between the block and the trough is μk, acceleration of the block down the plane is:

136616_074d03fab962460d803c37fc51cbe480.png

A
g(sinθ2μkcosθ)
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B
g(sinθ+2μkcosθ)
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C
g(sinθ2μkcosθ)
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D
g(sinθμkcosθ)
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Solution

The correct option is C g(sinθ2μkcosθ)
Here the frictional forces are perpendicular to the plane. As there are two surfaces,
So, frictional force ,f=2μkN
Here also mg downward from the plane. thus,
mgcosθN2=0....(1)
ma=mgsinθf=mgsinθ2μkN...(2)
using (1), from (1), ma=mgsinθ2μkmgcosθ2
or, a=g(sinθ2μkcosθ)

171515_136616_ans_284bc7f2a7ec405cb3948658874655dc.png

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