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Question

A body is sliding down a rough inclined plane of inclination θ, for which the coefficient of friction varies with distance x as μ(x)=kx, where k is a constant. Here x is the distance moved by body down the plane. The net force on the body down the plane will be zero at a distance x0 given by

A
tanθk
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B
ktanθ
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C
cotθk
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D
kcotθ
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Solution

The correct option is A tanθk
The net downward force along the inclined plane on the body at a
distance x is:
f(x)=mgsinθμmgcosθ
=mg(sinθμcosθ)
=mg(sinθkxcosθ)
f(x)=0 at a value of x=x0 given by
sinθkx0cosθ=0 x0=tanθk
130046_7836_ans_9b5f1dd0447d4811ac6b0febc976430a.png

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