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Question

A small mass slides down an inclined plane of inclination θ with the horizontal. The coefficent of friction is μ=μox where the x is the distance through which the mass slides down and μo is a positive constant. Then the distance covered by the mass before it stops is

A
2μotanθ
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B
4μotanθ
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C
12μotanθ
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D
1μotanθ
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Solution

The correct option is A 2μotanθ
Free body diagram:

From free body diagram,
Net force acting downwards parallel to the inclined plane is,

Fnet=mgsinθfk

Fnet=mgsinθμN

Fnet=mgsinθμmgcosθ

Thus, Fnet=mgsinθμ0xmgcosθ....(1)

From Newton's second law of motion, Fnet=ma

mgsinθμ0xmgcosθ=ma

a=gsinθμ0xgcosθ

Now, substituting a=vdvdx, we have,

vdvdx=gsinθμ0xgcosθ

Let the body stops at distance x.

Since, the intial and final velocities are zero on integrating we get,

vf=0vi=0vdv=x0(gsinθμ0xgcosθ)dx

0=gsinθ×xμ0gcosθ×x22

sinθ=μocosθ×x2

x=2μotanθ

Hence, option (a) is the correct answer.

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