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The upper portion of an inclined plane of inclination $α$ is smooth and the lower portion is rough. A particle slides down from rest from the top and just comes to rest at the foot. If the ratio of smooth length to rough length is m: n, find the coefficient of friction.

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$\frac{m + n}{n} \tan α$

$\frac{m + n}{n} \cos α$

$\frac{m + n}{n} \sin α$

$\frac{m}{n} \tan α$

answer is A.

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Detailed Solution

$2 a_{1} s_{1} = 2 a_{2} s_{2}$

$\frac{a_{1}}{a_{2}} = \frac{s_{2}}{s_{1}} = \frac{n}{m}$

$\frac{g \sin α}{μ g \cos α - g \sin α} = \frac{n}{m}$

Solving it, we get

$μ = (\frac{m + n}{n}) \tan α$

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