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There is an inclined plane of inclination 30° to the horizontal. The top $\frac{1}{4}$ th of the length is smooth and the lower 3/4th is rough. A block starts from rest from the top of the plane slides and comes to rest at the bottom of the inclined plane. The coefficient of sliding friction between block and inclined plane is

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$\frac{4}{3 \sqrt{3}}$

$\frac{2}{3 \sqrt{3}}$

$\frac{2 \sqrt{3}}{5}$

$\frac{\sqrt{3}}{2}$

answer is A.

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Detailed Solution

total energy of the body mgh is lost due to friction in travelling last $\frac{3}{4} t h$ of the inclined plane

mgh = $μ m g \cos θ \frac{3 l}{4}$

$h = μ \cos θ \frac{3 l}{4}$ $\Rightarrow$ $\frac{h}{l} = \frac{3}{4} μ c o s θ$

$\sin θ = \frac{3}{4} μ \cos θ$ $μ = \frac{4}{3} \tan θ = \frac{4}{3 \sqrt{3}}$

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