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A body is moving down along inclined plane of angle of inclination $θ$ . The coefficient of friction between the body and the plane varies as $μ$ = 0.5x, where x is the distance moved down the plane. The body will have the maximum velocity when it has travelled a distance x given by

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$x = \frac{\sqrt{2}}{\cot θ}$

$x = \frac{2}{\tan θ}$

$x = 2 \tan θ$

$x = \sqrt{2} \cot θ$

answer is A.

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Detailed Solution

$a = g (\sin θ - μ \cos θ)$

$\frac{d v}{d t} = g (\sin θ - \frac{x}{2} \cos θ)$

$\frac{d v}{d x} . \frac{d x}{d t} = g (\sin θ - \frac{x}{2} \cos θ)$

$v \frac{d v}{d x} = g (\sin θ - \frac{x}{2} \cos θ)$

For max velocity $\frac{d v}{d x} = 0$

$\sin θ = \frac{x}{2} \cos θ$

$x = 2 \tan θ$

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