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Q.

A given object takes $n$ times more time to slide down $45^{0}$ rough inclined plane as it takes to slide down a perfectly smooth $45^{0}$ incline of same dimensions. The coefficient of kinetic friction between the object and the incline is

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a

$\frac{1}{2 - n^{2}}$

b

$\sqrt{1 - \frac{1}{n^{2}}}$

c

$\sqrt{\frac{1}{1 - n^{2}}}$

d

$1 - \frac{1}{n^{2}}$

answer is C.

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Detailed Solution

From $s = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} a t^{2}, t = \sqrt{\frac{2 s}{a}}$
From smooth plane $a = g \sin θ$
For rough plane, $a^{'} = g (\sin θ - μ \cos θ)$
$∴ t^{'} = n t \Rightarrow \sqrt{\frac{2 s}{g (\sin θ - μ \cos θ)}} = n \sqrt{\frac{2 s}{g \sin θ}} ∴ n^{2} g (\sin θ - μ \cos θ) = g \sin θ$
When $θ = 45^{0}, \sin θ = \cos θ = \frac{1}{\sqrt{2}}$
Solving we get $μ = (1 - \frac{1}{n^{2}})$

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A given object takes n times more time to slide down 450 rough inclined plane as it takes to slide down a perfectly smooth 450 incline of same dimensions. The coefficient of kinetic friction between the object and the incline is