First slide

Friction on inclined plane of angle more than angle of repose

Question

A given object takes n times more time to slide down 45° rough inclined plane as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline is

Moderate

A

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

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

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

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

Solution

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

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