First slide

Friction on inclined plane of angle more than angle of repose

Question

A block released from rest from the top of a smooth inclined plane of inclination 45° takes time 't' to reach the bottom. The same block released from rest, from top of a rough inclined plane of same inclination, takes time '2t' to reach the bottom, coefficient of friction is:

Moderate

A

0.75
B

0.5
C

0.25
D

0.4

Solution

acceleration on smooth plane is $g \sin θ and for rough plane g (sinθ - μcosθ)$
time is inversely proportional to square root of acceleration
$\frac{t_{r}}{t_{s}} = \sqrt{\frac{g (\sin θ)}{g (\sin θ - μ \cos θ)}} = 2 \frac{1}{1 - μ} = 4 μ = 0.75$

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