Question 172: Laws of Motion

Question

When a body slides down from rest along a smooth inclined plane making an angle of 30$$^\circ$$ with the horizontal, it takes time T. When the same body slides down from the rest along a rough inclined plane making the same angle and through the same distance, it takes time $$\alpha$$T, where $$\alpha$$ is a constant greater than 1. The co-efficient of friction between the body and the rough plane is $${1 \over {\sqrt x }}\left( {{{{\alpha ^2} - 1} \over {{\alpha ^2}}}} \right)$$ where x = __________.

Accepted Answer

On smooth incline
a = g sin30$$^\circ$$
by S = ut + $${1 \over 2}$$at
2
S = $${1 \over 2}$$$${g \over 2}$$T
2
= $${g \over 4}$$T
2
........ (i)
For rough surface,
$$ma = mg\sin 30^\circ  - \mu mg\cos 30^\circ $$
$$a = g\sin 30^\circ  - \mu g\cos 30^\circ $$
Distance covered by the block on the rough surface in time $$\alpha$$T,
$$s = ut + {1 \over 2}a{t^2}$$
$$s = 0 + {1 \over 2}(g\sin 30^\circ  - \mu g\sin 30^\circ ){t^2}$$
$$s = {g \over 4}(1 - \sqrt 3 \mu ){(\alpha T)^2}$$ .... (ii)
Distance covered by the block is same for both the case,
$$ \Rightarrow {g \over 4}(1 - \sqrt 3 \mu ){(\alpha T)^2} = {g \over 4}{T^2}$$ [from Eq. (i) and Eq. (ii)]
$$ \Rightarrow 1 - \sqrt 3 \mu  = {1 \over {{\alpha ^2}}} \Rightarrow \mu  = \left( {{{{\alpha ^2} - 1} \over {{\alpha ^2}}}} \right){1 \over {\sqrt 3 }}$$
Comparing with $$\mu  = \left( {{{{\alpha ^2} - 1} \over {{\alpha ^2}}}} \right){1 \over {\sqrt x }}$$
The value of the x = 3.

Laws of Motion

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