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
\(\dfrac{1}{\sqrt{x}}\left(\dfrac{\alpha^2-1}{\alpha^2}\right)\)
where \(x\) = ____________.

Hint:

For the same distance from rest, \(a\,t^2\) is constant, so \(a_{rough} = a_{smooth}/\alpha^2\); then use \(a_{rough} = g(\sin30^\circ - \mu\cos30^\circ)\).

Answer 1

Correct Answer: 3

Smooth plane: \(a_1 = g\sin30^\circ = \dfrac{g}{2}\). Rough plane: \(a_2 = g(\sin30^\circ - \mu\cos30^\circ) = g\left(\dfrac12 - \dfrac{\sqrt3}{2}\mu\right)\).
Same distance from rest: \(s = \tfrac12 a_1T^2 = \tfrac12 a_2(\alpha T)^2\), so \(a_1 = a_2\alpha^2\).
\(\dfrac{g}{2} = \alpha^2 g\left(\dfrac12 - \dfrac{\sqrt3}{2}\mu\right)\) gives \(\dfrac{1}{\alpha^2} = 1 - \sqrt3\,\mu\).
\(\mu = \dfrac{1}{\sqrt3}\left(1 - \dfrac{1}{\alpha^2}\right) = \dfrac{1}{\sqrt3}\left(\dfrac{\alpha^2-1}{\alpha^2}\right)\).
Comparing with the given form, \(x = 3\).
Answer: 3