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 $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$ where $x=$, ...

Answer 1

On smooth incline
$a = g \sin 30^{\circ}$
by $S = ut +\frac{1}{2} at ^{2}$
$S =\frac{1}{2} \frac{ g }{2} T ^{2}=\frac{ g }{4} T ^{2} \ldots \ldots .$ (i)

On rough incline
$a = g \sin 30^{\circ}-\mu g \cos 30^{\circ}$
by $S = ut +\frac{1}{2} at ^{2}$
$S =\frac{1}{4} g (1-\sqrt{3} \mu)(\alpha T )^{2} \,\,\, ...(ii)$
By (i) and (ii)
$\frac{1}{4} g T ^{2}=\frac{1}{4} g (1-\sqrt{3} \mu) \alpha^{2} T ^{2}$
$\Rightarrow 1-\sqrt{3} g=\frac{1}{\alpha^{2}} $
$\Rightarrow g=\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right).\frac{1}{\sqrt{3}}$
$\Rightarrow x=3.00$