The time taken by a block to slide down a rou | vedclass

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Question

(A) For a frictionless inclined plane,the acceleration is $a_1 = g \sin 	heta$. The time taken to cover distance $s$ is $t_1 = \sqrt{\frac{2s}{g \sin

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{4}$

Vedclass · Answer

(A) For a frictionless inclined plane,the acceleration is $a_1 = g \sin 	heta$. The time taken to cover distance $s$ is $t_1 = \sqrt{\frac{2s}{g \sin 	heta}}$.For a rough inclined plane,the acceleration is $a_2 = g(\sin 	heta - \mu \cos 	heta)$. The time taken is $t_2 = \sqrt{\frac{2s}{g(\sin 	heta - \mu \cos 	heta)}}$.Given $t_2 = n t_1$,so $\frac{t_2}{t_1} = n$.Thus,$\frac{\sin 	heta}{\sin 	heta - \mu \cos 	heta} = n^2$.Rearranging for $\mu$: $\mu = 	an 	heta \left( 1 - \frac{1}{n^2} ight)$.Substituting $	heta = 30^{\circ}$ and $n = 2$:$\mu = 	an 30^{\circ} \left( 1 - \frac{1}{2^2} ight) = \frac{1}{\sqrt{3}} \left( 1 - \frac{1}{4} ight) = \frac{1}{\sqrt{3}} \left( \frac{3}{4} ight) = \frac{\sqrt{3}}{4}$.

The time taken by a block to slide down a rough inclined plane of angle $30^{\circ}$ is $n=2$ times the time taken to slide down a frictionless inclined plane of the same angle $30^{\circ}$. The coefficient of kinetic friction between the block and the plane is:

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