Starting from rest, the time taken by a body | vedclass

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(D) For a smooth inclined plane, the acceleration is $a_s = g \sin 	heta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin 	he

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(D) For a smooth inclined plane, the acceleration is $a_s = g \sin 	heta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin 	heta}}$.For a rough inclined plane, the acceleration is $a_r = g(\sin 	heta - \mu \cos 	heta)$. The time taken to cover the same distance $s$ is $t_r = \sqrt{\frac{2s}{g(\sin 	heta - \mu \cos 	heta)}}$.Given $t_r = n t_s$, where $n = 2$.Squaring both sides, we get $\frac{2s}{g(\sin 	heta - \mu \cos 	heta)} = n^2 \frac{2s}{g \sin 	heta}$.This simplifies to $\sin 	heta = n^2(\sin 	heta - \mu \cos 	heta)$.Rearranging for $\mu$, we get $\mu = 	an 	heta \left[1 - \frac{1}{n^2}ight]$.Substituting $	heta = 45^{\circ}$ and $n = 2$:$\mu = 	an 45^{\circ} \left[1 - \frac{1}{2^2}ight] = 1 	imes \left[1 - \frac{1}{4}ight] = \frac{3}{4} = 0.75$.

Starting from rest, the time taken by a body sliding down on a rough inclined plane at $45^{\circ}$ with the horizontal is twice the time taken to travel on a smooth plane of the same inclination and same distance. Then the coefficient of kinetic friction is:

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