Starting from rest,the time taken by a body s | 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 	het

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 is $t_r = \sqrt{\frac{2s}{g(\sin 	heta - \mu \cos 	heta)}}$.Given $t_r = n t_s$,where $n = 2$.Squaring both sides: $\frac{2s}{g(\sin 	heta - \mu \cos 	heta)} = n^2 \frac{2s}{g \sin 	heta}$.This simplifies to $\sin 	heta = n^2(\sin 	heta - \mu \cos 	heta)$.Rearranging for $\mu$: $\mu = 	an 	heta \left(1 - \frac{1}{n^2}ight)$.Substituting $	heta = 45^{\circ}$ and $n = 2$: $\mu = 	an 45^{\circ} \left(1 - \frac{1}{2^2}ight) = 1 	imes (1 - 0.25) = 0.75$.

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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