Starting from rest,a body slides down a 45^ci | vedclass

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Question

(C) Let the distance be $s$ and the angle of inclination be $	heta = 45^\circ$. In the absence of friction,the acceleration is $a_1 = g \sin 	heta$.

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(C) Let the distance be $s$ and the angle of inclination be $	heta = 45^\circ$. In the absence of friction,the acceleration is $a_1 = g \sin 	heta$. The time taken is $t_1 = \sqrt{\frac{2s}{g \sin 	heta}}$. With friction,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 $t_2 = 2t_1$,so $t_2^2 = 4t_1^2$. Substituting the expressions: $\frac{2s}{g(\sin 	heta - \mu \cos 	heta)} = 4 	imes \frac{2s}{g \sin 	heta}$. This simplifies to $\sin 	heta = 4(\sin 	heta - \mu \cos 	heta)$. $\sin 	heta = 4 \sin 	heta - 4 \mu \cos 	heta \implies 4 \mu \cos 	heta = 3 \sin 	heta$. $\mu = \frac{3}{4} 	an 	heta$. Since $	heta = 45^\circ$,$	an 45^\circ = 1$. $\mu = \frac{3}{4} 	imes 1 = 0.75$.

Starting from rest,a body slides down a $45^\circ$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is

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