A smooth block is released at rest on a 45^ci | vedclass

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(A) For a smooth incline,the acceleration is $a_1 = g \sin 	heta$. The time taken to cover distance $d$ is $t_1 = \sqrt{\frac{2d}{g \sin 	heta}}$.Fo

Vedclass · Accepted Answer

$\mu_k = 1 - \frac{1}{n^2}$

Vedclass · Answer

(A) For a smooth incline,the acceleration is $a_1 = g \sin 	heta$. The time taken to cover distance $d$ is $t_1 = \sqrt{\frac{2d}{g \sin 	heta}}$.For a rough incline,the acceleration is $a_2 = g \sin 	heta - \mu_k g \cos 	heta$. The time taken to cover distance $d$ is $t_2 = \sqrt{\frac{2d}{g \sin 	heta - \mu_k g \cos 	heta}}$.Given $t_2 = n t_1$,we have $\sqrt{\frac{2d}{g \sin 	heta - \mu_k g \cos 	heta}} = n \sqrt{\frac{2d}{g \sin 	heta}}$.Squaring both sides: $\frac{1}{g \sin 	heta - \mu_k g \cos 	heta} = \frac{n^2}{g \sin 	heta}$.Since $	heta = 45^\circ$,$\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$.Thus,$\frac{1}{g \sin 	heta (1 - \mu_k)} = \frac{n^2}{g \sin 	heta}$.This simplifies to $1 - \mu_k = \frac{1}{n^2}$,which gives $\mu_k = 1 - \frac{1}{n^2}$.

$A$ smooth block is released at rest on a $45^\circ$ incline and then slides a distance $d$. The time taken to slide is $n$ times as much to slide on a rough incline than on a smooth incline. The coefficient of kinetic friction is:

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