A block is released from rest on a 45^circ sm | vedclass

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

Vedclass · Accepted Answer

$\mu _k = 1 - \frac{1}{{{n^2}}}$

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(B) 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 $n = \frac{t_2}{t_1} = \sqrt{\frac{g \sin 	heta}{g \sin 	heta - \mu_k g \cos 	heta}} = \sqrt{\frac{\sin 	heta}{\sin 	heta - \mu_k \cos 	heta}}$.Squaring both sides: $n^2 = \frac{\sin 	heta}{\sin 	heta - \mu_k \cos 	heta}$.Since $	heta = 45^\circ$,$\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$.$n^2 = \frac{1/\sqrt{2}}{1/\sqrt{2} - \mu_k (1/\sqrt{2})} = \frac{1}{1 - \mu_k}$.Rearranging gives $1 - \mu_k = \frac{1}{n^2}$,so $\mu_k = 1 - \frac{1}{n^2}$.

$A$ block is released from rest on a $45^\circ$ smooth incline and slides a distance '$d$'. The time taken to slide the same distance '$d$' on a rough incline is '$n$' times the time taken on the smooth incline. The coefficient of kinetic friction is

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