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Question

(A) Let the length of the inclined plane be $s$. For the smooth inclined plane,the acceleration is $a_1 = g \sin 	heta$. Using $v^2 - u^2 = 2as$ with

Vedclass · Accepted Answer

$\mu = 	an 	heta \left( 1 - \frac{1}{n^2} ight)$

Vedclass · Answer

(A) Let the length of the inclined plane be $s$. For the smooth inclined plane,the acceleration is $a_1 = g \sin 	heta$. Using $v^2 - u^2 = 2as$ with $u=0$,we get $v^2 = 2(g \sin 	heta)s$ --- $(1)$. For the rough inclined plane,the acceleration is $a_2 = g(\sin 	heta - \mu \cos 	heta)$. The final velocity is $v' = v/n$. Using $v'^2 = 2a_2s$,we get $(v/n)^2 = 2g(\sin 	heta - \mu \cos 	heta)s$ --- $(2)$. Dividing equation $(1)$ by equation $(2)$,we get: $n^2 = \frac{\sin 	heta}{\sin 	heta - \mu \cos 	heta}$. Rearranging the terms: $n^2(\sin 	heta - \mu \cos 	heta) = \sin 	heta$. $n^2 \sin 	heta - n^2 \mu \cos 	heta = \sin 	heta$. $n^2 \mu \cos 	heta = n^2 \sin 	heta - \sin 	heta = \sin 	heta (n^2 - 1)$. $\mu = \frac{\sin 	heta (n^2 - 1)}{n^2 \cos 	heta} = 	an 	heta \left( \frac{n^2 - 1}{n^2} ight) = 	an 	heta \left( 1 - \frac{1}{n^2} ight)$.

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