The upper portion of an inclined plane of inc | vedclass

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(A) Let the length of the smooth portion be $L_1 = m$ and the length of the rough portion be $L_2 = n$.For the smooth portion,the acceleration is $a_1

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$\left[ \frac{m + n}{n} ight] 	an \alpha$

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(A) Let the length of the smooth portion be $L_1 = m$ and the length of the rough portion be $L_2 = n$.For the smooth portion,the acceleration is $a_1 = g \sin \alpha$.Using the equation of motion $v^2 = u^2 + 2a_1 L_1$,where $u = 0$:$v^2 = 0 + 2(g \sin \alpha)m = 2mg \sin \alpha$.For the rough portion,the acceleration is $a_2 = g \sin \alpha - \mu g \cos \alpha$.The particle comes to rest at the foot,so the final velocity $v_f = 0$.Using the equation of motion $v_f^2 = v^2 + 2a_2 L_2$:$0 = 2mg \sin \alpha + 2(g \sin \alpha - \mu g \cos \alpha)n$.Dividing by $2g$:$0 = m \sin \alpha + n \sin \alpha - n \mu \cos \alpha$.$n \mu \cos \alpha = (m + n) \sin \alpha$.$\mu = \left[ \frac{m + n}{n} ight] \frac{\sin \alpha}{\cos \alpha} = \left[ \frac{m + n}{n} ight] 	an \alpha$.

The upper portion of an inclined plane of inclination $\alpha$ is smooth and the lower portion is rough. $A$ particle slides down from rest from the top and just comes to rest at the foot. If the ratio of the smooth length to rough length is $m : n$,the coefficient of friction is

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