The upper half of an inclined plane of inclin | vedclass

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(D) Let the total length of the incline be $L$. The length of each half is $l = L/2$.For the smooth upper half,the acceleration is $a_1 = g \sin 	het

Vedclass · Accepted Answer

$\mu = 2 	an 	heta$

Vedclass · Answer

(D) Let the total length of the incline be $L$. The length of each half is $l = L/2$.For the smooth upper half,the acceleration is $a_1 = g \sin 	heta$. Starting from rest $(u=0)$,the velocity $v$ at the midpoint is given by:$v^2 = u^2 + 2 a_1 l = 0 + 2(g \sin 	heta)(L/2) = gL \sin 	heta$.For the rough lower half,the acceleration is $a_2 = g(\sin 	heta - \mu \cos 	heta)$. The block comes to rest at the bottom,so the final velocity is $0$. Using $v_f^2 = v^2 + 2 a_2 l$:$0 = gL \sin 	heta + 2[g(\sin 	heta - \mu \cos 	heta)](L/2)$.$0 = gL \sin 	heta + gL \sin 	heta - gL \mu \cos 	heta$.$2 \sin 	heta = \mu \cos 	heta$.Therefore,$\mu = 2 	an 	heta$.

The upper half of an inclined plane of inclination $\theta$ is smooth,while the lower half is rough. If a block released from the top comes to rest at the bottom,what is the coefficient of friction between the block and the rough surface?

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