The upper half of an inclined plane of inclin | vedclass

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(D) Let the total length of the inclined plane be $l$. The upper half has length $l/2$ and is smooth,while the lower half has length $l/2$ and is roug

Vedclass · Accepted Answer

$\mu = 2 	an 	heta$

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(D) Let the total length of the inclined plane be $l$. The upper half has length $l/2$ and is smooth,while the lower half has length $l/2$ and is rough with coefficient of friction $\mu$.For the upper half (smooth):The acceleration is $a_1 = g \sin 	heta$. Starting from rest $(u=0)$,the velocity $v$ at the midpoint is given by:$v^2 = u^2 + 2 a_1 (l/2) = 0 + 2(g \sin 	heta)(l/2) = gl \sin 	heta$.For the lower half (rough):The acceleration is $a_2 = g(\sin 	heta - \mu \cos 	heta)$. The body starts with velocity $v$ and comes to rest $(v_f = 0)$ at the bottom after covering distance $l/2$:$v_f^2 = v^2 + 2 a_2 (l/2) = 0$.Substituting $v^2 = gl \sin 	heta$ and $a_2 = g(\sin 	heta - \mu \cos 	heta)$:$gl \sin 	heta + 2[g(\sin 	heta - \mu \cos 	heta)](l/2) = 0$.$gl \sin 	heta + gl(\sin 	heta - \mu \cos 	heta) = 0$.$2 \sin 	heta - \mu \cos 	heta = 0$.$\mu \cos 	heta = 2 \sin 	heta$.$\mu = 2 	an 	heta$.

The upper half of an inclined plane of inclination $\theta$ is perfectly smooth,while the lower half is rough. $A$ body starting from rest at the top comes back to rest at the bottom. The coefficient of friction $\mu$ for the lower half is given by:

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