The upper half of an inclined plane of inclin | vedclass

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

Vedclass · Accepted Answer

$\mu = 2 	an 	heta$

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(D) Let the total length of the inclined plane be $2s$. The upper half has length $s$ and is smooth,while the lower half has length $s$ and is rough.According to the work-energy theorem,the total work done by all forces on the block is equal to the change in its kinetic energy.Since the block starts from rest at the top and comes to rest at the bottom,the change in kinetic energy is $\Delta KE = 0 - 0 = 0$.The forces doing work are gravity and friction.Work done by gravity over the total distance $2s$ is $W_g = (mg \sin 	heta)(2s)$.Work done by friction over the lower half of length $s$ is $W_f = -(\mu mg \cos 	heta)(s)$.Setting the total work to zero: $W_g + W_f = 0$.$(mg \sin 	heta)(2s) - (\mu mg \cos 	heta)(s) = 0$.$2mg \sin 	heta = \mu mg \cos 	heta$.Dividing both sides by $mg \cos 	heta$,we get $\mu = 2 \frac{\sin 	heta}{\cos 	heta} = 2 	an 	heta$.

The upper half of an inclined plane of inclination $\theta$ is perfectly smooth while the lower half is rough. $A$ block starting from rest at the top of the plane will again come to rest at the bottom,if the coefficient of friction between the block and the lower half of the plane is given by:

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