The upper half of an inclined plane of inclin | vedclass

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(C) Let $m$ be the mass of the block and $L$ be the total length of the inclined plane.According to the work-energy theorem,the total work done on the

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$\mu = 2	an	heta$

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(C) Let $m$ be the mass of the block and $L$ be the total length of the inclined plane.According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.Since the block starts from rest and comes to rest at the bottom,the change in kinetic energy is $\Delta K = 0$.Therefore,the total work done by all forces (gravity and friction) must be zero:$W_{	ext{gravity}} + W_{	ext{friction}} = 0$Work done by gravity over the entire length $L$ is $W_{	ext{gravity}} = mg \sin	heta \cdot L$.Work done by friction over the lower half length $L/2$ is $W_{	ext{friction}} = -f_k \cdot (L/2) = -(\mu mg \cos	heta) \cdot (L/2)$.Equating the sum to zero:$mg \sin	heta \cdot L - \mu mg \cos	heta \cdot \frac{L}{2} = 0$Dividing both sides by $mgL$:$\sin	heta - \frac{\mu}{2} \cos	heta = 0$$\sin	heta = \frac{\mu}{2} \cos	heta$$\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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