The upper half of an inclined plane of inclination $ heta$ is perfectly smooth while the lower half is rough.

Question

The upper half of an inclined plane of inclination $	heta$ 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 lower half of the plane is given by:

Accepted Answer

1. **Concept (Work-Energy Theorem):** The work-energy theorem states that the change in kinetic energy of a body is equal to the net work done by all the forces acting on it [NCERT Class 11, Physics Part I, Sec 5.5].
2. **Analysis:** 
   - The block starts from rest and comes to rest at the bottom, so the change in kinetic energy $\Delta K = K_f - K_i = 0$.
   - Let the total length of the inclined plane be $L$. The upper half ($L/2$) is smooth, and the lower half ($L/2$) is rough.
   - Forces doing work: Gravity acts over the entire length $L$. Friction acts only over the lower half $L/2$.
3. **Work Done Calculation:**
   - Work done by gravity ($W_g$): Force is $mg\sin	heta$ downwards along the plane. Displacement is $L$. 
     $$W_g = mg L \sin	heta$$
   - Work done by friction ($W_f$): Frictional force $f = \mu N = \mu mg \cos	heta$ acts upwards along the plane (opposing motion). Displacement is $L/2$.
     $$W_f = -f 	imes \frac{L}{2} = -(\mu mg \cos	heta) \frac{L}{2}$$
4. **Equation:**
   $$W_{net} = W_g + W_f = 0$$
   $$mg L \sin	heta - \frac{\mu mg L \cos	heta}{2} = 0$$
   $$mg L \sin	heta = \frac{\mu mg L \cos	heta}{2}$$
   Dividing both sides by $mgL \cos	heta$:
   $$\frac{\sin	heta}{\cos	heta} = \frac{\mu}{2}$$
   $$	an	heta = \frac{\mu}{2} \implies \mu = 2	an	heta$$

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