The upper half of an inclined plane with inclination `phi` is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

Options

A. `mu=(1)/(tantheta)`
B. `mu=(2)/(tantheta)`
C. `mu=2tantheta`
D. `mu=tantheta`

Answer

To solve the problem, we will use the work-energy theorem and analyze the forces acting on the body as it moves down the inclined plane. Here’s a step-by-step solution:

### Step 1: Understand the Setup
The inclined plane is divided into two halves: the upper half is smooth (frictionless) and the lower half is rough. The angle of inclination is given as `phi`. The body starts from rest at the top and comes to rest at the bottom.

### Step 2: Define the Distances
Let the total length of the inclined plane be `L`. Therefore, the length of the upper half (smooth) is `L/2` and the length of the lower half (rough) is also `L/2`.

### Step 3: Apply Work-Energy Theorem
According to the work-energy theorem, the total work done on the body is equal to the change in kinetic energy. Since the body starts from rest and comes to rest again, the initial and final kinetic energy are both zero. Thus, the total work done is also zero.

\[ \text{Total Work Done} = \text{Work Done by Gravity} + \text{Work Done by Friction} = 0 \]

### Step 4: Calculate Work Done by Gravity
The work done by gravity as the body moves down the incline can be calculated as follows:

- The component of gravitational force acting along the incline is \( mg \sin \phi \).
- The work done by gravity over the distance \( L \) is:

\[ W_{\text{gravity}} = mg \sin \phi \cdot L \]

### Step 5: Calculate Work Done by Friction
For the rough surface, the frictional force acts opposite to the direction of motion. The normal force \( N \) on the rough surface is given by:

\[ N = mg \cos \phi \]

The frictional force \( f \) can be expressed as:

\[ f = \mu N = \mu mg \cos \phi \]

The work done by friction over the distance \( L/2 \) is:

\[ W_{\text{friction}} = -f \cdot \frac{L}{2} = -\mu mg \cos \phi \cdot \frac{L}{2} \]

### Step 6: Set Up the Equation
Now, substituting the expressions for work done by gravity and friction into the work-energy equation:

\[ mg \sin \phi \cdot L - \mu mg \cos \phi \cdot \frac{L}{2} = 0 \]

### Step 7: Simplify the Equation
We can cancel \( mg \) and \( L \) from both sides (assuming \( m \neq 0 \) and \( L \neq 0 \)):

\[ \sin \phi - \frac{\mu}{2} \cos \phi = 0 \]

### Step 8: Solve for the Coefficient of Friction
Rearranging the equation gives:

\[ \sin \phi = \frac{\mu}{2} \cos \phi \]

Dividing both sides by \( \cos \phi \):

\[ \tan \phi = \frac{\mu}{2} \]

Thus, we find:

\[ \mu = 2 \tan \phi \]

### Conclusion
The coefficient of friction \( \mu \) for the lower half of the inclined plane must be:

\[ \mu = 2 \tan \phi \]