The upper half of an inclined plane with an angle of inclination theta, is smooth while the lower half is rough. A body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

Options

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

Answer

To solve the problem, we need to analyze the motion of a body on an inclined plane that has both smooth and rough sections. Here’s a step-by-step solution:

### Step 1: Understand the Problem
The inclined plane has two sections: the upper half is smooth, and the lower half is rough. A body starts from rest at the top and comes to rest at the bottom. We need to find the coefficient of friction (μ) for the lower half.

### Step 2: Define the Variables
- Let the angle of inclination be θ.
- Let the total distance of the inclined plane be d. Therefore, each half of the inclined plane is d/2.
- The initial velocity (u) at the top is 0, and the final velocity (v) at the bottom is also 0.

### Step 3: Calculate Velocity at the Middle Point
Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( u = 0 \)
- \( a = g \sin \theta \) (acceleration down the smooth incline)
- \( s = \frac{d}{2} \)

Substituting the values:
\[ v^2 = 0 + 2(g \sin \theta) \left(\frac{d}{2}\right) \]
\[ v^2 = g d \sin \theta \]

### Step 4: Analyze the Lower Half of the Incline
In the lower half, the forces acting on the body are:
- Gravitational force component down the incline: \( mg \sin \theta \)
- Frictional force acting up the incline: \( f = \mu mg \cos \theta \)

Using Newton's second law:
\[ m a' = mg \sin \theta - \mu mg \cos \theta \]
Dividing by m:
\[ a' = g \sin \theta - \mu g \cos \theta \]

### Step 5: Apply Kinematic Equation for the Lower Half
Using the kinematic equation again for the lower half:
\[ v'^2 = u'^2 + 2a's \]
where:
- \( v' = 0 \) (final velocity at the bottom)
- \( u' = v \) (initial velocity at the middle point)
- \( s = \frac{d}{2} \)

Substituting the values:
\[ 0 = v^2 + 2\left(g \sin \theta - \mu g \cos \theta\right)\left(\frac{d}{2}\right) \]
Substituting \( v^2 = g d \sin \theta \):
\[ 0 = g d \sin \theta + 2\left(g \sin \theta - \mu g \cos \theta\right)\left(\frac{d}{2}\right) \]

### Step 6: Simplify the Equation
Rearranging gives:
\[ 0 = g d \sin \theta + g d \sin \theta - \mu g d \cos \theta \]
\[ 0 = 2g d \sin \theta - \mu g d \cos \theta \]

### Step 7: Solve for μ
Dividing through by \( g d \) (assuming \( g \) and \( d \) are not zero):
\[ 0 = 2 \sin \theta - \mu \cos \theta \]
Rearranging gives:
\[ \mu \cos \theta = 2 \sin \theta \]
Thus:
\[ \mu = \frac{2 \sin \theta}{\cos \theta} \]
\[ \mu = 2 \tan \theta \]

### Final Answer
The coefficient of friction for the lower half of the inclined plane is:
\[ \mu = 2 \tan \theta \]

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