A small mass slides down an inclined plane of inclination `theta` with the horizontal. The co-efficient of friction is `mu=mu_(0)` x where x is the distance through which the mass slides down and `mu_(0)` a constant. Then, the distance covered by the mass before it stops is

Options

A. `(2)/(mu_(0)) tan theta`
B. `(4)/(mu_(0)) tan theta`
C. `(1)/(2mu_(0))tan theta`
D. `(1)/(mu_(0)) tan theta`

Answer

To solve the problem of a small mass sliding down an inclined plane with a varying coefficient of friction, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Mass**:
- The weight of the mass \( M \) acts vertically downward, which can be resolved into two components:
- Perpendicular to the incline: \( Mg \cos \theta \)
- Parallel to the incline: \( Mg \sin \theta \)

2. **Determine the Normal Force**:
- The normal force \( N \) acting on the mass is equal to the perpendicular component of the weight:
\[ N = Mg \cos \theta \]

3. **Calculate the Frictional Force**:
- The frictional force \( F_f \) opposes the motion and is given by:
\[ F_f = \mu N = \mu_0 x N = \mu_0 x (Mg \cos \theta) \]

4. **Write the Net Force Equation**:
- The net force acting on the mass as it slides down the incline is:
\[ F_{\text{net}} = Mg \sin \theta - F_f = Mg \sin \theta - \mu_0 x (Mg \cos \theta) \]

5. **Express the Net Force in Terms of Acceleration**:
- According to Newton's second law, the net force is also equal to mass times acceleration:
\[ F_{\text{net}} = M a \]
- Therefore, we have:
\[ M a = Mg \sin \theta - \mu_0 x (Mg \cos \theta) \]
- Dividing through by \( M \):
\[ a = g \sin \theta - \mu_0 x g \cos \theta \]

6. **Relate Acceleration to Velocity**:
- We can express acceleration \( a \) as:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]
- Thus, we can rewrite the equation:
\[ v \frac{dv}{dx} = g \sin \theta - \mu_0 x g \cos \theta \]

7. **Separate Variables and Integrate**:
- Rearranging gives:
\[ v \, dv = \left( g \sin \theta - \mu_0 x g \cos \theta \right) dx \]
- Integrating both sides from \( v = 0 \) to \( v \) and \( x = 0 \) to \( x \):
\[ \int_0^v v \, dv = \int_0^x \left( g \sin \theta - \mu_0 x g \cos \theta \right) dx \]

8. **Evaluate the Integrals**:
- The left side gives:
\[ \frac{v^2}{2} \]
- The right side becomes:
\[ g \sin \theta x - \mu_0 g \cos \theta \frac{x^2}{2} \]
- Therefore, we have:
\[ \frac{v^2}{2} = g \sin \theta x - \mu_0 g \cos \theta \frac{x^2}{2} \]

9. **Set Final Velocity to Zero**:
- When the mass comes to rest, \( v = 0 \):
\[ 0 = g \sin \theta x - \mu_0 g \cos \theta \frac{x^2}{2} \]
- Rearranging gives:
\[ g \sin \theta x = \mu_0 g \cos \theta \frac{x^2}{2} \]

10. **Solve for Distance \( x \)**:
- Cancel \( g \) (assuming \( g \neq 0 \)):
\[ \sin \theta x = \mu_0 \cos \theta \frac{x^2}{2} \]
- Rearranging gives:
\[ \mu_0 \cos \theta \frac{x^2}{2} - \sin \theta x = 0 \]
- Factoring out \( x \):
\[ x \left( \mu_0 \cos \theta \frac{x}{2} - \sin \theta \right) = 0 \]
- This gives:
\[ x = 0 \quad \text{or} \quad \mu_0 \cos \theta \frac{x}{2} = \sin \theta \]
- Solving for \( x \):
\[ x = \frac{2 \sin \theta}{\mu_0 \cos \theta} \]
- Thus, the distance covered by the mass before it stops is:
\[ x = \frac{2 \tan \theta}{\mu_0} \]

### Final Answer:
\[ x = \frac{2 \tan \theta}{\mu_0} \]