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 variable coefficient of friction, we can follow these steps:

### Step 1: Identify the Forces Acting on the Mass
When the mass \( m \) slides down the inclined plane, the forces acting on it are:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting perpendicular to the surface.
- The frictional force \( f \) acting opposite to the direction of motion.

### Step 2: Determine the Components of the Gravitational Force
The gravitational force can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

### Step 3: Express the Frictional Force
The frictional force \( f \) is given by:
\[ f = \mu N \]
where \( \mu = \mu_0 x \) (as given in the problem) and \( N = mg \cos \theta \). Therefore,
\[ f = \mu_0 x (mg \cos \theta) \]

### Step 4: Apply the Work-Energy Principle
According to the work-energy principle, the work done by all forces is equal to the change in kinetic energy. Initially, the mass starts from rest, so the initial kinetic energy is zero. The final kinetic energy will also be zero when the mass stops.

The work done by gravity is:
\[ W_g = mg \sin \theta \cdot d \]

The work done against friction is:
\[ W_f = -f \cdot d = -\mu_0 x (mg \cos \theta) \cdot d \]

### Step 5: Set Up the Equation
Using the work-energy principle:
\[ W_g + W_f = 0 \]
Substituting the expressions for work:
\[ mg \sin \theta \cdot d - \mu_0 x (mg \cos \theta) \cdot d = 0 \]

### Step 6: Simplify the Equation
We can cancel \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ \sin \theta \cdot d - \mu_0 x \cos \theta \cdot d = 0 \]
This simplifies to:
\[ d \sin \theta = \mu_0 x \cos \theta \cdot d \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ d \sin \theta = \mu_0 \cos \theta \cdot x \]

### Step 8: Integrate to Find Distance
Since \( \mu = \mu_0 x \), we can express the frictional work in terms of \( x \):
\[ \int_0^d \mu_0 x (mg \cos \theta) \, dx = \frac{\mu_0 mg \cos \theta}{2} d^2 \]
Now substituting back into the work-energy equation gives:
\[ mg \sin \theta \cdot d - \frac{\mu_0 mg \cos \theta}{2} d^2 = 0 \]

### Step 9: Solve for Distance \( d \)
Factoring out \( mg \):
\[ d \sin \theta = \frac{\mu_0 \cos \theta}{2} d^2 \]
Dividing both sides by \( d \) (assuming \( d \neq 0 \)):
\[ \sin \theta = \frac{\mu_0 \cos \theta}{2} d \]
Thus,
\[ d = \frac{2 \sin \theta}{\mu_0 \cos \theta} \]

### Final Answer
The distance covered by the mass before it stops is:
\[ d = \frac{2 \mu_0 \sin \theta}{\cos \theta} \]