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

Options

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

Explanation

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
The mass \( m \) experiences the following forces:
- Gravitational force \( mg \) acting downwards.
- Normal force \( N \) acting perpendicular to the inclined plane.
- Frictional force \( F_f \) acting up the incline, which is given by \( F_f = \mu N \).

### Step 2: Resolve 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: Write the expression for the normal force
The normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

### Step 4: Write the expression for the frictional force
Given that the coefficient of friction \( \mu = \mu_0 x \), the frictional force becomes:
\[ F_f = \mu N = \mu_0 x (mg \cos \theta) \]

### Step 5: Set up the equation of motion
Using Newton's second law, we can write the equation of motion along the incline:
\[ mg \sin \theta - F_f = ma \]
Substituting for \( F_f \):
\[ mg \sin \theta - \mu_0 x (mg \cos \theta) = ma \]
Dividing through by \( m \):
\[ g \sin \theta - \mu_0 x g \cos \theta = a \]

### Step 6: Express acceleration in terms of velocity and position
Using the relation \( a = v \frac{dv}{dx} \), we can rewrite the equation:
\[ v \frac{dv}{dx} = g \sin \theta - \mu_0 x g \cos \theta \]

### Step 7: Separate variables and integrate
Rearranging gives:
\[ \int v \, dv = \int \left(g \sin \theta - \mu_0 x g \cos \theta\right) dx \]
Integrating both sides:
\[ \frac{v^2}{2} = g \sin \theta \cdot x - \frac{\mu_0 g \cos \theta}{2} x^2 + C \]
Since the mass starts from rest, when \( x = 0 \), \( v = 0 \), thus \( C = 0 \).

### Step 8: Set the final condition when the mass stops
When the mass comes to a stop, \( v = 0 \):
\[ 0 = g \sin \theta \cdot x_m - \frac{\mu_0 g \cos \theta}{2} x_m^2 \]
Factoring out \( x_m \):
\[ x_m \left(g \sin \theta - \frac{\mu_0 g \cos \theta}{2} x_m\right) = 0 \]
Since \( x_m \neq 0 \), we can set the term in parentheses to zero:
\[ g \sin \theta = \frac{\mu_0 g \cos \theta}{2} x_m \]

### Step 9: Solve for \( x_m \)
Rearranging gives:
\[ x_m = \frac{2 g \sin \theta}{\mu_0 g \cos \theta} \]
Simplifying:
\[ x_m = \frac{2 \sin \theta}{\mu_0 \cos \theta} = \frac{2}{\mu_0} \tan \theta \]

### Final Result
Thus, the distance covered by the mass before it stops is:
\[ x_m = \frac{2}{\mu_0} \tan \theta \]

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