A body is sliding down a rough inclined plane of angle of inclination `theta`for which the coefficient of friction varies with distance x as `mu`(x) = kx, where k is a constant. Here x is the distance moved by the body down the plane. The net force on the body will be zero at a distance `x_(0)` given by :

Options

A. `(tan theta)/(k)`
B. `k tan theta `
C. `(cot theta)/(k)`
D. `k cot theta `

Answer

To solve the problem, we need to analyze the forces acting on a body sliding down a rough inclined plane where the coefficient of friction varies with distance. Let's break it down step by step.

### Step 1: Identify the Forces Acting on the Body
When a body is placed on an inclined plane, the following forces act on it:
- The gravitational force (weight) acting downwards, which can be resolved into two components:
- Perpendicular to the plane: \( mg \cos \theta \)
- Parallel to the plane: \( mg \sin \theta \)
- The normal force (\( N \)) acting perpendicular to the surface of the incline.
- The frictional force (\( f \)), which opposes the motion and is given by \( f = \mu N \).

### Step 2: Write the Expression for the Net Force
The net force (\( F_x \)) acting on the body along the inclined plane can be expressed as:
\[ F_x = mg \sin \theta - f \]
Since the frictional force \( f \) is given by \( f = \mu N \), we can substitute this into the equation:
\[ F_x = mg \sin \theta - \mu mg \cos \theta \]

### Step 3: Substitute the Coefficient of Friction
According to the problem, the coefficient of friction varies with distance \( x \) as:
\[ \mu(x) = kx \]
Substituting this into the net force equation gives:
\[ F_x = mg \sin \theta - kx \cdot mg \cos \theta \]
Factoring out \( mg \):
\[ F_x = mg \left( \sin \theta - kx \cos \theta \right) \]

### Step 4: Set the Net Force to Zero
For the body to be in equilibrium (net force is zero), we set \( F_x = 0 \):
\[ mg \left( \sin \theta - kx_0 \cos \theta \right) = 0 \]
This implies:
\[ \sin \theta - kx_0 \cos \theta = 0 \]

### Step 5: Solve for \( x_0 \)
Rearranging the equation gives:
\[ \sin \theta = kx_0 \cos \theta \]
Dividing both sides by \( \cos \theta \):
\[ \tan \theta = kx_0 \]
Finally, solving for \( x_0 \):
\[ x_0 = \frac{\tan \theta}{k} \]

### Conclusion
Thus, the distance \( x_0 \) at which the net force on the body will be zero is given by:
\[ x_0 = \frac{\tan \theta}{k} \]