A body starts from rest on a long inclined plane of slope `45^(@)` . The coefficient of friction between the body and the plane varies as `mu = 0.3 x` , where x is the distance travelled down the plane. The body will have maximum speed (for `g = 10 m//s^(2)` ) when x =

Options

A. 9.8 m
B. 12 m
C. 12 m
D. 3.33 m

Explanation

To solve the problem, we need to analyze the forces acting on the body as it moves down the inclined plane. Here’s a step-by-step solution:

### Step 1: Identify the forces acting on the body
The forces acting on the body are:
1. The gravitational force acting down the slope: \( F_{\text{gravity}} = mg \sin \theta \)
2. The normal force acting perpendicular to the slope: \( F_{\text{normal}} = mg \cos \theta \)
3. The frictional force acting up the slope: \( F_{\text{friction}} = \mu F_{\text{normal}} = \mu mg \cos \theta \)

### Step 2: Write the expression for the net force
The net force acting on the body as it moves down the incline can be expressed as:
\[ F_{\text{net}} = F_{\text{gravity}} - F_{\text{friction}} = mg \sin \theta - \mu mg \cos \theta \]

### Step 3: Substitute the values
Given that the angle of inclination \( \theta = 45^\circ \), we know:
- \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
- The coefficient of friction is given as \( \mu = 0.3x \)

Substituting these values into the net force equation:
\[ F_{\text{net}} = mg \left( \frac{1}{\sqrt{2}} \right) - (0.3x) mg \left( \frac{1}{\sqrt{2}} \right) \]
Factoring out \( mg \frac{1}{\sqrt{2}} \):
\[ F_{\text{net}} = mg \frac{1}{\sqrt{2}} \left( 1 - 0.3x \right) \]

### Step 4: Relate net force to acceleration
According to Newton's second law, \( F_{\text{net}} = ma \), where \( a \) is the acceleration of the body. Thus:
\[ ma = mg \frac{1}{\sqrt{2}} \left( 1 - 0.3x \right) \]
Dividing both sides by \( m \):
\[ a = g \frac{1}{\sqrt{2}} \left( 1 - 0.3x \right) \]

### Step 5: Set the acceleration to zero for maximum speed
The body will reach maximum speed when the acceleration becomes zero:
\[ 0 = g \frac{1}{\sqrt{2}} \left( 1 - 0.3x \right) \]
This implies:
\[ 1 - 0.3x = 0 \]
Solving for \( x \):
\[ 0.3x = 1 \implies x = \frac{1}{0.3} = \frac{10}{3} \approx 3.33 \text{ meters} \]

### Conclusion
The body will have maximum speed when \( x = 3.33 \) meters.