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
The forces acting on the body are:
1. The gravitational force \( mg \) acting downwards.
2. The normal force \( N \) acting perpendicular to the inclined plane.
3. The frictional force \( F_f \) acting up the incline, which depends on the coefficient of friction \( \mu \).

### Step 2: Resolve the Gravitational Force
Since the incline is at an angle of \( 45^\circ \), we can resolve the gravitational force into two components:
- The component parallel to the incline: \( F_{\parallel} = mg \sin(45^\circ) = mg \frac{1}{\sqrt{2}} \)
- The component perpendicular to the incline: \( F_{\perpendicular} = mg \cos(45^\circ) = mg \frac{1}{\sqrt{2}} \)

### Step 3: Calculate the Normal Force
The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = mg \cos(45^\circ) = mg \frac{1}{\sqrt{2}} \]

### Step 4: Calculate the Frictional Force
The frictional force \( F_f \) is given by:
\[ F_f = \mu N = \mu mg \cos(45^\circ) = (0.3x) \left( mg \frac{1}{\sqrt{2}} \right) \]
where \( \mu = 0.3x \).

### Step 5: Write the Equation of Motion
The net force acting on the body along the incline is given by:
\[ F_{\text{net}} = F_{\parallel} - F_f = mg \sin(45^\circ) - F_f \]
Substituting the expressions we found:
\[ F_{\text{net}} = mg \frac{1}{\sqrt{2}} - (0.3x) \left( mg \frac{1}{\sqrt{2}} \right) \]
Factoring out \( mg \frac{1}{\sqrt{2}} \):
\[ F_{\text{net}} = mg \frac{1}{\sqrt{2}} \left( 1 - 0.3x \right) \]

### Step 6: Relate Net Force to Acceleration
Using Newton's second law, \( F = ma \):
\[ ma = mg \frac{1}{\sqrt{2}} (1 - 0.3x) \]
Dividing both sides by \( m \):
\[ a = g \frac{1}{\sqrt{2}} (1 - 0.3x) \]

### Step 7: Find Maximum Speed Condition
The body will reach maximum speed when the acceleration becomes zero:
\[ 0 = g \frac{1}{\sqrt{2}} (1 - 0.3x) \]
Setting the term in parentheses to zero:
\[ 1 - 0.3x = 0 \]
Solving for \( x \):
\[ 0.3x = 1 \implies x = \frac{1}{0.3} = \frac{10}{3} \approx 3.33 \text{ meters} \]

### Final Answer
The body will have maximum speed when \( x \approx 3.33 \) meters.

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