A box of mass 8 kg is placed on a rough inclined plane of inclination `45^(@)`. Its downward motion c can be prevented by applying an upward pull F and it can be made to slide upwards by applying a force 2F. The coefficient of friction between the box and the inclined plane is

Options

A. `(1)/(2)`
B. `(1)/(sqrt2)`
C. `(1)/(2sqrt2)`
D. `(1)/(3)`

Explanation

To solve the problem, we will analyze the forces acting on the box placed on the inclined plane and use the equations of motion to find the coefficient of friction.

### Step 1: Identify the forces acting on the box
1. The weight of the box (W = mg), where m = 8 kg and g = 9.8 m/s².
2. The normal force (N) acting perpendicular to the inclined plane.
3. The frictional force (f) acting opposite to the direction of motion.
4. The applied force (F) and (2F) in different scenarios.

### Step 2: Calculate the weight of the box
\[ W = mg = 8 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 78.4 \, \text{N} \]

### Step 3: Resolve the weight into components
Since the inclined plane is at an angle of 45 degrees:
- The component of weight parallel to the incline (down the slope):
\[ W_{\parallel} = mg \sin(45^\circ) = 78.4 \, \text{N} \times \frac{1}{\sqrt{2}} = 39.2 \, \text{N} \]
- The component of weight perpendicular to the incline:
\[ W_{\perpendicular} = mg \cos(45^\circ) = 78.4 \, \text{N} \times \frac{1}{\sqrt{2}} = 39.2 \, \text{N} \]

### Step 4: Set up the equations for the two scenarios
1. **Preventing downward motion with force F:**
\[ F = W_{\parallel} - f \]
Where \( f = \mu N \) and \( N = W_{\perpendicular} \).
Thus, we can write:
\[ F = mg \sin(45^\circ) - \mu mg \cos(45^\circ) \]
Substituting the values:
\[ F = 39.2 \, \text{N} - \mu \times 39.2 \, \text{N} \]
\[ F = 39.2(1 - \mu) \]

2. **Sliding upwards with force 2F:**
\[ 2F = f + W_{\parallel} \]
Thus:
\[ 2F = \mu N + mg \sin(45^\circ) \]
Substituting the values:
\[ 2F = \mu \times 39.2 \, \text{N} + 39.2 \, \text{N} \]
\[ 2F = 39.2(\mu + 1) \]

### Step 5: Relate the two equations
From the two equations we have:
1. \( F = 39.2(1 - \mu) \)
2. \( 2F = 39.2(\mu + 1) \)

Substituting \( F \) from the first equation into the second:
\[ 2 \times 39.2(1 - \mu) = 39.2(\mu + 1) \]
Dividing both sides by 39.2:
\[ 2(1 - \mu) = \mu + 1 \]
Expanding and rearranging:
\[ 2 - 2\mu = \mu + 1 \]
\[ 2 - 1 = 3\mu \]
\[ 1 = 3\mu \]
\[ \mu = \frac{1}{3} \]

### Conclusion
The coefficient of friction between the box and the inclined plane is:
\[ \mu = \frac{1}{3} \]