A block of mass 2kg rests on a rough inclined plane making an angle of `30^@` with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

Options

A. 10.3N
B. 23.8N
C. 11.9N
D. 6.3N

Explanation

To find the frictional force acting on the block resting on a rough inclined plane, we can follow these steps:

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
- The weight of the block (mg) acting vertically downward.
- The normal force (N) acting perpendicular to the inclined plane.
- The frictional force (f) acting parallel to the inclined plane, opposing the motion.

### Step 2: Calculate the weight of the block
The weight of the block can be calculated using the formula:
\[ W = mg \]
Where:
- \( m = 2 \, \text{kg} \) (mass of the block)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \]

### Step 3: Resolve the weight into components
The weight can be resolved into two components:
- Perpendicular to the inclined plane: \( W_{\perp} = mg \cos \theta \)
- Parallel to the inclined plane: \( W_{\parallel} = mg \sin \theta \)

Here, \( \theta = 30^\circ \).

Calculating these components:
1. **Perpendicular Component**:
\[ W_{\perp} = 19.6 \, \text{N} \times \cos(30^\circ) = 19.6 \, \text{N} \times \frac{\sqrt{3}}{2} \approx 16.97 \, \text{N} \]

2. **Parallel Component**:
\[ W_{\parallel} = 19.6 \, \text{N} \times \sin(30^\circ) = 19.6 \, \text{N} \times \frac{1}{2} = 9.8 \, \text{N} \]

### Step 4: Calculate the normal force
The normal force (N) is equal to the perpendicular component of the weight:
\[ N = W_{\perp} = mg \cos(30^\circ) \approx 16.97 \, \text{N} \]

### Step 5: Calculate the maximum static frictional force
The maximum static frictional force (f_max) can be calculated using the formula:
\[ f_{\text{max}} = \mu_s N \]
Where \( \mu_s = 0.7 \) (coefficient of static friction).

Calculating the maximum static frictional force:
\[ f_{\text{max}} = 0.7 \times 16.97 \, \text{N} \approx 11.88 \, \text{N} \]

### Step 6: Determine the frictional force acting on the block
Since the block is at rest, the frictional force (f) will balance the parallel component of the weight:
\[ f = W_{\parallel} = 9.8 \, \text{N} \]

### Conclusion
The frictional force acting on the block is \( 9.8 \, \text{N} \).

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