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. 9.8 N
B. `0.7xx9.8xxsqrt(3) N`
C. `9.8xxsqrt(3) N`
D. `0.7xx9.8 N`

Explanation

To find the frictional force acting on a block of mass 2 kg resting on a rough inclined plane at an angle of 30 degrees with a coefficient of static friction of 0.7, we can follow these steps:

### Step 1: Calculate the gravitational force acting on the block.
The gravitational force (weight) acting on the block can be calculated using the formula:
\[ F_g = m \cdot g \]
where:
- \( m = 2 \, \text{kg} \) (mass of the block)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating this gives:
\[ F_g = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \]

### Step 2: Resolve the gravitational force into components.
The gravitational force can be resolved into two components:
1. Perpendicular to the inclined plane:
\[ F_{g\perp} = F_g \cdot \cos(\theta) \]
2. Parallel to the inclined plane:
\[ F_{g\parallel} = F_g \cdot \sin(\theta) \]

For \( \theta = 30^\circ \):
- \( F_{g\perp} = 19.6 \cdot \cos(30^\circ) \)
- \( F_{g\parallel} = 19.6 \cdot \sin(30^\circ) \)

Calculating these components:
\[ F_{g\perp} = 19.6 \cdot \frac{\sqrt{3}}{2} \approx 19.6 \cdot 0.866 \approx 16.97 \, \text{N} \]
\[ F_{g\parallel} = 19.6 \cdot \frac{1}{2} = 9.8 \, \text{N} \]

### Step 3: Calculate the maximum static frictional force.
The maximum static frictional force can be calculated using the formula:
\[ F_{f_{\text{max}}} = \mu_s \cdot F_{g\perp} \]
where:
- \( \mu_s = 0.7 \) (coefficient of static friction)

Calculating this gives:
\[ F_{f_{\text{max}}} = 0.7 \cdot 16.97 \approx 11.88 \, \text{N} \]

### Step 4: Compare the parallel gravitational force with the maximum static frictional force.
We compare \( F_{g\parallel} \) with \( F_{f_{\text{max}}} \):
- \( F_{g\parallel} = 9.8 \, \text{N} \)
- \( F_{f_{\text{max}}} = 11.88 \, \text{N} \)

Since \( F_{g\parallel} < F_{f_{\text{max}}} \), the block will not slide down the incline.

### Step 5: Determine the actual frictional force.
Since the block is at rest and not sliding, the actual frictional force will equal the parallel component of the gravitational force:
\[ F_f = F_{g\parallel} = 9.8 \, \text{N} \]

Thus, the frictional force acting on the block is **9.8 N**.

### Summary
The frictional force on the block is **9.8 N**.