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. 9.8N
B. `0.7xx9.8xxsqrt3N`
C. `9.8xxsqrt3N`
D. `0.7xx9.8N`

Explanation

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

### Step 1: Identify the forces acting on the block
The forces acting on the block include:
- The gravitational force (\(mg\)) acting downward.
- The normal force (\(N\)) acting perpendicular to the inclined plane.
- The frictional force (\(f_s\)) acting parallel to the inclined plane, opposing the motion.

### Step 2: Calculate the gravitational force
The gravitational force can be calculated using the formula:
\[ F_g = mg \]
Where:
- \(m = 2 \, \text{kg}\)
- \(g = 9.8 \, \text{m/s}^2\)

Calculating this gives:
\[ F_g = 2 \times 9.8 = 19.6 \, \text{N} \]

### Step 3: Resolve the gravitational force into components
The gravitational force can be resolved into two components:
1. Parallel to the incline: \(F_{\parallel} = mg \sin \theta\)
2. Perpendicular to the incline: \(F_{\perpendicular} = mg \cos \theta\)

For \(\theta = 30^\circ\):
\[ F_{\parallel} = 19.6 \sin(30^\circ) = 19.6 \times \frac{1}{2} = 9.8 \, \text{N} \]
\[ F_{\perpendicular} = 19.6 \cos(30^\circ) = 19.6 \times \frac{\sqrt{3}}{2} \approx 16.97 \, \text{N} \]

### Step 4: Calculate the normal force
The normal force (\(N\)) is equal to the perpendicular component of the gravitational force:
\[ N = F_{\perpendicular} = 16.97 \, \text{N} \]

### Step 5: Calculate the maximum static frictional force
The maximum static frictional force (\(f_{s,\text{max}}\)) can be calculated using the formula:
\[ f_{s,\text{max}} = \mu_s N \]
Where \(\mu_s = 0.7\):
\[ f_{s,\text{max}} = 0.7 \times 16.97 \approx 11.88 \, \text{N} \]

### Step 6: Compare the forces
Since the block is at rest, the static frictional force will be equal to the parallel component of the gravitational force, as long as it does not exceed the maximum static frictional force:
\[ f_s = F_{\parallel} = 9.8 \, \text{N} \]
Since \(9.8 \, \text{N} < 11.88 \, \text{N}\), the static frictional force will be \(9.8 \, \text{N}\).

### Final Answer
The frictional force on the block is:
\[ \boxed{9.8 \, \text{N}} \]