A block of mass 3 kg 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. `0.7 xx 9.8 xx sqrt(3)` N
C. `9.8 xx sqrt(3)` N
D. `0.7 xx 9.8` N

Explanation

To solve the problem step by step, we will analyze the forces acting on the block on the inclined plane and calculate the frictional force.

### Step 1: Identify the Forces Acting on the Block
The forces acting on the block are:
- The gravitational force (weight) acting downwards, \( F_g = mg \)
- 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
Given:
- Mass of the block, \( m = 3 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

The weight of the block is calculated as:
\[ F_g = mg = 3 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N} \]

### Step 3: Resolve the Weight into Components
The weight can be resolved into two components:
1. Perpendicular to the inclined plane: \( F_{g\perp} = mg \cos \theta \)
2. Parallel to the inclined plane: \( F_{g\parallel} = mg \sin \theta \)

Given \( \theta = 30^\circ \):
\[ F_{g\parallel} = mg \sin 30^\circ = 30 \, \text{N} \times \frac{1}{2} = 15 \, \text{N} \]
\[ F_{g\perp} = mg \cos 30^\circ = 30 \, \text{N} \times \frac{\sqrt{3}}{2} \approx 25.98 \, \text{N} \]

### Step 4: Calculate the Normal Force
The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = F_{g\perp} = mg \cos 30^\circ \approx 25.98 \, \text{N} \]

### Step 5: Calculate the Maximum Static Frictional Force
The maximum static frictional force \( f_{\text{max}} \) can be calculated using the coefficient of static friction \( \mu_s \):
\[ f_{\text{max}} = \mu_s \cdot N \]
Given \( \mu_s = 0.7 \):
\[ f_{\text{max}} = 0.7 \times 25.98 \, \text{N} \approx 18.19 \, \text{N} \]

### Step 6: Compare the Applied Force and the Maximum Static Frictional Force
The applied force (the component of weight parallel to the incline) is \( 15 \, \text{N} \), and the maximum static frictional force is \( 18.19 \, \text{N} \).

Since \( f_{\text{max}} > F_{g\parallel} \):
\[ 18.19 \, \text{N} > 15 \, \text{N} \]
The block will not move, and the frictional force will equal the applied force to prevent motion.

### Step 7: Conclusion
The frictional force on the block is equal to the applied force:
\[ \text{Frictional force} = 15 \, \text{N} \]