A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. If `mu_(S)= 0.6` , what is the frictional force on the block ?

Options

A. 9.8 N
B. 19.6N
C. 14.7N
D. 4.9N

Answer

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

### Step 1: Identify the forces acting on the block
The forces acting on the block include:
- The gravitational force (weight) acting downwards, which is \( mg \).
- 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 gravitational force
The gravitational force \( mg \) can be calculated as:
\[ mg = m \cdot g \]
where:
- \( m = 2 \, \text{kg} \) (mass of the block)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

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

### Step 3: Resolve the gravitational force into components
The gravitational force can be resolved into two components:
- Perpendicular to the inclined plane: \( mg \cos \theta \)
- Parallel to the inclined plane: \( mg \sin \theta \)

For \( \theta = 30° \):
\[ mg \cos 30° = 19.6 \cos 30° = 19.6 \cdot \frac{\sqrt{3}}{2} \approx 16.97 \, \text{N} \]
\[ mg \sin 30° = 19.6 \sin 30° = 19.6 \cdot \frac{1}{2} = 9.8 \, \text{N} \]

### Step 4: Calculate the maximum static frictional force
The maximum static frictional force \( f_s \) is given by:
\[ f_s = \mu_s \cdot N \]
where \( N \) is the normal force. Since the normal force is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta = 19.6 \cdot \cos 30° \approx 16.97 \, \text{N} \]
Now, substituting into the equation for static friction:
\[ f_s = 0.6 \cdot 16.97 \approx 10.18 \, \text{N} \]

### Step 5: Compare the frictional force with the component of weight
The component of the weight acting down the incline is \( mg \sin 30° = 9.8 \, \text{N} \). Since the maximum static friction (10.18 N) is greater than the component of the weight (9.8 N), the block will not slide down the incline.

### Conclusion
The frictional force acting on the block will be equal to the component of the weight acting down the incline, which is:
\[ f = mg \sin 30° = 9.8 \, \text{N} \]

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

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