A block rests on a rough inclined plane making an angle of `30^(@)` with horizontal. The coefficient of static friction between the block and inclined plane is `0.8` . If the frictional force on the block is `10 N`, the mass of the block in `kg` is `(g = 10 m//s^(2))`

Options

A. `(a)2.0`
B. `(b)4.0`
C. `(c)1.6`
D. `(d)2.5`

Explanation

To solve the problem step by step, we will analyze the forces acting on the block resting on the inclined plane and use the given information to find the mass of the block.

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

### Step 2: Resolve the Weight into Components
The weight of the block can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

Given that the angle \( \theta = 30^\circ \), we can write:
- \( mg \sin 30^\circ = \frac{1}{2} mg \)
- \( mg \cos 30^\circ = \frac{\sqrt{3}}{2} mg \)

### Step 3: Apply the Condition of Equilibrium
Since the block is at rest, the frictional force (f) is balancing the component of the weight acting down the incline:
\[ f = mg \sin 30^\circ \]

Given that the frictional force \( f = 10 \, \text{N} \), we can set up the equation:
\[ 10 = mg \sin 30^\circ \]

### Step 4: Substitute the Value of \( \sin 30^\circ \)
Substituting \( \sin 30^\circ = \frac{1}{2} \):
\[ 10 = mg \cdot \frac{1}{2} \]

### Step 5: Solve for Mass (m)
Rearranging the equation to solve for mass \( m \):
\[ 10 = \frac{1}{2} mg \implies mg = 20 \]
Now, substituting the value of \( g = 10 \, \text{m/s}^2 \):
\[ m \cdot 10 = 20 \implies m = \frac{20}{10} = 2 \, \text{kg} \]

### Final Answer
The mass of the block is \( 2 \, \text{kg} \).

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