A block 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 0.8. If the frictional force on the block is 10N, the mass of the block (in kg) is

Options

A. 1.6
B. `4.0`
C. `2.0`
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.

### Step 1: Identify the Forces Acting on the Block
The block experiences two main forces:
1. The gravitational force acting downwards, which can be expressed as \( mg \) (where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity, approximately \( 10 \, \text{m/s}^2 \)).
2. The frictional force acting parallel to the incline, which is given as \( 10 \, \text{N} \).

### Step 2: Resolve the Gravitational Force into Components
The gravitational force can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

Given that the angle \( \theta = 30^\circ \):
- The component of gravitational force parallel to the incline is \( mg \sin(30^\circ) \).
- The component of gravitational force perpendicular to the incline is \( mg \cos(30^\circ) \).

### Step 3: Use the Given Frictional Force
According to the problem, the frictional force \( f \) is equal to \( 10 \, \text{N} \). The frictional force can be expressed in terms of the normal force \( N \) and the coefficient of static friction \( \mu \):
\[ f = \mu N \]
Where \( \mu = 0.8 \).

### Step 4: Calculate the Normal Force
The normal force \( N \) is equal to the component of the gravitational force perpendicular to the incline:
\[ N = mg \cos(30^\circ) \]
Substituting this into the frictional force equation:
\[ 10 = 0.8 (mg \cos(30^\circ)) \]

### Step 5: Calculate \( mg \sin(30^\circ) \)
From the equilibrium condition, the frictional force must balance the component of the gravitational force acting down the incline:
\[ mg \sin(30^\circ) = 10 \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ mg \cdot \frac{1}{2} = 10 \]

### Step 6: Solve for Mass \( m \)
Rearranging the equation gives:
\[ mg = 10 \cdot 2 = 20 \]
Now substituting \( g = 10 \, \text{m/s}^2 \):
\[ m \cdot 10 = 20 \]
\[ m = \frac{20}{10} = 2 \, \text{kg} \]

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

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