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

Explanation

To find the mass of the block resting on a rough inclined plane, we can use the information provided about the angle of inclination, the coefficient of static friction, and the frictional force acting on the block. Here’s how to solve the problem step by step:

### Step 1: Understand the Forces Acting on the Block
The forces acting on the block on the inclined plane include:
1. The gravitational force (weight) acting downwards, which can be resolved into two components:
- Perpendicular to the incline: \( W_{\perpendicular} = mg \cos(\theta) \)
- Parallel to the incline: \( W_{\parallel} = mg \sin(\theta) \)
2. The frictional force acting up the incline, which is given as 10 N.

### Step 2: Write the Equation for Frictional Force
The maximum static frictional force can be calculated using the formula:
\[ f_s = \mu_s N \]
where:
- \( f_s \) is the frictional force (10 N in this case),
- \( \mu_s \) is the coefficient of static friction (0.8),
- \( N \) is the normal force.

The normal force \( N \) can be expressed as:
\[ N = mg \cos(\theta) \]

### Step 3: Substitute the Normal Force into the Friction Equation
Substituting for \( N \) in the frictional force equation gives:
\[ f_s = \mu_s (mg \cos(\theta)) \]
\[ 10 = 0.8 (mg \cos(30^\circ)) \]

### Step 4: Calculate \( \cos(30^\circ) \)
The value of \( \cos(30^\circ) \) is:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

### Step 5: Substitute \( \cos(30^\circ) \) into the Equation
Now substituting \( \cos(30^\circ) \) into the equation:
\[ 10 = 0.8 (mg \cdot \frac{\sqrt{3}}{2}) \]

### Step 6: Simplify the Equation
Now simplify the equation:
\[ 10 = 0.4 \sqrt{3} mg \]

### Step 7: Solve for Mass \( m \)
Rearranging the equation to solve for \( m \):
\[ m = \frac{10}{0.4 \sqrt{3} g} \]
Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ m = \frac{10}{0.4 \sqrt{3} \cdot 9.8} \]

### Step 8: Calculate the Value of \( m \)
Calculating the numerical value:
1. Calculate \( 0.4 \sqrt{3} \):
\[ 0.4 \sqrt{3} \approx 0.4 \cdot 1.732 \approx 0.6928 \]
2. Now substitute back:
\[ m = \frac{10}{0.6928 \cdot 9.8} \]
\[ m \approx \frac{10}{6.795} \approx 1.47 \, \text{kg} \]

### Final Answer
The mass of the block is approximately **1.47 kg**.

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