A block of mass 10kg pushed by a horizontal force F on a horizontal rough plane moves with an aceleration `5ms^(-2)`. When force is doubled, its acceleration becomes `18ms^(-2)`. The coefficient of friction is (g=10`ms^(-2)`)

Options

A. 0.8
B. 0.2
C. 0.4
D. 0.6

Answer

To solve the problem, we need to analyze the forces acting on the block and use Newton's second law of motion. Let's break it down step by step.

### Step 1: Identify the known values
- Mass of the block, \( m = 10 \, \text{kg} \)
- Initial acceleration, \( a_1 = 5 \, \text{m/s}^2 \)
- Final acceleration when force is doubled, \( a_2 = 18 \, \text{m/s}^2 \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the normal force
The normal force \( N \) acting on the block is equal to the weight of the block, which is given by:
\[ N = mg = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N} \]

### Step 3: Write the equations of motion for both cases
1. **First case (force \( F \) and acceleration \( a_1 \))**:
The net force acting on the block can be expressed as:
\[ F - f = ma_1 \]
where \( f \) is the frictional force. The frictional force can be expressed as:
\[ f = \mu N \]
Substituting \( N \):
\[ F - \mu N = ma_1 \implies F - \mu(100) = 10(5) \implies F - 100\mu = 50 \quad \text{(Equation 1)} \]

2. **Second case (force \( 2F \) and acceleration \( a_2 \))**:
The net force in this case is:
\[ 2F - f = ma_2 \]
Substituting for \( f \):
\[ 2F - \mu N = ma_2 \implies 2F - \mu(100) = 10(18) \implies 2F - 100\mu = 180 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( F - 100\mu = 50 \) (Equation 1)
2. \( 2F - 100\mu = 180 \) (Equation 2)

To eliminate \( \mu \), we can subtract Equation 1 from Equation 2:
\[ (2F - 100\mu) - (F - 100\mu) = 180 - 50 \]
This simplifies to:
\[ 2F - F = 130 \implies F = 130 \, \text{N} \]

### Step 5: Substitute \( F \) back to find \( \mu \)
Now substitute \( F = 130 \, \text{N} \) back into Equation 1:
\[ 130 - 100\mu = 50 \]
Rearranging gives:
\[ 100\mu = 130 - 50 \implies 100\mu = 80 \implies \mu = 0.8 \]

### Conclusion
The coefficient of friction \( \mu \) is:
\[ \mu = 0.8 \]