A block of weight `100N` is pushed by a force `F` on a horizantal rough plane moving with an acceleration `1m//s^(2)` when force is doubled its acceleration becomes `10m//s^(2)`. The coefficient of friction is `(10m//s^(-2))`

Options

A. `0.4`
B. `0.6`
C. `0.5`
D. `0.8`

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and use Newton's second law of motion.

### Given Data:
- Weight of the block, \( W = 100 \, \text{N} \)
- Acceleration, \( a_1 = 1 \, \text{m/s}^2 \) (when force \( F \) is applied)
- Acceleration, \( a_2 = 10 \, \text{m/s}^2 \) (when force \( 2F \) is applied)
- Coefficient of friction, \( \mu \) (to be calculated)

### Step 1: Calculate the mass of the block
Using the weight of the block:
\[ W = mg \implies m = \frac{W}{g} \]
Assuming \( g = 10 \, \text{m/s}^2 \):
\[ m = \frac{100 \, \text{N}}{10 \, \text{m/s}^2} = 10 \, \text{kg} \]

### Step 2: Write the equations of motion for both cases

#### Case 1: When force \( F \) is applied
Using Newton's second law:
\[ F - f_{\text{friction}} = ma_1 \]
Where \( f_{\text{friction}} = \mu N \) and \( N = W = 100 \, \text{N} \):
\[ F - \mu \cdot 100 = 10 \cdot 1 \]
This simplifies to:
\[ F - 100\mu = 10 \quad \text{(Equation 1)} \]

#### Case 2: When force \( 2F \) is applied
Using Newton's second law again:
\[ 2F - f_{\text{friction}} = ma_2 \]
\[ 2F - \mu \cdot 100 = 10 \cdot 10 \]
This simplifies to:
\[ 2F - 100\mu = 100 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations simultaneously

From Equation 1:
\[ F = 10 + 100\mu \quad \text{(1)} \]

Substituting \( F \) from Equation 1 into Equation 2:
\[ 2(10 + 100\mu) - 100\mu = 100 \]
Expanding this gives:
\[ 20 + 200\mu - 100\mu = 100 \]
Combining like terms:
\[ 20 + 100\mu = 100 \]
Subtracting 20 from both sides:
\[ 100\mu = 80 \]
Dividing by 100:
\[ \mu = 0.8 \]

### Final Answer:
The coefficient of friction \( \mu \) is \( 0.8 \).

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