A `30kg` block rests on a rough horizontal surface A force of `200N` is applied on the block The block acquires a speed of `4m//s` starting from rest in `2s` What is the value of coefficient of friction ? .

Options

A. `10 //sqrt( 3)`
B. ` sqrt( 3) //10`
C. `0.47`
D. 0.185

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given data
- Mass of the block, \( m = 30 \, \text{kg} \)
- Applied force, \( F = 200 \, \text{N} \)
- Final velocity, \( v = 4 \, \text{m/s} \)
- Initial velocity, \( u = 0 \, \text{m/s} \)
- Time, \( t = 2 \, \text{s} \)

### Step 2: Calculate the acceleration of the block
Using the formula for acceleration:
\[ a = \frac{v - u}{t} \]
Substituting the values:
\[ a = \frac{4 \, \text{m/s} - 0 \, \text{m/s}}{2 \, \text{s}} = \frac{4}{2} = 2 \, \text{m/s}^2 \]

### Step 3: Apply Newton's second law of motion
According to Newton's second law:
\[ F_{\text{net}} = m \cdot a \]
The net force acting on the block is the applied force minus the frictional force:
\[ F_{\text{net}} = F - F_{\text{friction}} = F - \mu \cdot N \]
Where \( N \) (normal force) is equal to \( m \cdot g \):
\[ N = m \cdot g = 30 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 300 \, \text{N} \]
Thus, the frictional force can be expressed as:
\[ F_{\text{friction}} = \mu \cdot N = \mu \cdot 300 \, \text{N} \]

### Step 4: Set up the equation for net force
Now we can set up the equation:
\[ F - \mu \cdot N = m \cdot a \]
Substituting the known values:
\[ 200 \, \text{N} - \mu \cdot 300 \, \text{N} = 30 \, \text{kg} \cdot 2 \, \text{m/s}^2 \]
This simplifies to:
\[ 200 - 300\mu = 60 \]

### Step 5: Solve for the coefficient of friction \( \mu \)
Rearranging the equation gives:
\[ 200 - 60 = 300\mu \]
\[ 140 = 300\mu \]
\[ \mu = \frac{140}{300} = \frac{14}{30} = \frac{7}{15} \approx 0.467 \]

### Step 6: Final result
The coefficient of friction \( \mu \) is approximately \( 0.47 \).