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. `sqrt3//10`
B. `10//3`
C. `0.47`
D. `0.185`

Explanation

To find the coefficient of friction for the given problem, we can follow these steps:

### Step 1: Identify the given data
- Mass of the block (m) = 30 kg
- Force applied (F) = 200 N
- Final velocity (v) = 4 m/s
- Initial velocity (u) = 0 m/s (starting from rest)
- Time (t) = 2 s

### Step 2: Calculate the acceleration of the block
Using the equation of motion:
\[ v = u + at \]
Substituting the known values:
\[ 4 = 0 + a \cdot 2 \]
\[ a = \frac{4}{2} = 2 \, \text{m/s}^2 \]

### Step 3: Apply Newton's second law
According to Newton's second law:
\[ F_{\text{net}} = m \cdot a \]
The net force acting on the block can be expressed as the applied force minus the frictional force:
\[ F_{\text{net}} = F - F_{\text{friction}} \]
Thus, we can write:
\[ F - F_{\text{friction}} = m \cdot a \]
Substituting the known values:
\[ 200 - F_{\text{friction}} = 30 \cdot 2 \]
\[ 200 - F_{\text{friction}} = 60 \]

### Step 4: Solve for the frictional force
Rearranging the equation gives:
\[ F_{\text{friction}} = 200 - 60 = 140 \, \text{N} \]

### Step 5: Relate frictional force to the coefficient of friction
The frictional force can also be expressed as:
\[ F_{\text{friction}} = \mu \cdot N \]
Where \( N \) is the normal force. For a horizontal surface:
\[ N = m \cdot g = 30 \cdot 10 = 300 \, \text{N} \]
Now substituting into the friction equation:
\[ 140 = \mu \cdot 300 \]

### Step 6: Solve for the coefficient of friction
Rearranging gives:
\[ \mu = \frac{140}{300} \]
\[ \mu = 0.4667 \]

### Step 7: Round to two decimal places
Thus, the coefficient of friction \( \mu \) is approximately:
\[ \mu \approx 0.47 \]

### Final Answer:
The coefficient of friction is \( \mu \approx 0.47 \).

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