A block of mass 0.18 kg is attached to a spring of force constant 2N/m. The coefficient of friction between the block and the floor is 0.1. Initially, the block is at rest and the spring is unstretched. An impulse is given to the block.
The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m/s is v=N/10. Then, N is

Options

A. 2
B. 3
C. 4
D. 6

Explanation

To solve the problem step by step, we will analyze the forces acting on the block, the energy transformations involved, and use the given data to find the initial velocity of the block.

### Step 1: Identify the given data
- Mass of the block, \( m = 0.18 \, \text{kg} \)
- Spring constant, \( k = 2 \, \text{N/m} \)
- Coefficient of friction, \( \mu = 0.1 \)
- Distance slid, \( x = 0.06 \, \text{m} \)

### Step 2: Calculate the gravitational force acting on the block
The gravitational force \( F_g \) acting on the block can be calculated using:
\[ F_g = m \cdot g \]
where \( g \approx 9.81 \, \text{m/s}^2 \).

Calculating:
\[ F_g = 0.18 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 1.764 \, \text{N} \]

### Step 3: Calculate the frictional force
The frictional force \( F_f \) can be calculated using:
\[ F_f = \mu \cdot F_g = \mu \cdot m \cdot g \]
Calculating:
\[ F_f = 0.1 \cdot 1.764 \, \text{N} = 0.1764 \, \text{N} \]

### Step 4: Calculate the work done against friction
The work done against friction \( W_f \) when the block slides a distance \( x \) is given by:
\[ W_f = F_f \cdot x \]
Calculating:
\[ W_f = 0.1764 \, \text{N} \cdot 0.06 \, \text{m} = 0.010584 \, \text{J} \]

### Step 5: Set up the energy conservation equation
The initial kinetic energy \( KE \) of the block when it is given an impulse is equal to the work done against friction plus the potential energy stored in the spring when it is compressed:
\[ \frac{1}{2} mv^2 = W_f + \frac{1}{2} k x^2 \]

### Step 6: Calculate the potential energy stored in the spring
The potential energy \( PE \) stored in the spring when compressed by \( x \) is given by:
\[ PE = \frac{1}{2} k x^2 \]
Calculating:
\[ PE = \frac{1}{2} \cdot 2 \, \text{N/m} \cdot (0.06 \, \text{m})^2 = \frac{1}{2} \cdot 2 \cdot 0.0036 = 0.0036 \, \text{J} \]

### Step 7: Substitute values into the energy equation
Substituting the values into the energy equation:
\[ \frac{1}{2} \cdot 0.18 \cdot v^2 = 0.010584 + 0.0036 \]
Calculating the right side:
\[ \frac{1}{2} \cdot 0.18 \cdot v^2 = 0.014184 \]

### Step 8: Solve for \( v^2 \)
Now, solving for \( v^2 \):
\[ 0.09 \cdot v^2 = 0.014184 \]
\[ v^2 = \frac{0.014184}{0.09} = 0.1576 \]

### Step 9: Calculate \( v \)
Taking the square root to find \( v \):
\[ v = \sqrt{0.1576} \approx 0.396 \, \text{m/s} \]

### Step 10: Relate \( v \) to \( N \)
Given that \( v = \frac{N}{10} \):
\[ 0.396 = \frac{N}{10} \]
\[ N = 0.396 \times 10 = 3.96 \]

### Conclusion
Since \( N \) must be an integer, we round \( N \) to the nearest integer:
\[ N \approx 4 \]

### Final Answer
Thus, the value of \( N \) is \( 4 \).