a block of mass 0.1 kg attached to a spring of spring constant` 400 N/m` pulled horizontally from `x=0` to` x_1 =10`mm. Find the work done by the spring force

Options

A. `2*10^-2`J
B. `2*10^-6`J
C. `3*10^-6`J
D. `4*10^-2`J

Answer

To solve the problem of finding the work done by the spring force when a block of mass 0.1 kg is attached to a spring with a spring constant of 400 N/m and is pulled from x = 0 to x1 = 10 mm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
We have a spring with a spring constant \( k = 400 \, \text{N/m} \) that is stretched from an initial position \( x = 0 \) to a final position \( x_1 = 10 \, \text{mm} \).

2. **Convert Units**:
Convert the displacement from millimeters to meters, since the spring constant is in N/m.
\[ x_1 = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m} = 0.01 \, \text{m} \]

3. **Use the Formula for Work Done by Spring Force**:
The work done by the spring force when it is stretched or compressed is given by:
\[ W = -\frac{1}{2} k x^2 \]
where \( W \) is the work done, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.

4. **Substitute the Values**:
Substitute \( k = 400 \, \text{N/m} \) and \( x = 0.01 \, \text{m} \) into the formula:
\[ W = -\frac{1}{2} \times 400 \times (0.01)^2 \]

5. **Calculate \( (0.01)^2 \)**:
\[ (0.01)^2 = 0.0001 \, \text{m}^2 \]

6. **Calculate the Work Done**:
Now substitute \( (0.01)^2 \) back into the equation:
\[ W = -\frac{1}{2} \times 400 \times 0.0001 \]
\[ W = -\frac{400}{2} \times 0.0001 = -200 \times 0.0001 = -0.02 \, \text{J} \]

7. **Final Result**:
The work done by the spring force is:
\[ W = -0.02 \, \text{J} = -2 \times 10^{-2} \, \text{J} \]

### Conclusion:
The work done by the spring force when the block is pulled from \( x = 0 \) to \( x_1 = 10 \, \text{mm} \) is \( -2 \times 10^{-2} \, \text{J} \).