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 find the work done by the spring force when a block is pulled from \( x = 0 \) to \( x_1 = 10 \) mm, we can use the formula for the work done by a spring:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Mass of the block, \( m = 0.1 \, \text{kg} \)
- Spring constant, \( k = 400 \, \text{N/m} \)
- Displacement, \( x_1 = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m} = 0.01 \, \text{m} \)

2. **Understand the Work Done by the Spring**:
The work done by the spring force when it is stretched or compressed is given by the formula:
\[ W = -\frac{1}{2} k x^2 \]
Here, the negative sign indicates that the work done by the spring force is in the opposite direction of the displacement.

3. **Calculate the Work Done**:
Substitute the values into the formula:
\[ W = -\frac{1}{2} \times 400 \, \text{N/m} \times (0.01 \, \text{m})^2 \]
\[ W = -\frac{1}{2} \times 400 \times 0.0001 \]
\[ W = -\frac{1}{2} \times 0.04 \]
\[ W = -0.02 \, \text{J} \]

4. **Final Result**:
The work done by the spring force when the block is pulled from \( x = 0 \) to \( x = 10 \) mm is:
\[ W = -0.02 \, \text{J} \quad \text{or} \quad W = -2 \times 10^{-2} \, \text{J} \]