Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring one each mass is

Options

A. `1/2kx^2`
B. `-1/2kx^2`
C. `1/4kx^2`
D. `-1/4kx^2`

Answer

To find the work done by the spring on each mass when two equal masses are attached to the ends of a spring and stretched symmetrically, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two equal masses (let's call them \( m_1 \) and \( m_2 \)) attached to a spring with spring constant \( k \).
- The spring is stretched symmetrically by a length \( x \) from its natural length.

2. **Work Done by the Spring**:
- The work done by the spring when it is stretched is related to the potential energy stored in the spring.
- The potential energy \( U \) stored in a spring when stretched by a length \( x \) is given by:
\[ U = \frac{1}{2} k x^2 \]

3. **Work Done on Each Mass**:
- Since the two masses are equal and the spring is stretched symmetrically, the work done by the spring on each mass will be half of the total potential energy stored in the spring.
- Therefore, the work done on each mass \( W \) is:
\[ W = \frac{1}{2} \left( \frac{1}{2} k x^2 \right) = \frac{1}{4} k x^2 \]

4. **Direction of Work Done**:
- The work done by the spring on each mass is negative because the spring force acts in the opposite direction to the displacement of the masses. Thus:
\[ W = -\frac{1}{4} k x^2 \]

### Final Answer:
The work done by the spring on each mass is:
\[ W = -\frac{1}{4} k x^2 \]