A long spring is stretched by 3 cm and its potential energy is V . If the spring is stretched by 6 cm ,its potential energy will be

Options

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

Answer

To solve the problem, we need to understand the relationship between the potential energy stored in a spring and its extension. The potential energy \( V \) stored in a spring is given by the formula:

\[ V = \frac{1}{2} k x^2 \]

where:
- \( V \) is the potential energy,
- \( k \) is the spring constant,
- \( x \) is the extension (or compression) of the spring.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- The spring is initially stretched by \( x_1 = 3 \) cm.
- The potential energy at this extension is given as \( V \).

2. **Write the potential energy for the initial extension**:
\[ V = \frac{1}{2} k (3)^2 = \frac{1}{2} k \cdot 9 \]

3. **Determine the new extension**:
- The spring is stretched by \( x_2 = 6 \) cm.

4. **Write the potential energy for the new extension**:
\[ V' = \frac{1}{2} k (6)^2 = \frac{1}{2} k \cdot 36 \]

5. **Relate the new potential energy to the initial potential energy**:
- We can express \( V' \) in terms of \( V \):
\[ V' = \frac{1}{2} k \cdot 36 = \frac{1}{2} k \cdot 4 \cdot 9 = 4 \left( \frac{1}{2} k \cdot 9 \right) = 4V \]

6. **Conclusion**:
- Therefore, when the spring is stretched by 6 cm, the potential energy will be \( 4V \).

### Final Answer:
The potential energy when the spring is stretched by 6 cm will be \( 4V \).