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 stretch. The potential energy (PE) stored in a spring can be expressed using Hooke's Law, which states:

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

where:
- \( PE \) is the potential energy,
- \( k \) is the spring constant,
- \( x \) is the amount of stretch in the spring.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The spring is initially stretched by \( x_1 = 3 \) cm, and the potential energy at this stretch is given as \( V \).
- Therefore, we can write:
\[ V = \frac{1}{2} k (3)^2 \]

2. **Calculate the Initial Potential Energy**:
- From the equation above, we can express \( V \) as:
\[ V = \frac{1}{2} k \cdot 9 \]
- Simplifying this, we get:
\[ V = \frac{9}{2} k \]

3. **Identify the New Conditions**:
- Now, the spring is stretched by \( x_2 = 6 \) cm. We need to find the new potential energy \( P_2 \) when the spring is stretched to this new length.
- Using the potential energy formula again:
\[ P_2 = \frac{1}{2} k (6)^2 \]

4. **Calculate the New Potential Energy**:
- Substituting \( x_2 = 6 \) cm into the potential energy formula:
\[ P_2 = \frac{1}{2} k \cdot 36 \]
- Simplifying this, we get:
\[ P_2 = 18k \]

5. **Relate the New Potential Energy to the Initial Potential Energy**:
- We already have \( V = \frac{9}{2} k \). Now, we can express \( P_2 \) in terms of \( V \):
\[ P_2 = 18k = 4 \cdot \frac{9}{2} k = 4V \]

6. **Final Result**:
- Therefore, the potential energy when the spring is stretched by 6 cm is:
\[ P_2 = 4V \]

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