When a long spring is stretched by 2 cm, its potential energy is V. If the spring is stretched by 10 cm, the potential energy in it will be

Options

A. 10 V
B. 25 V
C. `V/5`
D. 5V

Explanation

To solve the problem, we need to use the formula for the potential energy stored in a spring, which is given by:

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

where:
- \( PE \) is the potential energy,
- \( k \) is the spring constant,
- \( x \) is the displacement from the equilibrium position.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- When the spring is stretched by \( x_1 = 2 \) cm, the potential energy is \( V \).
- Thus, we can write:
\[ V = \frac{1}{2} k (2)^2 \]

2. **Calculate the potential energy for the second condition**:
- Now, when the spring is stretched by \( x_2 = 10 \) cm, we need to find the potential energy \( V_2 \):
\[ V_2 = \frac{1}{2} k (10)^2 \]

3. **Express \( V_2 \) in terms of \( V \)**:
- We can relate \( V_2 \) to \( V \) using the ratios of the displacements:
\[ V_2 = \frac{1}{2} k (10)^2 = \frac{1}{2} k (100) = 50k \]
- From the first condition, we have:
\[ V = \frac{1}{2} k (2)^2 = \frac{1}{2} k (4) = 2k \]

4. **Relate \( V_2 \) to \( V \)**:
- Now, we can express \( V_2 \) in terms of \( V \):
\[ V_2 = 50k = 25 \times 2k = 25V \]

5. **Conclusion**:
- Therefore, the potential energy when the spring is stretched by 10 cm is:
\[ V_2 = 25V \]

### Final Answer:
The potential energy in the spring when stretched by 10 cm is \( 25V \).

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