A long elastic spring is stretched by `2 cm` and its potential energy is `U`. If the spring is stretched by `10 cm`, the `PE` will be

Options

A. `(U)/(25)`
B. `(U)/(5)`
C. 5U
D. 25U

Explanation

To solve the problem, we need to determine the potential energy of a spring when it is stretched by different amounts. We will use the formula for the potential energy stored in a spring, which is given by:

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

where:
- \( U \) 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 stretch and potential energy**:
- The spring is initially stretched by \( x_1 = 2 \) cm, and its potential energy at this stretch is \( U \).

2. **Write the potential energy formula for the initial stretch**:
- Using the formula, we have:
\[ U = \frac{1}{2} k (x_1)^2 = \frac{1}{2} k (2 \text{ cm})^2 \]

3. **Calculate the potential energy for the new stretch**:
- Now, the spring is stretched by \( x_2 = 10 \) cm. We need to find the new potential energy \( U' \):
\[ U' = \frac{1}{2} k (x_2)^2 = \frac{1}{2} k (10 \text{ cm})^2 \]

4. **Relate the two potential energies**:
- Since both potential energies involve the same spring constant \( k \), we can set up a ratio:
\[ \frac{U'}{U} = \frac{\frac{1}{2} k (x_2)^2}{\frac{1}{2} k (x_1)^2} = \frac{(x_2)^2}{(x_1)^2} \]

5. **Substitute the values of \( x_1 \) and \( x_2 \)**:
- Plugging in the values:
\[ \frac{U'}{U} = \frac{(10 \text{ cm})^2}{(2 \text{ cm})^2} = \frac{100 \text{ cm}^2}{4 \text{ cm}^2} = 25 \]

6. **Express \( U' \) in terms of \( U \)**:
- This implies:
\[ U' = 25 U \]

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