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

Options

A. 25U
B. U/5
C. 5U
D. 10U

Explanation

To solve the problem, we need to understand the relationship between the potential energy stored in a spring and the displacement from its equilibrium position. The potential energy (U) stored in a spring is given by the formula:

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

where:
- \( U \) 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 condition**:
When the spring is stretched by 2 cm, the potential energy is given as \( U \).
\[ U = \frac{1}{2} k (2)^2 \]

2. **Calculate the potential energy for the initial stretch**:
Substitute \( x = 2 \) cm into the potential energy formula:
\[ U = \frac{1}{2} k (2^2) = \frac{1}{2} k (4) = 2k \]

3. **Identify the new condition**:
Now, we need to find the potential energy when the spring is stretched by 10 cm.
\[ U' = \frac{1}{2} k (10)^2 \]

4. **Calculate the potential energy for the new stretch**:
Substitute \( x = 10 \) cm into the potential energy formula:
\[ U' = \frac{1}{2} k (10^2) = \frac{1}{2} k (100) = 50k \]

5. **Find the ratio of the new potential energy to the initial potential energy**:
We have \( U = 2k \) and \( U' = 50k \). We can express \( U' \) in terms of \( U \):
\[ \frac{U'}{U} = \frac{50k}{2k} = \frac{50}{2} = 25 \]

6. **Express the new potential energy in terms of U**:
Therefore, the new potential energy \( U' \) is:
\[ U' = 25U \]

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