When a spring is stretched by 10 cm, the potential energy stored is E. When the spring is stretched by 10 cm more, the potential energy stored in the spring becomes

Options

A. 2E
B. 4E
C. 6E
D. 10E

Explanation

To solve the problem, we need to calculate the potential energy stored in a spring when it is stretched by different amounts. The potential energy (PE) stored in a spring is given by the formula:

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

where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

### Step-by-Step Solution:

1. **Initial Stretch of the Spring**:
- When the spring is stretched by 10 cm (which is 0.1 m), the potential energy stored is given as \( E \).
- Using the formula for potential energy:
\[ PE_1 = \frac{1}{2} k (0.1)^2 = E \]

2. **Convert the Potential Energy Expression**:
- We can express \( E \) in terms of \( k \):
\[ E = \frac{1}{2} k (0.1)^2 = \frac{1}{2} k \cdot 0.01 = \frac{k}{200} \]

3. **Stretching the Spring Further**:
- When the spring is stretched by an additional 10 cm, the total stretch becomes 20 cm (0.2 m).
- Now, we calculate the new potential energy:
\[ PE_2 = \frac{1}{2} k (0.2)^2 \]

4. **Calculate the New Potential Energy**:
- Substitute \( 0.2 \) into the potential energy formula:
\[ PE_2 = \frac{1}{2} k (0.2)^2 = \frac{1}{2} k \cdot 0.04 = \frac{k}{50} \]

5. **Relate the New Potential Energy to \( E \)**:
- We already expressed \( E \) as \( \frac{k}{200} \).
- Now, we can express \( PE_2 \) in terms of \( E \):
\[ PE_2 = \frac{k}{50} = \frac{k}{200} \cdot 4 = 4E \]

### Final Answer:
The potential energy stored in the spring when it is stretched by 20 cm is \( 4E \).

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