When a spring is stretched by 2 cm, it stores 100 J of energy. If it is further stretched by 2 cm, the stored energy will be increased by

Options

A. 100 J
B. 200 J
C. 300 J
D. 400 J

Explanation

To solve the problem, we need to understand the relationship between the extension of a spring and the potential energy stored in it. The potential energy (U) stored in a spring when it is stretched or compressed 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 natural length of the spring.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The spring is initially stretched by \( x_1 = 2 \) cm, and it stores \( U_1 = 100 \) J of energy.

2. **Calculate the Energy for the First Stretch**:
- Using the formula for potential energy:
\[ U_1 = \frac{1}{2} k x_1^2 \]
- We can rearrange this to find \( k \):
\[ 100 = \frac{1}{2} k (0.02)^2 \]
- Solving for \( k \):
\[ k = \frac{100 \times 2}{(0.02)^2} = \frac{200}{0.0004} = 500000 \, \text{N/m} \]

3. **Determine the New Stretch**:
- The spring is further stretched by another \( 2 \) cm, making the total stretch \( x_2 = 4 \) cm.

4. **Calculate the Energy for the Second Stretch**:
- Now, we calculate the potential energy when the spring is stretched to \( x_2 \):
\[ U_2 = \frac{1}{2} k x_2^2 \]
- Substituting \( x_2 = 0.04 \) m:
\[ U_2 = \frac{1}{2} (500000) (0.04)^2 = \frac{1}{2} (500000) (0.0016) = 400 \, \text{J} \]

5. **Calculate the Increase in Energy**:
- The increase in stored energy when the spring is stretched from \( x_1 \) to \( x_2 \) is:
\[ \Delta U = U_2 - U_1 = 400 \, \text{J} - 100 \, \text{J} = 300 \, \text{J} \]

### Final Answer:
The stored energy will be increased by **300 J**.