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,
- \( x \) is the amount of stretch in the spring.

### Step 1: Calculate the potential energy when the spring is stretched by 10 cm

When the spring is stretched by 10 cm (0.1 m), we can denote this potential energy as \( E \).

\[ E = \frac{1}{2} k (0.1)^2 \]

### Step 2: Calculate the potential energy when the spring is stretched by an additional 10 cm

Now, when the spring is stretched by an additional 10 cm, the total stretch becomes 20 cm (0.2 m). We denote this potential energy as \( E' \).

\[ E' = \frac{1}{2} k (0.2)^2 \]

### Step 3: Substitute the values into the potential energy formula

Now we can substitute the values into the formula for \( E' \):

\[ E' = \frac{1}{2} k (0.2)^2 = \frac{1}{2} k (0.04) = \frac{1}{2} k \cdot 0.04 \]

### Step 4: Relate \( E' \) to \( E \)

We know that:

\[ E = \frac{1}{2} k (0.1)^2 = \frac{1}{2} k (0.01) = \frac{1}{2} k \cdot 0.01 \]

Now, we can express \( E' \) in terms of \( E \):

\[ E' = \frac{1}{2} k \cdot 0.04 = 4 \left( \frac{1}{2} k \cdot 0.01 \right) = 4E \]

### Conclusion

Thus, when the spring is stretched by an additional 10 cm, the potential energy stored in the spring becomes:

\[ E' = 4E \]