The energy of a loaded wire is E. In doubling the load, what is the energy stored in the wire ?

Answer

To solve the problem of finding the energy stored in a wire when the load is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Stored in a Wire:**
The energy stored per unit volume in a wire is given by the formula:
\[ U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \]
where Stress is defined as force per unit area (σ = F/A) and Strain is the relative deformation (ε).

2. **Relate Stress and Strain:**
We know that:
\[ \text{Stress} = \sigma = \frac{F}{A} \]
and for a material, the relationship between stress and strain is given by:
\[ \sigma = E \times \epsilon \]
where E is the Young's modulus of the material.

3. **Express Strain in Terms of Stress:**
Rearranging the above equation gives:
\[ \epsilon = \frac{\sigma}{E} \]

4. **Substituting into the Energy Formula:**
Substituting the expression for strain into the energy formula, we get:
\[ U = \frac{1}{2} \times \sigma \times \left(\frac{\sigma}{E}\right) = \frac{\sigma^2}{2E} \]

5. **Initial Energy Stored:**
Let the initial load be \( W \). The initial stress is:
\[ \sigma = \frac{W}{A} \]
Therefore, the initial energy stored \( E \) can be expressed as:
\[ E = \frac{1}{2} \times \frac{W^2}{A^2} \times \frac{1}{E} \]

6. **Doubling the Load:**
When the load is doubled, the new load \( W' = 2W \). The new stress becomes:
\[ \sigma' = \frac{2W}{A} \]

7. **Calculate the New Energy Stored:**
The new energy stored \( E' \) is:
\[ E' = \frac{1}{2} \times \sigma'^2 \times \frac{1}{E} \]
Substituting \( \sigma' \):
\[ E' = \frac{1}{2} \times \left(\frac{2W}{A}\right)^2 \times \frac{1}{E} = \frac{1}{2} \times \frac{4W^2}{A^2} \times \frac{1}{E} = 4 \times \left(\frac{1}{2} \times \frac{W^2}{A^2} \times \frac{1}{E}\right) = 4E \]

8. **Final Result:**
Thus, when the load is doubled, the energy stored in the wire becomes:
\[ E' = 4E \]

### Conclusion:
When the load on the wire is doubled, the energy stored in the wire increases to four times the initial energy \( E \).