Energy of a stretched wire is

Options

A. half of load ` xx` strain
B. load `xx` strain
C. stress `xx` strain
D. half of stress ` xx` strain

Answer

To find the energy stored in a stretched wire, we can follow these steps:

### Step 1: Define the parameters
Let:
- \( L_0 \) = original length of the wire
- \( L \) = change in length of the wire
- \( A \) = cross-sectional area of the wire
- \( F \) = applied force
- \( Y \) = Young's modulus of the material

### Step 2: Relate stress and strain
Stress (\( \sigma \)) is defined as:
\[ \sigma = \frac{F}{A} \]
Strain (\( \epsilon \)) is defined as:
\[ \epsilon = \frac{L}{L_0} \]

### Step 3: Express force in terms of Young's modulus
Young's modulus (\( Y \)) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{L/L_0} \]
Rearranging gives:
\[ F = Y \cdot A \cdot \frac{L}{L_0} \]

### Step 4: Calculate the work done (energy stored)
The energy stored in the wire is equal to the work done to stretch it, which can be expressed as:
\[ \text{Energy} = \int F \, ds \]
In our case, \( ds \) can be replaced with \( dL \) (the change in length). Therefore:
\[ \text{Energy} = \int_0^L F \, dL \]
Substituting \( F \) from the previous step:
\[ \text{Energy} = \int_0^L \left( Y \cdot A \cdot \frac{L}{L_0} \right) dL \]

### Step 5: Simplify and evaluate the integral
Taking \( Y \cdot A / L_0 \) as a constant:
\[ \text{Energy} = \frac{Y \cdot A}{L_0} \int_0^L L \, dL \]
Evaluating the integral:
\[ \int_0^L L \, dL = \frac{L^2}{2} \]
Thus:
\[ \text{Energy} = \frac{Y \cdot A}{L_0} \cdot \frac{L^2}{2} \]

### Step 6: Final expression for energy stored
The energy stored in the stretched wire can be expressed as:
\[ \text{Energy} = \frac{1}{2} \cdot \frac{Y \cdot A}{L_0} \cdot L^2 \]

### Step 7: Relate to stress and strain
Using the definitions of stress and strain:
- Stress = \( \frac{F}{A} \)
- Strain = \( \frac{L}{L_0} \)

We can express the energy stored in terms of stress and strain:
\[ \text{Energy} = \frac{1}{2} \cdot \sigma \cdot \epsilon \cdot V \]
where \( V \) is the volume of the wire (\( A \cdot L_0 \)).

Thus, the final expression for the energy stored in a stretched wire is:
\[ \text{Energy} = \frac{1}{2} \cdot \sigma \cdot \epsilon \cdot V \]