Elastic Potential Energy Stretched Copper Wire

To find the elastic potential energy stored in the stretched copper wire, we can use the relation between potential energy density, Young's modulus, and longitudinal strain.

1. Given Data:

• Length of the wire, $L = 3\text{ m}$
• Elongation of the wire, $\Delta L = 3\text{ mm} = 3 \times 10^{-3}\text{ m}$
• Volume of the wire, $V = 600 \times 10^{-6}\text{ m}^3$
• Young's modulus of copper, $Y = 1.1 \times 10^{11}\text{ N/m}^2$

2. Calculation of Longitudinal Strain ($\epsilon$): $\epsilon = \frac{\Delta L}{L} = \frac{3 \times 10^{-3}\text{ m}}{3\text{ m}} = 10^{-3}$

3. Calculation of Energy Density ($u$): The elastic potential energy stored per unit volume (energy density) is given by: $u = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{1}{2} Y \epsilon^2$

Substituting the given values: $u = \frac{1}{2} \times \left(1.1 \times 10^{11}\text{ N/m}^2\right) \times \left(10^{-3}\right)^2$ $u = 0.55 \times 10^{11} \times 10^{-6}\text{ J/m}^3 = 5.5 \times 10^{4}\text{ J/m}^3$

4. Calculation of Total Elastic Potential Energy ($U$): The total elastic potential energy stored in the wire is: $U = u \times V$ $U = \left(5.5 \times 10^{4}\text{ J/m}^3\right) \times \left(600 \times 10^{-6}\text{ m}^3\right)$ $U = 3300 \times 10^{-2}\text{ J} = 33\text{ J}$

Thus, the elastic potential energy stored in the wire is 33 $\text{J}$.