The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length $100 ext{ cm}$

Question

The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length $100	ext{ cm}$ to stretch it by $1	ext{ mm}$ is: (given: Young's modulus of the wire $Y = 2.0 	imes 10^{11}	ext{ N/m}^2$)

Accepted Answer

1.  **Identify Given Values:**
    *   Length of wire ($L$) = $100	ext{ cm} = 1	ext{ m}$
    *   Extension ($\Delta L$) = $1	ext{ mm} = 10^{-3}	ext{ m}$
    *   Young's Modulus ($Y$) = $2.0 	imes 10^{11}	ext{ N/m}^2$
2.  **Calculate Strain:**
    $$	ext{Strain} = \frac{\Delta L}{L} = \frac{10^{-3}}{1} = 10^{-3}$$
3.  **Formula for Energy Density:** The elastic potential energy per unit volume ($u$) is given by:
    $$u = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain}$$
    Using Young's Modulus ($Y = 	ext{Stress} / 	ext{Strain}$), we can substitute Stress = $Y 	imes 	ext{Strain}$:
    $$u = \frac{1}{2} 	imes Y 	imes (	ext{Strain})^2$$
4.  **Calculation:**
    $$u = \frac{1}{2} 	imes (2.0 	imes 10^{11}) 	imes (10^{-3})^2$$
    $$u = 1.0 	imes 10^{11} 	imes 10^{-6}$$
    $$u = 10^5	ext{ J/m}^3$$

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