Properties of Matter Part 3

Concept: The elastic potential energy stored in a material when it is stretched is proportional to the Young's modulus, the square of the strain, and the volume of the material.Formula/Principle: The elastic potential energy ($U$) is given by $U = \frac{1}{2} Y E^2 V$, where $Y$ is Young's modulus, $E$ is the strain, and $V$ is the volume.Solution/Analysis:1. Identify the given parameters:

• Original Length ($L$) = $3$ m
• Extension ($\Delta L$) = $3$ mm = $3 \times 10^{-3}$ m
• Volume ($V$) = $600 \times 10^{-6} \mathrm{m}^3$
• Young's Modulus ($Y$) = $1.1 \times 10^{11} \mathrm{N/m}^2$

2. Calculate the strain ($E$): Strain is defined as the ratio of the change in length to the original length.$E = \frac{\Delta L}{L} = \frac{3 \times 10^{-3} \mathrm{m}}{3 \mathrm{m}} = 1 \times 10^{-3}$3. Use the formula for elastic potential energy ($U$):$U = \frac{1}{2} Y E^2 V$4. Substitute the calculated values into the formula:$U = \frac{1}{2} \times (1.1 \times 10^{11}) \times (1 \times 10^{-3})^2 \times (600 \times 10^{-6})$5. Simplify the expression:$U = \frac{1}{2} \times (1.1 \times 10^{11}) \times (1 \times 10^{-6}) \times (600 \times 10^{-6})$$U = \frac{1}{2} \times (1.1 \times 600) \times (10^{11} \times 10^{-6} \times 10^{-6})$$U = \frac{1}{2} \times 660 \times 10^{-1}$$U = 330 \times 10^{-1} = 33 \mathrm{J}$The elastic potential energy stored in the wire is $33 \mathrm{J}$.