Question 226: Properties of Matter

Question

A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is $600 	imes 10^{-6} \mathrm{~m}^3$. The elastic potential energy stored in the wire in stretched condition would be
$\_\_\_\_$ J.
(Given Young modulus of copper $=1.1 	imes 10^{11} \mathrm{~N} / \mathrm{m}^2$ )

Accepted Answer

Elastic potential energy $(U)$ is the work done in stretching a wire. For a wire with Young's modulus $Y$, length $L$, and cross-sectional area A , the force F required for an extension x is:
$$ F=\frac{Y A x}{L} $$
Work done (dW) for a small extension dx:
$$ \mathrm{dW}=\mathrm{F} \cdot \mathrm{dx}=\left(\frac{\mathrm{YA}}{\mathrm{~L}}ight) \mathrm{xdx} $$
To find the total work done in stretching the wire to $\Delta \mathrm{L}$ we need to integrate from 0 to total extension $\Delta \mathrm{L}$ :
$$ \mathrm{U}=\int_0^{\Delta \mathrm{L}} \frac{\mathrm{YAL}}{\mathrm{x}} \mathrm{dx}=\frac{\mathrm{YA}}{2 \mathrm{~L}}(\Delta \mathrm{~L})^2 $$
Since volume $\mathrm{V}=\mathrm{A} \cdot \mathrm{L}$, we can substitute $\mathrm{A}=\frac{\mathrm{V}}{\mathrm{L}}$ :
$$ \begin{gathered} \mathrm{U}=\frac{1}{2} \mathrm{Y}\left(\frac{\Delta \mathrm{~L}}{\mathrm{~L}}ight)^2 	imes \mathrm{V} \ \mathrm{U}=\frac{1}{2} 	imes 	ext { Young's Modulus } 	imes(	ext { Strain })^2 	imes 	ext { Volume } \end{gathered} $$
The length of wire is $\mathrm{L}=3 \mathrm{~m}$
The elongation in the length of wire is $\Delta \mathrm{L}=3 \mathrm{~mm}=3 	imes 10^{-3} \mathrm{~m}$
The volume of wire is $\mathrm{V}=600 	imes 10^{-6} \mathrm{~m}^3$
The Young's modulus of copper is $\mathrm{Y}=1.1 	imes 10^{11} \mathrm{~N} / \mathrm{m}^2$
The strain in wire due to elongation is
$$ \epsilon=\frac{\Delta \mathrm{L}}{\mathrm{~L}}=\frac{3 	imes 10^{-3}}{3}=10^{-3} $$
Now, substituting into the energy formula:
$$ U=\frac{1}{2} 	imes\left(1.1 	imes 10^{11}ight) 	imes\left(10^{-3}ight)^2 	imes\left(600 	imes 10^{-6}ight) $$
$\mathrm{U}=\frac{1}{2} 	imes 1.1 	imes 10^{11} 	imes 10^{-6} 	imes 600 	imes 10^{-6}$
$$ \begin{gathered} U=0.5 	imes 1.1 	imes 600 	imes 10^{11-6-6} \ U=0.5 	imes 1.1 	imes 600 	imes 10^{-1} \ U=33 \mathrm{~J} \end{gathered} $$
Therefore, the elastic potential energy stored is 33 J .

Properties of Matter

Integer / Numeric Answer Type