Calculate the work of stretching required to elongate

Calculate the work of stretching required to elongate a copper wire of length $L_0=1$ m and cross-sectional radius $r=2$ mm by $\Delta L=0.001$ m.

Step 1: Cross-sectional area

\[A=\pi r^2=\pi(0.002)^2=4\pi\times10^{-6}\ \text{m}^2\approx1.257\times10^{-5}\ \text{m}^2.\]

Step 2: Hooke's law for the force during stretching

At elongation $x\in[0,\Delta L]$, the tension is $F(x)=\dfrac{EA}{L_0}x$, where $E\approx120$ GPa is copper's Young's modulus.

Step 3: Work integral

\[W=\int_0^{\Delta L}F(x)\,dx=\frac{EA}{L_0}\int_0^{\Delta L}x\,dx=\frac{EA(\Delta L)^2}{2L_0}.\]

\[\boxed{W=\frac{EA(\Delta L)^2}{2L_0}}\]

Step 4: Numerical value

\[W=\frac{(120\times10^9)(4\pi\times10^{-6})(0.001)^2}{2(1)}=240\pi\times10^{-3}\approx0.754\ \text{J}.\]

Step 5: Cross-check via elastic energy density

The strain is $\epsilon=\Delta L/L_0=0.001$, well within copper's elastic limit. The elastic energy per unit volume is $u=\tfrac12E\epsilon^2$, so $W=u\cdot AL_0=\tfrac12E\epsilon^2AL_0=\dfrac{EA(\Delta L)^2}{2L_0}$, matching Step 3 exactly. Equivalently, $W=F_{\max}\Delta L/2\approx1508\times0.001/2\approx0.754$ J, since the average force during a linear ramp is half the maximum.