Question
A copper wire of diameter $2 \mathrm{mm}$ is $10 \mathrm{m}$ long and stretched out between two posts. The normal stress (pressure) $\sigma=E\left(L-L_{0}\right) / L_{0},$ depends on the length $L$ versus the unstretched length $L_{0}$ and Young's modulus $E=1.1 \times 10^{6} \mathrm{kPa}$. The force is $F=A \sigma$ and measured to be $110 \mathrm{N}$. How much longer is the wire and how much work was put in?
Step 1
The formula for the area of a circle is $A=\pi d^{2}/4$, where $d$ is the diameter of the wire. Substituting $d=0.002m$ into the formula, we get: \[A=\pi (0.002)^{2}/4 = 3.142 \times 10^{-6} m^{2}\] Show more…
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