Question

The figure shows the stress-strain plot for an aluminium wire, which is stretched by a machine pulling in opposite directions at the ends of the wire. The wire has an initial length of $0.8\text{m}$ and initial cross-sectional area of $2.0\times {10}^{-6}{\text{m}}^2$. How much work is done by the machine on the wire to produce a strain of $1.0\times {10}^{-3}$ $?$

• A. $54.4\text{mJ}$
• B. $56\text{mJ}$
• C. $108.8\text{mJ}$
• D. $128\text{mJ}$

Answer 1

The correct option is B $56\text{mJ}$

Given,
Initial length, $l=0.8\text{m}$,
Initial cross-sectional area, $A=2\times {10}^{-6}{\text{m}}^2$
Strain induced in the wire is $1\times {10}^{-3}$, which is shown by point $\text{Q}$ in figure.

Now, from the figure, stress induced in the wire for given strain is $7\times {10}^7{\text{N/m}}^2$.

The work done by the force from the machine will be stored as elastic potential energy in the wire.

$\text{Elastic potential energy}=\text{Area of stress-strain curve}\times \text{Volume of the wire}$

$\Delta U=\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}$

$\Rightarrow \Delta U=\frac{1}{2}\times (7\times {10}^7)\times (1\times {10}^{-3})\times (2\times {10}^{-6}\times 0.8)$

$\Rightarrow \Delta U=0.056\text{J}$

$\therefore \Delta U=56\text{mJ}$

Hence, option $\left(b\right)$ is correct answer.