Question

Figure 12-59 shows the stress versus strain plot for an aluminum wire that is stretched by a machine pulling in opposite directions at the two ends of the wire. The scale of the stress axis is set by $\mathit{s}\mathbf{=}\mathbf{7}\mathbf{.0}$, in units of${\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathit{N}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$. The wire has an initial length of$\mathbf{0}\mathbf{.800}\mathit{m}$and an initial cross-sectional area of$\mathbf{2}\mathbf{.00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}{\mathit{m}}^{\mathbf{2}}$. How much work does the force from the machine do on the wire to produce a strain of$\mathbf{1}\mathbf{.00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$?

Figure:

Short answer

Force from the machine does work on the wire to produce a strain of$1.00\times {10}^{-3}$ .

Step-by-step solution

Determine the given quantities

A stress versus strain plot for an aluminum wire is given and the scale of the stress axis is

$s=7.0,inunitsof{10}^7\text{\hspace{0.17em}N}/m^2$

The initial length of is $0.800m$and initial cross-sectional area of$2.00\times {10}^{-6}{m}^2$

Determine the concept of force, stress and strain

The stress on the object is equal to force per unit area. The strain is equal to change in the length per original length.

Using formula for stress and strain, find how much work does the force from the machine do on the wire to produce a strain of

Consider the formula for the work done as:

$w=\int Fdx$

Determine the work on the wire by the force to produce a strain of 1.00×10-3

Consider the formula for the work done:

$w=\int Fdx$

Consider the equation for the stress as:

$F=stress\times area$

Consider the formula for the differential length as:

$dx=strain\times length$

Substitute the values in the formula for the work done and solve as:

$\begin{array}{c}w=\int stress\times A\times strain\times L\\ =AL\int stress\times strain\\ =wirearea\times lengthofwire\times areaunderthecurveofgraph\end{array}$

$\begin{array}{c}Areaunderthecurveofgraph=\frac{1}{2}base\times height\\ =\frac{1}{2}(1.0\times {10}^{-3}\text{\hspace{0.17em}N}/m^2)(7.0\times {10}^7\text{\hspace{0.17em}N}/m^2)\\ =35000\text{N}/m^2\end{array}$

Therefore, solve for the work done on the wire as:

$\begin{array}{c}K=W\\ =AL(grapharea)\\ =(2.0\times {10}^{-6}{\text{\hspace{0.17em}m}}^{\text{2}})\left(0.800\text{\hspace{0.17em}m}\right)(35000\text{\hspace{0.17em}\hspace{0.17em}N}/{\text{m}}^{\text{2}})\\ =0.0560\text{J}\end{array}$

Force from the machine does $0.0560J$work on the wire to produce a strain of$1.00\times {10}^{-3}$