91Ó°ÊÓ

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

$s=7.0,inunitsof{10}^7\text{ N}/m^2$

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

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:

$w=∫Fdx$

Consider the formula for the work done:

$w=∫Fdx$

Consider the equation for the stress as:

$F=stress×area$

Consider the formula for the differential length as:

$dx=strain×length$

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

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

$\begin{array}{c}Area\text{ }under\text{ }the\text{ }curve\text{ }of\text{ }graph=\frac{1}{2}base×height\\ =\frac{1}{2}(1.0×{10}^{−3}\text{ N}/m^2)(7.0×{10}^7\text{ N}/m^2)\\ =35000\text{N}/m^2\end{array}$

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

$\begin{array}{c}K=W\\ =AL(grapharea)\\ =(2.0×{10}^{−6}{\text{″¾}}^{\text{2}})\left(0.800\text{″¾}\right)(35000\text{ â¶Ä‰N}/{\text{m}}^{\text{2}})\\ =0.0560\text{J}\end{array}$

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