Question 218: Properties of Matter

Question

A steel wire of length 2 m and Young's modulus $2.0 	imes 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$ is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and $10^{-3}$ respectively, then the elastic potential energy density of the wire is __________ $	imes 10^5$ (in SI units).

Accepted Answer

To find the elastic potential energy density of the steel wire, we need to use the given information and formulae for strain and energy density.
Given:
The length of the wire, $ \ell = 2 \, 	ext{m} $
Young's modulus, $ Y = 2.0 	imes 10^{11} \, 	ext{N/m}^2 $
Poisson's ratio, $ \mu = 0.2 $
Transverse strain, $ \frac{\Delta r}{r} = 10^{-3} $
The formula for Poisson's ratio is:
$ \mu = -\frac{\left(\frac{\Delta r}{r}ight)}{\left(\frac{\Delta \ell}{\ell}ight)} $
From this, we solve for the longitudinal strain $\frac{\Delta \ell}{\ell}$:
$ \frac{\Delta \ell}{\ell} = \frac{1}{\mu} 	imes \left(\frac{\Delta r}{r}ight) $
Substitute the given values:
$ \frac{\Delta \ell}{\ell} = \frac{1}{0.2} 	imes 10^{-3} = 5 	imes 10^{-3} $
The elastic potential energy density $ u $ is given by:
$ u = \frac{1}{2} Y \varepsilon_{\ell}^2 $
where $ \varepsilon_{\ell} = \frac{\Delta \ell}{\ell} $. Plug in the values:
$ u = \frac{1}{2} 	imes 2 	imes 10^{11} 	imes \left(5 	imes 10^{-3}ight)^2 $
Simplify further:
$ u = \frac{1}{2} 	imes 2 	imes 10^{11} 	imes 25 	imes 10^{-6} $
$ u = 25 	imes 10^5 \, 	ext{(in SI units)} $
Thus, the elastic potential energy density of the wire is $ 25 	imes 10^5 $ SI units.

Properties of Matter

A steel wire of length 2 m and Young's modulus 2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2} is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10^{-3} respectively, then the elastic potential energy density of the wire is __________ \times 10^5 (in SI units).

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