A wire of length 0.89 m and cross-sectional area 1.7 cm2 is stretched elastically by an amount 1.2 cm. By Hooke’s law, the restoring force is $−k\Delta L$.

Calculate the work done in stretching this wire by an amount $\Delta L$. Young's modulus is 1.3 $\times$ 1010 N/m2

I don't really know where to start.


closed as off-topic by ja72, Danu, Kyle Kanos, ACuriousMind, Qmechanic Apr 12 '15 at 9:40

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In terms of the Young's modulus the Hook's law (up to the overall sign) is written as $$F=ES \frac{\Delta L}{L_0}=kx, $$ the corresponding elastic energy (or the work that has be done to stretch a wire) is $$W=ES \frac{\Delta L^2}{2L_0}, $$ here you know the Young's modulus $E=1.3 \times 10^{10}$, the cross area $S=1.7cm^2$ and the initial length $L_0=0.89m$ and the amount of stretching $\Delta L=0.012 m$. Do not forget to transform the cross area into square meters!

  • 1
    $\begingroup$ Check your units. Force and Energy shouldn't have the same units. If $k=\frac{E S}{L_0}$ then $W=\frac{1}{2}k x^2$ and not $W=k x$ $\endgroup$ – ja72 Apr 12 '15 at 7:01
  • $\begingroup$ thank you for giving me minuses because of a typo, great job! $\endgroup$ – Yuri Apr 13 '15 at 1:53
  • $\begingroup$ It ain't just a typo when the answer was totally wrong. Thanks for correcting. $\endgroup$ – ja72 Apr 13 '15 at 2:57

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