An aluminium wire (Y = 7.0 × 10¹⁰ Pa) and a rubber cord (Y = 4.0 × 10⁶ Pa) are each under a stress…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

An aluminium wire ( $Y = 7.0 \times 10^{10}\ \text{Pa}$ ) and a rubber cord ( $Y = 4.0 \times 10^{6}\ \text{Pa}$ ) are each under a stress of $2.0 \times 10^{6}\ \text{Pa}$ . Find the elastic energy in each cubic metre of each, and the ratio. Which stores more?

Need a reminder? The key ideaShow

Elastic energy: $U = \tfrac12 F\,\Delta L = \tfrac12 k(\Delta L)^2$ . Per unit volume, $u = \tfrac12\,\text{stress} \times \text{strain} = \dfrac{\text{stress}^2}{2Y}$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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