Elastic potential energy

Work goes in, energy is stored

To stretch a wire you must pull, and the pull grows as the wire gets longer: from zero at the start to $F$ at the end. The work done is the average force times the stretch:

That is the area of the triangle under the force-stretch line. Within the elastic limit, the energy is stored, like the energy of a stretched spring, and it comes back when the wire is let go.

Since $F = \dfrac{YA}{L}\Delta L$, the wire is a spring with $k = \dfrac{YA}{L}$, and $U = \tfrac12 k(\Delta L)^2$.

You should find that twice the stretch stores four times the energy, and that at the same stretch the stiffer steel stores more.

Energy in each cubic metre

Divide $U$ by the volume $AL$:

Watch out. For the same stretch, a stiff wire stores more energy. For the same force, a soft one does: $U = \dfrac{F^2L}{2AY}$.

Rubber and lost energy

Stretch rubber and let it go, and the graph on the way back lies below the graph on the way out. The area between them is energy turned into heat in each cycle. This is why a car tyre warms up on a long drive, and why rubber is good at soaking up shocks and vibrations.

A worked example

A steel guy-wire, $5.0\ \text{m}$ long and $1.6\ \text{mm}$ across, holds up a mast and is stretched by $3.0\ \text{mm}$. How much energy does it store? ($Y = 2.0 \times 10^{11}\ \text{Pa}$.)

1. Area. $A = \pi(0.80 \times 10^{-3}\ \text{m})^2 = 2.01 \times 10^{-6}\ \text{m}^2$.
2. Force. $F = YA \times \dfrac{\Delta L}{L}$, with $YA = 4.02 \times 10^{5}\ \text{N}$ and $\dfrac{\Delta L}{L} = 6.0 \times 10^{-4}$, so $F = 241\ \text{N}$.
3. Energy. $U = \tfrac12 \times 241\ \text{N} \times 3.0 \times 10^{-3}\ \text{m} = 0.362\ \text{J}$.

Check. Per cubic metre: stress $1.2 \times 10^{8}\ \text{Pa}$, strain $6.0 \times 10^{-4}$, so $u = 3.6 \times 10^{4}\ \text{J/m}^3$; times the volume, $1.0 \times 10^{-5}\ \text{m}^3$, gives the same $0.36\ \text{J}$.

Practise with new numbers

Try it in Python

The program works out the guy-wire both ways.

Remember these

• $U = \tfrac12 F\,\Delta L = \tfrac12 k(\Delta L)^2$, with $k = \dfrac{YA}{L}$.
• Per unit volume: $u = \tfrac12\,\text{stress} \times \text{strain} = \dfrac{\text{stress}^2}{2Y}$.
• Rubber's loading and unloading graphs make a loop; its area is heat.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A force rising to $60\ \text{N}$ stretches a wire by $2.5\ \text{mm}$. How much energy is stored?

2. Why is the energy $\tfrac12 F\,\Delta L$ and not $F\,\Delta L$?

3. Two wires of the same metal and thickness are stretched by the same amount, but one is twice as long. Which stores more energy?

4. Why does a rubber ball not bounce back to the height it was dropped from?