Elastic potential energy: the ½ in ½Fx and ½kx²

8 minAS Level6.2.36.2.4

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By the end of this lesson

By the end you can derive $E_P = \dfrac{1}{2}Fx = \dfrac{1}{2}kx^2$ from the area under a straight-line force–extension graph, and use it to find an energy, a force, an extension or a spring constant.

Where the stored energy comes from

You have just seen that the work done stretching an object is the area under its force–extension graph.

If the deformation is elastic, the object gives that work back when it returns to its original shape. Until then, the energy is stored in the stretched bonds between its atoms.

That stored energy is called elastic potential energy, $E_P$ . It is also called strain energy.

So the rule is: area under the graph = work done = elastic potential energy stored (provided the deformation is elastic).

Definition

Elastic potential energy

Elastic potential energy is the energy stored in an object when it is deformed elastically, which is released when the object returns to its original shape.

For deformation within the limit of proportionality, it is equal to the area under the force–extension graph.

Deriving ½Fx — no memorising needed

Within the limit of proportionality, the graph is a straight line through the origin. The area underneath is a triangle.

The area of any triangle is

$\text{area} = \dfrac{1}{2} \times \text{base} \times \text{height}$

On this graph the base is the extension $x$ and the height is the final force $F$ . So

$E_P = \dfrac{1}{2}Fx$

Now use Hooke's law, $F = kx$ , to replace $F$ :

$E_P = \dfrac{1}{2}(kx)x = \dfrac{1}{2}kx^2$

Both forms are on the syllabus. The first is easiest when you know the final force; the second when you know the spring constant.

Why it is not just Fx

The force is not $F$ for the whole stretch. It starts at zero and rises steadily to $F$ .

The average force is therefore $\dfrac{0 + F}{2} = \dfrac{F}{2}$ , and the work done is average force × extension:

$W = \dfrac{F}{2} \times x = \dfrac{1}{2}Fx$

Multiplying the final force by the extension would pretend you pulled with the full force from the very first millimetre. That is why the ½ appears.

Worked example: Worked example: energy from a spring constant

A spring has a spring constant of $250\ \text{N m}^{-1}$ . It is stretched by $0.12\ \text{m}$ within its limit of proportionality.

Calculate the elastic potential energy stored.

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Quick check

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