Elastic energy from force–extension graphs

A stretched bow, a drawn catapult and a compressed car spring all store energy that they release when they spring back. This lesson compares elastic potential energy with the two kinds of field potential energy, and shows how to find it from the area under a force–extension graph.

Three kinds of potential energy

Potential energy is energy stored because of the position or shape of things, not their motion.

Gravitational and electric potential energy change when a field force does work. Elastic potential energy is different: it is stored in a body that has been stretched, squashed or bent, by the forces between its particles.

Elastic potential energy from a force–extension graph

When a spring or wire is stretched slowly by a force F, the work done over a small extra extension Δ x is FΔ x, the area of a thin strip under the force–extension graph. Adding all the strips gives the work done in stretching it: the area under the force–extension graph. For an ideal elastic material with negligible dissipation, this work is stored as elastic potential energy and recovered during unloading. A material can regain its original shape while dissipating energy; compare its loading and unloading curves.

For a spring that obeys Hooke’s law, F = kx, the graph is a straight line through the origin and the area is a triangle:

For a spring stretched further, from extension x₁ to x₂, subtract the energies stored at each extension:

This is not (1/2)k(x₂-x₁)², because the spring was already exerting a force at the start. When the graph is not a straight line, split the area into triangles, rectangles and trapezia.

Worked example 1

Energy from a two-part graph

Problem

A material is stretched slowly from its natural length and follows the force–extension graph below. Assume elastic deformation with negligible dissipation over this range. The graph joins (0,0), (0.10 m,20 N) and (0.20 m,50 N) with straight lines. Find the elastic potential energy stored at an extension of 0.20 m.

Piecewise elastic loading

Applied force versus extension, with sample points [(0, 0), (0.1, 20), (0.2, 50)]. The stated straight segments define the loading path.

Scroll across the figure to read all labels.

Open full-size graph

What to noticeIdeal reversible material, loaded slowly from natural length with negligible dissipation. The constructed graph is exactly piecewise linear.

View figure data

Figure data: Piecewise elastic loading

Extension, x (m)	Loading force
0.0	0
0.1	20
0.2	50

Figure data: Piecewise elastic loading

Extension, x (m)	Loading force
0.0	0
0.1	20
0.2	50

Show full solution

1. First region: a triangle

   Method

   Find the area from 0 to 0.10 m, where the force rises steadily to 20 N.

   Reason

   The energy stored is the area under the graph, and this part of it is a triangle.

   Working

   Answer

   (1/2)(0.10)(20) = 1.0 J

2. Second region: a trapezium

   Reason

   The force rises from 20 N to 50 N over a width of 0.10 m.

   Working

   Answer

   (1/2)(20 + 50)(0.10) = 3.5 J

3. Total

   Working

   Answer

   Eₚ = 1.0 + 3.5 = 4.5 J

Quick check

Using (1/2)Fx with the final point, (1/2)(50)(0.20) = 5.0 J, would be wrong: the graph is not one straight line.

Guided practice 2

Energy from a second graph

About 4 min

Problem

Another material is stretched with negligible dissipation along this force–extension graph. The graph joins (0,0), (0.10 m,30 N) and (0.16 m,42 N) with straight lines. Find the elastic potential energy stored at an extension of 0.16 m.

Read the two graph regions

Applied force versus extension, with sample points [(0, 0), (0.1, 30), (0.16, 42)]. The stated straight segments define the loading path.

Scroll across the figure to read all labels.

Open full-size graph

What to noticeIdeal reversible material with negligible dissipation; straight segments define the complete loading path for this constructed exercise.

View figure data

Figure data: Read the two graph regions

Extension, x (m)	Loading force
0.00	0
0.10	30
0.16	42

Figure data: Read the two graph regions

Extension, x (m)	Loading force
0.00	0
0.10	30
0.16	42

Find the energy stored

Show the steps Optional working

Step 1: Area from 0 to 0.10 m

JUnit: J

Step 2: Area from 0.10 to 0.16 m

JUnit: J

Elastic potential energy

JUnit: J

Check answer

Hints

Hint 1

two regions

Where does the gradient change, and what shapes lie under each section?

Show full solution

1. Add the two areas

   Working

   Answer

   (1/2)(0.10)(30) + (1/2)(30 + 42)(0.06) = 1.50 + 2.16 = 3.66 J, or 3.7 J to two significant figures

Only elastic deformation stores recoverable energy

The area under the loading graph is always the work done in stretching the material. It is all recoverable as elastic potential energy when deformation is elastic and dissipation is negligible. Beyond the elastic limit, part of the work goes into permanently deforming the material and ends up as internal energy.

Mind stretcher 1 Different loading and unloading pathsExtension

Some materials, such as rubber, return to their original length but follow a lower curve as they are unloaded. They give back less work than was put in.

Different loading and unloading paths

Loading follows force values 0, 8 and 16 N at extensions 0, 0.05 and 0.10 m. Unloading follows 16, 5 and 0 N as extension decreases through the same positions. The unloading path lies below the loading path between their endpoints.

Scroll across the figure to read all labels.

Open full-size graph

What to noticeOptional constructed cycle. The material returns to the same extension but dissipates energy; this is not an ideal Hooke’s-law spring. Paths are shown against increasing extension, even though unloading proceeds from right to left.

View figure data

Figure data: Different loading and unloading paths

Extension, x (m)	Loading (left to right)	Unloading (right to left)
0.00	0	0
0.05	8	5
0.10	16	16

Figure data: Different loading and unloading paths

Extension, x (m)	Loading (left to right)	Unloading (right to left)
0.00	0	0
0.05	8	5
0.10	16	16

The area under the loading curve is 0.80 J and the area under the unloading curve is 0.65 J. Find the energy dissipated in one complete cycle.

Energy dissipated per cycle

JUnit: J

Check answer

Show answer

The difference, 0.80-0.65 = 0.15 J, is the area between the curves. It becomes internal energy of the material, which is why a rubber band that is stretched and released many times gets warm. Original shape is restored, but not all the input work is recovered.

Common mistakes

• Using the length of the spring instead of its extension.
• Using (1/2)Fx for a graph that is not a straight line through the origin.
• Using (1/2)k(x₂-x₁)² for a spring that is already stretched.
• Assuming all the work done is stored when the material has passed its elastic limit.

Before you move on