Elastic energy from force–extension graphs

A stretched bow, a wound-up spring and a squashed ball all store energy that can be released to make something move. This elastic potential energy comes from deforming a material, rather than from the position of a mass or a charge in a field. The work needed to stretch a material is the area under its force–extension graph, and for a material that springs back fully, that area is also the energy it stores.

Three kinds of potential energy

Potential energy is energy stored because of the position or arrangement of things that interact.

Gravitational and electric potential energy change when an object moves through a field. Elastic potential energy changes when a material changes shape, even if nothing moves up, down or through a field.

Work as the area under a force–extension graph

When a material is stretched, the force needed increases as the extension increases, so you cannot simply multiply one force by the extension. Instead, split the extension into many small steps Δ x. Over each step the force is almost constant, so the work done is about FΔ x: the area of a thin strip under the graph. Adding all the strips gives the total: the work done in stretching is the area under the force–extension graph.

If the material returns to its original shape when released, and no energy is lost to internal energy while it is stretched, all this work is stored as elastic potential energy.

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

where F is the final force and x the final extension. The average force during stretching is half the final force, which is where the 1/2 comes from. For any other shape of graph, find the actual area.

Worked example 1

Energy from a graph with two straight parts

Problem

A material is stretched slowly from its natural length and stores all the loading work as elastic potential energy, with no heating. Its force–extension graph is made of the straight lines shown. Find the 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 part: a triangle

   Method

   Find the area from 0 to 0.10 m.

   Reason

   The force rises in a straight line from zero.

   Working

   Answer

   1/2 × 0.10 × 20 = 1.0 J

2. Second part: a trapezium

   Method

   Find the area from 0.10 to 0.20 m.

   Reason

   Here the force starts at 20 N, not zero, and rises to 50 N.

   Working

   Answer

   1/2 × (20 + 50) × 0.10 = 3.5 J

3. Total

   Working

   Answer

   1.0 + 3.5 = 4.5 J

Quick check

Using (1/2)Fx = 1/2 × 50 × 0.20 = 5.0 J would be wrong: the graph is not one straight line through the origin, so the material does not obey Hooke’s law all the way.

Guided practice 2

Energy stored at 0.16 m

About 4 min

Problem

Another material stores all the loading work as elastic potential energy, with no heating, and has the force–extension graph shown. Find the 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

Split the area into two shapes

First area

JUnit: J

Show the steps Optional working

Step 1: Mean force on the second segment

NUnit: N

Step 2: Extension change on the second segment

mUnit: m

Second area

JUnit: J

Show the steps Optional working

Step 1: First area

JUnit: J

Step 2: Second area

JUnit: J

Energy stored

JUnit: J

Check answer

Hints

Hint 1

shapes

Split the area into a triangle up to 0.10 m and a trapezium from 0.10 to 0.16 m.

Hint 2

trapezium

The trapezium’s parallel sides are 30 N and 42 N, and its width is 0.06 m.

Show solution step by step

1. The two areas

   Method

   Find each area.

   Reason

   The total work is the total area.

   Working

   Answer

   Triangle: 1/2 × 0.10 × 30 = 1.50 J. Trapezium: 1/2 × (30 + 42) × 0.06 = 2.16 J.

2. Total

   Working

   Answer

   1.50 + 2.16 = 3.66 J, or 3.7 J to two significant figures.

Common mistakes

• Multiplying the final force by the extension, when the force rose from zero.
• Using (1/2)Fx for a graph that is not a straight line through the origin.
• Using the change in extension in (1/2)Fx when the spring was already stretched. Find each stored energy from the natural length, or the area between the two extensions.
• Confusing elastic potential energy with gravitational potential energy: stretching a spring sideways changes the first but not the second.