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Energy and Fields overview

Topic 5 of 6

Elastic energy from a force-extension graph

The stretching force usually changes as a material extends. Find the work from the area under its force-extension graph, then state when that work is stored as recoverable elastic energy.

Extension x is the increase from the unloaded reference length. For a small extra extension, work is approximately stretching force × extra extension. Adding these small areas gives the loading work.

Use a settled, slow-loading model with negligible kinetic-energy change, reversible deformation and negligible dissipative heating. Under those conditions:

Increase in elastic potential energy = area under the stretching-force/extension graph

The graph shows the applied stretching-force magnitude. The material's restoring force on the load points the opposite way. Do not change the stored energy's sign because these forces have different recipients and directions.

A linear spring

For a spring obeying Hooke's law, F = kx. Take k = 75 N/m and x = 0.040 m. The final stretching force is 75 × 0.040 = 3.0 N. The line begins at zero extension and zero applied load.

Eelastic = ½ × base × height
= ½(0.040)(3.0) = 0.060 J

Substituting F = kx into the triangular area gives Eelastic = ½kx2. This expression follows from the linear relation through the unloaded origin. It is not a rule that every force-extension curve encloses a triangle.

A force-extension model with changing gradient

The following supplied model joins each pair of neighbouring points with a straight segment. Assume the same slow, reversible, negligible-loss conditions. These are model values rather than recorded experimental observations.

Supplied extension and stretching force
Extension x / mForce F / N
0.0000.0
0.0201.0
0.0403.0
0.0606.0

Elastic energy is an area, not final force times extension

Loading is slow, so kinetic-energy changes are negligible. Deformation is reversible and dissipative heating is negligible, so the stretching work becomes recoverable elastic potential energy. Both graphs use the same axis scales.

Linear spring: k = 75 N/m

A triangular force-extension area stores 0.060 joulesThe horizontal axis is extension in metres, zero to 0.060. The vertical axis is positive stretching-force magnitude in newtons, zero to six. Loading is slow with negligible kinetic-energy change; deformation is reversible with negligible dissipative heating. The line joins the unloaded origin to extension 0.040 metres and force three newtons. The triangular shaded area is one half times 0.040 times three, or 0.060 joules.0.0000.0200.0400.0600123456Stretching force F / NExtension x / m0.060 J

Area = ½ × 0.040 m × 3.0 N = 0.060 J. This triangular formula follows from the linear relation through the unloaded origin.

A different model with three straight segments

Three exact trapezium areas store 0.140 joulesThe horizontal axis is extension in metres, zero to 0.060. The vertical axis is positive stretching-force magnitude in newtons, zero to six. Loading is slow with negligible kinetic-energy change; deformation is reversible with negligible dissipative heating. The specified graph joins extension-force points zero and zero, 0.020 and one, 0.040 and three, and 0.060 and six with straight segments. Areas A1, A2 and A3 are 0.010, 0.040 and 0.090 joules. The last two, shaded green, total 0.130 joules between extensions 0.020 and 0.060 metres. All three total 0.140 joules. Because these segments are the supplied model itself, the trapezium sum is exact, unlike an estimate from samples of an unknown smooth curve.0.0000.0200.0400.0600123456Stretching force F / NExtension x / mA1A2A3

A1 = 0.010 J, A2 = 0.040 J and A3 = 0.090 J. Total = 0.140 J; the increase from 0.020 to 0.060 m is A2 + A3 = 0.130 J. Do not replace this specified piecewise graph with one endpoint triangle.

Each shaded force-extension area is an energy in N m = J. The second model uses its actual straight segments, so the endpoint triangle is not its area.

For each segment, the trapezium area is its mean endpoint force multiplied by its extension interval:

  1. 0 to 0.020 m: (0 + 1.0)/2 × 0.020 = 0.010 J.
  2. 0.020 to 0.040 m: (1.0 + 3.0)/2 × 0.020 = 0.040 J.
  3. 0.040 to 0.060 m: (3.0 + 6.0)/2 × 0.020 = 0.090 J.

Total energy stored from zero extension is 0.140 J. The additional energy from 0.020 m to 0.060 m is 0.040 + 0.090 = 0.130 J. Equivalently, subtract the energy already stored at 0.020 m from the final total.

Using the endpoint triangle would give ½(0.060)(6.0) = 0.180 J; using the endpoint rectangle would give 0.060 × 6.0 = 0.360 J. Neither follows the actual graph. The trapezium sum is exact for its stated straight segments. Joining sparse points sampled from a smooth curve would instead give an estimate.

Check the area units. With force in N and extension in m, the numerical area is in J. If the horizontal axis is in cm, convert the length scale before interpreting the area as joules.

Optional check A reversible force-extension model joins (0.020 m, 1.0 N), (0.040 m, 3.0 N) and (0.060 m, 6.0 N) with straight segments. Loading is slow and dissipation is negligible. What additional elastic energy is stored from 0.020 m to 0.060 m?
A reversible force-extension model joins (0.020 m, 1.0 N), (0.040 m, 3.0 N) and (0.060 m, 6.0 N) with straight segments. Loading is slow and dissipation is negligible. What additional elastic energy is stored from 0.020 m to 0.060 m?

What would the measurements establish?

Measure the unloaded reference, then use settled length readings to calculate extension as load increases. Keep the force and extension axes and their units explicit. The spring investigation's force readings and extension graph can then determine loading work from area.

Compare loading and unloading behaviour before claiming that all that work is recoverable. In a material that dissipates energy or remains permanently deformed, not all loading work is recovered as useful mechanical output. Reaching the same final extension is not enough evidence of an ideal spring model.

If the elastic energy is later transferred to a body's kinetic energy, an equality between the two requires negligible losses and no other changing stores. The presence of a spring alone does not establish those conditions.