Why is elastic energy ½Fx rather than Fx?

The force starts at zero and rises linearly to F, so the average force during the stretch is F/2. The work is that average force times the extension — the triangle area.

What is the difference between the limit of proportionality and the elastic limit?

The limit of proportionality is where the graph stops being straight. The elastic limit is where deformation stops being reversible. A material can be non-proportional yet still elastic between them.

Does Hooke’s law apply to compression?

Yes, for springs and many solids, within the limit of proportionality. The same stiffness usually applies.

Why do parallel springs make a stiffer combination?

The load is shared between them, so each extends less for a given total force — which is what greater stiffness means.

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53, 55, 58MechanicsCore

Hooke’s Law and Elastic Energy

Stretch a spring gently and it returns to its original length; stretch it too far and it stays deformed. $\textbf{Hooke’s law}$ describes the gentle region, where extension is proportional to force, and the work you do stretching it is stored as $\textbf{elastic strain energy}$ .

35 min Topic 4 · Materials

What you'll be able to do

State Hooke’s law and identify the limit of proportionality
Use $F = k\Delta x$ and combine springs in series and parallel
Distinguish elastic from plastic deformation
Calculate elastic strain energy as the area under a force–extension graph

Hooke’s law

For many materials, the extension is proportional to the applied force, provided the $\textbf{limit of proportionality}$ is not exceeded: $F = k\Delta x$ . The constant $k$ is the $\textbf{stiffness}$ (spring constant), in $\mathrm{N\,m^{-1}}$ .

On a force–extension graph this is a straight line through the origin with gradient $k$ . A stiffer spring has a steeper line — it takes more force for the same extension.

Beyond the limit of proportionality the graph curves. The material may still return to its original length up to the $\textbf{elastic limit}$ , but past that point it is permanently deformed.

F = k\Delta x

Hooke’s law, up to the limit of proportionality.

Example 1A spring extends by 4.0 cm under a load of 6.0 N. Find its stiffness, and its extension under 9.0 N.

k = \dfrac{6.0}{0.040} = 150\ \mathrm{N\,m^{-1}}

\Delta x = \dfrac{9.0}{150}

Answer

k = 150\ \mathrm{N\,m^{-1}}

; extension

0.060

m, i.e. 6.0 cm.

Tip — Extension is the change in length, not the total length. Subtract the original length before substituting.

Springs in series and parallel

Two identical springs side by side in $\textbf{parallel}$ share the load, so each extends half as much and the combination is twice as stiff: $k = k_{1} + k_{2}$ .

Two springs end to end in $\textbf{series}$ each carry the full load, so each extends fully and the total extension adds. The combination is less stiff: $\dfrac{1}{k} = \dfrac{1}{k_{1}} + \dfrac{1}{k_{2}}$ .

Note these rules are the reverse of those for resistors — parallel springs add directly, series springs add reciprocally.

k_{\text{parallel}} = k_{1} + k_{2}

Shared load: stiffer.

\dfrac{1}{k_{\text{series}}} = \dfrac{1}{k_{1}} + \dfrac{1}{k_{2}}

Full load on each: less stiff.

Example 1Springs of stiffness 200 and 300

\mathrm{N\,m^{-1}}

are joined in series. Find the combined stiffness.

\dfrac{1}{k} = \dfrac{1}{200} + \dfrac{1}{300} = \dfrac{5}{600}

k = 120\ \mathrm{N\,m^{-1}}

Answer

120\ \mathrm{N\,m^{-1}}

— less than either spring.

Elastic and plastic deformation

$\textbf{Elastic}$ deformation is fully reversible: remove the load and the material returns to its original shape. At the atomic scale, bonds stretch but no atoms change place.

$\textbf{Plastic}$ deformation is permanent. Beyond the elastic limit, layers of atoms slide past one another, so removing the load leaves a permanent extension.

On a loading–unloading graph, elastic behaviour retraces the same line; plastic behaviour unloads along a line parallel to the original but offset, meeting the extension axis at the permanent extension.

Elastic strain energy

Work done stretching a spring is stored as $\textbf{elastic strain energy}$ . Because the force increases from zero, the work is the $\textbf{area under the force–extension graph}$ , not simply force times extension.

Within Hooke’s law that area is a triangle of base $\Delta x$ and height $F$ , giving $E = \tfrac12 F\Delta x$ . Substituting $F = k\Delta x$ gives the alternative $E = \tfrac12 k\Delta x^{2}$ .

Beyond the limit of proportionality the graph is curved, so the formula no longer applies — the area must be estimated, for instance by counting squares.

E_{\text{el}} = \tfrac{1}{2}F\Delta x

Triangle area.

E_{\text{el}} = \tfrac{1}{2}k\Delta x^{2}

Using Hooke’s law.

Example 1A spring of stiffness

400\ \mathrm{N\,m^{-1}}

is stretched by 5.0 cm. Find the energy stored.

E = \tfrac12 \times 400 \times 0.050^{2}

= 200 \times 0.0025

Answer

0.50\ \mathrm{J}

Tip — Energy scales with the square of extension. Doubling the stretch stores four times the energy — useful for checking answers.

Equation recap

F = k\Delta x

Hooke’s law.

k_{p} = k_{1}+k_{2}

Parallel.

\tfrac1{k_{s}} = \tfrac1{k_{1}}+\tfrac1{k_{2}}

Series.

E = \tfrac12 F\Delta x = \tfrac12 k\Delta x^{2}

Elastic strain energy.

Common mistakes to avoid

Using

E = F\Delta x

The force grows from zero, so the energy is the triangle area

\tfrac12 F\Delta x

Substituting total length for extension.

Extension is new length minus original length.

Adding spring constants in series.

Series springs combine reciprocally; parallel springs add.

Treating the elastic limit and limit of proportionality as the same.

Proportionality can end before elasticity does.

Key takeaways

Hooke’s law $F = k\Delta x$ holds up to the limit of proportionality.
Parallel springs add stiffness; series springs combine reciprocally.
Elastic deformation is reversible; plastic deformation leaves a permanent extension.
Elastic strain energy is the area under a force–extension graph.
Within Hooke’s law, $E = \tfrac12 F\Delta x = \tfrac12 k\Delta x^{2}$ .

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