Force-Extension Graphs for Different Materials

Force-extension graphs are among the most important diagrams in the materials topic. By examining the shape of the graph, you can identify the material, determine its stiffness, find the energy stored, and understand how the material will behave under different loading conditions. In the Edexcel A-level exam, you must be able to draw, interpret, and compare these graphs for metals, rubber, and polythene.

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Metal Wire (e.g., Copper or Steel)

Loading

The force-extension graph for a metal wire shows:

1. A straight-line region through the origin — the wire obeys Hooke’s law. The gradient equals the spring constant k = F/x.
2. At the limit of proportionality, the line begins to curve upward (the wire still extends, but no longer proportionally to the force).
3. At the elastic limit, further loading causes permanent deformation.
4. Beyond the elastic limit, the wire extends significantly with smaller increases in force (plastic region).
5. The wire eventually fractures at the breaking point.

Unloading (within elastic limit)

The unloading path is the same straight line as the loading path. All elastic potential energy is recovered. No energy is dissipated.

Unloading (beyond elastic limit)

The unloading path is a straight line parallel to the original loading line, but shifted to the right. The gradient is the same (same spring constant / Young modulus) because the material itself has not changed — only its permanent length has. The x-intercept of the unloading line gives the permanent extension.

Energy

• Within the elastic limit: energy stored = area under the loading curve = ½Fx = ½kx².
• Beyond the elastic limit: energy stored elastically = area under the unloading line. Energy dissipated plastically = area between the loading curve and the unloading line.

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Rubber Band

Loading

The force-extension graph for a rubber band is distinctly non-linear and has an S-shape (sigmoidal curve):

1. Initially, the rubber is relatively stiff — the curve rises steeply.
2. At intermediate extensions, the rubber becomes much more compliant — the curve flattens and extends easily with little additional force. This is where the tangled polymer chains are uncoiling and aligning.
3. At very large extensions (approaching the limit of chain extension), the rubber stiffens again dramatically — the curve rises steeply once more.

Rubber can sustain enormous strains — extensions of 300–500% are common.

Unloading

The unloading curve follows a different path from the loading curve, lying below it. However, the rubber returns to its original length — the extension returns to zero when the force is removed.

This means:

• The deformation is elastic (the rubber recovers).
• But energy is dissipated during the cycle — the rubber warms up.

Hysteresis Loop

The area enclosed between the loading and unloading curves is the hysteresis loop. The area of this loop equals the energy dissipated as internal energy (heat) per loading-unloading cycle.

The greater the area of the hysteresis loop, the more energy is dissipated per cycle. Natural rubber has a relatively large hysteresis loop; silicone rubber has a smaller one.

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Polythene Strip

Loading

A polythene strip has a very distinctive force-extension graph:

1. An initial steep region where the polythene resists extension (relatively stiff).
2. A yield point where the polythene suddenly “gives” — the graph flattens dramatically.
3. An extended region where the polythene stretches enormously with very little additional force. In this region, the molecular chains are being pulled apart and sliding past each other. The polythene may visibly “neck” — a thin section forms that gradually extends along the strip.
4. At very large extensions, the graph may steepen slightly as the aligned chains resist further extension, until the strip eventually breaks.

Unloading

The polythene does not return to its original length when unloaded. The extension is almost entirely plastic (permanent). The unloading curve drops steeply to zero force, but the polythene remains at roughly its extended length.

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Comparing the Three Materials

Feature	Metal Wire	Rubber Band	Polythene Strip
Hooke’s law region	Yes (linear)	No (non-linear)	Brief initial region only
Maximum strain	Small (< 1%)	Very large (300–500%)	Very large (100–300%)
Returns to original length?	Yes (if within elastic limit)	Yes (with hysteresis)	No (plastic deformation)
Energy dissipated on unloading	None (if elastic)	Yes (hysteresis loop area)	Most energy dissipated (permanently deformed)
Stiffness (initial)	High	Moderate (variable)	Moderate then very low

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Identifying Materials from Graph Shape

graph TD
    A["Examine the force-extension\ngraph shape"] --> B{"Straight line\nthrough origin?"}
    B -- Yes --> C["METAL WIRE\n(within elastic limit)\nGradient = spring constant k"]
    B -- No --> D{"S-shaped curve?"}
    D -- Yes --> E{"Returns to origin\non unloading?"}
    E -- Yes --> F["RUBBER BAND\n(hysteresis loop present)"]
    E -- No --> G["Rubber stretched\nbeyond elastic limit"]
    D -- No --> H{"Steep start then\nsudden yield?"}
    H -- Yes --> I["POLYTHENE\n(large plastic extension)"]
    H -- No --> J["Metal wire loaded\nbeyond elastic limit\n(curved after P)"]

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Area Under Force–Extension Graphs

The area under a force-extension graph has a specific physical meaning:

$\text{Area under graph} = \text{Work done} = \text{Energy transferred}$Area under graph=Work done=Energy transferred

• For loading: area under the loading curve = total work done on the material.
• For unloading: area under the unloading curve = energy recovered (returned as kinetic energy, stored energy, etc.).
• For a complete loading-unloading cycle: area between the curves = energy dissipated as internal energy.

For a metal wire within its elastic limit: loading area = unloading area → no energy dissipated. For rubber: unloading area < loading area → the difference (hysteresis loop area) = energy dissipated as heat. For polythene: almost all the loading energy is dissipated → the material is permanently deformed and very little energy is recovered.

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Worked Calculations Using Graph Areas

Worked example 1: A metal wire is loaded to 80 N with an extension of 1.6 mm (within elastic limit). Calculate the energy stored.

E = ½Fx = ½ × 80 × 1.6 × 10⁻³ = 0.064 J

This is the area of the triangle under the straight-line loading graph.

Worked example 2: A rubber band is stretched with loading area = 2.4 J and unloading area = 1.6 J. If this cycle is repeated 200 times per minute, what is the power dissipated as heat?

Energy per cycle = 2.4 − 1.6 = 0.8 J Power = energy/time = 0.8 × 200 / 60 = 2.67 W

This is why rubber in high-frequency applications (car tyres, engine mounts) must be designed for low hysteresis to avoid overheating.

Worked example 3: A metal wire is loaded to 150 N at an extension of 3.0 mm (beyond elastic limit). On unloading, the permanent extension is 1.2 mm. Estimate the energy dissipated plastically.

Elastic extension recovered = 3.0 − 1.2 = 1.8 mm Elastic energy recovered = ½ × 150 × 1.8 × 10⁻³ = 0.135 J Total work done during loading (approximate, assuming roughly linear up to plastic region): ≈ ½ × 150 × 3.0 × 10⁻³ = 0.225 J Energy dissipated ≈ 0.225 − 0.135 = 0.090 J

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Practical Work: Comparing Materials

A common A-level practical involves stretching a rubber band and a polythene strip with the same masses and plotting their force-extension graphs.

Method:

1. Measure the original length of each sample.
2. Add masses in equal increments, recording extension after each addition.
3. Remove masses in the same increments, recording extension during unloading.
4. Plot both loading and unloading curves for each material on separate axes.

Key observations:

• The rubber band returns to approximately its original length; the polythene does not.
• The rubber band shows a clear hysteresis loop; the polythene shows a huge gap between loading and unloading.
• The rubber band’s loading curve is S-shaped; the polythene’s loading curve shows a sudden yield point.

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Molecular Explanations Summary

Material	Loading behaviour	Molecular explanation
Metal wire	Linear then plastic	Bonds stretch elastically; atoms slip along planes in plastic region
Rubber band	S-shaped elastic	Coiled polymer chains uncoil at moderate extensions; covalent bonds resist at high extensions
Polythene	Steep then sudden yield	Chains initially resist through entanglement; chains slide past each other after yield point

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Exam Tips

When comparing force-extension graphs:

1. Identify the material from the shape of the curve (linear = metal, S-shaped = rubber, yield then plateau = polythene).
2. State the stiffness by comparing gradients (steeper = stiffer).
3. Calculate energy from the area under the curve (count squares if on graph paper).
4. Describe recovery — does the material return to its original length?
5. Explain hysteresis — is there an enclosed area between loading and unloading curves?

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Common Exam Mistakes

1. Drawing rubber as a straight line: Rubber is always non-linear. Never draw it with a constant gradient.
2. Forgetting the unloading curve: Many exam questions specifically ask about loading AND unloading. Always consider both.
3. Claiming polythene is elastic: Polythene undergoes permanent plastic deformation — it does not return to its original length.
4. Confusing the hysteresis loop with the plastic deformation gap: For rubber, the loop closes (returns to origin). For a plastically deformed metal, the unloading line hits the x-axis at a positive extension (permanent deformation).

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A-Level Deep Dive: Force-Extension for Different Materials

Spec mapping

Edexcel 9PH0 Topic 4 — Materials, sub-topic on force-extension behaviour for different materials requires students to interpret and sketch force-extension graphs for metal wires, rubber, and polymers; identify the limit of proportionality, elastic limit and yield point on a metal graph; recognise loading and unloading curves; and use the area under the graph to determine work done or elastic strain energy stored (refer to the official Pearson Edexcel specification document for exact wording). It is the synoptic capstone of the deformation strand: pulling together Hooke's law (linearity), elastic vs plastic deformation (recovery), and the energy-area relation, then extending them to non-Hookean materials. Examined in 9PH0 Paper 1 with 6-mark "compare and explain" questions. The formula booklet provides $E_{\text{el}} = \tfrac{1}{2} F \Delta x$Eel​=21​FΔx for the linear region only — for non-linear materials, the area must be computed by counting squares.

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Worked example with full mark scheme

Question (8 marks):

The force-extension behaviour of three materials is investigated using identical loading apparatus: a steel wire, a copper wire, and a rubber band. Loading proceeds in equal force increments until the sample reaches its limit, then masses are removed in the same increments to give an unloading curve.

(a) Sketch on a single set of axes the qualitative force-extension graphs for the three materials, labelling the limit of proportionality and yield point on the metal curves and the loading/unloading branches on the rubber. (4)

(b) The rubber band has a loading-curve area of 0.85 J and an unloading-curve area of 0.52 J between extensions of 0 and 12 cm. Calculate the energy dissipated per loading-unloading cycle and the average power dissipated if the cycle repeats at 4.0 Hz. (4)

Solution with mark scheme:

(a) Step 1 — sketch the steel wire. Steep straight line from the origin (high stiffness, obeys Hooke's law) up to a clearly marked limit of proportionality; a short curved continuation up to the yield point where the gradient drops sharply; further extension beyond the yield with low gradient (plastic flow); fracture marked.

M1 — steel drawn as a straight line through the origin with limit of proportionality and yield point both labelled.

Step 2 — sketch the copper wire. Less steep straight line than steel (lower Young modulus) up to a lower limit of proportionality; substantially longer plastic region than steel (copper is more ductile); fracture at a larger extension than steel.

M1 — copper drawn with smaller initial gradient than steel and a longer plastic region.

Step 3 — sketch the rubber band. S-shaped loading curve from the origin: initially shallow (uncoiling of polymer chains is easy), then steepening as chains align and bonds resist further extension. Unloading curve sits below the loading curve, returning to the origin; the enclosed region is the hysteresis loop.

M1 — rubber drawn as an S-shape with loading curve above unloading curve, both starting and ending at the origin.

A1 — all three curves on a single axis with consistent labelling of branches; limit of proportionality and yield point only on the metals (not on rubber); hysteresis loop visible on rubber.

(b) Step 1 — energy dissipated per cycle.

The area enclosed between the loading and unloading curves represents the energy dissipated as internal energy (heating) per cycle:

$\Delta E = E_{\text{loading}} - E_{\text{unloading}} = 0.85 - 0.52 = 0.33\ \text{J}$ΔE=Eloading​−Eunloading​=0.85−0.52=0.33 J

M1 — recognising that dissipated energy = loading area − unloading area.

A1 — value 0.33 J quoted to 2 sf, consistent with input data.

Step 2 — average power.

At 4.0 Hz, the cycle repeats 4.0 times per second, so:

$P = \Delta E \times f = 0.33 \times 4.0 = 1.32\ \text{W}$P=ΔE×f=0.33×4.0=1.32 W

M1 — applying $P = \Delta E / T = \Delta E \times f$P=ΔE/T=ΔE×f.

A1 — 1.3 W to 2 sf.

Total: 8 marks (M5 A3, split as shown).

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Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A student stretches a glass fibre, a steel wire, and a polythene strip — all of equal initial cross-sectional area and length — and plots force-extension graphs.

(a) Compare the three force-extension curves, identifying for each material whether plastic deformation occurs before fracture, and whether the material is elastic or plastic at the point of fracture. (3)

(b) The steel wire has a limit of proportionality at $F_p = 60$Fp​=60 N, $\Delta x_p = 0.40$Δxp​=0.40 mm. Calculate the elastic strain energy stored at this point and explain how the energy stored up to fracture would compare to this value (no calculation required for the second part). (3)

Mark scheme decomposition by AO:

(a)

• B1 (AO1.1a) — glass: linear up to fracture, no plastic region (brittle); fractures while still elastic.
• B1 (AO1.1a) — steel: linear up to limit of proportionality, then plastic region with substantial extension before fracture; ductile.
• B1 (AO1.1a) — polythene: short steep region followed by a sudden yield, then a long low-gradient plastic plateau; not elastic — recovers very little extension.

(b)

• M1 (AO2.1) — applying $E_{\text{el}} = \tfrac{1}{2} F \Delta x = \tfrac{1}{2}(60)(0.40 \times 10^{-3})$Eel​=21​FΔx=21​(60)(0.40×10−3).
• A1 (AO1.1b) — $E_{\text{el}} = 1.2 \times 10^{-2}$Eel​=1.2×10−2 J $= 12$=12 mJ.
• B1 (AO2.5) — energy up to fracture would be considerably greater than 12 mJ because the plastic region adds a large area beneath the curve, even at lower gradient; the comparison illustrates why ductile materials absorb more energy than brittle ones for the same fracture force.

Total: 6 marks split AO1 = 4, AO2 = 2. This is a compare-and-explain question that rewards graph reasoning over numerical fluency.

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