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Density and Hooke's law

Two ideas open the study of materials. Density is how much mass a material has per cubic metre, and Hooke's law says that extension is proportional to force, but only up to a limit. Where that law stops holding matters as much as where it holds.

Year 12AQA 3.4.2.1 CIE 4.3.1, 6.1.1, 6.1.2, 6.1.3, 6.1.4, 6.2.1, 6.2.2, 6.2.3, 6.2.4 OCR A 3.2.4(a), 3.4.1(a), 3.4.1(b), 3.4.1(c), 3.4.1(d)(i), 3.4.2(a), 3.4.2(b), 3.4.2(f)EDEXCEL 49, 53, 55, 58

IN THIS TOPIC

Use $\rho = m/V$ , and get the g cm⁻³ to kg m⁻³ conversion the right way round.
Apply $F = k\Delta L$ while it holds, and say what the spring constant belongs to.
Separate the limit of proportionality from the elastic limit.
Find the stored energy from the area under a force-extension graph, and account for the energy that never comes back after plastic deformation.

COMMON MISCONCEPTION

Hooke's law holds right up until the spring snaps.

Density

The density of a material is its mass per unit volume.

\rho = \frac{m}{V}

ON THE AQA DATA SHEET

Density is measured in kg m⁻³, and for a homogeneous sample at a given temperature, pressure and state it belongs to the material and not to any particular lump of it. Cut a brick in half and mass and volume both halve, leaving the ratio exactly where it was. Water sits at 1000 kg m⁻³, a benchmark worth carrying everywhere, since a fully immersed lump of anything denser than the fluid around it sinks unless something else holds it up.

Handbooks quote densities in g cm⁻³, and the conversion is a factor students misplace constantly. 1 g cm⁻³ = 1000 kg m⁻³, because a kilogram is a thousand grams while a cubic metre is a million cubic centimetres. Convert before substituting, never after.

WORKED EXAMPLE

Identifying a metal

A solid metal cube of side 2.0 cm has a mass of 63 g. Find its density and suggest what the metal might be.

Work in SI from the first line. Side 0.020 m, mass 0.063 kg, so the volume is (0.020)³ = 8.0 × 10⁻⁶ m³.

ρ = m/V = 0.063/(8.0 × 10⁻⁶) = 7.9 × 10³ kg m⁻³.

Iron or steel, on any density table. The step to watch is cubing the side; skip it and the answer lands three powers of ten adrift, which a glance at the table catches at once.

Hooke's law, and where it stops

Load a spring or a wire and it extends. For modest loads the response obeys Hooke's law, which says the force is proportional to the extension.

F = k\Delta L

ON THE AQA DATA SHEET

Here k is the spring constant, or stiffness, measured in N m⁻¹. It belongs to that particular object, not to the material it is made from. Hooke's law covers compression as well as stretching.

FIG. 1Force against extension. The straight section is Hooke's law; the graph leaves the line at the limit of proportionality and reaches the elastic limit a little beyond. The shaded area is the energy stored.

The law fails in two stages, and the exam separates them. The limit of proportionality is where the graph stops being straight. The elastic limit, usually a little beyond it, is where deformation stops being reversible. Load past that second point and the object no longer returns to its original length when you take the load off.

The energy stored

Stretching a spring transfers energy into it, the elastic strain energy, equal to the area under the force-extension graph. While the graph is straight that area is a triangle.

\text{energy stored} = \tfrac{1}{2}F\Delta L

ON THE AQA DATA SHEET

Substitute $F = k\Delta L$ into that and the same triangle becomes $E = \tfrac{1}{2}k(\Delta L)^{2}$ . Only the first version is printed for you, so the second is genuine recall, and it is worth knowing because it makes the square visible. Double the extension and the stored energy goes up four times. Once the graph curves, the half-formula no longer applies while the area still does, so count squares under the curve instead.

That stored energy is fully recoverable while the deformation is elastic, and energy conservation follows it around. Release a catapult and the strain energy becomes kinetic energy of the projectile; fire it upwards and the kinetic energy becomes gravitational potential energy in turn.

GUIDED PRACTICE

A spring constant from data

A spring extends by 0.080 m under a 3.2 N load, within its limit of proportionality. Find k, then the energy stored at an extension of 0.15 m, choosing the right energy expression.

Show the working

k = F/ΔL = 3.2/0.080 = 40 N m⁻¹.

At 0.15 m, use E = ½k(ΔL)² = ½ × 40 × 0.15² = 0.45 J. The ½ is the triangle under the force-extension line; leaving it out claims the force was at full strength throughout the stretch.

INDEPENDENT PRACTICE

Energy from the area

A different spring's force-extension line passes through the origin and reaches 6.0 N at 0.12 m. Find the stored energy from the area under the line, and check it against ½FΔL.

Show the working

The area is a triangle: ½ × base × height = ½ × 0.12 × 6.0 = 0.36 J.

½FΔL gives ½ × 6.0 × 0.12 = 0.36 J, the same number, because the formula is that triangle written out. Bend the line and only the area survives.

Past the elastic limit

Load beyond the elastic limit and the material deforms plastically, meaning its internal structure rearranges for good. Unload it and, in the idealised metal model these graphs draw, the contraction follows a line parallel to the original elastic section but shifted sideways, meeting the extension axis some distance from the origin.

FIG. 2Loading first climbs the straight elastic line, then turns and creeps along the shallow plastic path. This schematic draws the limit of proportionality and the elastic limit merged at the marked point; on a real wire, as the previous figure separates them, the line bends slightly, while still elastic, between the two. Unloading runs straight back down, parallel to the elastic line in this idealised model, but lands to the right of where it began: the sample keeps a permanent set. The wire on the left acts it out, stretching further and never quite returning.

That intercept is the permanent extension. The energy accounting shifts too. Work done during loading now exceeds the energy recovered during unloading, and the difference, which is the area trapped between the two curves, went into rearranging the material and warming it. Energy spent on plastic deformation is not recovered as mechanical work; it ends up mainly as internal energy.

That one-way loss is useful as well as instructive. A ductile part built to yield a little on every large swing of an oscillation loses the same energy each time, so the oscillation's amplitude falls, which is how a sacrificial steel damper protects a tall building. Forced vibrations and resonance, over in periodic motion, picks the idea up as a damping mechanism.

ASSESSMENT FOCUS

1 g cm⁻³ = 1000 kg m⁻³. This conversion appears constantly and is missed constantly; a density answer of 2.7 when the data sheet world works in thousands should ring an alarm.
Limit of proportionality and elastic limit are different points: the first is where the graph stops being straight, the second where deformation stops being reversible. Exam questions test the distinction directly.
Strain energy is the area under the force-extension graph. Use $\tfrac{1}{2}F\Delta L$ only while the line is straight, and count squares once it curves.
Read a permanent extension where the unloading line meets the extension axis, and expect it to run parallel to the original straight section.
Volumes convert with the cube, so 1 cm³ is $10^{−6}$ m³. Convert the length first and cube afterwards.
$\rho = m/V$ , $F = k\Delta L$ and $\tfrac{1}{2}F\Delta L$ are all printed for you. $\tfrac{1}{2}k(\Delta L)^{2}$ is not, and that one you must carry in.

CHECK YOURSELF

A spring with spring constant 25 N m⁻¹ is stretched by 0.20 m, within its limit of proportionality. Find (a) the force applied and (b) the energy stored. (c) If the extension were doubled, still within the limit, what would the stored energy become?

Show a hint

Energy depends on the square of the extension.

Show the answer

(a) $F = k\Delta L$ = 25 × 0.20 = 5.0 N.

(b) $E = \tfrac{1}{2}k(\Delta L)^{2}$ = ½ × 25 × 0.20² = 0.50 J.

(c) The square quadruples it, giving 2.0 J. Picture the triangle under the force-extension line. Doubling the extension doubles its base and its height at once, so the area goes up four times.

The energy stored is the area under the force-extension graph.

½FΔL is that area only while the line is straight.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

18 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the density and hooke's law questions page.

10 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Gradients and areas under graphs is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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Use $\rho = m/V$ , and get the g cm⁻³ to kg m⁻³ conversion the right way round.
Apply $F = k\Delta L$ while it holds, and say what the spring constant belongs to.
Separate the limit of proportionality from the elastic limit.
Find the stored energy from the area under a force-extension graph, and account for the energy that never comes back after plastic deformation.

Open the full revision checklist to track your progress across the whole unit.

Density and Hooke's law

Density

Hooke's law, and where it stops

The energy stored

Past the elastic limit

Question workbook

Question workbook

Answer book