Forces and elasticity & RP6 — AQA GCSE Physics

SPEC 4.5.3 • FORCES — ELASTICITY & RP6

Forces and Elasticity

Springs, Hooke's law, and Required Practical 6 — the most-examined practical on Paper 2. Two equations, one graph, one experiment that appears every series.

Stretching, Bending & Compressing

Forces can change the shape of an object — by stretching it (springs, rubber bands), bending it (a ruler held at both ends), or compressing it (a foam block, a closed coil spring). All three are covered by the same physics.

Spec point — more than one force required

"To change the shape of an object (by stretching, bending or compressing), more than one force has to be applied — this is limited to stationary objects only." If only one force acted, the object would accelerate (move) rather than change shape. Two (or more) equal-and-opposite forces are needed.

Elastic vs Inelastic Deformation

Elastic deformation

A change in shape that is reversed when the deforming force is removed. The object returns to its original shape.

Examples: a spring stretched by small forces; a rubber band; a tennis ball squashed momentarily.

Inelastic deformation

A change in shape that remains after the deforming force is removed. The object does not return to its original shape.

Examples: a paperclip bent out of shape; a spring stretched too far; modelling clay squashed.

Mark-scheme wording trap

Mark schemes require the word "elastically deformed" when describing a spring's behaviour, not just "elastic". Saying a spring "is elastic" is insufficient (Jun 2022 P2 Q05.4 rejected this exact phrasing). Use "the spring is elastically deformed" or "the spring deforms elastically".

Hooke's Law — F ∝ e

Hooke's Law (spec wording)

The extension of an elastic object is directly proportional to the force applied, provided that the limit of proportionality is not exceeded.

In symbols: F ∝ e (force is proportional to extension). Doubling the force doubles the extension. Tripling the force triples the extension. The two quantities are linked by a constant of proportionality called the spring constant (k).

Compression too

The same relationship applies when an elastic object is compressed — in that case e is the compression (how much shorter the spring becomes). The spec specifically says this.

Equation 1 — F = k × e

F = k × e

F (force, N) · k (spring constant, N/m) · e (extension, m)

Spring constant (k)

The force per unit extension of a spring, in newtons per metre (N/m). A larger k means a stiffer spring — it needs more force to stretch by 1 m.

On the RECALL list

F = k e is on the recall list — memorise it. Spec phrase: "Students should be able to recall and apply this equation." Also written as "force = spring constant × extension" in mark-scheme word form.

Worked Example — Calculate the Force

A spring with a spring constant of 40 N/m is stretched by 5.0 cm.

Calculate the force applied to the spring. 3 marks

CONV

Convert cm to m: 5.0 cm ÷ 100 = 0.050 m(the trap step)

E

Equation: F = k × e

S

Substitute: F = 40 × 0.050

U

Unit: newtons (N)

A

Answer: F = 2.0 N

Three forms to memorise

For any F = k × e question, you might find any one of F, k, or e from the other two:

Find F: F = k × e

Find k: k = F ÷ e

Find e: e = F ÷ k

The Force-Extension Graph

If you plot force (y-axis) against extension (x-axis), Hooke's law gives a straight line through the origin — until you reach the limit of proportionality. Beyond that point, the line curves.

Mark-scheme phrasing

When asked "How does the graph show the spring obeys Hooke's law?", the mark scheme requires BOTH: "the line is straight AND passes through the origin". Each is a separate mark point. Saying only one earns half the marks.

When the equations stop applying

Both equations (F = ke and Eₑ = ½ke²) only apply in the linear region — up to the limit of proportionality. Beyond that point, the actual force-extension relationship curves, and the linear equation underestimates the energy stored (the real area under a curving F-e graph is larger than ½ke²).

RP6 — The Setup

Apparatus checklist

Clamp stand (with bench-clamp for stability)

Spring (with hook at the bottom)

Vertical ruler (mm resolution) alongside the spring

Set of slotted masses (typically 100 g each)

Eye protection (in case the spring breaks)

RP6 — IV, DV, Controls, Method

Independent variableForce applied to the spring (varied by hanging different known masses; F = m g)

Dependent variableExtension of the spring (e = new length − original length)

ControlsSame spring throughout (so k stays constant) Original (unstretched) length measured once at the start Allow spring to stop oscillating before each reading

Method (numbered)

Clamp the spring vertically; record its original length with the ruler

Add a known mass to the hook; allow oscillations to settle; record the new length

Calculate extension = new length − original length

Repeat for at least 5 different masses

Repeat each reading on the way down (removing masses) to confirm elastic behaviour (spring returns to original length)

Plot a graph of force (y-axis) against extension (x-axis)

RP6 accuracy tips — examiner-evidenced

Ruler must be vertical (parallel to the spring) — so it measures the actual length change. A tilted ruler reads more than the real length. ("Explain why the ruler must be vertical" is a 2-mark question pattern.)

Read at eye level to avoid parallax error

Repeat readings on the way up and back down to check for anomalous results and confirm elastic behaviour

Risk assessment — for 6-mark method questions

If a 6-mark question asks you to "describe the method", mark schemes credit hazards and precautions as part of the answer:

Hazard: clamp stand could topple → Precaution: use a bench-clamp to secure it

Hazard: spring may break or fly off → Precaution: wear eye protection

Hazard: masses may fall on feet → Precaution: stand to one side

Finding k from the Graph

For a force-extension graph in the linear region, k = gradient = ΔF ÷ Δe. Pick two clear points on the straight line (one ideally at the origin), calculate the rise over the run, and that's your spring constant.

From a force-extension graph for a spring, two points on the straight line are read off: (0, 0) and (0.125 m, 5.0 N).

Calculate the spring constant. 3 marks

E

Equation: F = k × e → k = F ÷ e

S

Substitute: k = 5.0 ÷ 0.125

U

Unit: newtons per metre (N/m)

A

Answer: k = 40 N/m

Mark scheme tip

The mark scheme allows any correct pair of values from the graph, as long as your F and e values are read off the SAME point on the line. If you misread a value, you can still earn marks for correct method using your incorrect value (error carried forward).

Equation 2 — Elastic Potential Energy

When a force stretches or compresses a spring, the force does work on the spring (from lesson 5.04). That work is stored as elastic potential energy:

Eₑ = ½ × k × e²

Eₑ (elastic PE, J) · k (spring constant, N/m) · e (extension, m)

On the EQUATION SHEET

Unlike F = ke, this equation is provided on the Physics Equations Sheet — you don't need to memorise it. But you DO need to apply it correctly, and the squared on extension is where most marks are lost.

Spec link — work done = energy stored

"A force that stretches (or compresses) a spring does work and elastic potential energy is stored in the spring. Provided the spring is not inelastically deformed, the work done on the spring and the elastic potential energy stored are equal." So in the elastic region: W = Eₑ.

Spec qualifier: the equation Eₑ = ½ke² only applies up to the limit of proportionality. Beyond that point the relationship is non-linear and the equation no longer gives the correct stored energy.

The squared makes things grow fast

Because e is squared, doubling the extension quadruples the elastic PE (×4). Tripling extension stores nine times the energy (×9). Same quadratic behaviour as KE = ½ m v². This is why the cm² unit error is so devastating — squaring amplifies it.

Compression too

The same equation works for compression — just use the compression as e. Real examples: a closed-coil spring in a car suspension, the pen-click spring in a ballpoint, a foam mattress under your weight. All of them store elastic PE when squashed.

Worked Example — Elastic PE

A spring with spring constant 25 N/m is stretched by 8.0 cm.

Calculate the elastic potential energy stored. 3 marks

CONV

Convert cm to m FIRST: 8.0 cm ÷ 100 = 0.080 m(critical — must convert BEFORE squaring)

E

Equation: Eₑ = ½ × k × e²

S

Substitute: Eₑ = 0.5 × 25 × (0.080)²

= 0.5 × 25 × 0.0064

U

Unit: joules (J)

A

Answer: Eₑ = 0.080 J

Why this calculation traps students

Substituting 8.0 (cm, not converted) and then squaring gives Eₑ = 0.5 × 25 × 64 = 800 J — that's 10 000 times too big. The squared amplifies the unit error. Always convert e to metres first, then square.

Examiner Alert — Three Calculation Traps

Trap 1: forgetting the squared

Eₑ = ½ k e² — the extension is squared. Writing Eₑ = ½ k e (no square) is a recurring multi-year error. The squared isn't optional — it's an exponent, not a typo. Forgetting it loses 2 of 3 marks.

Trap 2: substituting e in cm before squaring

The squared amplifies unit errors:

e in cm: (8.0)² = 64

e in m: (0.080)² = 0.0064

Ratio: 64 / 0.0064 = 10 000

Substituting cm without converting gives an answer 10 000× too large. Convert to metres at the substitution line — before you square.

Trap 3: forgetting the ½ (half)

Eₑ = ½ × k × e² — the half at the front is easy to drop. Writing Eₑ = k × e² gives an answer 2× too big. The half comes from the energy being the area under a triangular force-extension graph (½ × base × height). Don't lose it.

Examiner Alert — Precise Definitions

"Elastic" alone is insufficient — use "elastically deformed"

Jun 2022 P2 Q05.4 mark scheme rejected "the spring is elastic" — the credited wording is "the spring is elastically deformed" or "the spring deforms elastically". The adverb form ties the deformation to the behaviour. Same for inelastic: write "inelastically deformed", not just "inelastic".

Spring constant definition

Spring constant is force per unit extension, not "the force needed to stretch the spring". The "per unit" phrasing matters — k has units of N per metre, so it's force divided by extension, not force on its own.

Mark-scheme phrasing for Hooke's law confirmation

If asked how a graph shows Hooke's law applies: write both halves — "the line is straight AND passes through the origin". Each half is a separate mark point. "Constant gradient" is also credited as the "straight" half.

Practice Question

A spring has a spring constant of 200 N/m. It is compressed by 2.0 cm.

Calculate the elastic potential energy stored in the spring. 3 marks

CONV

Convert cm to m FIRST: 2.0 cm ÷ 100 = 0.020 m

E

Equation: Eₑ = ½ × k × e²

S

Substitute: Eₑ = 0.5 × 200 × (0.020)² = 0.5 × 200 × 0.0004

U

Unit: joules (J)

A

Answer: Eₑ = 0.040 J

Compression too

The equations F = ke and Eₑ = ½ke² work for both stretching AND compression. The spec specifically states this. For a compressed spring, e is just the compression (how much shorter the spring becomes).

Weak vs Strong Answers

Q: A spring of spring constant 50 N/m is stretched by 4.0 cm. Calculate the elastic potential energy stored. (3 marks)

✗ Weak (1/3)

Eₑ = ½ × k × e²

Eₑ = 0.5 × 50 × 4.0²

Eₑ = 0.5 × 50 × 16

Eₑ = 400 J

Correct equation and squaring (1 mark) but no unit conversion. Answer 10 000× too big. Two marks lost.

✓ Strong (3/3)

4.0 cm = 0.040 m

Eₑ = ½ × k × e²

Eₑ = 0.5 × 50 × (0.040)²

Eₑ = 0.5 × 50 × 0.0016

Eₑ = 0.040 J

Conversion shown first, equation, squared correctly, full marks.

The double-trap habit

For every Eₑ calculation: (1) convert e to metres on its own line; (2) show the squared explicitly when substituting. These two habits together secure full marks every time.

Common Errors Recap

Students often writeCorrect versionWhy it matters

F = k + eF = k × eOperation error

Eₑ = ½ k e (no square)Eₑ = ½ k e²Loses 2 of 3 marks

e in cm substituted into Eₑ = ½ke²Convert to m FIRST, then square10 000× error

"The spring is elastic""The spring is elastically deformed "Jun 2022 P2 Q05.4 — verbatim wording required

"Spring constant is the force needed to stretch the spring"Spring constant is force per unit extensionDefinition mark error

"Line is straight" alone (for Hooke's confirmation)"Line is straight AND passes through origin"Two mark points required

Summary & Checkpoint

Core content

Two equations: F = k × e (recall) · Eₑ = ½ × k × e² (equation sheet)

Hooke's law: extension ∝ force, up to the limit of proportionality

Elastic vs inelastic: elastic = returns to original shape; inelastic = doesn't

Spec wording: "elastically deformed" — never just "elastic"

Two forces needed to change shape (one would just accelerate the object)

RP6 — Investigate force-extension for a spring

IV: force applied (different masses, F = m g)

DV: extension (new length − original length)

Controls: same spring, original length once, settled readings

Method: measure new length per mass, plot F (y) vs e (x), gradient = k

Hooke's law confirmation phrase: "line is straight AND through origin"

The exam-winning habits

Convert e to metres BEFORE squaring in Eₑ

Show the squared explicitly when substituting

Use " elastically deformed ", not "elastic"

For Hooke's law graph: state BOTH "straight" AND "through origin"

Work done on spring = elastic PE stored (provided elastic deformation)