Home / IGCSE 0625 / Motion, forces, energy / Hooke’s law
IGCSE 0625 / Section 1.5 / Core

A spring keeps its promise, up to a point.

Hang a load on a spring and it stretches. Double the load and it stretches twice as far, a tidy straight-line relationship, but only up to a point. Past the limit of proportionality, the spring stops playing fair.

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The key idea

Up to the limit of proportionality, the extension of a spring is directly proportional to the load: F = kx, where k is the spring constant in N/m. Beyond that limit, extension is no longer proportional to load and the graph curves.

Definition · learn the exact words

Up to the limit of proportionality, the extension of a spring is directly proportional to the load applied.

F = k x

k is the spring constant, in N/m. x is the extension, the increase in length, not the total stretched length.

Section 01

Load it up, watch it stretch.

Add loads one at a time and watch the extension grow in equal steps, the signature of a proportional relationship. Keep going past the limit of proportionality and the steps start to grow unevenly.

Stage 1 · Learn

Check Hooke’s law

Four quick checks on proportionality, extension, and the spring constant. Each correct answer earns XP and lights this skill on your star map.

Quick check+10 XP

Up to the limit of proportionality, the extension of a spring is...

Quick check+10 XP

The point where a load-extension graph stops being a straight line is called the...

Quick check+10 XP

The extension of a spring means...

Quick check+10 XP

A spring obeys Hooke law. If the load on it is doubled, the extension...

Section 02

Straight, then curved.

extension load limit of proportionality
Load against extension: a straight line through the origin while F = kx holds, then it curves once the load passes the limit of proportionality.
Worked example

A spring extends by 0.060 m when a load of 9.0 N is hung from it, within the limit of proportionality. Find the spring constant.

Step 1 · Rearrange F = kxTo find the constant, k = F / x.
Step 2 · Substitutek = 9.0 / 0.060 = 150 N/m
Step 3 · AnswerThe spring constant is 150 N/m, meaning 150 N would be needed for each metre of extension.
Examiner trap

The point where the straight line ends is the limit of proportionality. Avoid writing elastic limit, which is a different point on the graph and is not the term the syllabus wants here. When a load-extension graph stops being a straight line, the quantity you name is the limit of proportionality.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written in the style of Paper 1.

Finish the four checks above to unlock the exam questions
Exam style+20 XP

A load of 6.0 N stretches a spring by 0.030 m, within the limit of proportionality. The spring constant is...

Exam style+20 XP

A spring has a spring constant of 50 N/m. A 10 N load, within the limit, produces an extension of...

Exam style+20 XP

Beyond the limit of proportionality, doubling the load on a spring...

Skill unlocked

Hooke’s law, mastered.

This skill is now lit gold on your star map. Keep the chain going.

-Rank -Level -Score -Topics
Go deeper · practice
Six original Cambridge-style questions
Proportionality up to the limit, the meaning of extension, the spring constant, and F = kx calculations both ways. Attempt each, then reveal the worked solution.
Stage 3 · master the unit
Motion, forces and energy challenge
Mixed questions across the whole unit, each one worth XP. Start this only when you feel confident across every topic in the unit, not just Hooke’s law.
Start the unit challenge →