IGCSE CP2 – Extension | Edexcel International GCSE Physics

Theory — Hooke's Law

The extension of a spring is directly proportional to the force applied — up to the elastic limit.

Hooke's Law

When a force is applied to a spring it stretches. Hooke's Law states that the extension is directly proportional to the force applied, provided the elastic limit has not been exceeded:

F = k × e

F = force applied (N) · k = spring constant (N/m or N/cm) · e = extension from natural length (m or cm)

The spring constant k tells you how stiff the spring is. A large k means a stiff spring — a large force produces only a small extension.

Elastic and Plastic Deformation

Below the elastic limit: the spring returns to its original length when the force is removed. This is elastic deformation — no permanent change.

Beyond the elastic limit: the spring is permanently deformed — it does not return to its original length. This is plastic deformation. On the F vs e graph, the line curves beyond the elastic limit.

The F vs Extension Graph

Straight line through origin — Hooke's Law obeyed, extension proportional to force
Gradient = k — the spring constant in N/cm or N/m
Curve beyond elastic limit — Hooke's Law no longer applies, more extension per newton

Real Lab vs This Simulation

In a real school lab, standard springs are designed to stay within their elastic limit using the masses provided (typically 6 × 100g). The experiment focuses on verifying the linear relationship — getting a straight F vs e graph and calculating k from the gradient.

In this simulation you can deliberately go past the elastic limit to observe plastic deformation and the curve on the graph — something that would damage real equipment. Use this to understand the concept, not as an expectation of what happens in the lab.

Procedure

Hanging slotted masses on a spring and measuring extension at each load.

Equipment

Spiral spring · Clamp stand, boss and clamp · Slotted masses (100g) · Mass hanger · Metre ruler · Pointer (wire or sticky label) · Safety mat

1

Set up and record natural length

Hang the spring from the clamp. Attach a pointer at the bottom. Record the ruler reading — all extensions are measured from this reference position.

💡 Extension = new reading − original reading. You do not need to know the actual length of the spring.

2

Add masses one at a time

Add 100g masses one at a time. Wait for the spring to stop oscillating before reading the ruler. Calculate extension e = new reading − original reading.

💡 Calculate force using F = mg where g = 10 N/kg. A 100g mass has weight F = 0.1 × 10 = 1 N.

3

Take readings up to the maximum safe load

Keep adding masses up to the maximum provided (usually 6 × 100g = 600g). In a real school lab, the spring will stay within its elastic limit throughout — the F vs e graph will remain a straight line. This is the expected result.

💡 In practice, school springs are designed to stay elastic within the load range provided. The elastic limit and permanent deformation are real phenomena — they just require loads beyond what is safe to use in a standard lab session.

4

Remove masses — check spring returns to natural length

Remove masses one at a time. The spring should return to its original natural length after each mass is removed — confirming elastic behaviour throughout. If the spring does not return to its original length, the elastic limit was exceeded and permanent deformation has occurred.

💡 In the simulation you can explore permanent deformation deliberately by using the soft spring and adding many masses. In a real lab this would damage the equipment — so this is one advantage of a simulation.

5

Plot F vs extension

Plot F (y-axis, N) against extension e (x-axis, cm). Draw a best-fit line through the straight-line section. Gradient = k. Mark where the graph begins to curve — this is the elastic limit.

      🌀 Add 100g masses one at a time — watch the spring stretch against the ruler. In a real lab the spring stays elastic throughout. In this simulation you can explore past the elastic limit using extra virtual masses.
    

Spring Type

Add / Remove Mass

Live Readings

0.0

Extension / cm

Mass added0 g

Force F = mg0.0 N

Natural length8.0 cm

Stretched length8.0 cm

Extension e0.0 cm

Hooke's Law✓ obeyed

Readings recorded: 0

Data Table

Force vs extension. Red rows are beyond the elastic limit — notice how the extension per newton increases.

#	Mass / g	Force F / N	Extension e / cm	k = F/e / N/cm	Hooke's Law
No readings yet.

Graph — Force vs Extension

Straight line through origin confirms Hooke's Law. Gradient = spring constant k. Curve beyond elastic limit.

Spring constant k

—N/cm from graph gradient

Results

Gradient (= k)—

R² (linear)—

Elastic limit—

True k—

% error—

Interpretation

Record at least 5 readings to see analysis.

Questions

Question 1

A spring has a natural length of 12 cm. A force of 4 N is applied and it stretches to 20 cm. Calculate the spring constant k. Give the correct unit.

Extension e = 20 − 12 = 8 cm. Using F = ke: k = F/e = 4/8 = 0.5 N/cm. To convert: 0.5 N/cm = 50 N/m (multiply by 100). The spring constant is 0.5 N/cm or 50 N/m. A higher k means a stiffer spring that needs more force to produce the same extension.

Question 2

Explain what happens to the spring at the elastic limit. What is the difference between elastic deformation and plastic deformation?

At the elastic limit: the F vs extension graph stops being a straight line and begins to curve — the spring extends more and more for each extra newton of force. Below the elastic limit: elastic deformation — the spring returns to its original length when the force is removed. No permanent change occurs. Beyond the elastic limit: plastic deformation — the spring is permanently stretched. Even when all the masses are removed, the spring is longer than it was originally. It has been permanently deformed.

Question 3

A student plots an F vs extension graph. The straight line has a gradient of 2.5 N/cm. What is the spring constant in N/m? How far will the spring extend if a force of 6 N is applied (assuming Hooke's Law applies)?

k = 2.5 N/cm × 100 = 250 N/m. (Multiply by 100 because 1 m = 100 cm.) Extension e = F/k = 6/2.5 = 2.4 cm (using k in N/cm). Or: e = F/k = 6/250 = 0.024 m = 2.4 cm. The spring extends by 2.4 cm.