Hooke's law, explained with examples

By The STEM-Tutor team · Updated October 2026

Hooke's law says that the extension of a spring is proportional to the force stretching it, up to a limit. It explains how spring balances and car suspensions work, and it is the basis of a GCSE required practical.

The law

F = k e

• F is the force applied, in newtons (N);
• k is the spring constant, in newtons per metre (N/m);
• e is the extension, the increase in length, in metres (m).

The spring constant measures stiffness. A stiff spring, like a car suspension spring, has a large k and needs a big force to stretch it a little. A soft spring, like one in a pen, has a small k. The same law applies to compression: e is then the decrease in length.

Extension means the change in length, not the total length. If a 10 cm spring stretches to 14 cm, the extension is 4 cm.

The limit of proportionality

Hooke's law only holds up to a point called the limit of proportionality. Up to that point, a graph of force against extension is a straight line through the origin, and its gradient is k. Beyond it, the line curves: the spring stretches more easily, and extension is no longer proportional to force.

Elastic and inelastic deformation

An object is elastically deformed if it returns to its original shape and size when the force is removed. If it does not, it has been inelastically (plastically) deformed. A spring stretched only a little behaves elastically. Stretched too far, it becomes permanently longer. To stretch, squash or bend an object, you need more than one force acting on it; a single force would just make it accelerate.

Worked examples

Example 1: finding the force

A spring has k = 40 N/m. What force stretches it by 0.05 m?

• F = ke = 40 × 0.05 = 2 N.

Example 2: finding the spring constant

A 3 N weight stretches a spring from 12 cm to 18 cm. Find k.

• Extension = 18 − 12 = 6 cm = 0.06 m.
• k = F/e = 3/0.06 = 50 N/m.

Example 3: finding extension

A spring with k = 250 N/m supports a 0.5 kg mass. Find its extension. (g = 9.8 N/kg)

• Weight = mg = 0.5 × 9.8 = 4.9 N.
• e = F/k = 4.9/250 = 0.0196 m ≈ 2.0 cm.

Example 4: reading a graph

On a force-extension graph, a spring extends 8 cm under 4 N, and the line is straight. What force gives an extension of 5 cm?

• k = 4/0.08 = 50 N/m.
• F = 50 × 0.05 = 2.5 N.

Elastic potential energy

Stretching a spring does work on it, and that energy is stored in its elastic potential energy store:

E_e = ½ k e² (valid up to the limit of proportionality)

This is the area under the force-extension graph, a triangle with base e and height ke. The work done to stretch a spring equals the energy it stores, as long as it is not stretched beyond the limit of proportionality.

Example 5: energy stored

A spring with k = 200 N/m is stretched by 0.1 m. How much energy does it store?

• E_e = ½ × 200 × 0.1² = 1 J.

Example 6: doubling the extension

If the extension in Example 5 is doubled, what happens to the energy stored?

• E ∝ e², so doubling e multiplies the energy by 4: 4 J.

Hooke's law in everyday life

A spring balance, or newton meter, works because extension is proportional to force: the scale is evenly spaced, so each newton moves the pointer the same distance. Car suspension springs compress under bumps and release the energy gradually. Trampolines, mattresses and bungee cords all store elastic potential energy. Engineers choose materials with the right stiffness, and they design structures so normal loads stay well below the limit of proportionality, so nothing is permanently bent.

Combining springs

Two identical springs side by side (in parallel) share the load, so each stretches half as much: the pair behaves like one spring with twice the stiffness, 2k. Two identical springs end to end (in series) each feel the full load, so the total extension doubles: the pair behaves like one spring of stiffness k/2. You may meet this as a practical extension question.

The force and extension required practical

1. Hang a spring from a clamp stand, with a ruler clamped vertically beside it.
2. Record the position of the bottom of the spring with no load. Use a pointer attached to the spring to read the ruler precisely.
3. Add masses one at a time, for example 100 g each, which adds about 1 N of force each time.
4. After each addition, record the new position and calculate the extension from the original position.
5. Plot force against extension and find k from the gradient of the straight part.

For accuracy, read the ruler at eye level to avoid parallax error, and wait until the spring stops moving. For safety, wear eye protection in case the spring snaps, and keep feet clear of the masses. If you go past the limit of proportionality, the spring will not return to its original length, so the last few points will bend away from the line.

Common mistakes

• Using the total length instead of the extension.
• Forgetting to convert centimetres to metres.
• Using F = ke beyond the limit of proportionality.
• Writing E = ke² and forgetting the ½.
• Using mass in kg as the force instead of weight in N.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Is Hooke's law in Combined Science?

Yes. Forces and elasticity, F = ke and the required practical are in both Combined Science and separate Physics.

Is F = ke on the equation sheet?

For GCSE exams in 2025, 2026 and 2027, an equations sheet is provided with the Physics and Combined Science papers, including the equations students previously had to recall. Check the arrangements for your exam year, and learn the equation anyway, because using it fluently saves time.

Key takeaways

• F = ke, with e the extension in metres.
• k is the gradient of the straight part of a force-extension graph.
• Hooke's law holds only up to the limit of proportionality.
• E_e = ½ke² is the area under the graph.
• Elastic deformation is reversible; inelastic is permanent.

Related guides

• GCSE Physics forces
• Kinetic energy
• GCSE Physics required practicals