Forces and motion · GCSE Physics
Hooke’s law
Teacher-written GCSE Physics revision on Hooke’s law: F = ke up to the limit of proportionality, spring constant in N/m, force–extension graphs, and elastic versus inelastic deformation.
Hooke’s law: F = ke, up to the limit of proportionality. The gradient of a force–extension graph is k. Beyond the limit the spring is permanently deformed.
The important bits
What you need to know
- 1
Extension e is the increase in length: stretched length minus original length. Units: metres (m). Do not use the full length in F = ke.
- 2
Hooke’s law: F = ke, where F is the applied force in newtons and k is the spring constant in N/m. k is large for a stiff spring.
- 3
The law holds up to the limit of proportionality, where F is no longer proportional to e. That point is where a force–extension graph stops being a straight line through the origin.
- 4
Elastic deformation: the object returns to its original length when the force is removed. Inelastic (plastic) deformation: it does not; it is permanently stretched.
- 5
The elastic potential energy stored (while Hooke’s law holds) is E = ½ke², also equal to the area under the F–e graph, ½Fe.
- 6
Required practical: hang masses, measure extension with a ruler, plot F against e, find k from the gradient of the straight section. Use a safety pointer and a catch for falling masses.
- 7
Two springs in series share the force but add extensions, so the combination is less stiff (smaller k). Two springs in parallel share the load, so the combination is stiffer.
- 8
A rubber band is often non-linear: the graph curves, and loading and unloading may give a hysteresis loop as energy is transferred to a thermal store.
Quotations worth analysing
Short evidence. Real method.
“Force = spring constant × extension”
Extension, not total length. Convert centimetres to metres if k is in N/m. 5.0 cm is 0.050 m.
“The limit of proportionality is the point beyond which force is no longer proportional to extension.”
It is a graph skill: the straight line through the origin ends. Past that, do not use F = ke blindly.
“Elastic potential energy E = ½ke²”
Only while Hooke’s law applies. The area under a straight F–e graph is the same energy: a triangle of area ½Fe.
Go deeper
Gradient is k; area is energy
Plot force up and extension along. A straight line through the origin is Hooke’s law. Gradient = rise/run = ΔF/Δe = k, in N/m. Do not invert it. The area under that straight section is the energy transferred to the elastic store: a triangle, ½ × F × e, which matches ½ke² because F = ke. Past the limit of proportionality the graph bends; k is no longer constant and the energy is not a simple triangle if you include the bent part without care. In the practical, measure original length with no load, then length with each mass, subtract, convert to metres, and repeat on unloading if you are looking for elastic behaviour. A graph that does not return to the origin has gone inelastic.
Go deeper
Stiff is large k, not “strong”
A spring with k = 40 N/m needs 40 N to extend by 1.0 m, or 2.0 N to extend by 0.050 m. A stiffer spring has a steeper graph. “Strong” is a vague everyday word; in the exam say large spring constant or small extension for a given force. Series springs: each feels the same force, extensions add, effective k falls. Parallel: extensions match, forces add, effective k rises. That is the same logic as resistors, which can help you remember, but say it in spring language. Rubber bands and polythene strips are there to show that not every material is Hookean. Describe the shape; do not force F = ke onto a curve.
See the idea in action
A spring’s original length is 12.0 cm. With a 3.0 N load it is 18.0 cm. Extension e = 6.0 cm = 0.060 m. Spring constant k = F/e = 3.0 / 0.060 = 50 N/m. Energy stored E = ½ke² = 0.5 × 50 × (0.060)² = 0.090 J, which matches ½Fe = 0.5 × 3.0 × 0.060 = 0.090 J. If a 6.0 N load produced 13.0 cm of extension (0.130 m) instead of 0.120 m, the extra stretch shows the limit of proportionality has been passed and F = ke with k = 50 N/m no longer holds.
Exam technique
Turn knowledge into marks
Use extension, not total length. Convert to metres. Find k from the straight-line gradient. Quote the limit of proportionality as the end of that straight line. Use ½ke² only on the Hookean section.
Common mistakes
Do not give these marks away
- 01
Putting total length into F = ke instead of extension, or leaving e in centimetres.
- 02
Calling the limit of proportionality the “breaking point”, or using F = ke on the curved part of the graph.
- 03
Reading the gradient as e/F, which gives 1/k.
A spring with k = 25 N/m extends 0.080 m. What force is applied, if Hooke’s law holds?
A0.32 N
B2.0 N
C3.1 N
D200 N
Show the answer
2.0 N. F = ke = 25 × 0.080 = 2.0 N. 200 N used 8.0 cm as if it were 8.0 m. 0.32 N is k × e² or a mix-up with energy.
Quick questions
If this is the bit you searched
What is the spring constant?
k in F = ke, measured in N/m. It is the force needed per metre of extension, and the gradient of a linear force–extension graph.
What is the limit of proportionality?
The maximum extension (or force) at which F is still proportional to e. Beyond it the F–e graph is no longer a straight line through the origin.
How do you find the energy stored in a stretched spring?
If Hooke’s law holds, E = ½ke², which is also the area of the triangle under the force–extension graph.
What is the difference between elastic and inelastic deformation?
Elastic: the object returns to its original size when the force is removed. Inelastic: it is permanently deformed.