Hooke's Law

When you stretch a spring by pulling on it, the spring pulls back. The more you stretch it, the harder it pulls. Robert Hooke first described this behaviour in 1676 and it remains one of the most useful relationships in physics.

Hooke’s Law

Hooke’s law states that the extension of a spring is directly proportional to the applied force, provided the limit of proportionality is not exceeded:

$F = kx$

where:

F is the applied force in N (newtons)
k is the spring constant (also called stiffness constant) in N m⁻¹
x is the extension (or compression) in m — this is the change in length from the natural (unstretched) length

The spring constant k tells you how stiff the spring is. A large k means you need a large force to produce a small extension — the spring is stiff. A small k means the spring extends easily — it is compliant.

Worked example 1: A spring has a natural length of 15.0 cm. When a force of 4.0 N is applied, the spring extends to 21.0 cm. What is the spring constant?

Extension x = 21.0 − 15.0 = 6.0 cm = 0.060 m k = F/x = 4.0 / 0.060 = 66.7 N m⁻¹

Force–Extension Graphs and Key Regions

A force-extension graph plots force (y-axis) against extension (x-axis).

For a spring obeying Hooke’s law, the graph is a straight line through the origin. The gradient of this line equals the spring constant k.

graph LR
    A["Apply Force F"] --> B{"Is F < F at limit\nof proportionality?"}
    B -- Yes --> C["Linear region:\nF = kx\nGradient = k"]
    B -- No --> D{"Is F < F at\nelastic limit?"}
    D -- Yes --> E["Non-linear elastic:\nStill returns to\noriginal length"]
    D -- No --> F["Plastic region:\nPermanent\ndeformation"]
    F --> G{"Is F < F at\nbreaking point?"}
    G -- Yes --> H["Material stretches\nplastically"]
    G -- No --> I["Fracture"]

As the force increases beyond a certain point, the graph begins to curve. Two important points on this graph are:

Limit of proportionality (P): The point beyond which force and extension are no longer directly proportional. The graph starts to curve.
Elastic limit (E): The point beyond which the spring will not return to its original length when the force is removed. Permanent deformation has occurred.

The limit of proportionality always occurs at or before the elastic limit. Below the elastic limit, deformation is elastic (reversible). Above the elastic limit, deformation is plastic (permanent).

Springs in Series

When two springs with spring constants k₁ and k₂ are connected end to end (in series), the same force acts on each spring. Each spring extends independently.

Total extension = x₁ + x₂ = F/k₁ + F/k₂

The effective spring constant for springs in series is:

$\frac{1}{k_{\text{total}}} = \frac{1}{k_1} + \frac{1}{k_2}$

The combined spring is always less stiff than either individual spring. This is analogous to resistors in parallel in electricity.

Worked example 2: Two springs with k₁ = 40 N m⁻¹ and k₂ = 60 N m⁻¹ are connected in series. A 12 N force is applied. Find the effective spring constant and the total extension.

1/k_total = 1/40 + 1/60 = 3/120 + 2/120 = 5/120 k_total = 120/5 = 24 N m⁻¹

Total extension = F/k_total = 12/24 = 0.50 m

Check: x₁ = 12/40 = 0.30 m, x₂ = 12/60 = 0.20 m, total = 0.50 m ✔

Springs in Parallel

When two springs with spring constants k₁ and k₂ are connected side by side (in parallel), they share the applied force and have the same extension.

The effective spring constant for springs in parallel is:

$k_{\text{total}} = k_1 + k_2$

The combined spring is stiffer than either individual spring. This is analogous to resistors in series in electricity.

Worked example 3: Two springs with k₁ = 40 N m⁻¹ and k₂ = 60 N m⁻¹ are connected in parallel. A 12 N force is applied.

k_total = 40 + 60 = 100 N m⁻¹ Extension = F/k_total = 12/100 = 0.12 m

Summary: Series vs Parallel Springs

Property	Series	Parallel
Force on each spring	Same (F)	Shared (F = F₁ + F₂)
Extension	Different; total = x₁ + x₂	Same for each spring
Effective k	1/k = 1/k₁ + 1/k₂ (softer)	k = k₁ + k₂ (stiffer)
Analogy	Resistors in parallel	Resistors in series

Elastic Potential Energy

When you stretch a spring, you do work against the restoring force. This work is stored as elastic potential energy in the spring.

For a spring obeying Hooke’s law, the elastic potential energy is:

$E = \frac{1}{2}kx^2 = \frac{1}{2}Fx$

This can be derived from the force-extension graph: the energy stored equals the area under the graph. For a straight line through the origin, the area is a triangle with base x and height F = kx, giving area = ½ × x × kx = ½kx².

Worked example 4: A spring with k = 80 N m⁻¹ is stretched by 0.15 m. How much elastic potential energy is stored?

E = ½kx² = ½ × 80 × (0.15)² = ½ × 80 × 0.0225 = 0.90 J

This energy is fully recoverable if the spring has not been stretched beyond its elastic limit. If it has, some energy is dissipated in permanently deforming the material.

Worked example 5: A spring with k = 200 N m⁻¹ is initially extended by 0.10 m and then further extended to 0.25 m. How much additional work is done?

This is a common exam trap. The additional work is NOT ½k(Δx)²:

ΔE = ½k(x₂² − x₁²) = ½ × 200 × (0.25² − 0.10²) = ½ × 200 × (0.0625 − 0.010) = ½ × 200 × 0.0525 = 5.25 J

If you incorrectly used ½k(Δx)² = ½ × 200 × 0.15² = 2.25 J, you would be wrong because the force is not constant during the extension.

Hooke’s Law Beyond Springs

Hooke’s law applies to any elastic material within its limit of proportionality. This includes:

Metal wires under tension
Rubber bands (approximately, over a limited range)
Bones (within physiological loading ranges)
The suspension system of a car

Any time you see a linear region on a force-extension graph, Hooke’s law is being obeyed and the spring constant (or stiffness) can be calculated from the gradient.

Loading and Unloading

When a spring that has been stretched within its elastic limit is unloaded, it returns along the same path on the force-extension graph. No energy is lost — all the elastic potential energy is recovered as the spring returns to its natural length.

If the spring is stretched beyond its elastic limit, the unloading curve does not return to the origin. The difference between the loading and unloading paths represents energy dissipated in permanently deforming the material. The spring has a new, longer natural length — it has been permanently stretched.

Experimental Determination of a Spring Constant

To find k experimentally:

Hang the spring vertically from a clamp stand.
Measure and record the natural length L₀ using a ruler.
Add masses in equal increments (e.g., 50 g), recording the total length after each addition.
Calculate extension x = L − L₀ for each mass.
Plot a graph of force F (= mg) against extension x.
The gradient of the straight-line region equals k.

Typical data table:

Mass / g	Force / N	Length / cm	Extension / cm
0	0	15.0	0
50	0.49	16.2	1.2
100	0.98	17.4	2.4
150	1.47	18.5	3.5
200	1.96	19.7	4.7
250	2.45	20.8	5.8
300	2.94	22.5	7.5

The first six readings are approximately proportional (gradient ≈ 0.49/0.012 = 40.8 N m⁻¹). At 300 g the extension is 7.5 cm rather than the expected 7.0 cm, suggesting the limit of proportionality has been exceeded between 250 g and 300 g.

Common Exam Mistakes

Using total length instead of extension: x is the change in length from the natural length, NOT the total length.
Additional work = ½k(Δx)²: This is WRONG. The correct formula for work done between two extensions is ΔE = ½k(x₂² − x₁²).
Assuming Hooke’s law always applies: It only applies up to the limit of proportionality. Check whether the question states the spring is within this limit.
Confusing spring constant with Young modulus: The spring constant (k) depends on dimensions; the Young modulus (E) is a material property. Two identical materials can have very different spring constants if their dimensions differ.

A-Level Deep Dive: Hooke's Law

Spec mapping

Edexcel 9PH0 Topic 4 — Materials, sub-topic on Hooke's law and elastic energy requires students to recall and use $F = k\Delta x$ within the limit of proportionality, to interpret force–extension graphs (gradient gives $k$ , area gives work done), and to compute elastic strain energy as $E = \tfrac{1}{2}F\Delta x = \tfrac{1}{2}k(\Delta x)^2$ (refer to the official Pearson Edexcel specification document for exact wording). The topic is genuinely synoptic: it is examined in 9PH0 Paper 1 (spring–mass systems in equilibrium and dynamics), in Paper 2 (Topic 6 oscillations, where SHM rests on $F = -kx$ ), and in the Practical Skills paper (CP2 Young modulus — the linear region of the force–extension graph is the Hooke's-law region whose gradient gives $k$ ). The formula booklet provides $F = k\Delta x$ and $E = \tfrac{1}{2}F\Delta x$ , but $E = \tfrac{1}{2}k(\Delta x)^2$ must be reconstructed by substitution.

Worked example with full mark scheme

Question (8 marks):

Two identical springs, each with spring constant $k = 30$ N m⁻¹ and natural length 0.20 m, both obey Hooke's law within their limit of proportionality.

(a) The two springs are connected end-to-end (in series) and a load of 6.0 N is hung from the lower end. Calculate (i) the effective spring constant of the combination and (ii) the total extension. (3)

(b) The same two springs are connected side-by-side (in parallel) and the same 6.0 N load is hung from a light bar joining their lower ends so that each spring extends by the same amount. Calculate (i) the effective spring constant and (ii) the extension of each spring. (3)

Solution with mark scheme:

(a) Step 1 — derive the series combination.

Each spring carries the full load 6.0 N (the lower spring's tension equals the load; the upper spring's tension equals the load plus the weight of the lower spring, taken as negligible). Each extends by $\Delta x_1 = F/k = 6.0/30 = 0.20$ m.

Total extension $= \Delta x_1 + \Delta x_2 = 0.40$ m. The effective spring constant is defined by $F = k_{\text{eff}} \Delta x_{\text{total}}$ :

$k_{\text{series}} = \dfrac{6.0}{0.40} = 15\ \text{N m}^{-1}$

Equivalently, $1/k_{\text{series}} = 1/k_1 + 1/k_2$ .

M1 — recognising the same force acts on each spring in series and adding extensions, or using $1/k_{\text{series}} = 1/k_1 + 1/k_2$ .

A1 — $k_{\text{series}} = 15$ N m⁻¹.

A1 — total extension $= 0.40$ m.

(b) Step 1 — derive the parallel combination.

In parallel, each spring extends by the same amount and shares the load. Let each carry force $F_1$ . Then $2F_1 = 6.0$ N, so $F_1 = 3.0$ N. Each extends by $\Delta x = F_1/k = 3.0/30 = 0.10$ m.

The effective spring constant: $F = k_{\text{eff}} \Delta x \Rightarrow k_{\text{parallel}} = 6.0/0.10 = 60$ N m⁻¹. Equivalently, $k_{\text{parallel}} = k_1 + k_2$ .

M1 — recognising the load is shared between parallel springs and that each extends equally.

A1 — $k_{\text{parallel}} = 60$ N m⁻¹.

A1 — extension of each $= 0.10$ m.

Each spring stores $E_1 = \tfrac{1}{2}(30)(0.10)^2 = 0.15$ J. Two springs in parallel store $E_{\text{total}} = 2 \times 0.15 = 0.30$ J.

M1 — correct use of $E = \tfrac{1}{2}k(\Delta x)^2$ for one spring.

A1 — total elastic potential energy $= 0.30$ J.

Cross-check using the combined spring constant: $E = \tfrac{1}{2}k_{\text{parallel}}(\Delta x)^2 = \tfrac{1}{2}(60)(0.10)^2 = 0.30$ J. The two methods agree, as they must.

Total: 8 marks (M3 A5, split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A spring of natural length 0.250 m is suspended from a fixed support. A student adds masses in 100 g increments and records the new length of the spring after each addition.

Mass added / g	New length / m
0	0.250
100	0.272
200	0.295
300	0.317
400	0.340
500	0.395

(a) Plot in your mind a force–extension graph and identify the load at which the spring first stops obeying Hooke's law. Justify your answer. (3)

(b) Calculate the spring constant for the linear region. State an appropriate unit. (2)

(c) Estimate the elastic potential energy stored in the spring at an extension of 0.090 m within the Hooke's-law region. (1)

Mark scheme decomposition by AO:

(a)

M1 (AO2.1) — converting masses to forces using $F = mg$ with $g = 9.81$ N kg⁻¹, and computing extensions $x = L - L_0$ . The first five extensions are 0.022, 0.045, 0.067, 0.090 m for forces 0.981, 1.962, 2.943, 3.924 N — each force-per-extension ratio is $\approx 43.6$ N m⁻¹, indicating linearity.
A1 (AO3.2a) — at 500 g, force $= 4.905$ N gives extension 0.145 m, breaking the constant ratio. The Hooke's-law region ends between 400 g and 500 g.
B1 (AO2.5) — explicit justification: "the ratio $F/x$ is constant within the first five points and rises sharply at the sixth, so the limit of proportionality lies between 4.0 N and 4.9 N".

(b)

M1 (AO1.1b) — gradient of best-fit line through the linear region, e.g. $k = (3.924 - 0)/(0.090 - 0) = 43.6$ N m⁻¹.
A1 (AO1.1b) — $k \approx 44$ N m⁻¹ with unit N m⁻¹.

(c)

B1 (AO1.1b) — $E = \tfrac{1}{2}k x^2 = \tfrac{1}{2}(44)(0.090)^2 = 0.18$ J.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. This is a classic data-handling question where AO2/AO3 marks reward the candidate who interrogates the table (constant-ratio test) rather than just plugging the last value into a formula.

Synoptic links

Connects to:

Topic 4 — Young modulus: Hooke's law for a wire is $F = k\Delta x$ , but the modulus $E = \sigma/\varepsilon$ is material-specific. Comparing the two gives $k = EA/L$ — thicker wires are stiffer and longer wires are softer for the same material.
Topic 3 — Energy conservation: the area under a force–extension graph is work done on the spring, equal to the elastic potential energy stored. For Hooke's-law materials this is the triangle $\tfrac{1}{2}F\Delta x$ ; for non-Hookean materials, integrate or count squares.