1.8 Hooke's law and elastic strain energy
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Unit 1 Mechanics and Materials, AS content. This page is about objects (a particular spring, wire or band); stress, strain and the Young modulus, which describe the material, are on the next page (1.9).
What this page covers
7 points
- Hooke's law, \(\Delta F = k\Delta x\), and the stiffness \(k\) of an object
- Measuring the stiffness of a spring from a force–extension graph
- Springs in series and in parallel
- Force–extension and force–compression graphs: limit of proportionality, elastic limit, yield point
- Elastic and plastic deformation; loading and unloading a metal wire and a rubber band; hysteresis
- Elastic strain energy from \(\Delta E_{el} = \tfrac12F\Delta x\) and from the area under a graph
- Estimating the area under a curved graph by counting squares or adding strips
In one minute
6 points
- Hooke's law says the extension of an object is proportional to the force applied, up to the limit of proportionality, so \(\Delta F = k\Delta x\) and the gradient of the straight part of a force–extension graph is the stiffness.
- \(k\) is the stiffness (N m⁻¹), the force per unit extension. It belongs to that object: its material, length, thickness and shape.
- Series: \(1/k = 1/k_1 + 1/k_2\) (less stiff). Parallel: \(k = k_1 + k_2\) (stiffer).
- Elastic deformation disappears when the force is removed; plastic deformation is permanent. The elastic limit separates them.
- Rubber returns to its length but its unloading curve lies below its loading curve; the hysteresis loop is energy dissipated as heat.
- Elastic strain energy = area under the force–extension graph; for a straight line through the origin, \(\Delta E_{el} = \tfrac12F\Delta x\).
Contents
13 sections
- What is Hooke's law?
- How do you measure the stiffness of a spring?
- How do springs combine in series and in parallel?
- What happens when a wire is stretched beyond the straight line?
- Why does rubber behave differently?
- How much energy does a stretched object store?
- How do you find the energy when the graph is curved?
- Exam tips for 1.8
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is Hooke's law?
Pull on a spring and it gets longer; pull twice as hard and, for a while, it gets twice as much longer. That simple proportionality is Hooke's law, and it holds for springs, metal wires and rods as long as the deformation stays small.
- Extension = stretched length − original length. A compression is the reduction in length when a force squashes the object.
- Stiffness \(k\) is the force needed per unit extension. A stiff object has a large \(k\) and stretches little.
- \(k\) is a property of the object, not of the material alone: a longer spring of the same wire has a smaller \(k\), a thicker one a larger \(k\).
- The law applies to tension (stretching) and to compression (squashing). Springs, metal wires and rods obey it for small deformations; rubber largely does not.
- In SI base units, 1 N m⁻¹ = 1 kg s⁻².
How do you measure the stiffness of a spring?
The standard experiment hangs masses on a spring and measures how far it stretches. The method matters because the gradient you draw is only as good as the readings behind it.
- Hang the spring from a clamp, with a metre rule clamped vertically beside it. Fix a horizontal pointer to the bottom of the spring.
- Record the pointer reading with no load (or a small hanger to straighten the coils). Read at eye level to avoid parallax.
- Add masses in equal steps (for example 100 g) and record each new reading. Extension = reading − original reading; load = \(mg\).
- Remove the masses one at a time and read again. If the readings return to the originals, the spring was not taken past its elastic limit.
- Plot force on the y-axis against extension on the x-axis; the gradient of the straight section is \(k\).
- Safety: clamp the stand to the bench, keep feet clear of the masses, and wear eye protection in case the spring flies off.
Use the graph above to (a) determine the stiffness of the spring, (b) state the force at the limit of proportionality and (c) calculate the extension produced by a force of 4.0 N.
(a) Gradient of the straight section from the large triangle, with cm converted to m: \(k = \Delta F/\Delta x = 4.8 \div 0.150\) → k = 32 N m⁻¹.
(b) The last point on the straight line is the 600 g load: \(F = 0.600\times9.81 = 5.886\ \text{N}\) → 5.9 N.
(c) 4.0 N is below the limit, so Hooke's law applies: \(\Delta x = F/k = 4.0 \div 32 = 0.125\ \text{m}\) → 12.5 cm.
If a graph has extension on the y-axis and force on the x-axis, its gradient is \(1/k\), not \(k\). Check the axes before you call a gradient "the stiffness".
How do springs combine in series and in parallel?
Series: one after the other
Each spring carries the whole load \(F\), so each stretches by its own amount, \(F/k_1\) and \(F/k_2\). The extensions add: \(\Delta x = F/k_1 + F/k_2\). Dividing by \(F\) gives the combined stiffness.
Parallel: side by side
Joined by a bar that stays level, both springs stretch by the same \(\Delta x\) and the load is shared: \(F = k_1\Delta x + k_2\Delta x = (k_1 + k_2)\Delta x\). The stiffer spring carries the larger share.
Spring A has stiffness 120 N m⁻¹ and spring B has stiffness 80 N m⁻¹. A load of 3.6 N is hung from them (a) in series and (b) in parallel, joined by a light bar that stays horizontal. Calculate the total extension in each case, and the force on each spring in (b).
(a) Each spring carries the full 3.6 N: \(\Delta x_A = 3.6\div120 = 0.030\ \text{m}\), \(\Delta x_B = 3.6\div80 = 0.045\ \text{m}\); total → 0.075 m. Check: \(k_{\text{series}} = 1/(1/120 + 1/80) = 48\ \text{N m}^{-1}\) and \(3.6\div48 = 0.075\ \text{m}\).
(b) \(k_{\text{parallel}} = 120 + 80 = 200\ \text{N m}^{-1}\), so \(\Delta x = 3.6\div200\) → 0.018 m. Same extension for both: \(F_A = 120\times0.018\) → 2.16 N; \(F_B = 80\times0.018\) → 1.44 N.
Check: \(2.16 + 1.44 = 3.6\ \text{N}\), the whole load. The series arrangement stretches 4.2 times as far.
Adding stiffnesses for springs in series. Joining springs end to end makes the arrangement easier to stretch, so the combined stiffness must be smaller than either one: a quick check on any answer.
What happens when a wire is stretched beyond the straight line?
| Point or region | What it means |
|---|---|
| Limit of proportionality (P) | the end of the straight section: beyond it, force and extension are no longer proportional |
| Elastic limit (E) | the greatest force for which the object still returns to its original length when the force is removed |
| Yield point (Y) | where the material starts to stretch a lot for little or no increase in force: the start of large plastic flow |
| Elastic deformation | a change of shape that is fully reversed when the force is removed |
| Plastic deformation | a permanent change of shape: the object does not return to its original length |
P and E are often close together, but they are different points: between them the graph curves yet the deformation is still elastic. Beyond E, layers of atoms in a metal slip past one another, so part of the extension remains after the force is removed.
When a metal is unloaded from beyond E, the unloading line is (nearly) parallel to the original straight section, because the stiffness of the atomic bonds is unchanged. Where that unloading line meets the extension axis gives the permanent extension.
The graph above is for a metal wire. The wire is stretched to an extension of 10.0 mm, where the force is 105 N, and then the force is removed. (a) Determine the stiffness of the wire on the straight section. (b) Calculate the permanent extension of the wire.
(a) Gradient of OP, with mm converted to m: \(k = 80 \div (2.0\times10^{-3})\) → k = 4.0 × 10⁴ N m⁻¹.
(b) The unloading line has the same gradient, so the extension recovered is \(\Delta x = F/k = 105 \div (4.0\times10^{4}) = 2.63\times10^{-3}\ \text{m} = 2.63\ \text{mm}\). Permanent extension = total − recovered = \(10.0 - 2.63\) → 7.4 mm.
Compression graphs are read the same way. For a compression spring the line stays straight until the coils touch; after that the force rises very steeply (see the pop-up toy graph below).
When you label a graph, put the limit of proportionality exactly where the straight line ends. Do not put the elastic limit at the same point unless the question says they coincide, and never place the yield point on the straight section.
Why does rubber behave differently?
A rubber band does not obey Hooke's law: its force–extension graph is curved almost from the start. Rubber is made of long, tangled polymer chains. A small force straightens and lines up the chains easily; once they are nearly straight, further stretching pulls on the bonds and the band becomes stiffer. Rubber can stretch to several times its length and still return to (almost) its original length: a large elastic deformation.
The unloading curve lies below the loading curve: at each extension the band pulls back with less force than was needed to stretch it. This is hysteresis. The area under the loading curve is the work done on the band; the area under the unloading curve is the energy returned. The loop area is the energy dissipated, mainly as thermal energy in the rubber, which is why a band stretched and released repeatedly gets warm.
| Metal wire past its elastic limit | Rubber band | |
|---|---|---|
| After unloading | permanent extension | returns to (almost) its original length |
| Unloading curve | straight, parallel to the loading line | curved, below the loading curve |
| Energy not returned | area between loading curve and unloading line: work done in permanently deforming the wire | loop area: dissipated as thermal energy |
Describing rubber after hysteresis as "plastically deformed". The band goes back to its original length, so the deformation is elastic; the energy loss comes from the loading and unloading curves being different, not from a permanent change of shape.
How much energy does a stretched object store?
Stretching an object needs work. While the deformation is elastic, that work is stored as elastic strain energy and can be recovered. Work = force × distance, but here the force grows as the extension grows, so the work is the area under the force–extension graph. For a straight line through the origin that area is a triangle:
Doubling the extension doubles the force, so the stored energy goes up four times (\(\Delta E \propto \Delta x^2\)). The energy added between two extensions is the trapezium under the line between them, not \(\tfrac12F\Delta x\) using the change in extension.
The spring of worked example 1 has stiffness 32 N m⁻¹. Calculate (a) the elastic strain energy stored when the force is 5.0 N and (b) the extra energy stored when the force is increased from 3.0 N to 5.0 N.
(a) \(\Delta x = 5.0\div32 = 0.1562\ \text{m}\), then \(\Delta E_{el} = \tfrac12\times5.0\times0.1562 = 0.391\ \text{J}\) → 0.39 J.
(b) At 3.0 N, \(\Delta x = 3.0\div32 = 0.0938\ \text{m}\). Trapezium: \(\tfrac12\times(3.0 + 5.0)\times(0.1562 - 0.0938)\) → 0.25 J.
Check: \(\tfrac12k(\Delta x)^2\) at each force gives \(0.391 - 0.141 = 0.25\ \text{J}\).
A toy of mass 25 g sits on a spring of stiffness 450 N m⁻¹ (graph above). The toy is pressed down, compressing the spring by 30 mm, and held by a suction cup. When the cup lets go, the toy jumps vertically. \(g = 9.81\ \text{N kg}^{-1}\). (a) Calculate the elastic strain energy stored. (b) Calculate the maximum height the toy could rise. (c) Suggest why the real height is less.
(a) \(F = 450\times0.030 = 13.5\ \text{N}\), so \(\Delta E_{el} = \tfrac12\times13.5\times0.030 = 0.2025\ \text{J}\) → 0.20 J.
(b) All of it becomes gravitational potential energy at the top: \(mg\Delta h = 0.2025\), so \(\Delta h = 0.2025\div(0.025\times9.81)\) → 0.83 m.
(c) Not all the stored energy ends up as gravitational potential energy: some becomes kinetic energy of the spring itself, thermal energy in the spring, and work against air resistance.
Convert millimetres and centimetres to metres before you use \(\tfrac12F\Delta x\). N × mm gives millijoules, so an answer 1000 times too big is the usual sign of a missed conversion.
How do you find the energy when the graph is curved?
The energy is still the area under the graph up to the extension reached, but \(\tfrac12F\Delta x\) no longer applies. Two methods work:
- Counting squares: find the energy of one grid square (its width in m × its height in N), count whole squares under the curve, count squares that are about half or more as whole and ignore the rest, then multiply.
- Strips: split the area into vertical strips of equal width \(\Delta x\), treat each as a trapezium of area \(\tfrac12(F_1 + F_2)\Delta x\), and add them.
For a curve that bends over (gets less steep), each straight strip top lies slightly below the curve, so the strip method slightly underestimates; narrower strips give a better estimate.
A nylon cord is stretched in steps of 0.10 m. The force readings at 0, 0.10, 0.20, 0.30, 0.40 and 0.50 m are 0.0, 59.3, 99.1, 125.8, 143.7 and 155.6 N. (a) Estimate the elastic strain energy stored at an extension of 0.50 m, assuming the cord is still elastic. (b) Explain why \(\tfrac12F\Delta x\) would give the wrong answer.
(a) Area of each strip = \(\tfrac12(F_1 + F_2)\times0.10\): 2.965 + 7.920 + 11.245 + 13.475 + 14.965 J. Adding: \(\Delta E_{el} \approx 50.6\ \text{J}\) → ≈ 51 J.
(b) \(\tfrac12F\Delta x\) is the area of a triangle under a straight line: \(\tfrac12\times155.6\times0.50 = 38.9\ \text{J}\). This is too small, because the curve lies above the straight line from the origin.
The exact area under the curve drawn is 51.1 J: the strip estimate is slightly low, as expected for a curve that bends over.
Using the force at the end × the extension (\(F\Delta x = 77.8\ \text{J}\) here). That is the area of the whole rectangle, which includes the region above the curve, and overestimates the energy.
Exam tips for 1.8
Hooke's law and \(\tfrac12F\Delta x\) use the extension (or compression). If a question gives lengths, subtract the original length first.
A statement of Hooke's law needs the condition "provided the limit of proportionality is not exceeded". A definition of the elastic limit must mention returning to the original length when the force is removed.
Use \(\tfrac12F\Delta x\) only on the proportional part. For a curved graph, or for energy added between two extensions, find the area of the region you actually need.
Ask two questions: does each spring carry the whole load (series) or a share of it (parallel)? Do the extensions add (series) or are they equal (parallel)? Then use \(\Delta F = k\Delta x\) for each spring.
When asked to describe a force–extension graph, name the regions in order (straight section, curve, plastic region) and give the values where they change. When asked to explain, link each feature to what is happening: proportional, elastic, permanent.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| The limit of proportionality and the elastic limit are the same point. | An object can still return to its original length after the graph has started to curve. | P is where the line stops being straight; E is where permanent deformation starts. E is at or beyond P. |
| A spring obeys Hooke's law because it returns to its original length. | Returning to the original length shows elastic behaviour; rubber does this without obeying Hooke's law. | Hooke's law means force ∝ extension: a straight line through the origin. |
| Stiffness is a property of the material. | Two springs of the same steel can have very different \(k\). | \(k\) depends on the object's material, length, thickness and shape; the Young modulus describes the material. |
| Springs in series are stiffer. | Each spring carries the full load, so the extensions add. | In series the combined stiffness is less than either spring; in parallel it is the sum. |
| Doubling the extension doubles the stored energy. | The force also doubles, and the energy is ½ × force × extension. | \(\Delta E_{el} \propto (\Delta x)^2\): double the extension, four times the energy. |
| All the work done stretching rubber is stored and returned. | The unloading curve is below the loading curve. | The hysteresis loop area is dissipated as thermal energy each cycle. |
How this topic connects
1.9 Stress, strain and the Young modulus
Dividing force by area and extension by length removes the size of the object: the stress–strain graph describes the material, and its gradient (the Young modulus) replaces the stiffness \(k\).
1.5 Work and energy
The area under a force–extension graph is the work done, and the stored elastic strain energy can become kinetic or gravitational potential energy, as in a catapult or a trampoline.
Oscillations (Unit 5) and practical skills (3.3)
A mass on a spring that obeys Hooke's law oscillates with simple harmonic motion, with a period that depends on \(k\). The spring experiment is also a model for choosing equal steps, reading scales at eye level and taking a gradient from a large triangle.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(\Delta F = k\Delta x\) | Hooke's law; stiffness from force and extension (\(k\) in N m⁻¹) | yes |
| \(\Delta E_{el} = \tfrac12F\Delta x\) | elastic strain energy, straight-line region (J) | yes |
| \(\Delta E_{el} = \tfrac12k(\Delta x)^2\) | strain energy from stiffness and extension | no — derive it, substituting \(F = k\Delta x\) |
| \(1/k = 1/k_1 + 1/k_2\) | springs in series | no — derive it |
| \(k = k_1 + k_2\) | springs in parallel | no — derive it |
| energy = area under \(F\)–\(\Delta x\) graph | any graph, including curves (squares or strips) | no — learn it |
Check your understanding
1. A spring of natural length 0.250 m and stiffness 60 N m⁻¹ supports a load of 4.5 N. What is its new length?
Extension = \(4.5\div60 = 0.075\ \text{m}\), so length = \(0.250 + 0.075 = 0.325\ \text{m}\) (assuming the limit of proportionality is not passed).
2. Two identical springs are joined first in series and then in parallel, with the same load. Compare the extensions.
Series stiffness = \(k/2\), parallel stiffness = \(2k\), so the series extension is 4 times the parallel extension.
3. Explain the difference between the limit of proportionality and the elastic limit.
Beyond the limit of proportionality the extension is no longer proportional to the force (the graph curves). Beyond the elastic limit the object no longer returns to its original length when the force is removed (it is permanently deformed).
4. A spring of stiffness 800 N m⁻¹ is compressed by 5.0 cm. How much elastic strain energy does it store?
\(\tfrac12k(\Delta x)^2 = \tfrac12\times800\times0.050^2 = 1.0\ \text{J}\).
5. Why does a rubber band become warm when it is stretched and released many times?
Its unloading curve lies below its loading curve. The area of the hysteresis loop is energy that is not returned on unloading; it is dissipated as thermal energy in the rubber every cycle.
6. A wire with stiffness 4.0 × 10⁴ N m⁻¹ on its straight section is loaded beyond its elastic limit to 8.0 mm at 100 N, then unloaded. Find the permanent extension.
Recovered extension = \(100\div(4.0\times10^{4}) = 2.5\times10^{-3}\ \text{m} = 2.5\ \text{mm}\), so permanent extension = \(8.0 - 2.5 = 5.5\ \text{mm}\).
Specification checklist
- (27) I can state Hooke's law with its condition and use \(\Delta F = k\Delta x\), where \(k\) is the stiffness of the object.
- (27) I can find the combined stiffness and the extensions of springs joined in series and in parallel.
- (29) I can plot, sketch and interpret force–extension and force–compression graphs, and find a stiffness from a gradient.
- (29) I can mark and explain the limit of proportionality, elastic limit and yield point, and tell elastic from plastic deformation.
- (29) I can compare the loading and unloading curves of a metal wire and of rubber, including permanent extension and the hysteresis loop.
- (32) I can calculate elastic strain energy from \(\Delta E_{el} = \tfrac12F\Delta x\) and from the area under a force–extension graph.
- (32) I can estimate the area under a curved force–extension graph by counting squares or adding strips.