Hooke's law, ΔF = kΔx, 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 ΔEₑₗ = ½FΔx and from the area under a graph

Estimating the area under a curved graph by counting squares or adding strips

1. Hooke's law says the extension of an object is proportional to the force applied, up to the limit of proportionality, so ΔF = kΔx and the gradient of the straight part of a force–extension graph is the stiffness.

2. k is the stiffness (N m⁻¹), the force per unit extension. It belongs to that object: its material, length, thickness and shape.

3. Series: 1/k = 1/k₁ + 1/k₂ (less stiff). Parallel: k = k₁ + k₂ (stiffer).

4. Elastic deformation disappears when the force is removed; plastic deformation is permanent. The elastic limit separates them.

5. Rubber returns to its length but its unloading curve lies below its loading curve; the hysteresis loop is energy dissipated as heat.

6. Elastic strain energy = area under the force–extension graph; for a straight line through the origin, ΔEₑₗ = ½FΔx.

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.

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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) Gradient of the straight section from the large triangle, with cm converted to m: k = ΔF/Δx = 4.8 ÷0.150 → k = 32 N m⁻¹.

(b) The last point on the straight line is the 600 g load: F = 0.600×9.81 = 5.886 N → 5.9 N.

(c) 4.0 N is below the limit, so Hooke's law applies: Δx = F/k = 4.0 ÷32 = 0.125 m → 12.5 cm.

Series: one after the other

Each spring carries the whole load F, so each stretches by its own amount, F/k₁ and F/k₂. The extensions add: Δx = F/k₁ + F/k₂. Dividing by F gives the combined stiffness.

Parallel: side by side

Joined by a bar that stays level, both springs stretch by the same Δx and the load is shared: F = k₁Δx + k₂Δx = (k₁ + k₂)Δx. The stiffer spring carries the larger share.

WORKED EXAMPLE · QUESTION

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).

SOLUTION

(a) Each spring carries the full 3.6 N: Δx_A = 3.6÷120 = 0.030 m, Δx_B = 3.6÷80 = 0.045 m; total → 0.075 m. Check: kₛₑᵣᵢₑₛ = 1/(1/120 + 1/80) = 48 N m⁻¹ and 3.6÷48 = 0.075 m.

(b) kₚₐᵣₐₗₗₑₗ = 120 + 80 = 200 N m⁻¹, so Δx = 3.6÷200 → 0.018 m. Same extension for both: F_A = 120×0.018 → 2.16 N; F_B = 80×0.018 → 1.44 N.

Check: 2.16 + 1.44 = 3.6 N, the whole load. The series arrangement stretches 4.2 times as far.

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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) Gradient of OP, with mm converted to m: k = 80 ÷(2.0×10⁻³) → k = 4.0 × 10⁴ N m⁻¹.

(b) The unloading line has the same gradient, so the extension recovered is Δx = F/k = 105 ÷(4.0×10⁴) = 2.63×10⁻³ m = 2.63 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).

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.

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 (ΔE ∝Δx²). The energy added between two extensions is the trapezium under the line between them, not ½FΔx using the change in extension.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) Δx = 5.0÷32 = 0.1562 m, then ΔEₑₗ = ½×5.0×0.1562 = 0.391 J → 0.39 J.

(b) At 3.0 N, Δx = 3.0÷32 = 0.0938 m. Trapezium: ½×(3.0 + 5.0)×(0.1562 - 0.0938) → 0.25 J.

Check: ½k(Δx)² at each force gives 0.391 - 0.141 = 0.25 J.

WORKED EXAMPLE · QUESTION

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 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.

SOLUTION

(a) F = 450×0.030 = 13.5 N, so ΔEₑₗ = ½×13.5×0.030 = 0.2025 J → 0.20 J.

(b) All of it becomes gravitational potential energy at the top: mgΔh = 0.2025, so Δh = 0.2025÷(0.025×9.81) → 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.

The energy is still the area under the graph up to the extension reached, but ½FΔx 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 Δx, treat each as a trapezium of area ½(F₁ + F₂)Δx, 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.

WORKED EXAMPLE · QUESTION

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 ½FΔx would give the wrong answer.

SOLUTION

(a) Area of each strip = ½(F₁ + F₂)×0.10: 2.965 + 7.920 + 11.245 + 13.475 + 14.965 J. Adding: ΔEₑₗ ≈50.6 J → ≈ 51 J.

(b) ½FΔx is the area of a triangle under a straight line: ½×155.6×0.50 = 38.9 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.

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.

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.

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?

2. Two identical springs are joined first in series and then in parallel, with the same load. Compare the extensions.

3. Explain the difference between the limit of proportionality and the elastic limit.

4. A spring of stiffness 800 N m⁻¹ is compressed by 5.0 cm. How much elastic strain energy does it store?

5. Why does a rubber band become warm when it is stretched and released many times?

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.

1. Extension = 4.5÷60 = 0.075 m, so length = 0.250 + 0.075 = 0.325 m (assuming the limit of proportionality is not passed).

2. Series stiffness = k/2, parallel stiffness = 2k, so the series extension is 4 times the parallel extension.

3. 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. ½k(Δx)² = ½×800×0.050² = 1.0 J.

5. 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. Recovered extension = 100÷(4.0×10⁴) = 2.5×10⁻³ m = 2.5 mm, so permanent extension = 8.0 - 2.5 = 5.5 mm.