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FractionRush AQA A-Level Physics 4

Properties of Materials

Investigate elastic and plastic deformation, Hooke's law, elastic potential energy, and how force-extension graphs reveal material properties.

AQA A-Level Physics · Unit 4: Mechanics and Materials
🔩Hooke's law
State and apply F = kx; identify the limit of proportionality
↩️Elastic deformation
Describe elastic behaviour and the elastic limit
🔨Plastic deformation
Explain permanent deformation beyond the elastic limit
🌀Springs in series/parallel
Derive and apply effective spring constants for combinations
Elastic potential energy
Calculate EPE = ½kx² = ½Fx from force-extension graphs
📈Force-extension graphs
Interpret different regions and areas under F-x curves

Hooke's Law

When a spring (or elastic material) is stretched or compressed, the force required is proportional to the extension, provided the extension is not too large. This is Hooke's law:

F = kx
SymbolQuantityUnit
FForce applied (load)N
kSpring constant (stiffness)N m⁻¹
xExtension (or compression) from natural lengthm

The spring constant k measures the stiffness of the spring: a larger k means a stiffer spring requiring more force for the same extension.

Limit of proportionality: The point beyond which force and extension are no longer proportional. Above this point, a force-extension graph curves away from its initial straight line.
Elastic limit: The maximum force/extension from which the material will still return to its original length when the load is removed. The elastic limit is usually slightly beyond the limit of proportionality.

Hooke's law applies not just to springs but to many elastic materials under small deformations, including rubber bands (over a limited range), wires, and beams.

The limit of proportionality and the elastic limit are different points. Hooke's law fails at the limit of proportionality; elastic behaviour fails at the elastic limit (which is slightly higher).

Elastic and Plastic Deformation

Elastic deformation: Deformation that is fully recovered when the load is removed. The material returns to its original shape and size. The atoms are displaced from their equilibrium positions but return to them when the force is removed.
Plastic deformation: Permanent deformation that remains after the load is removed. Atoms are displaced permanently to new equilibrium positions. The material does not return to its original shape.

The transition from elastic to plastic behaviour occurs at the elastic limit. Beyond this point, the material is permanently deformed regardless of whether the load is removed.

Different materials show different deformation characteristics:

During plastic deformation, energy is not fully recovered on unloading — the unloading curve does not follow the loading curve. The area between the loading and unloading curves represents the energy dissipated as heat.

Springs in Series and Parallel

Springs in series (end to end): The same force acts on each spring, and the total extension is the sum of individual extensions.

1/k_eff = 1/k₁ + 1/k₂

The effective spring constant of springs in series is always less than either individual spring constant — the combination is less stiff.

Springs in parallel (side by side): They share the load, and both extend by the same amount.

k_eff = k₁ + k₂

The effective spring constant of parallel springs is the sum — the combination is stiffer than either spring alone.

This is analogous to electrical resistors: springs in series combine like resistors in parallel (reciprocals add), and springs in parallel combine like resistors in series (values add). Remember: the analogy is inverted!

Springs in series: softer (smaller k_eff). Springs in parallel: stiffer (larger k_eff).

Elastic Potential Energy and Force-Extension Graphs

When a spring obeys Hooke's law, the work done stretching it is stored as elastic potential energy. Since the force increases linearly with extension, the average force is ½F = ½kx:

E_elastic = ½kx² = ½Fx

On a force-extension graph, the elastic potential energy stored equals the area under the F-x curve. For a Hookean spring this is a triangle of area ½ × F × x = ½kx².

Key features of a force-extension graph for a metal wire:

For a rubber band, the loading and unloading curves are different (hysteresis). The area enclosed between the curves represents energy dissipated as thermal energy during each cycle of loading and unloading.

For a spring: E = ½kx² (area of triangle under F-x graph). For non-Hookean materials, count squares or use the formula for the area of the specific shape under the graph.
A spring has a spring constant of 250 N m⁻¹. A 3.0 kg mass is attached. Find the extension and the elastic potential energy stored.
1Force = weight = mg = 3.0 × 9.81 = 29.43 N
2Extension: x = F/k = 29.43 / 250 = 0.1177 m
3EPE = ½kx² = ½ × 250 × (0.1177)² = 125 × 0.01386 = 1.73 J
Extension = 0.118 m; Elastic PE = 1.73 J
Two springs of spring constants 200 N m⁻¹ and 300 N m⁻¹ are connected (a) in series and (b) in parallel. Find the effective spring constant in each case.
1In series: 1/k_eff = 1/200 + 1/300 = 3/600 + 2/600 = 5/600
2k_eff (series) = 600/5 = 120 N m⁻¹
3In parallel: k_eff = 200 + 300 = 500 N m⁻¹
Series: k_eff = 120 N m⁻¹; Parallel: k_eff = 500 N m⁻¹
A force of 60 N extends a spring by 0.12 m. (a) Is Hooke's law obeyed? (b) Calculate the spring constant. (c) Calculate the elastic PE stored.
1Assuming this is within the linear region, Hooke's law is obeyed.
2k = F/x = 60 / 0.12 = 500 N m⁻¹
3EPE = ½Fx = ½ × 60 × 0.12 = 3.6 J
k = 500 N m⁻¹; EPE = 3.6 J
A spring (k = 400 N m⁻¹) is compressed by 0.05 m and used to launch a 0.2 kg ball horizontally from a table. What speed does the ball have just after launch?
1EPE stored = ½kx² = ½ × 400 × (0.05)² = ½ × 400 × 0.0025 = 0.5 J
2All EPE converts to KE: ½mv² = 0.5
3v² = 2 × 0.5 / 0.2 = 5.0 → v = √5.0 = 2.24 m s⁻¹
Launch speed = 2.24 m s⁻¹

Q1. A spring obeys Hooke's law up to a load of 20 N, with an extension of 0.04 m. What is the spring constant?

Q2. What is the difference between the limit of proportionality and the elastic limit?

Q3. A spring of k = 150 N m⁻¹ is stretched by 0.06 m. What is the elastic potential energy stored?

Q4. Three identical springs (k = 60 N m⁻¹ each) are arranged: two in parallel, and that combination in series with the third. Find the effective spring constant.

Q5. A rubber band is loaded and then unloaded. The area under the loading curve is 0.80 J and the area under the unloading curve is 0.55 J. How much energy is dissipated as heat per cycle?

Challenge Q1. A vertical spring (k = 800 N m⁻¹) hangs from a fixed support. A 2.0 kg mass is attached and released from the spring's natural length. Find the maximum extension and the position of equilibrium. Explain the energy transfers.

Challenge Q2. On a force-extension graph for a metal wire, sketch and label (a) the limit of proportionality, (b) the elastic limit, (c) the yield point, and (d) the breaking point. Describe the type of deformation in each region.

Challenge Q3. A spring catapult fires a 0.05 kg ball. The spring (k = 1200 N m⁻¹) is compressed by 0.08 m. If 15% of the stored elastic energy is lost to friction, calculate the speed of the ball at launch.