3.4.2.1MechanicsCore

Bulk Properties of Solids

Before we look at what happens inside a stretched material, we describe how solids behave in bulk: how tightly their mass is packed ( $density$ ), how they stretch under load ( $Hooke’s law$ ), and how much energy a stretched sample stores. These ideas underpin everything from spring systems to the design of bridges.

45 min Video by Science Shorts 3.4.2 Materials

Springs & Hooke’s Law — GCSE & A-Level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

Define density and use ρ = m/V
State Hooke’s law and identify the limit of proportionality
Use the spring constant and combine springs in series and parallel
Distinguish elastic from plastic deformation
Calculate the elastic strain energy stored in a stretched sample

Density

$Density$ is the mass per unit volume of a material. It is a property of the substance itself, not of the size of the sample, and is measured in kilograms per cubic metre (kg m⁻³).

Density is what determines whether an object floats and is a first clue to a material’s identity — lead is dense, cork is not, and water sits at a convenient $1000$ kg m⁻³.

ρ = \frac{m}{V}

Density = mass ÷ volume (kg m⁻³).

1Volume

= 0.050 \times 0.040 \times 0.050 = 1.0 \times 1 0^{- 4}

m³.

ρ = m / V = 0.27/ (1.0 \times 1 0^{- 4}) = 2.7 \times 1 0^{3}

kg m⁻³.

3That matches the density of aluminium (

2.7 \times 1 0^{3}

kg m⁻³).

Answer

ρ = 2.7 \times 1 0^{3}

kg m⁻³ (aluminium)

Tip — Convert lengths to metres before finding a volume in m³: 1 cm = 0.01 m, so 1 cm³ = 1 × 10⁻⁶ m³.

Hooke’s law

$Hooke’s law$ says that the extension of a spring (or wire) is proportional to the force stretching it, provided you do not stretch it too far. The constant of proportionality is the $spring constant$ $k$ , measured in N m⁻¹ — a stiffer spring has a larger $k$ .

The proportionality holds only up to the $limit of proportionality$ . Beyond it the force–extension line curves, and beyond the nearby $elastic limit$ the material no longer returns to its original length.

F = k Δ L

Hooke’s law: force = spring constant × extension.

1Convert the extension:

Δ L = 4.0

= 0.040

2Rearrange Hooke’s law:

k = F /Δ L = 2.0/0.040

k = 50

N m⁻¹.

Answer

k = 50

N m⁻¹

Combining springs

Identical springs $in parallel$ share the load, so the combination is stiffer: the effective spring constant adds, $k_{total} = k_{1} + k_{2}$ . The same force produces a smaller extension.

Springs $in series$ each feel the full force and their extensions add, so the combination is more compliant: $1/ k_{total} = 1/ k_{1} + 1/ k_{2}$ . Two identical springs in series are half as stiff as one.

k_{parallel} = k_{1} + k_{2}

Springs in parallel — stiffer.

\frac{1}{k _{series}} = \frac{1}{k _{1}} + \frac{1}{k _{2}}

Springs in series — more compliant.

Tip — Sanity check: parallel springs share the load so they stretch less (bigger k); series springs pass the load along and stretch more (smaller k).

Elastic strain energy

Stretching a spring stores energy in its $elastic (strain) store$ . On a force–extension graph the energy stored is the $area under the line$ . While Hooke’s law holds this area is a triangle, giving $E = \frac{1}{2} F Δ L$ , which can also be written $\frac{1}{2} k Δ L^{2}$ .

If the material stays within its elastic limit, all of this energy is returned when the force is removed. If it is stretched into the plastic region, some energy is not recovered — it has gone into permanently rearranging the material.

E = \frac{1}{2} F Δ L = \frac{1}{2} k Δ L^{2}

Elastic strain energy = area under the force–extension graph (Hooke’s law region).

1Use

E = \frac{1}{2} k Δ L^{2}

E = \frac{1}{2} \times 50 \times (0.040)^{2}

E = 25 \times 1.6 \times 1 0^{- 3} = 0.040

Answer

E = 0.040

Equation recap

ρ = \frac{m}{V}

Density = mass ÷ volume.

F = k Δ L

Hooke’s law (up to the limit of proportionality).

k_{∥} = k_{1} + k_{2}, \frac{1}{k _{series}} = \frac{1}{k _{1}} + \frac{1}{k _{2}}

Combining springs.

E = \frac{1}{2} F Δ L = \frac{1}{2} k Δ L^{2}

Elastic strain energy stored.

Common mistakes to avoid

Leaving lengths in centimetres when calculating density or spring constant.

Convert to metres first: 1 cm = 0.01 m and 1 cm³ = 1 × 10⁻⁶ m³.

Assuming Hooke’s law always applies, however far you stretch.

F = kΔL only holds up to the limit of proportionality; beyond it the graph curves.

Swapping the rules for series and parallel springs.

Parallel springs add their spring constants (stiffer); series springs add their compliances (softer).

Using E = FΔL for the strain energy.

The area under a straight force–extension line is a triangle, so E = ½FΔL, not FΔL.

Key takeaways

Density ρ = m/V is a property of the material, in kg m⁻³.
Hooke’s law F = kΔL holds up to the limit of proportionality; k is the stiffness.
Parallel springs add k (stiffer); series springs add 1/k (softer).
Elastic strain energy = area under the force–extension graph = ½FΔL = ½kΔL².

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