Hooke's law and the spring constant, stress, strain and the Young modulus, elastic strain energy, and the behaviour of materials from stress-strain graphs.

What this dot point is asking

CCEA wants you to state and apply Hooke's law, define stress, strain and the Young modulus, calculate the elastic strain energy stored in a stretched material, and interpret stress-strain graphs for different materials. Expect a numerical Young modulus calculation and a graph interpretation in every series.

The answer

Hooke's law and the spring constant

Beyond the elastic limit the material no longer returns to its original length when the force is removed; it is permanently (plastically) deformed. Springs in series share the same force and add extensions, so the combined constant is smaller; springs in parallel share the extension and add forces, so the combined constant is larger.

Stress, strain and the Young modulus

The Young modulus measures stiffness: steel ($\approx 2 \times 10^{11}\ \text{Pa}$) is far stiffer than rubber. A common CCEA experiment measures it with a long, thin wire (Searle's apparatus): a long wire gives a measurable extension, and a thin wire raises the stress, both reducing percentage uncertainty.

Strain energy and stress-strain graphs

The elastic strain energy stored in a stretched material is the area under the force-extension graph. For a material obeying Hooke's law,

A stress-strain graph shows the limit of proportionality, the elastic limit, the yield point and the breaking stress. A brittle material (such as glass) breaks soon after the elastic limit with little plastic deformation; a ductile material (such as copper) deforms plastically a long way before breaking; a polymeric material (such as rubber) shows a curved loading and unloading path enclosing a hysteresis loop, the area of which is energy lost as heat.

Worked example: energy stored in a stretched wire

Examples in context

Example 1. A suspension-bridge cable. A steel cable supporting part of a footbridge carries a tension of $1.2 \times 10^{5}\ \text{N}$ over a cross-section of $8.0 \times 10^{-4}\ \text{m}^2$. The stress is $\sigma = F/A = 1.5 \times 10^{8}\ \text{Pa}$, well below steel's breaking stress of around $4 \times 10^{8}\ \text{Pa}$, so the cable has a safety factor of roughly $2.7$. Engineers keep the working stress below the elastic limit so the cable always returns to length when the load eases.

Example 2. A climbing rope absorbing a fall. Climbing ropes are deliberately made of polymer fibres with a large area under the force-extension curve, so they store and then dissipate a fall's kinetic energy gradually rather than transmitting a sudden force. The hysteresis loop means the rope warms slightly on each fall; the energy not returned is the area between the loading and unloading curves.

Try this

Q1. A wire of length $2.0\ \text{m}$ and cross-sectional area $1.0 \times 10^{-6}\ \text{m}^2$ extends by $1.0\ \text{mm}$ under a force of $50\ \text{N}$. Find the Young modulus. [3 marks]

• Cue. $E = \frac{FL}{Ax} = \frac{50 \times 2.0}{(1.0 \times 10^{-6})(1.0 \times 10^{-3})} = 1.0 \times 10^{11}\ \text{Pa}$.

Q2. A spring of spring constant $200\ \text{N m}^{-1}$ is stretched by $0.10\ \text{m}$. Find the energy stored. [2 marks]

• Cue. $E = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.10)^2 = 1.0\ \text{J}$.

Q3. Explain why a long thin wire is used when measuring the Young modulus experimentally. [2 marks]

• Cue. A long wire gives a larger, more measurable extension; a thin wire raises the stress for a given load, so both reduce percentage uncertainty.