Hooke's law and force-extension graphsEdexcel A-Level Physics: Revision notes
Section 1
Hooke's law and stiffness
Hooke's law states that the extension of a spring or wire is directly proportional to the force applied, provided the limit of proportionality is not exceeded:
where F is the force in N, x the extension in m and k the stiffness (spring constant) in N m⁻¹. A larger k means a stiffer spring. The same equation applies to compression, where x is the decrease in length.
Worked example: a force of 4.0 N extends a spring by 25 mm, so k = 4.0 / 0.025 = 160 N m⁻¹. A 10 N force gives x = 10 / 160 = 0.0625 m = 62.5 mm.
Using the extension in mm in F = kx with k in N m⁻¹. Convert to metres first.
Section 2
Force–extension graphs
A graph of force (y-axis) against extension (x-axis) for a spring or wire that obeys Hooke's law is a straight line through the origin. The gradient is the stiffness k. For compression the same straight line continues into negative force and negative extension, if the material behaves in the same way.
Beyond the limit of proportionality the graph curves and its gradient falls, so the extension grows faster than the force.
Section 3
Elastic and plastic deformation
In elastic deformation the material returns to its original length when the force is removed. The elastic limit is the greatest force (or extension) from which this can happen. It is at, or very slightly beyond, the limit of proportionality.
Beyond the elastic limit the deformation is plastic: atoms move to new positions and some extension is permanent. If a wire is unloaded from the plastic region, the unloading graph is a line parallel to the original straight line and ends at a non-zero permanent extension.
Some ductile materials reach a yield point, where the material extends a large amount for little or no increase in force, because layers of atoms slide over each other.
Evidence for plastic deformation: after the force is removed the material is longer than at the start.
Section 4
Using the graph: a worked example
A wire gives a straight line up to 12 N at an extension of 1.5 mm, so k = 12 / 1.5 × 10⁻³ = 8.0 × 10³ N m⁻¹.
A car spring with k = 4.5 × 10⁴ N m⁻¹ that obeys Hooke's law up to 0.18 m can support at most F = kx = 4.5 × 10⁴ × 0.18 = 8100 N, which is a mass of 826 kg (g = 9.81 N kg⁻¹). Designers keep working loads below the limit of proportionality so that the material behaves predictably and returns to its original length.
Must Know
- F = kx; k in N m⁻¹; gradient of the F–x graph = k
- Hooke's law holds up to the limit of proportionality
- Elastic: returns to original length; plastic: permanent extension
- Yield point: large extension for little extra force
- Unloading from the plastic region follows a line parallel to the original straight line
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Hooke's law and force-extension graphs
- A helical spring obeys Hooke's law. A tensile force of 4.0 N extends it by 25 mm.The spring is instead compressed by 15 mm. Calculate the force needed, assuming Hooke's law is obeyed in compression.2 marks
- A metal wire is loaded in steps and the extension is measured. The graph of force against extension is a straight line through the origin up to a force of 12 N, where the extension is 1.5 mm. Beyond 12 N the line curves so that its gradient decreases. When the wire is unloaded from a force of 12 N it returns to its original length. When it is unloaded from a force of 14 N it is found to be 0.30 mm longer than at the start.Explain what the 0.30 mm permanent extension shows about the wire.2 marks
- A car suspension spring has a stiffness of 4.5 × 10⁴ N m⁻¹. When the car is parked, the spring supports one quarter of the car's mass of 1200 kg. The spring obeys Hooke's law up to a compression of 0.18 m. Take g = 9.81 N kg⁻¹.Calculate the compression of the spring when the car is parked.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).