Ch 3 · 6 Deformation of Solids · A Level Physics 9702
Topic 3 of 3 in Forces, Energy and Materials — A Level Physics 9702 — 58 lessons.
Introduction
Bridges, cables, springs and bones all stretch or squash under load. Engineers need to know how much a material deforms, whether it springs back, and how much energy it stores. This topic gives you the quantities that describe materials, not just objects. (Syllabus topic 6: sections 6.1–6.2.)
Key ideas — 6.1 (a) Tension, compression and the terms you need
Deformation is caused by tensile forces (pulling, stretching) or compressive forces (pushing, squashing). (Forces and deformations are in one dimension only.)
🧑🏫 Teacher's explanation
Imagine you are playing with a thick rubber band. When you pull it, you are applying a 'load'. The amount it stretches is the 'extension'. If you pull it gently, it behaves predictably: double the pull, double the stretch. This predictable behavior is what we call Hooke's Law. However, if you pull it too hard, it might stretch and never return to its original shape—this is the 'limit of proportionality'.
📖 Definitions
🔍 Worked example — Easy
A spring has an original length of 0.15 m. When a load of 12 N is applied, the length becomes 0.18 m. Calculate the extension x.
🔍 Worked example — Standard
A metal wire obeys Hooke's law with a spring constant k = 450 N/m. Calculate the extension produced by a load of 15 N.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 6.1 (b) Hooke's law and the spring constant
Hooke's law: the extension of an object is directly proportional to the load applied, provided the limit of proportionality is not exceeded.
🧑🏫 Teacher's explanation
Imagine you are stretching a thick rubber band versus a thin hair tie. The hair tie is 'stretchy' (low spring constant), while the rubber band is 'stiff' (high spring constant). When you pull on them, they resist your pull. This resistance is the force required to cause a specific change in length.
📖 Definitions
🔍 Worked example — Easy
A spring has a spring constant k = 250 N m^-1. Calculate the extension produced when a load of 15 N is applied.
🔍 Worked example — Standard
A force-extension graph for a metal wire shows a straight line passing through the origin. At a load of 40 N, the extension is 2.0 mm. Determine the spring constant k in N m^-1.
🔍 Worked example — Exam-style
A spring with k = 120 N m^-1 is stretched by 0.05 m. Calculate the elastic potential energy stored in the spring.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 6.1 (c) Stress, strain and the Young modulus
The spring constant depends on the object's size. To describe the material, use:
🧑🏫 Teacher's explanation
Imagine you are pulling on a thick rubber band versus a thin sewing thread. Even if you apply the same force, the thin thread stretches much more. This is because the thickness (cross-sectional area) matters. To compare materials fairly, we don't just look at how much they stretch (extension), but how much they stretch *relative* to their original length and thickness.
📖 Definitions
🔍 Worked example — Easy
A steel wire of length 2.0 m and cross-sectional area 0.5 mm^2 is subjected to a load of 100 N. Calculate the stress in the wire.
🔍 Worked example — Standard
A copper wire has an original length of 5.0 m and a cross-sectional area of 1.0 × 10^-6 m^2. When a force of 50 N is applied, the extension is 2.5 × 10^-3 m. Calculate the Young modulus.
🔍 Worked example — Exam-style
A metal rod of length 1.0 m and area 2.0 × 10^-4 m^2 is stretched by 0.5 mm by a force of 400 N. Determine the Young modulus of the metal.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 6.2 (a) Elastic and plastic behaviour
🧑🏫 Teacher's explanation
Imagine you are stretching a thick rubber band. If you pull it slightly and let go, it snaps back to its original shape—this is elastic behavior. If you pull it so hard that it stays stretched out and looks 'baggy', you have permanently changed its structure. This is plastic behavior.
📖 Definitions
🔍 Worked example — Easy
A spring has a spring constant k = 150 N m^-1. Calculate the extension produced by a load of 30 N.
🔍 Worked example — Standard
A force-extension graph for a metal wire is a straight line passing through the origin. The wire extends 2.0 mm when a load of 40 N is applied. Calculate the elastic potential energy stored in the wire when it has an extension of 5.0 mm.
🔍 Worked example — Exam-style
A wire is subjected to a tensile force. The stress σ is 1.2 × 10^8 Pa and the strain epsilon is 0.004. Calculate the Young Modulus E.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the tensile stress in the wire. [2] (b) Calculate the tensile strain. [2] (c) Determine the Young Modulus of the wire. [2] [6 marks]
Key ideas — 6.2 (b) Work done and elastic potential energy
The area under a force–extension graph is the work done in deforming the object.
🧑🏫 Teacher's explanation
Imagine you are stretching a thick rubber band. To pull it, you have to exert a force over a certain distance. You are essentially 'doing work' on the rubber band. This work doesn't just disappear; it is stored inside the rubber band as energy, waiting to be released when you let go.
📖 Definitions
🔍 Worked example — Easy
A spring with a spring constant k = 250 N m^-1 is stretched by 0.08 m. Calculate the elastic potential energy stored in the spring.
🔍 Worked example — Standard
A force-extension graph for a metal wire shows a straight line passing through the origin. A force of 40 N results in an extension of 2.0 mm. Calculate the work done in stretching the wire to this extension.
🔍 Worked example — Exam-style
A material is stretched from x = 0 to x = 5.0 mm. The force-extension graph is non-linear. The area under the graph from 0 to 2 mm is 0.001 J. The area from 2 mm to 5 mm is 0.004 J. Calculate the total work done and state if the energy is entirely stored as elastic potential energy if the material has reached its limit of proportionality at 2 mm.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the force applied to the spring. [1] (b) Calculate the elastic potential energy stored in the spring. [2] [3 marks]
Practical notes
Precautions: wear eye protection (the wire may snap); start with a small load to straighten the wire; the diameter measurement has the biggest percentage uncertainty (it is squared).
Worked example
Example 1. A steel wire 2.0 m long, with cross-sectional area 1.0 × 10^-7 m², stretches by 1.0 mm under a load of 10 N. Find the stress, strain and Young modulus.
🔍 Worked example — Easy
A copper wire of length L = 3.0 m and cross-sectional area A = 2.0 × 10^-7 m^2 is subjected to a load of 20 N. Calculate the stress σ and the strain \varepsilon if the extension x is 1.5 mm.
🔍 Worked example — Exam-style
A spring with a spring constant k = 600 N m^-1 is stretched by 12 cm. Calculate the elastic potential energy stored in the spring.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) Stress = force ÷ cross-sectional area; strain = extension ÷ original length; Young modulus = stress ÷ strain. (2) Extension is proportional to load, provided the limit of proportionality is not exceeded. (3) A = π(0.40 × 10^-3)^2 = 5.03 × 10^-7 m²; E = dfrac40 × 3.05.03 × 10^-7 × 1.5 × 10^-3 = 1.6 × 10^11 Pa. (4) Elastic: returns to its original shape when the load is removed; plastic: permanently deformed. (5) A long wire gives a larger extension for the same strain, and a thin wire gives a larger stress for the same load; both make the extension large enough to measure with a small percentage uncertainty.