OCR A-Level Physics B H557
Planning
Implementing
Analysis
Evaluation
Practical Skills
Use of Apparatus and Techniques
Fundamental Data Analysis
Imaging and Signalling
Sensing
Mechanical Properties of Materials
Waves and Quantum Behaviour
Space, Time and Motion
Creating Models
Out Into Space
Our Place in the Universe
Matter: Very Simple
Matter: Hot or Cold
Electromagnetism
Charge and Field
Probing Deep Into Matter
Ionising Radiation and Risk
Mechanical Properties of Materials
Elastic strain energy
- 3 things to know
- Components 01, 02 and 03
Overview
Recall questions on elastic strain energy, the area under a force–extension graph, E = ½kx² and why it is ½Fx rather than Fx, calculations for springs, bungee cords and catapults, doubling the extension, materials that do not obey Hooke's law, energy beyond the elastic limit, and why a stretched rubber band warms.
What you need to know
Answer each one before you reveal it.
Explain why the energy stored is ½Fx rather than Fx.
A spring of stiffness 80 N m⁻¹ is stretched by 50 mm. Calculate the energy stored.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Define elastic strain energy.
Model answer
Elastic strain energy is the energy stored in a material as it is elastically deformed, which is recovered when the load is removed.
- 2
What does the area under a force–extension graph represent?
Model answer
It represents the work done in stretching the specimen, which is the energy stored if the deformation is elastic.
- 3
State the relationship for the energy stored in a specimen obeying Hooke's law.
Model answer
E = ½kx², where k is the stiffness and x the extension.
- 4
Explain why the energy stored is ½Fx rather than Fx.
Model answer
The force grows steadily from zero to F as the specimen extends, so the mean force is F/2 and the work done is the mean force multiplied by the extension.
- 5
A spring of stiffness 80 N m⁻¹ is stretched by 50 mm. Calculate the energy stored.
Model answer
E = ½ × 80 × 0.050² = 0.10 J.
- 6
A force of 4.0 N stretches a spring obeying Hooke's law by 50 mm. Calculate the energy stored.
Model answer
E = ½Fx = ½ × 4.0 × 0.050 = 0.10 J.
Other ways to work this
The same content, four other ways.
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