Eduqas A-Level Physics A420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Conduction of Electricity
Resistance
D.C. Circuits
Capacitance
Solids Under Stress
Electrostatic and Gravitational Fields of Force
Using Radiation to Investigate Stars
Orbits and the Wider Universe
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Nuclear Decay
Particles and Nuclear Structure
Nuclear Energy
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Solids Under Stress
Work done in deforming a solid
- 3 things to know
- Component 2
Overview
Recall questions on the area under a force–extension graph, the equation ½Fx and its equivalent ½kx², why the factor of a half appears, the energy stored elastically, and what the area of a loading and unloading loop represents.
What you need to know
Answer each one before you reveal it.
State the expression for the work done in stretching a solid that obeys Hooke's law.
Why is the work done in stretching a Hookean spring ½Fx rather than Fx?
Questions from this subtopic
Each one with the model answer underneath.
- 1
How is the work done in deforming a solid found from a force–extension graph?
Model answer
The work done is the area between the force–extension curve and the extension axis.
- 2
State the expression for the work done in stretching a solid that obeys Hooke's law.
Model answer
The work done is ½Fx, where F is the final force in newtons and x is the final extension in metres.
- 3
Why is the work done in stretching a Hookean spring ½Fx rather than Fx?
Model answer
The force rises steadily from zero to F as the extension grows, so the average force acting over the extension is only ½F.
- 4
Give the expression for the work done in stretching a spring in terms of its spring constant.
Model answer
Substituting F = kx into ½Fx gives a work done of ½kx².
- 5
What is meant by elastic strain energy?
Model answer
Elastic strain energy is the energy stored in a deformed solid that is recovered when the deforming force is removed.
- 6
A spring of spring constant 50 N m⁻¹ is stretched by 4.0 cm. Calculate the elastic strain energy stored.
Model answer
The energy is ½kx² = 0.5 × 50 × (0.040)² = 0.040 J. The force at that extension is 2.0 N, and ½Fx = 0.5 × 2.0 × 0.040 gives the same 0.040 J.
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