Force-Extension Graph and Elastic Potential Energy
PhysicsChapter 7: Physics of Solids
Force-Extension Graph and Elastic Potential Energy
Explanation level
Overview
Within the elastic limit, work done stretching a material equals the area under its force-extension graph and is stored as elastic potential or strain energy: U = ½Fx = ½kx² = ½(YA/L)x².
Learning Outcomes
Area under a force-extension graph gives work done.
For linear loading, average force is F/2.
Elastic energy is U = ½Fx.
Using F = kx gives U = ½kx².
For a wire, k = YA/L and U = ½(YA/L)x².
Energy is recoverable only while deformation remains elastic.
Intermediate Level
Work from the Graph
As load increases from zero, an elastic wire extends. Work done by a variable force equals the area under the force-extension curve. In the linear Hooke-law region, this area is a triangle.
Key equation
W=area under the F-x graph
Key equation
Favg=0+F2=F2
Key equation
W=Favgx=12Fx
Elastic Potential Energy
When deformation stays within the elastic limit, the work done is stored in the material and can be recovered. This stored energy is elastic potential energy or strain energy.
Key equation
U=12Fx
Hooke-Law Form
Key equation
F=kx
Key equation
U=12kx2
Young-Modulus Form for a Wire
Key equation
F=YALx
Key equation
U=12(YAL)x2
Meaning of k for a Wire
Key equation
k=YAL
Larger Y makes the wire stiffer.
Larger A makes the wire stiffer.
Greater L makes the wire less stiff.
Textbook Chandelier Example
Energy Stored in a Steel Wire
A 1 m steel wire of diameter 0.004 m supports 980 N. Take Y = 2.0 × 10¹¹ Pa. Find extension and strain energy.
1
Given
L = 1 m
d = 0.004 m
F = 980 N
Y = 2.0 × 10¹¹ Pa
2
Working
A = π(0.002)² = 1.257 × 10⁻⁵ m²
3
Step 2
x = FL/(YA) = 3.90 × 10⁻⁴ m
4
Step 3
U = ½Fx
5
Step 4
U = ½(980)(3.90 × 10⁻⁴)
Final answer:x = 3.90 × 10⁻⁴ m and U ≈ 0.191 J
Energy Density
Dividing U by wire volume AL gives the elastic energy stored per unit volume.
Key equation
UAL=12(FA)(xL)=12(stress)(strain)
Common Mistakes
Keep force, length, area and extension in SI units.
Use the original wire length in the modulus formula.
Work and stored energy are equal only for deformation within the elastic limit when energy is recoverable.
Short-Question Preparation
Q. How is work found from a force-extension graph?
A. It equals the area under the graph.
Q. Derive U = ½Fx.
A. For linear loading, average force is 0 + F2 = F/2, so work = Favg x = ½Fx.
Q. Write three forms of elastic energy.
A.U = ½Fx = ½kx² = ½(YA/L)x².
Q. What is strain energy?
A. Elastic potential energy stored in a deformed material.
Q. Write energy density in stress-strain form.
A.u = ½ × stress × strain.
Sticky Notes
Area under a force-extension graph gives work done.
For linear loading, average force is F/2.
Elastic energy is U = ½Fx.
Using F = kx gives U = ½kx².
For a wire, k = YA/L and U = ½(YA/L)x².
Energy is recoverable only while deformation remains elastic.
60-Second Revision
Within the elastic limit, work done stretching a material equals the area under its force-extension graph and is stored as elastic potential or strain energy: U = ½Fx = ½kx² = ½(YA/L)x².
Summary
Within the elastic limit, work done stretching a material equals the area under its force-extension graph and is stored as elastic potential or strain energy: U = ½Fx = ½kx² = ½(YA/L)x².
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Exam preparation
Exam Focus
Important points
Area under a force-extension graph gives work done.
For linear loading, average force is F/2.
Elastic energy is U = ½Fx.
Using F = kx gives U = ½kx².
For a wire, k = YA/L and U = ½(YA/L)x².
Energy is recoverable only while deformation remains elastic.
Common mistakes
Use the elastic-limit conditions.
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.
Force-Extension Graph and Elastic Potential Energy
Sticky Notes · Quick Revision
1.
Key Point 1
Area under a force-extension graph gives work done.
2.
Key Point 2
For linear loading, average force is F/2.
3.
Key Point 3
Elastic energy is U = ½Fx.
4.
Key Point 4
Using F = kx gives U = ½kx².
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6 concept questions
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Explanation after each answer
Exam Practice
Important Question
Q1. How is work found from a force-extension graph?
