Topic 7.3.29 min read
Force-Extension Graph and Elastic Potential Energy
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.
Choose Your Level
Beginner Level
Stretching an elastic wire stores energy. Because force rises from zero to its final value, the average force is half the final force, giving stored energy equal to one-half force times extension.
Recommended for Class 11
Force-Extension Graph and Elastic Potential Energy
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.
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.
Hooke-Law Form
Young-Modulus Form for a Wire
Meaning of k for a Wire
- 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.
- L = 1 m
- d = 0.004 m
- F = 980 N
- Y = 2.0 × 10¹¹ Pa
A = π(0.002)² = 1.257 × 10⁻⁵ m²
x = FL/(YA) = 3.90 × 10⁻⁴ m
U = ½Fx
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.
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.
Ask-the-Tutor Modes
- Explain the setup simply.
- Derive the formula.
- Solve the textbook-style numerical.
- Explain the graph.
- Test me with an MCQ.
Exam Focus
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 exam traps
- Wrong axes or units
Long question structure
Setup → measurements → observation → formula → calculation → precautions.
Common Mistakes
Common Mistakes
- Mistake
Use the elastic-limit conditions.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
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.
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.
- Definition
- 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.
Did you understand this topic?
Your answer is saved with your learning record. It does not mark the topic as mastered.
Ready to test yourself?
You've completed the learning section.
Exam Practice
Important Question
Q1. How is work found from a force-extension graph?
Important Question
Q2. Derive U = ½Fx.
Important Question
Q3. Write three forms of elastic energy.
Important Question
Q4. What is strain energy?
Important Question
Q5. Write energy density in stress-strain form.