Elasticity: Exploring Stress, Strain, and Hooke’s Law in A-Level Science
What Is Elasticity?
Elasticity refers to a material’s ability to return to its original shape after being deformed by an external force.
Key Concepts in Elasticity
Stress (σ\sigmaσ)
Stress measures the force per unit area applied to a material:
σ=FA\sigma = \frac{F}{A}σ=AF
Where:
- σ\sigmaσ: Stress (N/m2\text{N/m}^2N/m2).
- FFF: Force (NNN).
- AAA: Cross-sectional area (m2\text{m}^2m2).
Strain (ϵ\epsilonϵ)
Strain is the ratio of extension to original length:
ϵ=ΔLL\epsilon = \frac{\Delta L}{L}ϵ=LΔL
Where:
- ϵ\epsilonϵ: Strain (dimensionless).
- ΔL\Delta LΔL: Extension (mmm).
- LLL: Original length (mmm).
Hooke’s Law
Within the elastic limit, stress is proportional to strain:
σ=E⋅ϵ\sigma = E \cdot \epsilonσ=E⋅ϵ
Where EEE: Young’s modulus (N/m2\text{N/m}^2N/m2).
Applications of Elasticity
Engineering
Elastic properties ensure structural integrity in bridges and buildings.
Sports Equipment
Elasticity determines the performance of materials in racquets, balls, and tracks.
Medical Devices
Elastomers are used in prosthetics and medical implants.
Example Problem
A metal wire of length 2 m2 \, \text{m}2m and cross-sectional area 0.0001 m20.0001 \, \text{m}^20.0001m2 is stretched by 0.01 m0.01 \, \text{m}0.01m under a force of 200 N200 \, \text{N}200N. Find the stress, strain, and Young’s modulus.
- Stress (σ\sigmaσ):
σ=FA=2000.0001=2×106 N/m2\sigma = \frac{F}{A} = \frac{200}{0.0001} = 2 \times 10^6 \, \text{N/m}^2σ=AF=0.0001200=2×106N/m2
- Strain (ϵ\epsilonϵ):
ϵ=ΔLL=0.012=0.005\epsilon = \frac{\Delta L}{L} = \frac{0.01}{2} = 0.005ϵ=LΔL=20.01=0.005
- Young’s Modulus (EEE):
E=σϵ=2×1060.005=4×108 N/m2E = \frac{\sigma}{\epsilon} = \frac{2 \times 10^6}{0.005} = 4 \times 10^8 \, \text{N/m}^2E=ϵσ=0.0052×106=4×108N/m2
Common Mistakes in Elasticity Calculations
- Forgetting to use consistent units for stress and strain.
- Exceeding the elastic limit when applying Hooke’s law.
- Misinterpreting strain as a percentage instead of a ratio.
Practice Questions
- A wire of length 1.5 m1.5 \, \text{m}1.5m extends by 0.02 m0.02 \, \text{m}0.02m under a force of 100 N100 \, \text{N}100N. If the cross-sectional area is 0.00005 m20.00005 \, \text{m}^20.00005m2, calculate the stress, strain, and Young’s modulus.
- Explain the significance of Young’s modulus in material selection.
- Describe one application of elasticity in sports engineering.