determine the elastic potential energy of a material deformed within its limit of proportionality from the area under the force–extension graph

1. Learning Objective

By the end of this lesson you will be able to determine the elastic potential energy of a material deformed within its limit of proportionality by calculating the area under its force–extension graph, and to relate this energy to stress, strain and Young’s modulus as required by Cambridge IGCSE/A‑Level Physics (9702).

2. Key Definitions (Cambridge Syllabus)

3. Force–Extension Graph

The typical force–extension graph for a uniform specimen is shown below. The straight‑line segment from the origin to point A represents the linear (elastic) region up to the LoP. Beyond A the curve deviates, indicating the onset of plastic deformation and the elastic limit.

4. From Force–Extension to Stress–Strain

• Force = σA → σ = F/A
• Extension = ΔL → Strain = ε = ΔL / L₀
• Because A and L₀ are constants for a given specimen, the force–extension graph is a scaled version of the stress–strain graph (scaled by A and L₀).
• In the linear region σ ∝ ε; the slope of the stress–strain graph is Young’s modulus:

  \[

  Y = \frac{σ}{ε} = \frac{F L₀}{A ΔL} = \frac{k L₀}{A}

  \]

5. Elastic Potential Energy (EPE)

Elastic potential energy (Uₑ) is the energy stored when a material is stretched or compressed elastically. It equals the work done in deforming the material from zero extension to a final extension x_max that lies within the LoP.

Required formulae (Cambridge 6.2)

• \(U{e}= \displaystyle\int{0}^{x_{\max}} F\,dx\)
• For a linear region (Hooke’s law) the graph is a right‑angled triangle, giving

  \[

  U{e}= \tfrac12 F{\max} x{\max}= \tfrac12 k x{\max}^{2}

  \]

• Relation to Young’s modulus:

  \[

  U{e}= \tfrac12 \frac{Y A}{L{0}}\,x_{\max}^{2}

  \]

Why only the area up to the LoP?

Beyond the LoP the material yields plastically; the work done is dissipated as heat and micro‑structural change, not stored as recoverable elastic energy. Therefore the elastic potential energy is the area under the curve only up to the limit of proportionality.

6. Worked Example – Steel Wire (Force–Extension Method)

1. Given: \(F{\max}=120\ \text{N}\) at an extension \(x{\max}=0.015\ \text{m}\) (within LoP).
2. Triangle‑area method:

   \[

   U{e}= \tfrac12 F{\max} x_{\max}

   = \tfrac12 (120\ \text{N})(0.015\ \text{m})

   = 0.90\ \text{J}

   \]

3. Spring‑constant method:

   \[

   k = \frac{F{\max}}{x{\max}} = \frac{120\ \text{N}}{0.015\ \text{m}} = 8.0\times10^{3}\ \text{N m}^{-1}

   \]

   \[

   U{e}= \tfrac12 k x{\max}^{2}

   = \tfrac12 (8.0\times10^{3}\ \text{N m}^{-1})(0.015\ \text{m})^{2}

   = 0.90\ \text{J}

   \]

7. Example – Determining Young’s Modulus from a Stress–Strain Graph

A metal rod of original length \(L_{0}=1.00\ \text{m}\) and cross‑sectional area \(A=2.0\times10^{-4}\ \text{m}^{2}\) is loaded. The stress–strain graph is linear up to point B where \(\sigma = 2.0\times10^{8}\ \text{Pa}\) and \(\varepsilon = 0.0010\).

• Young’s modulus from the graph:

  \[

  Y = \frac{σ}{ε} = \frac{2.0\times10^{8}\ \text{Pa}}{0.0010}=2.0\times10^{11}\ \text{Pa}

  \]

• Corresponding spring constant:

  \[

  k = \frac{Y A}{L_{0}} = \frac{(2.0\times10^{11}\ \text{Pa})(2.0\times10^{-4}\ \text{m}^{2})}{1.00\ \text{m}}

  = 4.0\times10^{7}\ \text{N m}^{-1}

  \]

• If the rod is stretched to \(x_{\max}=0.002\ \text{m}\) (within LoP), the elastic energy is

  \[

  U{e}= \tfrac12 k x{\max}^{2}

  = \tfrac12 (4.0\times10^{7})(0.002)^{2}

  = 80\ \text{J}

  \]

8. Experiment to Determine Young’s Modulus (Cambridge 6.1)

Apparatus: Uniform wire (known length \(L_{0}\) and cross‑sectional area \(A\)), fixed clamp, set of calibrated masses, metre ruler or vernier, optional force sensor.

1. Measure the original length \(L_{0}\) and cross‑sectional area \(A\) of the wire.
2. Hang the wire vertically from the clamp.
3. Gradually add masses, recording the total load \(F\) (or mass \(m\) where \(F=mg\)) and the corresponding extension \(\Delta L\) each time.
4. Plot Force (N) versus Extension (m). The straight‑line portion gives the spring constant:

   \[

   k = \frac{F}{\Delta L}

   \]

5. Convert each point to stress and strain:

   • σ = F / A
   • ε = ΔL / L₀

6. Plot Stress (Pa) versus Strain (dimensionless). The slope of the linear region is Young’s modulus:

   \[

   Y = \frac{σ}{ε}= \frac{k L_{0}}{A}

   \]

7. Ensure that all data points used lie within the LoP; points beyond give a lower apparent Y because of plastic deformation.

9. Energy Loss During Plastic Deformation

When the material yields beyond the elastic limit, part of the work done is converted into heat, creates dislocations, and may cause micro‑cracks. This energy is not recoverable; the material does not return to its original shape, and the extra area under the force–extension curve represents energy loss rather than stored elastic potential energy.

10. Common Misconceptions (and Exam Impact)

• “All deformation stores energy.” Only elastic deformation (within the LoP) stores recoverable energy. Plastic work is dissipated. AO2 mark loss: using the whole area under the curve.
• “The area under any part of the graph gives elastic energy.” The area must be taken up to the limit of proportionality. AO1: wrong answer.
• “The spring constant is the same for any extension.” k is constant only in the linear region. AO2: incorrect calculation of Uₑ.
• “Stress and force are the same.” Stress = force per unit area. Two wires of different cross‑section experience different stresses even with the same force. AO1: wrong units.
• “The elastic limit is the same as the LoP.” For real materials the elastic limit may occur slightly after the LoP; the graph’s first deviation from linearity marks the elastic limit. AO2: mis‑identifying point A.

11. Summary Checklist (Exam‑Ready)

1. Identify the linear (Hooke’s law) region of the force–extension graph and locate the limit of proportionality (point A).
2. Read the maximum force \(F{\max}\) and extension \(x{\max}\) at that point.
3. Calculate elastic energy:

   • Triangle method: \(U{e}= \tfrac12 F{\max} x_{\max}\)
   • Spring‑constant method: \(k = F{\max}/x{\max}\), \(U{e}= \tfrac12 k x{\max}^{2}\)

4. If cross‑sectional area \(A\) and original length \(L_{0}\) are known, convert to stress and strain and obtain Young’s modulus:

   \[

   Y = \frac{F L{0}}{A \Delta L}= \frac{k L{0}}{A}

   \]

5. Remember that any area beyond the LoP does not represent elastic potential energy.

12. Real‑World Application

In a car suspension system, the springs are designed to operate well within their LoP so that the kinetic energy of road bumps is stored as elastic potential energy and then released, providing a smooth ride. If a spring were to be overloaded into the plastic region, the suspension would lose its ability to recover, leading to permanent sag and unsafe handling.

13. Practice Questions

1. A rubber band obeys Hooke’s law up to a force of 30 N, at which point its extension is 0.050 m**. Calculate the elastic potential energy stored.
2. A metal rod has a spring constant of \(2.5\times10^{4}\ \text{N m}^{-1}\). If it is stretched 2.0 mm within the LoP, what is the energy stored?
3. Explain why the area under the force–extension curve beyond the limit of proportionality does not represent elastic potential energy.
4. A wire of length 1.20 m and cross‑sectional area \(1.0\times10^{-6}\ \text{m}^{2}\) is loaded with a force of 60 N, producing an extension of 0.012 m. Determine:

   • Stress and strain.
   • Young’s modulus of the material.
   • The elastic potential energy stored (verify that the extension is within the LoP).

5. Using the stress–strain data below, calculate Young’s modulus and then the elastic energy stored when the specimen is stretched to an extension of 0.003 m.

   
   Data: \(σ = 1.8\times10^{8}\ \text{Pa}\) at \(ε = 0.0009\); original length \(L_{0}=0.80\ \text{m}\); cross‑sectional area \(A=5.0\times10^{-5}\ \text{m}^{2}\).