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Edexcel A Level (IAL) Physics-1.32 Elastic Strain Energy- Study Notes- New Syllabus

Edexcel A Level (IAL) Physics -1.32 Elastic Strain Energy- Study Notes- New syllabus

Edexcel A Level (IAL) Physics -1.32 Elastic Strain Energy- Study Notes -Edexcel A level Physics – per latest Syllabus.

Key Concepts:

  • 1.32 be able to calculate the elastic strain energy Eel in a deformed material sample, using the equation ∆Eel =1/2 F∆x , and from the area under the force-extension graph
    The estimation of area and hence energy change for both linear and non-linear force-extension graphs is expected.

Edexcel A level Physics-Study Notes- All Topics

Elastic Strain Energy \( \Delta E_{\text{el}} = \tfrac{1}{2}F\,\Delta x \) and Area Under Force–Extension Graph

Elastic strain energy is the work done on a material to deform it elastically (i.e., it returns to its original shape when the force is removed).

Elastic Strain Energy Formula

For materials obeying Hooke’s Law:

$ \Delta E_{\text{el}} = \frac{1}{2} F\,\Delta x $

  • \( \Delta E_{\text{el}} \) = elastic strain energy (J)
  • \( F \) = applied force (N)
  • \( \Delta x \) = extension (m)

This equation is valid only in the linear region where Hooke’s Law holds.

Energy from Spring Constant

If the spring constant \( k \) is known:

$ \Delta E_{\text{el}} = \frac{1}{2}k(\Delta x)^2 $

  • More useful when a force-extension graph is not provided.

 Energy from the Area Under the Force–Extension Graph

The work done in stretching a material is:

$ \Delta E_{\text{el}} = \text{Area under the } F\text{–}x \text{ graph} $

  • This works for linear and non-linear materials.
  • Linear graph → area is a triangle → use \( \frac{1}{2}F\Delta x \).
  • Non-linear graph → area must be estimated (trapeziums, counting squares, integration).

Linear (Hookean) Materials

  • Graph is a straight line.
  • Energy stored is triangular area under the line.
  • Accurate and simple to calculate.

$ \Delta E_{\text{el}} = \frac{1}{2}F\Delta x $

Non-Linear Materials (Rubber, Polymers, Plastics)

  • Curve is not straight → no simple formula.
  • Estimate area using:
    • small trapeziums beneath the curve
    • counting squares on graph paper
    • digital integration (A-level: trapezium rule)
  • The area still equals the energy stored.

 Important Notes

  • Elastic strain energy is only recoverable if deformation is elastic.
  • In the plastic region, work done is not fully recovered → some becomes internal energy.
  • Energy stored in real materials often shows hysteresis (energy loss during unloading).

Example (Easy)

A spring is stretched by \( 0.10\, \mathrm{m} \) under a force of \( 6\, \mathrm{N} \). Find the elastic strain energy stored.

▶️ Answer / Explanation

\( \Delta E_{\text{el}} = \frac{1}{2}F\Delta x = \frac{1}{2}(6)(0.10) = 0.3\, \mathrm{J} \)

Example (Medium)

A material obeys Hooke’s Law with a spring constant \( k = 200\, \mathrm{N\,m^{-1}} \). Find the elastic energy stored when it is stretched by \( 0.08\, \mathrm{m} \).

▶️ Answer / Explanation

\( \Delta E_{\text{el}} = \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}(200)(0.08)^2 = 0.64\, \mathrm{J} \)

Example (Hard)

The force–extension graph for a rubber band is non-linear. The force increases from 0 to \( 12\, \mathrm{N} \) over an extension of \( 0.20\, \mathrm{m} \). Estimate the elastic energy stored using the trapezium rule (take four strips).

▶️ Answer / Explanation

Divide extension range into 4 equal strips of width \( h = 0.05\,\mathrm{m} \).

Suppose measured forces at intervals are:

  • 0.00 m → 0 N
  • 0.05 m → 4 N
  • 0.10 m → 7 N
  • 0.15 m → 10 N
  • 0.20 m → 12 N

Use trapezium rule:

$ E \approx \frac{h}{2}\left(F_0 + 2(F_1 + F_2 + F_3) + F_4\right) $

$ E = \frac{0.05}{2}\left[0 + 2(4 + 7 + 10) + 12\right] = 0.025(2 \times 21 + 12) $

\( E = 0.025(54) = 1.35\, \mathrm{J} \)

Estimated elastic energy stored ≈ \( 1.35\, \mathrm{J} \)

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