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General physics
🎯 Learning Objectives
  • Determine the elastic potential energy from the area under a force–extension graph within the limit of proportionality
  • Recall and apply (E_P = tfrac{1}{2} F x) and (E_P = tfrac{1}{2} k x^2) for elastic materials
  • Interpret and sketch force–extension graphs for springs and wires
  • Calculate spring constant (k = frac{F}{x}) and energy stored for given extensions
🗣️ Language Objectives
  • Use terms “elastic potential energy,” “limit of proportionality,” “force–extension graph” correctly in English
  • Describe graph shapes (“linear,” “curved,” “area under the curve”) accurately
  • Explain calculations and reasoning in clear, academic English
  • Interpret problem statements and express results using precise physics language
📚 Key Terms and Translations
English TermRussianKazakh
Elastic potential energyУпругая потенциальная энергияСерпімді потенциалдық энергия
Limit of proportionalityПредел пропорциональностиПропорционалдылық шегі
Force–extension graphГрафик сила–удлинениеКүш–ұзарту графигі
Spring constant ((k))Жёсткость пружиныПружина тұрақтылығы
Extension ((x))УдлинениеҰзарту
Load ((F))НагрузкаЖүк
🃏 Vocabulary Study Cards

Elastic potential energy

Definition: Energy stored when a material is deformed elastically

Formula: (E_P = tfrac{1}{2}Fx = tfrac{1}{2}kx^2)

Limit of proportionality

Definition: Maximum load where (F propto x) holds

Note: Beyond this, material yields permanently

Force–extension graph

Use: Graphical representation of (F) vs (x)

Key: Area under line = energy stored

Spring constant ((k))

Definition: (k = frac{F}{x}), stiffness measure

Units: N/m

📖 Glossary of Terms

Elastic potential energy

Energy stored in a material when it is deformed elastically, equal to the work done to deform it.

Translation
Russian: Энергия, накопленная при упругой деформации, равная работе, проделанной для деформации.
Kazakh: Серпімді деформация кезінде жинақталған энергия, оны деформациялау үшін жасалған жұмысқа тең.
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Limit of proportionality

The maximum force for which the extension of the material remains directly proportional to the force applied.

Translation
Russian: Максимальная сила, при которой удлинение остается прямо пропорциональным приложенной силе.
Kazakh: Қолданылатын күшке ұзарту тура пропорционалды болатын ең үлкен күш.
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🔬 Theory: Energy of Deformation

Area under Force–Extension Graph

For an elastic material within its limit of proportionality, the graph of force (F) versus extension (x) is a straight line from the origin to ((x, F)). The elastic potential energy (E_P) stored is the area under this line:

(E_P = text{area} = tfrac{1}{2} times F times x)

Using Hooke’s Law

Because (F = k x) in the linear region, substitute to get:

(E_P = tfrac{1}{2} k x^2)

Translation
Russian: В линейной области (F = kx), поэтому энергия (E_P = tfrac{1}{2} k x^2).
Kazakh: Сызықтық аймақта (F = kx), сондықтан энергия (E_P = tfrac{1}{2} k x^2).
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Theory Questions

Easy: How do you calculate (E_P) from the area under the force–extension line?
Answer
Compute (tfrac{1}{2} times F times x), the triangular area under the line.
Medium: Show algebraically that (E_P = tfrac12 k x^2) using (F = kx).
Answer
Substitute (F) in (tfrac12 F x): (tfrac12 (kx)x = tfrac12 k x^2).
Medium: A spring extends by 0.10 m under 15 N. Calculate energy stored.
Answer
(E_P = tfrac12 F x = tfrac12 (15)(0.10) = 0.75text{ J}.)
Hard (Critical Thinking): Discuss why the triangular area method fails beyond the limit of proportionality.
Answer
Beyond the limit, the graph is no longer linear: area is not a simple triangle, and energy cannot be found by (tfrac12 Fx). One must integrate the actual curve or account for plastic deformation.
💪 Memorization Exercises

Match the Formula

  1. Energy from area under graph: ______
  2. Energy using Hooke’s law: ______
  3. Spring constant: ______
  4. Limit beyond which Hooke’s law fails: ______

Answer
1. (tfrac12 F x)
2. (tfrac12 k x^2)
3. (k = tfrac{F}{x})
4. Limit of proportionality
🎥 Video Lesson
📐 Worked Examples

Example 1: Area Method

A wire is extended by 0.08 m under 10 N. Find (E_P).

Force-extension graph

Solution
Answer
Area = (tfrac12 times 10 times 0.08 = 0.40text{ J}.)

Example 2: Using (k)

A spring has (k=200) N/m and is stretched 0.05 m. Calculate (E_P).

Spring calculation

Solution
Answer
(E_P = tfrac12 k x^2 = tfrac12 (200)(0.05)^2 = 0.25text{ J}.)
🧪 Interactive Investigation

Use the PhET “Hooke’s Law” simulation to measure energy stored:


Investigation Answers
Record force and extension, plot graph, calculate area or use (tfrac12 k x^2). Compare results.
👥 Collaborative Group Activity

In pairs, design and carry out an experiment to verify (E_P = tfrac12 k x^2). Share results and discuss deviations.

📝 Individual Assessment

Solve these structured questions:

  1. Derive (E_P = tfrac12 k x^2) by integrating (F(x)) from 0 to (x).
  2. A spring with constant 150 N/m is compressed 0.04 m. Find energy stored.
  3. Explain why area under a non-linear graph requires calculus beyond proportionality limit.
  4. Compare energy stored in two springs of different (k) but same extension.
  5. Design a real-world application where elastic potential energy is crucial; calculate required parameters.
🤔 Lesson Reflection
  • Which method (area vs formula) did you find clearer for calculating (E_P)?
  • What challenges arise when the graph is non-linear?
  • How will you ensure accurate measurements of (k) and (x) in experiments?
  • Describe an everyday device that stores elastic potential energy.