Elastic Potential Energy and Energy Transfers

In this lesson you will learn how to calculate elastic potential energy stored in a stretched or compressed spring, understand Hooke's law, and describe energy transfers involving elastic objects. This is part of AQA GCSE Combined Science Trilogy (8464), Section 6.1.

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Hooke's Law

When a spring is stretched or compressed by a force, it extends or compresses. Hooke's law states:

$F = k \, e$F=ke

Where:

• $F$F = force applied in newtons (N)
• $k$k = spring constant in newtons per metre (N/m)
• $e$e = extension (or compression) in metres (m)

This relationship is linear up to the limit of proportionality. Beyond this point, the spring no longer obeys Hooke's law.

Exam Tip: The limit of proportionality is the point where the force–extension graph stops being a straight line. Beyond this point, the spring may also pass its elastic limit — the point after which it will not return to its original length.

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Elastic Potential Energy

When a spring is stretched or compressed within its limit of proportionality, it stores elastic potential energy.

The Equation

$E_e = \frac{1}{2} k \, e^2$Ee​=21​ke2

Where:

• $E_e$Ee​ = elastic potential energy in joules (J)
• $k$k = spring constant in N/m
• $e$e = extension in metres (m)

Important: This equation is only valid when the spring has not exceeded the limit of proportionality.

Worked Example 1

A spring with a spring constant of 80 N/m is extended by 0.15 m. Calculate the elastic potential energy stored.

$E_e = \frac{1}{2} \times 80 \times 0.15^2 = \frac{1}{2} \times 80 \times 0.0225 = 0.9 \text{ J}$Ee​=21​×80×0.152=21​×80×0.0225=0.9 J

Worked Example 2

A spring stores 4.5 J of energy when extended by 0.3 m. What is the spring constant?

$k = \frac{2 E_e}{e^2} = \frac{2 \times 4.5}{0.3^2} = \frac{9}{0.09} = 100 \text{ N/m}$k=e22Ee​​=0.322×4.5​=0.099​=100 N/m

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Force–Extension Graphs

A force–extension graph for a spring that obeys Hooke's law is a straight line through the origin.

• The gradient of the linear section equals the spring constant $k$k.
• The area under the graph up to a given extension equals the elastic potential energy stored.

graph TD
    subgraph "Force–Extension Graph Interpretation"
        A["Straight line from origin"] --> B["Gradient = spring constant k"]
        A --> C["Area under line = elastic PE stored"]
        D["Line curves"] --> E["Limit of proportionality exceeded"]
    end

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Energy Transfers Involving Springs

When a compressed spring is released, elastic potential energy is transferred to kinetic energy of the object attached.

Worked Example 3

A toy car is attached to a compressed spring ($k = 200$k=200 N/m, compression = 0.05 m). When released, what speed does the 0.1 kg car reach? (Ignore friction.)

Step 1: Calculate elastic PE stored. $E_e = \frac{1}{2} \times 200 \times 0.05^2 = 0.25 \text{ J}$Ee​=21​×200×0.052=0.25 J

Step 2: Equate to kinetic energy. $E_k = E_e = 0.25 \text{ J}$Ek​=Ee​=0.25 J

Step 3: Find speed. $v = \sqrt{\frac{2 E_k}{m}} = \sqrt{\frac{2 \times 0.25}{0.1}} = \sqrt{5} \approx 2.24 \text{ m/s}$v=m2Ek​​​=0.12×0.25​​=5​≈2.24 m/s

graph LR
    A["Elastic potential store\n(compressed spring)"] -->|"Mechanically"| B["Kinetic store\n(toy car)"]
    B -->|"By heating / friction"| C["Internal store\n(surroundings)"]
    style C fill:#ffcccc,stroke:#cc0000

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Energy Dissipation

In any real energy transfer, some energy is dissipated — transferred to the internal (thermal) energy store of the surroundings. Dissipated energy is sometimes called "wasted" energy because it is spread out and cannot easily be recovered.

Reducing Dissipation

Method	How It Helps
Lubrication	Reduces friction between moving parts
Streamlining	Reduces air resistance on moving objects
Insulation	Reduces energy transfer by heating to the surroundings

Exam Tip: AQA may ask you to suggest ways to reduce unwanted energy transfers. Always link your answer to a specific mechanism: e.g., "Lubricating the moving parts reduces friction, so less energy is dissipated to the internal (thermal) store of the surroundings."

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Common Mistakes

Mistake	Correction
Using $E_e = \frac{1}{2}ke$Ee​=21​ke (forgetting to square $e$e)	The equation is $E_e = \frac{1}{2}ke^2$Ee​=21​ke2
Applying the equation beyond the limit of proportionality	The equation only works in the linear region
Confusing spring constant with extension	$k$k is a property of the spring; $e$e is how much it stretches
Forgetting to convert cm to m	Extension must be in metres for the equation to give joules

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Summary

• Hooke's law: $F = ke$F=ke, valid up to the limit of proportionality.
• Elastic potential energy: $E_e = \frac{1}{2}ke^2$Ee​=21​ke2.
• The gradient of a force–extension graph gives the spring constant; the area gives the energy stored.
• When a spring is released, elastic PE transfers to kinetic energy (and some is dissipated).
• Dissipation can be reduced by lubrication, streamlining, or insulation.
• Assessed in AQA GCSE Combined Science Trilogy (8464), Section 6.1.

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Extended Worked Examples

Worked Example 4 — Finding Extension

A spring stores 1.8 J of elastic potential energy. Its spring constant is 144 N/m. What is the extension?

$e = \sqrt{\frac{2E_e}{k}} = \sqrt{\frac{2 \times 1.8}{144}} = \sqrt{0.025} = 0.158 \text{ m} \approx 15.8 \text{ cm}$e=k2Ee​​​=1442×1.8​​=0.025​=0.158 m≈15.8 cm

Worked Example 5 — Toy Launcher

A toy launcher has a spring ($k = 320$k=320 N/m) compressed by 0.04 m. The ball it fires has mass 25 g. Ignoring friction, how fast is the ball launched, and what maximum height could it reach vertically? ($g = 10$g=10 N/kg.)

Elastic PE:

$E_e = \tfrac{1}{2} \times 320 \times 0.04^2 = 0.256 \text{ J}$Ee​=21​×320×0.042=0.256 J

All converted to kinetic energy:

$v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 0.256}{0.025}} = \sqrt{20.48} \approx 4.53 \text{ m/s}$v=m2Ek​​​=0.0252×0.256​​=20.48​≈4.53 m/s

Maximum height (converting $E_k \to E_p$Ek​→Ep​):

$h = \frac{E_k}{mg} = \frac{0.256}{0.025 \times 10} = 1.02 \text{ m}$h=mgEk​​=0.025×100.256​=1.02 m

Worked Example 6 — Combining Stretched Springs

Two identical springs (each $k = 60$k=60 N/m) are connected in parallel and stretched together by 0.2 m. How much total elastic PE is stored?

Each spring stores:

$E_e = \tfrac{1}{2} \times 60 \times 0.2^2 = 1.2 \text{ J}$Ee​=21​×60×0.22=1.2 J

Total: 2.4 J. (Parallel arrangement doubles the effective spring constant; same extension means double the energy.)

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Comparison Table — Elastic vs Gravitational vs Kinetic