Elastic Potential Energy

This lesson covers the elastic potential energy (EPE) store and the equation $EPE = \frac{1}{2}ke^2$EPE=21​ke2, as required by the Edexcel GCSE Combined Science specification (1SC0). You will also learn about Hooke's law and how springs store and release energy.

---

What Is Elastic Potential Energy?

Elastic potential energy (EPE) is the energy stored in an object when it is stretched or compressed. When the force is removed and the object returns to its original shape, the stored energy is released.

Examples of objects that store EPE:

• A stretched spring
• A compressed spring
• A stretched rubber band
• A bent diving board
• A drawn bow (archery)

---

Hooke's Law

Before studying EPE, you need to understand Hooke's law, which describes how springs behave when a force is applied.

$F = ke$F=ke

where:

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

Quantity	Unit	Symbol
Force	newton	N
Spring constant	newtons per metre	N/m
Extension	metre	m

Key Points

• The spring constant (k) tells you how stiff a spring is. A large k means a stiff spring that is hard to stretch.
• Extension (e) is the difference between the stretched length and the natural (original) length: $e = \text{stretched length} - \text{natural length}$e=stretched length−natural length
• Hooke's law applies only up to the limit of proportionality — beyond this, the extension is no longer proportional to the force.

Exam Tip: In the exam you may be given a force–extension graph. A straight line through the origin shows Hooke's law is obeyed. The gradient of this straight section equals the spring constant k.

---

The EPE Equation

The energy stored in a stretched or compressed spring (provided it has not exceeded its limit of proportionality) is:

$EPE = \frac{1}{2}ke^2$EPE=21​ke2

where:

• EPE = elastic potential energy in joules (J)
• k = spring constant in N/m
• e = extension (or compression) in metres (m)

Exam Tip: Notice the similarity to the kinetic energy equation — both have ½ and a squared term. In KE it is speed squared; in EPE it is extension squared. Doubling the extension quadruples the EPE.

---

Worked Examples

Worked Example 1

A spring has a spring constant of 40 N/m and is stretched by 0.15 m. Calculate the EPE stored.

1. Write the equation: $EPE = \frac{1}{2}ke^2$EPE=21​ke2
2. Substitute: $EPE = \frac{1}{2} \times 40 \times 0.15^2$EPE=21​×40×0.152
3. $0.15^2 = 0.0225$0.152=0.0225
4. $EPE = \frac{1}{2} \times 40 \times 0.0225 = 0.45 \text{ J}$EPE=21​×40×0.0225=0.45 J
5. Answer: EPE = 0.45 J

Worked Example 2

A spring with k = 200 N/m stores 1.6 J of EPE. What is the extension?

1. Rearrange: $e^2 = \frac{2 \times EPE}{k}$e2=k2×EPE​
2. Substitute: $e^2 = \frac{2 \times 1.6}{200} = 0.016$e2=2002×1.6​=0.016
3. $e = \sqrt{0.016} = 0.126 \text{ m}$e=0.016​=0.126 m
4. Answer: e ≈ 0.13 m (or 13 cm)

Worked Example 3

A catapult has a spring constant of 80 N/m. It is pulled back by 0.2 m and releases a 0.01 kg stone. Assuming all EPE is transferred to KE, find the speed of the stone.

1. Calculate EPE: $EPE = \frac{1}{2} \times 80 \times 0.2^2 = \frac{1}{2} \times 80 \times 0.04 = 1.6 \text{ J}$EPE=21​×80×0.22=21​×80×0.04=1.6 J
2. EPE = KE: $1.6 = \frac{1}{2} \times 0.01 \times v^2$1.6=21​×0.01×v2
3. $v^2 = \frac{2 \times 1.6}{0.01} = 320$v2=0.012×1.6​=320
4. $v = \sqrt{320} = 17.9 \text{ m/s}$v=320​=17.9 m/s
5. Answer: v ≈ 17.9 m/s

---

Force–Extension Graphs

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

Feature	Meaning
Straight line through origin	Hooke's law is obeyed (F ∝ e)
Gradient of straight section	Equals the spring constant k
Curve at higher forces	Limit of proportionality exceeded
Area under the graph	Equals the EPE stored

flowchart LR
    A["Force applied\n(N)"] -->|"Hooke’s law region\n(straight line)"| B["Extension\n(m)"]
    B -->|"Beyond limit"| C["Non-linear region\n(curve)"]

The Area Under the Graph

The area under a force–extension graph equals the work done on the spring, which equals the EPE stored.

For the straight-line (Hooke's law) region, this area is a triangle:

$\text{Area} = \frac{1}{2} \times F \times e = \frac{1}{2} \times ke \times e = \frac{1}{2}ke^2$Area=21​×F×e=21​×ke×e=21​ke2

This is how the EPE equation is derived.

Exam Tip: If the graph is not a straight line (the spring has gone beyond the limit of proportionality), you cannot use the equation $EPE = \frac{1}{2}ke^2$EPE=21​ke2. Instead, you would need to count squares or estimate the area under the curve.

---

Elastic and Inelastic Deformation

Type	Description	Example
Elastic deformation	Object returns to its original shape when the force is removed	A spring stretched within its limit
Inelastic deformation	Object does not return to its original shape	A spring stretched beyond its elastic limit

• The elastic limit is the maximum force (or extension) at which the object will still return to its original shape.
• Beyond the elastic limit, the deformation is permanent.

---

Summary

• Elastic potential energy is stored in objects that are stretched or compressed.
• Hooke's law: $F = ke$F=ke (force is proportional to extension up to the limit of proportionality).
• The spring constant (k) measures stiffness — a high k means a stiff spring.
• EPE is calculated using $EPE = \frac{1}{2}ke^2$EPE=21​ke2.
• Doubling the extension quadruples the EPE.
• The area under a force–extension graph equals the EPE stored.
• Elastic deformation is reversible; inelastic deformation is permanent.

---

Extended Worked Examples

Worked Example 4 — Trampoline

A trampoline mat behaves like a spring with effective spring constant 6000 N/m. A gymnast pushes the mat down by 0.30 m. Calculate the elastic potential energy stored.