When a spring or elastic band is stretched within its elastic limit, it stores elastic potential energy. The equation is: Eₑ = ½ke², where Eₑ is energy in joules, k is the spring constant in N/m, and e is the extension in metres. A stiffer spring (higher k) stores more energy for the same extension.
What is elastic potential energy?
Elastic potential energy (also called elastic strain energy) is energy stored in an object that has been stretched, compressed, or deformed — as long as the deformation is within the object's elastic limit. When released, this energy converts back to kinetic energy, returning the object to its original shape.
Examples:
- A stretched spring in a wind-up toy
- A compressed spring in a door hinge
- A drawn bow
- A catapult or slingshot
- A trampoline at the moment of maximum depression
The key condition is the elastic limit (or limit of proportionality): if a spring is stretched beyond this point, it deforms permanently and does not store or release energy elastically — it becomes permanently deformed (plastic deformation).
What is the elastic potential energy equation?
$$E_e = \frac{1}{2}ke^2$$
Where:
- Eₑ = elastic potential energy stored (J)
- k = spring constant (N/m) — a measure of the spring's stiffness; a larger k means a stiffer spring
- e = extension or compression (m) — the change from the natural (unstretched) length
This equation is given on the AQA GCSE physics equation sheet, so you do not need to memorise it, but you must be able to use and rearrange it.
How does elastic potential energy link to Hooke's law?
Hooke's law states that the force needed to extend (or compress) a spring is proportional to the extension, provided the elastic limit is not exceeded:
$$F = ke$$
Where F is the force in newtons (N), k is the spring constant (N/m), and e is the extension (m).
The elastic potential energy stored equals the area under the force-extension graph (a straight line through the origin for a Hookean spring). This area is a triangle with base e and height F = ke:
$$E_e = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times e \times ke = \frac{1}{2}ke^2$$
This confirms that Eₑ = ½ke² and shows that the energy stored in a spring is equal to the work done in stretching it (work = average force × distance = ½F × e = ½ke²).
How do you calculate elastic potential energy — worked examples?
Worked example 1 — finding energy stored:
A spring has spring constant k = 200 N/m. It is extended by 0.04 m. Calculate the elastic potential energy stored.
- Eₑ = ½ × 200 × (0.04)²
- Eₑ = ½ × 200 × 0.0016
- Eₑ = 100 × 0.0016 = 0.16 J
Worked example 2 — finding extension:
A spring with k = 500 N/m stores 0.25 J of elastic potential energy. What is the extension?
- 0.25 = ½ × 500 × e²
- 0.25 = 250 × e²
- e² = 0.25 / 250 = 0.001
- e = √0.001 = 0.0316 m (≈ 3.2 cm)
Worked example 3 — finding spring constant:
An elastic band stores 0.40 J of energy when stretched by 0.10 m. What is its spring constant?
- 0.40 = ½ × k × (0.10)²
- 0.40 = ½ × k × 0.01
- 0.40 = 0.005k
- k = 0.40 / 0.005 = 80 N/m
Summary of rearrangements
| Given | Find | Rearranged formula |
|---|---|---|
| k and e | Eₑ | Eₑ = ½ke² |
| Eₑ and k | e | e = √(2Eₑ / k) |
| Eₑ and e | k | k = 2Eₑ / e² |
How does stiffer spring constant affect energy stored?
For the same extension, a stiffer spring (larger k) stores more energy — because a stiffer spring requires a larger force to produce the same extension, and force × distance = more work done on the spring.
For the same force applied, a stiffer spring extends less, and the energy stored is actually less (not more) — because the small extension more than outweighs the larger k.
Compare:
- Spring A: k = 100 N/m, stretched by 0.1 m → Eₑ = ½ × 100 × 0.01 = 0.5 J
- Spring B: k = 400 N/m, stretched by 0.1 m → Eₑ = ½ × 400 × 0.01 = 2.0 J
Four times stiffer, same extension: four times more energy stored. The e² dependence means that doubling the extension (not the k) quadruples the energy stored for the same spring.
How does elastic potential energy convert to kinetic energy?
When a stretched spring is released (assuming no friction or air resistance), the elastic potential energy converts entirely to kinetic energy of the released object. By conservation of energy:
Eₑ = Eₖ (at the point of release)
½ke² = ½mv²
This allows you to find the speed v of a mass m launched by a spring:
$$v = e\sqrt{\frac{k}{m}}$$
Example: A spring with k = 800 N/m is compressed by 0.05 m. It launches a ball of mass 0.02 kg. What is the ball's speed at the moment of release?
- Eₑ = ½ × 800 × (0.05)² = ½ × 800 × 0.0025 = 1.0 J
- Eₖ = ½mv² → 1.0 = ½ × 0.02 × v²
- v² = 1.0 / 0.01 = 100
- v = 10 m/s
Frequently asked questions
What is the elastic limit and why does it matter for the equation?
The elastic limit is the maximum extension beyond which a material does not return to its original length — it has been permanently deformed. The equation Eₑ = ½ke² is only valid within the elastic limit, where Hooke's law applies and the force-extension graph is a straight line. Beyond the elastic limit, k is no longer constant, the graph curves, and the equation cannot be used. In exam questions, you can assume the elastic limit has not been exceeded unless told otherwise.
Why does the energy equation contain e² rather than just e?
Because Hooke's law relates force to extension (F = ke), but work done is force × distance. As you stretch a spring, the force required increases proportionally. The work done is the area under the force-extension graph — a triangle with area ½ × base × height = ½ × e × ke = ½ke². The squared term comes from the fact that both the force and the distance increase together as you stretch the spring further.
How is elastic potential energy different from gravitational potential energy?
Both are forms of potential energy — energy stored due to position or configuration. Gravitational potential energy (GPE = mgh) is stored due to an object's height in a gravitational field and depends on mass, gravitational field strength, and height. Elastic potential energy is stored due to deformation of a material (stretching or compressing) and depends on the spring constant and the square of the extension. GPE converts to KE as an object falls; EPE converts to KE when a spring is released.
Can elastic potential energy be stored in compression as well as extension?
Yes. The equation Eₑ = ½ke² applies to both extension and compression, provided the material obeys Hooke's law (is within its elastic limit). The variable e represents the magnitude of the deformation from natural length — whether the spring is stretched or compressed. For example, the spring in a shock absorber stores energy as it is compressed when a vehicle hits a bump, then releases it as it returns to its natural length.
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