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v0.1.1065
NotesPhysics HLTopic 1.3Elastic potential energy
Back to Physics HL Topics
1.3.42 min read

Elastic potential energy

IB Physics • Unit 1

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Contents

  • What elastic potential energy is
  • Working out the energy stored
  • Exam-style question
The big idea: Elastic potential energy is the energy stored in a spring (or any springy material) when you stretch or squash it.

It depends on the spring's stiffness and on the stretch squared: E_H = ½kΔx².

Let go, and that stored energy is released again. Unit: the joule (J).
Two words to know first: Spring constant (k): how stiff a spring is — the force needed per metre of stretch (unit N m⁻¹). A bigger k means a harder spring to pull.

Extension (Δx): how far the spring is stretched or squashed from its natural length (in metres).
Spot it — the stretch is SQUARED: Because of the Δx², the stretch matters a lot: double the stretch ⇒ four times the stored energy.

A spring pulled twice as far stores 4× the energy — and shoots back with 4× the energy to give away.

The elastic-energy formula is given in the data booklet. The spring constant k comes from Hooke's law (F = kΔx) — it links the force to the stretch — but for the stored energy you use the ½kΔx² formula directly.

Elastic potential energy — given in the data booklet (topic A.3). It is sometimes written E_p = ½kx² — same formula, same meaning.
elastic potential energy — energy stored in the stretched/squashed spring (J, joules)
spring constant — how stiff the spring is (N m⁻¹: newtons per metre)
extension or compression — how far the spring is stretched or squashed from its natural length (m)
Hooke's law (also given) — defines the spring constant k. The minus sign means the force pulls back toward the natural length. Use F = kΔx to find k from a force and a stretch.
the spring's restoring force — it pulls/pushes back toward the natural length (N)
spring constant — the stiffness (N m⁻¹)
extension or compression from the natural length (m)
Don't forget to square the stretch — and use metres: The most common slip is forgetting the Δx². Square the extension first, then multiply by k and halve.

And convert any centimetres to metres before squaring (e.g. 4 cm = 0.04 m).

Worked example — energy stored in a stretched spring

A spring has a spring constant of 250 N m⁻¹. It is stretched by 0.080 m. Find the elastic potential energy stored.

Solution

  1. Start with the given formula:
  2. Put in the numbers (k = 250, Δx = 0.080) — square the extension first:
  3. Work it out (0.080² = 0.0064) — keep the unit:

Final answer

EH = 0.80 J.

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How this is tested: Elastic energy is the link between forces and energy conservation.

- Paper 1A: a quick EH calculation, or finding the average power as a spring releases its stored energy (power = energy ÷ time). - Paper 2: the classic spring-coupled collision — a moving cart hits a spring on a second cart, and at the instant they move together you find the elastic energy stored using energy conservation.

Classic trap: when the carts momentarily move together, the spring stores the kinetic energy that has 'gone missing'. Find the total kinetic energy before, the kinetic energy of the combined motion, and the difference is the elastic energy stored.
Energy conservation with a spring: Energy is never lost — it just changes form. When a spring is squashed during a collision:

E_H stored = kinetic energy before − kinetic energy of the combined motion.

Later the spring pushes the carts apart and gives that energy back.
Find momentum first, then energies: At the instant the carts move together they share one common speed. Get it from conservation of momentum (total momentum before = total momentum after), then compare kinetic energies.

IB-style question — (a) common speed of the carts

A 2.0 kg cart moves at 6.0 m s⁻¹ toward a stationary 4.0 kg cart that carries a spring bumper. At the instant the spring is most compressed, the two carts move together with one common speed. Find that common speed.

Solution

  1. Momentum is conserved. Total momentum before = moving cart only:
  2. After, the carts move together as one mass (2.0 + 4.0 = 6.0 kg) at speed v:
  3. Set them equal and solve for v — keep the unit:

Final answer

common speed = 2.0 m s⁻¹.

IB-style question — (b) elastic energy stored in the spring

Using the common speed from part (a), find the elastic energy stored in the spring at the instant the carts move together.

Solution

  1. Kinetic energy before (only the moving cart) — use Ek = ½mv²:
  2. Kinetic energy of the combined motion (6.0 kg at 2.0 m s⁻¹):
  3. The 'missing' kinetic energy is now stored in the spring:

Final answer

elastic energy stored = 24 J — the kinetic energy lost from the motion is stored in the squashed spring (it isn't destroyed).

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what the area under a force–extension graph for a spring represents. [1 mark]

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1.1.1Velocity and displacement
1.1.2Acceleration
1.1.3Displacement from a velocity–time graph
1.1.4The suvat equations
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