Energy stored in a spring
You will be able to: Relate elastic energy to deformation and the area under a linear spring-force graph.
Why does doubling a spring’s stretch store four times the energy?
Stretch a light spring slowly from zero extension to 0.10 m. The required holding force grows as you pull, so energy builds up faster during the later centimeters than during the first ones.
A useful starting point: Hooke’s law →
Words and symbols before equations
- Deformation x
- Signed stretch or compression from relaxed length, in m.
- Stiffness k
- Spring constant, in N/m.
- Elastic energy U_s
- Recoverable energy stored by ideal spring deformation, in J.
What this picture assumes
Ideal massless spring, k=200 N/m, linear elastic range. U=0 at relaxed length. Signed spring force is −kx; elastic energy is ½kx².
Connect the picture to the physics
For an ideal spring, U_s=½kx², taking zero energy at its relaxed length. Stretching by +x and compressing by −x store the same positive energy, because x is squared.
A slow external pull increases from 0 to kx. Its force–extension graph is triangular, so the work stored is ½(x)(kx). The spring’s own work has the opposite sign: W_s=−ΔU_s.
Use deformation, not total length. An ideal massless spring remains in its linear elastic range. A hanging mass at equilibrium can still stretch a spring; relaxed length and the hanging system’s equilibrium position are different reference ideas.
A worked example, step by step
A spring has k=200 N/m. Find its energy when stretched 0.10 m and 0.20 m. Calculate the extra work needed for the second stage of a slow stretch.
- At 0.10 m: U_s=½(200)(0.10²)=1 J.
- At 0.20 m: U_s=½(200)(0.20²)=4 J.
- Extra external work from 0.10 to 0.20 m is ΔU=4−1=3 J, assuming negligible kinetic-energy change.
- The first 0.10 m required only 1 J; the second requires 3 J because the force is larger.
kx is a force; ½kx² is an energy. They have different units.
Does a compressed spring have negative elastic energy?
Compare with an explanation
No. With the relaxed state assigned zero, ideal elastic energy is ½kx² and is nonnegative.
Predict. Change one thing. Explain.
Keep k=200 N/m. Compare deformations −0.10, 0, +0.10 and +0.20 m. Explain the symmetry of energy and the opposite signs of the spring force.
U_s=1 J; signed spring force=-20 N. Equal stretch and compression magnitudes store equal energy.
Ideal massless spring, k=200 N/m, linear elastic range. U=0 at relaxed length. Signed spring force is −kx; elastic energy is ½kx².
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a work, energy or power relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA spring’s holding force rises from 0 to 12 N while extension increases from 0 to 0.06 m. (a) Find k. (b) Find stored energy using graph area. (c) Predict energy at 0.12 m. (d) Give one experimental check before trusting the model.
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Compare with the answer and four-point rubric
- 1 point: k=12/0.06=200 N/m.
- 1 point: Triangle area ½×12×0.06=0.36 J.
- 1 point: 1.44 J.
- 1 point: Repeat measurements and check that holding force remains linear in extension and the spring returns to relaxed length.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What length is squared?
Deformation from relaxed length.
RECALL 2Why a factor of one half?
Force grows linearly from zero; its graph area is triangular.
RECALL 3What happens to U when x changes sign?
It stays the same for equal deformation magnitudes.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Energy stored in a spring
- U_s=½kx² with zero at relaxed length.
- W_s=−ΔU_s; slow external work = +ΔU_s if no other energy changes.
Remember: kx is a force; ½kx² is an energy. They have different units.
Conditions: Ideal massless spring, k=200 N/m, linear elastic range. U=0 at relaxed length. Signed spring force is −kx; elastic energy is ½kx².
Refresh Kid · Unit 3 · Objectives 3.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.3, objectives 3.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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