Physics I: Mechanics › Work & energy › full formula sheet

Us = ½kx²

Say it: “spring potential energy equals one-half k x squared”

Spring potential energy

The energy coiled in a spring — why doubling the stretch quadruples the stored energy, and where the ½ comes from.

k is the spring constant in N/m (stiffness), x the stretch or compression from the relaxed length in meters. Us is in joules and never negative.

Before this lesson: Work (variable force)

Where it comes from

Stretch a spring slowly and your hand does work — but the spring doesn't move anywhere or heat up. Where did the work go? It's coiled inside the spring, ready to fling back. That stored work is spring potential energy.

Before reading on: stretching a spring 10 cm stores 1 J. Stretch it 20 cm instead — is the stored energy 2 J, 3 J, or 4 J? The force isn't constant, so think carefully.

4 J. Hooke's law says the force grows with stretch (F = kx), so the second 10 cm fights twice the average force of the first 10 cm. Double the distance and double the average force: 2 × 2 = 4× the energy. Another square law.

Us = ½kx²the work of stretching, banked in the coil — quadratic in x, like kinetic energy is in vSay it: “spring potential equals one-half k x squared”

Derivation

The stretching force varies (F = kx), so this is variable-force work — the area of a triangle:

W
=
∫₀x kx dx
Step 1 — set up. Stretch from 0 to x against the spring force kx. (See work by a variable force.)
=
½kx²
Step 2 — integrate. ∫ kx dx = kx²/2. Or read the triangle: base x, height kx, area ½·x·kx.
Us
=
½kx²
Step 3 — name it. The work you did stretching is stored as spring potential energy. The spring can return exactly this much (ideally). ∎

Why x is squared: stretching twice as far means twice the distance against twice the average force. And the ½ is the triangle's — the force ramps 0 → kx, averaging kx/2. Same ½, same square, same story as kinetic energy.

How to use it

The procedure, every time:

  1. Measure x from the relaxed length. Not from the floor, not from your hand — from where the spring is natural. Stretch and compression both count (x² erases the sign).
  2. Check k's units. N/m. A k in N/cm needs ×100 first.
  3. Compute ½kx². Square x before multiplying by k; halve after.
  4. For a change of stretch, use the difference. x₁ → x₂: ΔUs = ½k(x₂² − x₁²) — not ½k(x₂ − x₁)².
  5. Work by the spring is −ΔUs. The spring pushing outward does positive work as it relaxes (Us drops).
Common mistake: measuring x from the wrong origin — e.g. a spring hanging with a mass already stretches it x₀; further stretching x uses the total stretch in ½kx², or the difference formula for the change.

Worked examples

Four problems, easiest first. Mind the square and the half.

Example 1 — basic: k = 150 N/m, x = 0.2 m

  1. Square x. x² = 0.04.
  2. Assemble. Us = ½ × 150 × 0.04 = 3 J.
  3. Sanity check. Positive, joules. A firm spring stretched 20 cm holding a few joules — plausible. ✓
Common mistake: U = kx² = 6 J — dropping the ½. The force ramps 0 → kx; only the triangle's area counts.
Your turn — k = 100 N/m, x = 0.3 m. Us = ?

Answer: 4.5 J. ½ × 100 × 0.09 = 4.5 J.

Example 2 — the square bites: k = 200 N/m, x = 0.1 m vs 0.2 m

  1. At 0.1 m. Us = ½ × 200 × 0.01 = 1 J.
  2. At 0.2 m. Us = ½ × 200 × 0.04 = 4 J.
  3. Ratio: 4×. Twice the stretch, four times the energy — the second 10 cm cost 3 J, the first only 1 J.
Common mistake: guessing 2 J (“twice as far, twice the energy”). Squares punish linear intuition — the same trap as kinetic energy's v².
Your turn — k = 80 N/m, x = 0.5 m. Us = ?

Answer: 10 J. ½ × 80 × 0.25 = 10 J.

Example 3 — solve for stretch: Us = 8 J, k = 400 N/m

  1. Rearrange. x² = 2Us/k.
  2. Compute. x² = 16/400 = 0.04, so x = 0.2 m.
  3. Check. ½ × 400 × 0.04 = 8 J. ✓
Common mistake: x = 0.04 m — forgetting the square root. The formula stores x²; the question asks for x.
Your turn — Us = 18 J, k = 100 N/m. x = ?

Answer: 0.6 m. x² = 36/100 = 0.36; x = 0.6 m.

Example 4 — work done BY the spring: k = 60 N/m relaxing 0.3 m → 0

  1. Potential drop. ΔUs = 0 − ½ × 60 × 0.09 = −2.7 J.
  2. Spring's work. Wby = −ΔUs = +2.7 J — the spring does positive work as it relaxes, flinging the mass.
  3. Your work to stretch it was the reverse: you did +2.7 J (Us rose), the spring did −2.7 J on you.
Common mistake: sign confusion — saying the relaxing spring does −2.7 J. Relaxing releases energy: the spring's work is positive, the potential change negative. W = −ΔU, always.
Your turn — k = 120 N/m spring relaxes from 0.2 m to 0. Work by the spring?

Answer: +2.4 J. ΔUs = −½×120×0.04 = −2.4 J; W = +2.4 J.

Memorization tips

  • Say it aloud: “spring potential equals one-half k x squared” — and hear the twin of “one-half m v squared.”
  • The KE twin: ½kx² mirrors ½mv² — stiffness plays mass's role, stretch plays speed's. Learn one, get the other free.
  • x is from relaxed. Every spring problem's first question: “where is natural length?” Answer it before computing.
  • The doubling rule: 2× stretch → 4× energy. Same square-law reflex as kinetic energy.
  • Difference, not difference-squared: x₁→x₂ uses ½k(x₂²−x₁²). Squaring the difference is the trap.
  • W = −ΔU: relaxing spring does positive work; stretching (by you) banks positive potential. The minus is the exchange rate.

Final challenge

Five mixed questions — the square, the half, and sign traps. Score 5/5 and spring potential is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Why is it ½kx² and not kx²?

Because the force ramps 0 → kx as you stretch, averaging kx/2 — the triangle's area, not the rectangle's. ∫₀x kx dx = kx²/2. The ½ is the average of the ramp.

Does compression store energy too?

Yes — x² erases the sign, so compressing 0.1 m stores exactly what stretching 0.1 m stores. The spring doesn't care which way you deform it.

What is k, physically?

The spring constant: newtons of force per meter of stretch (N/m). Big k = stiff spring = more energy per meter. It's measured by hanging a known weight and reading the stretch: k = F/x.

Where is x = 0?

At the spring's natural (relaxed) length — not the floor, not your hand. If a mass already hangs on the spring, that pre-stretch counts in x.

How does spring potential become motion?

Release the spring and it does positive work W = −ΔUs on the mass, converting Us into kinetic energy. With no losses, ½kx² = ½mv² at the moment of release.

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