Physics I: Mechanics › Forces › full formula sheet

F = −kxSay it: “the spring force equals negative k x — it pulls back toward equilibrium”

Spring force

Hooke's law as a force — the restoring pull behind every bounce and vibration.

Notation on this page: F is the spring's force in N, k is the spring constant in N/m, x is the displacement from equilibrium in m (signed!).

Before this lesson: Newton's second law

Where it comes from

In 1678 Robert Hooke published the observation ut tensio, sic vis — “as the extension, so the force.” Hang weights on a spring and each added newton stretches it by the same extra amount. Double the pull, double the stretch. The spring's resistance grows in exact proportion to how far you deform it.

Before reading on: a spring stretches 0.1 m under a 10 N pull. You pull with 30 N instead. How far does it stretch — and which way does the spring pull back?

The proportionality answers the first half: 30 N is three times the pull, so the stretch is three times 0.1 m = 0.3 m. The second half is the famous minus sign: the spring never pushes you away — it always fights the deformation, pulling back toward its natural length. In symbols, with x measured from equilibrium:

F = −kxstretch it +x and it pulls −x; compress it −x and it pushes +x — always homewardSay it: “the spring force equals negative k x”

That homeward habit is why springs oscillate: release a stretched spring and it accelerates back, overshoots equilibrium, gets yanked back again — the restoring force is the engine of every vibration you will meet.

Derivation

Hooke's law is an empirical law — it comes from plotting measurements, not from Newton's axioms. Here is the experimental reasoning that produces it:

F ∝ x
Step 1 — the experiment. Plot measured force against stretch: the points fall on a straight line through the origin. Force is proportional to displacement.
|F|
=
k · |x|
Step 2 — name the slope. The line's slope is the spring constant k (N/m): force per metre of stretch. Stiffer spring, steeper line.
F
=
−kx
Step 3 — restore the direction. The force always opposes the displacement, so the signed form needs the minus: F and x have opposite signs, always. ∎

Two caveats ride along. First, the law holds only within the elastic limit — stretch too far and the spring deforms permanently, and the straight line bends. Second, x is measured from the natural length (equilibrium), not from wherever the spring happens to be.

How to use it

The procedure, every time:

  1. Set x = 0 at the natural length. x is the displacement from equilibrium — not from the floor, not from your hand.
  2. Sign x carefully. Pick +x as the stretch direction: stretched → x > 0, compressed → x < 0.
  3. Compute F = −kx (signed). Or compute the magnitude kx and state the direction yourself (“toward equilibrium”).
  4. Check the sign. F and x must have opposite signs. If they match, you dropped the minus.

Finding k

Given any force–stretch pair, k = |F|/|x|. A 12 N pull stretching a spring 0.04 m gives k = 12/0.04 = 300 N/m. Once you have k, the spring is fully characterized — every later question is arithmetic.

Common mistake: dropping the minus sign and reporting that a stretched spring pushes outward. The minus is the physics — without it, springs would explode instead of oscillate.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: k = 200 N/m, stretched x = 0.15 m

  1. Identify. k = 200 N/m, x = +0.15 m (stretched).
  2. Multiply with the sign. F = −kx = −200 × 0.15 = −30 N.
  3. Interpret. −30 N means 30 N in the −x direction — pulling back toward equilibrium. The spring resists the stretch.
Common mistake: answering +30 N (“the spring pushes outward”). Stretched springs pull in — the minus sign is doing real work.
Your turn — k = 150 N/m, stretched 0.2 m. Spring force?

Answer: −30 N (30 N toward equilibrium). F = −150 × 0.2 = −30 N.

Before reading on: a 12 N force stretches a spring by 0.04 m. Is this a stiff spring or a soft one — and what single number captures that?

Example 2 — finding k: 12 N stretches it 0.04 m

  1. Rearrange. k = |F|/|x|.
  2. Divide. k = 12/0.04 = 300 N/m — fairly stiff (a car suspension spring is ~30,000 N/m; a pen spring ~100 N/m).
  3. Use it. Now any stretch is known: 0.1 m of this spring pulls 30 N.
Common mistake: reporting k in N or N·m. k is force per stretch: N/m. Units first, always.
Your turn — an 8 N force stretches a spring 0.05 m. Find k.

Answer: 160 N/m. k = 8/0.05 = 160 N/m.

Example 3 — compression: k = 250 N/m, x = −0.1 m

  1. Sign x. Compressed 0.1 m → x = −0.1 m.
  2. Multiply. F = −kx = −250 × (−0.1) = +25 N.
  3. Interpret. +25 N: the spring pushes outward, back toward equilibrium. Compressed springs push; stretched springs pull — both homeward.
Common mistake: treating compression like a stretch (“F = −25 N”). The two minuses cancel — trust the algebra, then sanity-check the direction.
Your turn — k = 400 N/m, compressed 0.05 m. Spring force?

Answer: +20 N (20 N pushing outward). F = −400 × (−0.05) = +20 N.

Example 4 — hanging mass: 0.5 kg stretches it 0.02 m at rest

  1. Equilibrium means forces balance. Spring pulls up, weight pulls down: kx = mg.
  2. Solve for k. k = mg/x = (0.5 × 9.8)/0.02 = 4.9/0.02 = 245 N/m.
  3. Check: a quarter-kilonewton per metre — a plausible lab spring.
  4. The lesson: hanging a known mass is the standard way to measure k.
Common mistake: forgetting that at equilibrium the spring force equals the weight — then trying to use F = ma with a = 0 and getting 0 = 0. Statics first: ΣF = 0.
Your turn — a 1.2 kg mass stretches a spring 0.06 m at rest. Find k.

Answer: 196 N/m. k = (1.2 × 9.8)/0.06 = 11.76/0.06 = 196 N/m.

Memorization tips

  • Say it aloud: “spring force equals negative k x.” Stress the negative — it is the only hard part.
  • The homeward rule: whatever x does, F does the opposite. Stretched? Pulls in. Compressed? Pushes out. Always homeward.
  • Sign check: F and x must have opposite signs. If your answer shows them matching, the minus got dropped.
  • k in N/m: force per metre of stretch. Bigger k, stiffer spring, steeper F–x line.
  • Measure from equilibrium: x = 0 is the natural length. Every spring problem starts by finding where that is.
  • Oscillation seed: restoring + proportional-to-x = simple harmonic motion. When you meet T = 2π√(m/k), this page is its birthplace.

Final challenge

Five mixed questions — signs, stiffness, and the traps, all in one. Score 5/5 and the spring force 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

What is the spring force formula?

F = −kx: the force exerted by an ideal spring equals the negative of the spring constant times the displacement from equilibrium. The minus sign makes it a restoring force.

What does the minus sign mean in F = −kx?

It means the force always points back toward equilibrium (x = 0): stretch the spring in the +x direction and it pulls in −x; compress it and it pushes in +x.

What is the spring constant k?

A measure of stiffness in N/m: the force per metre of stretch. A bigger k means a stiffer spring. Find it from k = F/x using any measured force–stretch pair.

When does Hooke's law break down?

Beyond the elastic limit: stretch a spring too far and it deforms permanently, so force is no longer proportional to stretch. F = −kx only holds for modest displacements.

Why does a mass on a spring oscillate?

Because the force is restoring (always toward equilibrium) and grows with displacement: release a stretched spring and it accelerates back, overshoots, and gets pulled back again — simple harmonic motion.

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