Physics I: Mechanics › Forces › full formula sheet

fk = μkNSay it: “kinetic friction equals mu-k times the normal force, opposing the motion”

Kinetic friction

The drag of sliding — what takes over the instant the static grip breaks.

Notation on this page: fk is kinetic friction in N, μk (dimensionless, usually < μs) is the coefficient of kinetic friction, N is the normal force in N.

Before this lesson: Newton's second law

Where it comes from

Static friction is the grip before things move. But the instant the surfaces start sliding past each other, the microscopic gears shear and a new regime takes over: the bumps now collide, break, and reform thousands of times per second as the surfaces grind past. The resistance becomes a steady drag instead of a self-adjusting grip.

Before reading on: a box slides across the floor at 2 m/s, then at 4 m/s. Does the kinetic friction double, stay the same, or halve?

The experiments (Amontons, 1699; Coulomb, 1785) found something beautifully simple: once sliding, the drag is proportional to the squeeze and roughly independent of speed — the 4 m/s box feels about the same friction as the 2 m/s box. In symbols:

fk = μkNan equality, not an inequality — while sliding, friction is what it isSay it: “f-k equals mu-k times N”

Two contrasts with static friction matter: (1) this is an =, not a ≤ — no self-adjustment; (2) μk is usually smaller than μs, which is why the stuck box lurches when it breaks free — the resistance suddenly drops.

Derivation

Like its static sibling, kinetic friction is an empirical law — measured, not proved. The derivation is really a justification of its form from Amontons' two experimental laws:

fk ∝ N
Step 1 — Amontons' first law. The sliding drag is proportional to the normal load. Twice the weight on the sled, twice the drag.
fk
=
μk · N
Step 2 — name the constant. μk, the coefficient of kinetic friction: dimensionless, a property of the surface pair.
fk
≈
constant in v
Step 3 — Amontons' observation on speed. At ordinary speeds the drag barely changes with sliding speed — the microscopic collisions average out. So no v appears in the formula. ∎

Direction needs care: kinetic friction opposes the relative sliding motion, i.e. it points opposite the velocity. A block sliding right feels friction to the left — even if you keep pushing it right. The push and the friction then fight in the net-force sum, exactly like Example 4 of the second-law lesson.

Energy footnote: friction does negative work, W = −fkd, turning kinetic energy into heat and sound. That is why the sliding box slows down — and why its stopping distance is finite. (Full story in the Work & Energy unit.)

How to use it

The procedure, every time:

  1. Confirm it is sliding. If nothing moves, you want static friction. Kinetic friction exists only during relative motion.
  2. Find N. Flat: N = mg. Incline: N = mg cos θ.
  3. Compute fk = μkN and point it opposite the velocity.
  4. Net it. Fnet = (other forces) − fk along the motion axis, then a = Fnet/m. With no other horizontal forces, a = −μkg — a handy shortcut.
Common mistake: using μs for a sliding object. The coefficients switch the instant motion starts — check which regime you're in first.

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: 10 kg block sliding, μk = 0.25

  1. Normal force. N = mg = 10 × 9.8 = 98 N.
  2. Friction. fk = μkN = 0.25 × 98 = 24.5 N, opposing the slide.
  3. Meaning: whatever the speed (2 m/s or 20 m/s), the drag is ~24.5 N.
Common mistake: scaling friction with speed (“faster means more drag”). In this model it doesn't — that instinct belongs to air resistance, a different force.
Your turn — 4 kg block sliding, μk = 0.3. Kinetic friction?

Answer: about 11.8 N. N = 4 × 9.8 = 39.2 N; fk = 0.3 × 39.2 = 11.76 ≈ 11.8 N.

Before reading on: the block above is given a shove and left to slide with no other pushes. Heavier block or lighter block — which decelerates faster?

Example 2 — coasting to a stop: no pushes, just friction

  1. Net force. Only horizontal force is friction: Fnet = −24.5 N.
  2. Accelerate. a = Fnet/m = −24.5/10 = −2.45 m/s².
  3. The shortcut: a = −μkg = −0.25 × 9.8 = −2.45 m/s² — mass cancelled! Heavy and light blocks decelerate identically.
Common mistake: giving friction the wrong sign. It opposes the velocity — if the block moves right, a is negative. Get the sign right or the kinematics that follow will lie.
Your turn — μk = 0.4, no pushes. Deceleration?

Answer: −3.92 m/s² (≈ −3.9 m/s²). a = −μkg = −0.4 × 9.8 = −3.92 m/s², for any mass.

Example 3 — pushed while sliding: 60 N push, 10 kg, μk = 0.25

  1. Friction. fk = 24.5 N (from Example 1), opposing the motion.
  2. Net force. Fnet = 60 − 24.5 = 35.5 N in the push direction.
  3. Accelerate. a = 35.5/10 = 3.55 m/s².
  4. Notice: without friction it would be 6 m/s² — friction eats a fixed 24.5 N off the top.
Common mistake: adding friction to the push (60 + 24.5). Friction opposes — it subtracts.
Your turn — 50 N push, 5 kg block, μk = 0.2. Acceleration?

Answer: about 8.0 m/s². fk = 0.2 × 49 = 9.8 N; Fnet = 50 − 9.8 = 40.2 N; a = 40.2/5 = 8.04 ≈ 8.0 m/s².

Example 4 — stopping distance: v0 = 6 m/s, μk = 0.3

  1. Deceleration. a = −μkg = −0.3 × 9.8 = −2.94 m/s².
  2. Kinematics. v² = v0² + 2ad with v = 0: d = v0²/(2μkg) = 36/(2 × 0.3 × 9.8).
  3. Compute. d = 36/5.88 ≈ 6.1 m.
  4. The lesson: d ∝ v0² — double the speed, quadruple the stopping distance. Braking physics in one line.
Common mistake: thinking stopping distance doubles with speed. The v² in the kinematics makes it quadratic — the reason speeding kills.
Your turn — v0 = 4 m/s, μk = 0.2. Stopping distance?

Answer: about 4.1 m. d = 16/(2 × 0.2 × 9.8) = 16/3.92 ≈ 4.08 ≈ 4.1 m.

Memorization tips

  • Say it aloud: “kinetic friction equals mu-k N, against the motion.” The direction half is the memory.
  • = not ≤: static gets the inequality (self-adjusting); kinetic gets the equals sign (steady drag). The symbol tells you the regime.
  • μk < μs: the lurch rule — breaking free is harder than sliding. If a problem gives both, match each to its regime.
  • The −μkg shortcut: a freely sliding block decelerates at μkg regardless of mass. One line, no FBD needed.
  • Stopping distance squared: d = v0²/(2μkg). Speed doubles → distance quadruples. This is the most exam-used consequence of the formula.
  • Energy echo: friction eats fk × d of kinetic energy as heat. Force view and energy view must agree — use one to check the other.

Final challenge

Five mixed questions — directions, stopping distances, and the traps, all in one. Score 5/5 and kinetic friction 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 kinetic friction?

Kinetic friction is the resistance between two surfaces that are sliding past each other: fk = μkN, always opposing the direction of sliding. It switches on the instant motion starts.

What is the difference between static and kinetic friction?

Static friction (fs ≤ μsN) prevents motion and self-adjusts up to a maximum. Kinetic friction (fk = μkN) acts during sliding at a roughly constant value — and μk is usually smaller than μs, so starting is harder than sliding.

Which way does kinetic friction point?

Opposite the relative sliding motion — against the velocity, not against the applied push. A block sliding right feels kinetic friction to the left, even if you are still pushing it right.

Does kinetic friction depend on speed?

In the standard model, no: fk = μkN is roughly constant at ordinary speeds. (At very high speeds or with lubrication the simple model breaks down.)

How do you find how far a sliding object travels before stopping?

Friction gives a = −μkg; then v² = v0² + 2ad with v = 0 gives d = v0²/(2μkg). Double the speed and the stopping distance quadruples.

More from the codex

Live

Physics I formula sheet

All the mechanics formulas — kinematics, forces, energy — printable and quiz-ready.

Open sheet → Live

Formula Sheet Builder

Mix and match any sections into your own printable sheet.

Open tool → Live

Prompt Simulator

Practice prompt engineering with deterministic scoring.

Open tool →