Coefficient of Kinetic Friction Formula: μk = fk / N

Q: What is the formula for coefficient of kinetic friction?

μk = fk / N: sliding friction over the normal reaction. On level ground with only friction slowing an object, it also equals a / g.

Q: Does the coefficient of kinetic friction have a unit?

No. Newtons on top and newtons underneath cancel, leaving a pure number.

Q: Is μk always less than μs?

For nearly all ordinary pairs of surfaces, yes. A few special materials have the two nearly equal, but μk is never meant to exceed μs in school problems.

Q: Does kinetic friction depend on speed?

For dry surfaces at everyday speeds it is nearly constant, which is the assumption used in exams. At very high speeds or with lubricants it can change.

The coefficient of kinetic friction formula is $\mu_k = \frac{f_k}{N}$ : divide the friction force on a sliding object by the normal force between the surfaces. A 25 kg box dragged across a floor at steady speed by a 70 N horizontal rope has $\mu_k = \frac{70}{245} \approx 0.29$ .

Kinetic friction, also called sliding or dynamic friction, acts only once surfaces are moving relative to each other. Its coefficient is usually a bit lower than the static one for the same pair, which is why it is easier to keep a heavy box moving than to start it. This page shows three ways to find μk, all built on the friction force formula and the laws of motion.

Quick facts

Formula
$\mu_k = \frac{f_k}{N}$
From braking on level ground
$\mu_k = \frac{a}{g}$
Steady slide down a slope
$\mu_k = \tan\theta$
Accelerating down a slope
$\mu_k = \tan\theta - \frac{a}{g\cos\theta}$
Unit
None
Compared with static
Usually $\mu_k < \mu_s$

Kinetic coefficient of friction formula and its parts

Terms in μk = fk / N
Symbol	Meaning	Unit
$\mu_k$	Coefficient of kinetic friction	none
$f_k$	Friction while sliding	N
$N$	Perpendicular contact force	N

Once sliding starts, $f_k = \mu_k N$ stays roughly constant for ordinary speeds. That makes kinetic friction easy to handle: it is a fixed backward force, so the acceleration it causes is constant and the equations of motion apply.

Three ways to apply the coefficient of kinetic friction formula

Steady pull. Drag the object at constant velocity. The pull then equals friction, so $\mu_k = \frac{F_{\text{pull}}}{N}$ .
Sliding to a stop. Give the object a push on a level surface and let it slide. Friction is the only horizontal force, so $\mu_k mg = ma$ and $\mu_k = \frac{a}{g}$ . Find a from $v^2 = u^2 - 2as$ .
Incline. Tilt a ramp until a nudged block slides down at constant speed. Then $mg\sin\theta = \mu_k mg\cos\theta$ , so $\mu_k = \tan\theta$ . If the block speeds up with acceleration a, use $\mu_k = \tan\theta - \frac{a}{g\cos\theta}$ .

Worked examples for the formula for coefficient of kinetic friction

Example 1: steady drag

A 25 kg box is dragged steadily over a level floor by a 70 N horizontal force. Find μk (g = 9.8 m/s²).

N = 25 × 9.8 = 245 N. At constant speed, fk = 70 N.

$\mu_k = \frac{70}{245} = 0.286 \approx 0.29$

Example 2: a puck sliding to rest

An ice hockey puck leaves a stick at 6.0 m/s and slides 30 m before stopping. Find μk.

Deceleration: $a = \frac{u^2}{2s} = \frac{6.0^2}{2 \times 30} = \frac{36}{60} = 0.60\ \text{m/s}^2$ .

$\mu_k = \frac{a}{g} = \frac{0.60}{9.8} = 0.061$

The mass of the puck is not needed, because it cancels.

Example 3: a block speeding up on a 35° ramp

A block slides down a 35° ramp with an acceleration of 3.0 m/s². Find μk.

$\tan 35^\circ = 0.700$ and $g\cos 35^\circ = 9.8 \times 0.819 = 8.03\ \text{m/s}^2$ .

$\mu_k = 0.700 - \frac{3.0}{8.03} = 0.700 - 0.374 = 0.326 \approx 0.33$

Why kinetic friction is usually smaller

At the microscopic level, surfaces touch only at tiny high points. When they sit still, those contact points settle and bond more firmly. Once sliding, they break and remake contact constantly and never get as firm a hold. The result is that, for most pairs, $\mu_k$ is noticeably smaller than $\mu_s$ , though the size of the gap depends on the materials. The coefficient of static friction formula page covers the gripping case, and the general coefficient of friction formula page compares both.

Energy view of sliding friction

The work done by kinetic friction over a sliding distance d is $W = -\mu_k N d$ . On level ground with no other horizontal force, all the starting kinetic energy is used up: $\frac{1}{2}mu^2 = \mu_k mg d$ , so $d = \frac{u^2}{2\mu_k g}$ . Check with Example 2: $\frac{36}{2 \times 0.061 \times 9.8} \approx 30\ \text{m}$ . The work done by friction page develops this method.

Mistakes with μk

Using an accelerating pull as if it equalled friction. Only a constant velocity pull does.
Forgetting that N = mg cos θ on a slope.
Quoting μk with a unit.
Swapping μk and μs, so the moving case seems to have more grip.

If μk comes out negative on an incline, the block must be accelerating faster than a frictionless slope allows. Recheck the angle or the acceleration.

Frequently asked questions

What is the formula for coefficient of kinetic friction?

μk = fk / N: sliding friction over the normal reaction. On level ground with only friction slowing an object, it also equals a / g.

Does the coefficient of kinetic friction have a unit?

No. Newtons on top and newtons underneath cancel, leaving a pure number.

Is μk always less than μs?

For nearly all ordinary pairs of surfaces, yes. A few special materials have the two nearly equal, but μk is never meant to exceed μs in school problems.

Does kinetic friction depend on speed?

For dry surfaces at everyday speeds it is nearly constant, which is the assumption used in exams. At very high speeds or with lubricants it can change.