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The coefficient of kinetic friction is the dimensionless quantity ${\mu }_k=|f_k|/N$ for sliding contact with positive normal force $N$ in the ideal kinetic friction model. It characterizes the pair of contacting surfaces and their condition. A constant ${\mu }_k$ means the model neglects dependence on contact slip speed; unlike the coefficient of static friction, it specifies a sliding force magnitude rather than a no-slip upper bound.

// Target: classical-mechanics.bigb

Contact slip is the tangential relative velocity between the two material points instantaneously in contact. For a nonrotating box on a belt it is $w=v-U$; for a rigid body, use the velocity of its contact point, including the contribution from angular velocity. Zero contact slip defines rolling without slipping at a rolling contact, even when the center of mass moves.

Friction is a contact force tangent to the surfaces. Kinetic friction opposes contact slip; static friction maintains contact without contact slip, including rolling without slipping. The perpendicular contact component is the normal force. These two components answer different questions: how strongly the surfaces press together, and how strongly they resist sliding past one another.

The ideal Coulomb dry-contact model combines a bound on static friction with a sliding law for kinetic friction. Its parameters are the coefficient of static friction and the coefficient of kinetic friction; constant coefficients are an empirical approximation, rather than a consequence of Newton's laws of motion. Surface condition matters, and the static coefficient need not universally exceed the kinetic coefficient. ^[ref]

The model's historical setting was practical machinery: Coulomb's Théorie des machines simples investigated friction and the stiffness of ropes. The Académie's bibliography records its 1781 prize and its appearance in volume 10, pages 161–332, of the Recueil des savants étrangers. ^[ref]

In the ideal kinetic friction model, the loss of combined mechanical energy of the two contacting bodies becomes thermal internal energy at rateThis follows by adding the contact powers $f_{k}v$ and $-f_{k}U$ from Newton's third law. It depends on contact slip, whereas the contact power received by either body separately depends on its velocity in the chosen inertial frame. Thus positive frictional heating is compatible with increasing kinetic energy of one body, as the moving-belt example in kinetic friction shows. Heating need not cause a temperature rise if thermal energy escapes by heat transfer.

Kinetic friction is friction during nonzero contact slip. For one-dimensional motion along a surface, choose right as positive and write $v$ for the body's contact-point velocity and $U$ for the surface's contact-point velocity, both measured in the same inertial frame. The signed relative velocity is $w=v-U$. The ideal Coulomb law for the tangential force on the body isHere $N\ge 0$ is the magnitude of the normal force, ${\mu }_k\ge 0$ is the coefficient of kinetic friction, and $\mathrm{sgn}$ is the sign function. Thus kinetic friction opposes contact slip, which can differ from the body's direction of motion. In several tangential directions the same law reads ${\mathbf{f}}_k=-{\mu }_{k}N\mathbf{w}/|\mathbf{w}|$ for a nonzero tangential relative velocity $\mathbf{w}$. The magnitude law is documented in Ling, Sanny and Moebs, OpenStax University Physics 1, §6.2: static and kinetic friction.

For a stipulated toy model, take a nonrotating box with mass $m=1\mathrm{kg}$, normal force $N=10\mathrm{N}$, coefficient of static friction ${\mu }_s=0.5$, and coefficient of kinetic friction ${\mu }_k=0.3$. These are assumed model parameters, rather than measured properties of any particular pair of materials; the prescribed normal force is consistent with horizontal support under $g=10m/{\mathrm{s}}^2$ and no other vertical forces. On stationary ground, a box already sliding right has $f_k=-3\mathrm{N}$. With an applied horizontal force $F=6\mathrm{N}$ to the right, Newton's second law gives $a=(F+f_k)/m=3m/{\mathrm{s}}^2$, for as long as rightward contact slip persists.

Direction is clearer by comparing two instantaneous states of this same model. If the box has $v=1m/s$ on stationary ground, $U=0$ and $w=+1m/s$, so kinetic friction acts left with magnitude $3\mathrm{N}$. If the box still has $v=1m/s$ but a belt beneath it has $U=2m/s$, then $w=-1m/s$: kinetic friction acts right with magnitude $3\mathrm{N}$, even while the box moves right.

In the belt case, the power delivered to the box is $P_{\mathrm{box}}=f_{k}v=+3\mathrm{W}$. With no other horizontal force, its kinetic energy therefore increases by the work-energy theorem. By Newton's third law, the box exerts $-f_k$ on the belt, so the contact power delivered to the belt is $P_{\mathrm{belt}}=-f_{k}U=-6\mathrm{W}$. Their sum is $-3\mathrm{W}$: frictional heating produces thermal internal energy at $3\mathrm{W}$. An ideal motor maintaining the prescribed belt velocity supplies $+6\mathrm{W}$ against this contact load, neglecting other losses, so its input divides into $3\mathrm{W}$ of box kinetic energy and $3\mathrm{W}$ of frictional heating. These are instantaneous rates, since the box's velocity subsequently changes. This balance follows directly from power as force times velocity and conservation of energy; the power law is given in Ling, Sanny and Moebs, OpenStax University Physics 1, §7.4: power.

When contact slip vanishes, including a transition to rolling without slipping, the nonzero-slip law no longer applies. Determine the required static friction from the no-slip condition and the equations of motion, then test its bound. Substituting $\mathrm{sgn}(0)=0$ into the sliding law would incorrectly exclude a nonzero static friction.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal Coulomb dry-contact model, the allowable magnitude is bounded bywhere $N\ge 0$ is the magnitude of the normal force and ${\mu }_s\ge 0$ is the coefficient of static friction. The actual static friction is whatever the equations of motion require to maintain zero contact slip, provided it lies within this bound. Equality specifies limiting static friction; a smaller load generally requires a smaller contact force. ^[ref]

Use the stipulated toy model from kinetic friction: mass $m=1\mathrm{kg}$, normal force $N=10\mathrm{N}$, coefficient of static friction ${\mu }_s=0.5$, and coefficient of kinetic friction ${\mu }_k=0.3$. A box initially at rest on stationary horizontal ground is pushed right by $F\ge 0$. To remain at rest, Newton's second law requires $F+f_s=ma=0$, hence $f_s=-F$. For $F=4\mathrm{N}$, this gives $f_s=-4\mathrm{N}$ and $a=0$, within the $5\mathrm{N}$ bound. Automatically setting $|f_s|={\mu }_{s}N$ would predict the wrong force and acceleration.

For $F=6\mathrm{N}$, maintaining rest would require $|f_s|=6\mathrm{N}>5\mathrm{N}$, so rest cannot persist. Once the box slides right, kinetic friction is $f_k=-3\mathrm{N}$ and the acceleration is $a=(6-3)/1=3m/{\mathrm{s}}^2$. At exactly $F=5\mathrm{N}$, the inequality still permits rest with $f_s=-5\mathrm{N}$. An already right-sliding box under the same applied force instead has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. Applied force alone therefore does not specify the friction branch: the contact state and motion history matter.

The diagram compares conditional states, rather than specifying a force-ramp trajectory. The static branch includes its $5\mathrm{N}$ endpoint; the sliding branch also exists below that load when rightward contact slip is already present. To solve a new problem, first impose zero contact slip, find the required static friction and normal force from the equations of motion, and check the inequality. If that contact condition fails, use kinetic friction with the actual slip direction.

This distinction makes static friction useful for traction. For an ideal driven tire undergoing rolling without slipping on stationary ground, the road can exert a forward static friction that accelerates the vehicle while the tire's contact point has zero velocity. Thus static friction can point along the center of mass motion; its role is to prevent contact slip, rather than always to oppose the motion of the whole body.