= Kinetic friction
{title2=$|F_{\mathrm{fr}}|=\mu N$}
<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 is
$$
f_k=-\mu_kN\operatorname{sgn}(v-U),\qquad v-U\ne0.
$$
Here $N\ge0$ is the magnitude of the <normal force>, $\mu_k\ge0$ is the <coefficient of kinetic friction>, and $\operatorname{sgn}$ 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 $\mathbf f_k=-\mu_kN\mathbf w/|\mathbf w|$ for a nonzero tangential <relative velocity> $\mathbf w$. The magnitude law is documented in https://openstax.org/books/university-physics-volume-1/pages/6-2-friction[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 $g=10\,\mathrm{m/s^2}$ 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 $a=(F+f_k)/m=3\,\mathrm{m/s^2}$, for as long as rightward <contact slip> persists.
Direction is clearer by comparing two instantaneous states of this same model. If the box has $v=1\,\mathrm{m/s}$ on stationary ground, $U=0$ and $w=+1\,\mathrm{m/s}$, so <kinetic friction> acts left with magnitude $3\,\mathrm N$. If the box still has $v=1\,\mathrm{m/s}$ but a belt beneath it has $U=2\,\mathrm{m/s}$, then $w=-1\,\mathrm{m/s}$: <kinetic friction> acts right with magnitude $3\,\mathrm N$, even while the box moves right.
\Image[/kinetic-friction-relative-motion.png]
{title=The same rightward box velocity gives opposite friction directions on stationary ground and a faster moving belt}
In the belt case, the <power> delivered to the box is $P_{\mathrm{box}}=f_kv=+3\,\mathrm W$. 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 $P_{\mathrm{belt}}=-f_kU=-6\,\mathrm W$. 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 https://openstax.org/books/university-physics-volume-1/pages/7-4-power[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 $\operatorname{sgn}(0)=0$ into the sliding law would incorrectly exclude a nonzero <static friction>.
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