Irodov 1.127: Friction Sets Its Own Speed Limit – Amit & Deepali Sinha Foundation

Problem Irodov 1.127

A small body of mass $m$ starts at a point $O$ on a horizontal plane with an initial horizontal speed $v_0$ and slides to rest under kinetic friction alone. Find:

the mean power delivered by the friction force over the whole time the body is moving, for a constant friction coefficient $k = 0.27$ , mass $m = 1.0$ kg and initial speed $v_0 = 1.5$ m/s
the maximum magnitude of the instantaneous power delivered by the friction force, if instead the friction coefficient grows with distance traveled as $k = \alpha x$ , where $\alpha$ is a given constant and $x$ is the distance from $O$

The body at O, sliding with initial velocity v_0 under its weight m g, the normal N and kinetic friction f, the only horizontal force in either part of the problem. — The body at $O$ , sliding with initial velocity $\vect v_0$ under its weight $m\vect g$ , the normal $\vect N$ and kinetic friction $\vect f$ , the only horizontal force in either part of the problem.

Given

$m$
$v_0$
$k$ (part a, constant)
$k=\alpha x$ (part b)

Find

(a)

\langle P\rangle

for

k=0.27

m=1.0

kg,

v_0=1.5

m/s; (b)

P_{\mathrm{max}}

for

k=\alpha x

Solution

Step 1Total work friction does (part a)

On the horizontal plane, friction of constant magnitude $f=kmg$ is the only horizontal force, so the work-energy theorem ties its total work directly to the kinetic energy lost between the start and the moment the body stops.

W_f = 0-\frac12 mv_0^2

W_f = -\frac12 mv_0^2

(1)

Step 2Time taken to stop (part a)

Friction produces a constant deceleration $a=kg$ , so the velocity falls linearly with time; setting it to zero locates the stopping instant $t_1$ .

0 = v_0-kg\,t_1 \;\Longrightarrow\; t_1=\frac{v_0}{kg} \tag{2}

Step 3Mean power over the whole motion

Dividing the total work (1) by the stopping time (2) gives the mean power, then the given numbers evaluate it.

\begin{align*}\langle P\rangle &= \frac{W_f}{t_1} \\ &= \frac{-\frac12 mv_0^2}{v_0/(kg)}\end{align*}

\langle P\rangle = -\dfrac12 kmgv_0 = -2.0\ \mathrm W

(3)

Step 4Speed as a function of distance (part b)

With $k=\alpha x$ , friction’s magnitude grows linearly with distance traveled, so Newton’s second law along the direction of motion becomes a differential relation between $v$ and $x$ rather than between $v$ and $t$ .

mv\frac{dv}{dx} = -\alpha xmg

Separating variables and integrating from the start, where $v=v_0$ at $x=0$ , to a general distance $x$ ,

\int_{v_0}^{v}v'\,dv' = -\alpha g\int_0^{x}x'\,dx' \;\Longrightarrow\; \frac{v^2-v_0^2}{2} = -\frac{\alpha g x^2}{2}

v^2 = v_0^2-\alpha g x^2

(4)

Step 5Instantaneous power as a function of distance

The instantaneous power friction delivers is the product of its magnitude $\alpha xmg$ and the speed (4), carrying a negative sign because friction always opposes the motion it acts through.

P(x) = -\alpha xmg\,v(x) = -\alpha mg\,x\sqrt{v_0^2-\alpha gx^2} \tag{5}

Step 6Locating the extremum

$P(x)$ is negative throughout the motion, so its magnitude is greatest where the factor $x\sqrt{v_0^2-\alpha gx^2}$ is greatest; differentiating that factor and setting the derivative to zero locates the critical distance $x_m$ .

\begin{align*}\frac{d}{dx}\left[x\sqrt{v_0^2-\alpha gx^2}\right] &= \sqrt{v_0^2-\alpha gx^2}-\frac{\alpha gx^2}{\sqrt{v_0^2-\alpha gx^2}} \\ &= \frac{v_0^2-2\alpha gx^2}{\sqrt{v_0^2-\alpha gx^2}}\end{align*}

The denominator stays positive throughout the motion, so the derivative vanishes only where the numerator does; setting it to zero and solving for $x$ gives the critical distance.

\begin{align*}v_0^2-2\alpha gx_m^2 &= 0 \\ x_m^2 &= \frac{v_0^2}{2\alpha g}\end{align*}

x_m = \dfrac{v_0}{\sqrt{2\alpha g}}

(6)

Step 7Maximum magnitude of the power

Substituting $x_m$ (6) into the speed (4) gives the speed $v_m$ at the critical distance, and both together into the power (5) give the extremal value.

\begin{align*}v_m^2 &= v_0^2-\alpha g\left(\frac{v_0}{\sqrt{2\alpha g}}\right)^2 \\ &= \frac{v_0^2}{2}\end{align*}

With $v_m=v_0/\sqrt2$ ,

\begin{align*}P_{\max} &= -\alpha mg\,x_mv_m \\ &= -\alpha mg\cdot\frac{v_0}{\sqrt{2\alpha g}}\cdot\frac{v_0}{\sqrt2}\end{align*}

P_{\max} = -\dfrac12 mv_0^2\sqrt{\alpha g}

(7)

Check

Dimensionally, $[k][m][g][v_0]=(\text{dimensionless})(\mathrm{kg})(\mathrm{m/s^2})(\mathrm{m/s})=\mathrm W$ in part (a), and $[\alpha g]=(\mathrm{m^{-1}})(\mathrm{m/s^2})=\mathrm{s^{-2}}$ , so $\sqrt{\alpha g}$ is an inverse time and $mv_0^2\sqrt{\alpha g}$ is again a watt, as required in part (b). Numerically, part (a) gives $\frac12(0.27)(1.0)(9.8)(1.5)\approx1.98\ \mathrm W$ , matching the boxed $-2.0$ W. As a limiting check on part (b), at $x=0$ and at the stopping distance ( $v=0$ ) the power vanishes, consistent with $P(x)$ being the product of two factors that each start or end at zero.

Remark

Part (a)’s mean power is exactly half of $-kmgv_0$ , the power friction would deliver at the instant of release: because the speed falls linearly with time under constant deceleration, the power (proportional to speed) falls linearly too, and the mean of a linear quantity over an interval is its value at the midpoint, here the average of the initial value and zero. Part (b) has no such shortcut because $k$ itself changes with $x$ , which is exactly why the maximum there sits at a specific interior point, $v_m=v_0/\sqrt2$ , rather than at either end of the motion.