A small disc is projected on a horizontal floor with speed $u$. Coefficient of friction between the disc and floor varies as $\mu = \mu_0+ kx$, where $x$ is the distance covered. Find distance slid by the disc on the floor.

I have correctly solved this by using two methods :

  1. By integrating the kinematic equation $\displaystyle a=v\frac{dv}{dx}=-(\mu_0+kx)g$ and
  2. by considering the work done by frictional force ($\int_0^x (\mu_0+kx)mgdx$) to dissipate the initial kinetic energy of the disc.

Now I notice that the equation of motion, $a=-(\mu_0 +kx)g$, is kinda SHM-like, but with a non-conservative force instead. I conjecture that as soon as the disc reaches $v=0$, the force of friction, and hence the acceleration, drop to 0. Now, I think that this is pretty similar (except the $a=0$ at $v=0$ part; in SHM, from my knowledge, $a$ and $v$ are separated by a phase angle of $\frac π2$) to the first $\frac 14$ of an SHM cycle, i.e. starting from the mean position till it reaches the amplitude.

Now my calculations for the third method: The maximum speed obtained is at mean position: $v_{max}=u=A\omega$ where $\omega$ is the angular frequency of oscillation. Also, $\omega ^2=kg$ and so maximum distance travelled is equal to the amplitude, i.e, when velocity becomes 0. So $$ A=\frac{u}{\sqrt{kg}}. $$ However, this is not the correct answer.

What’s my error in concept?

EDIT: I understood where I got confused, thanks to user @J.Murray. But I’d like to see this question solved (or get the easily solvable equation $kgx^2+2\mu_0gx-u^2=0$) with concepts borrowed from SHM, and not directly the two methods I have listed above. No big deal if it is complicated.

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    $\begingroup$ You're right that the equation of motion is kinda SHM-like, but your analysis seems to proceed as though it is exactly SHM. What happened to the $\mu_0$ term? $\endgroup$
    – J. Murray
    Jun 25, 2022 at 17:58
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    $\begingroup$ Uh… @J.Murray why would that affect any calculations? It doesn't change $\omega$ $\endgroup$ Jun 25, 2022 at 18:01
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    $\begingroup$ Ah….I see… At the expected mean position the acceleration is not 0 (as it should be in SHM). Or is it? Because we start from there. So at x=0, is a=0? @J.Murray can you please see if the question be solved via some SHM-concepts? $\endgroup$ Jun 25, 2022 at 18:03
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    $\begingroup$ When $x=0$, we have $a = -(\mu_0 + k\cdot 0 )g = -\mu_0 g\neq 0$. The equilibrium point will be at $x= -k/\mu_0$. You could use some concepts from SHM, but it would not be quite as simple as just doing the integral. $\endgroup$
    – J. Murray
    Jun 25, 2022 at 20:03
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    $\begingroup$ @J.Murray I may come off as annoying, but pleaseeee confirm that even when the disc has started from a point (say A) with nonzero initial speed, there may be acceleration at A? I just feel silly asking this, but please confirm this for me. $\endgroup$ Jun 25, 2022 at 20:15

2 Answers 2


Consider a mass dangling from a vertical spring. The equation of motion is

$$ ma = mg -k(x-x_0) $$

where $x_0$ is the position of the end of the unstretched spring, and I’m using a downward-positive coordinate system. This system has an equilibrium at

$$ x-x_0= \frac{mg}k $$

The potential energy for a mass on a vertical spring is

$$ U= -mgx +\frac12 k(x-x_0)^2 =\frac12 k \left( x-\left(x_0+\frac{mg}k\right) \right)^2 + \text{constant} $$

That is, the potential energy is the sum of a linear gravitational term and a quadratic term from the spring. However, you may recall from algebra that a quadratic function plus a linear function is just a quadratic function with the same curvature but a different minimum. If you enjoy algebra and completing the square, you can find the second equality.

So oscillations of a dangling spring about its effective equilibrium are described by simple harmonic motion with the same frequency $\omega^2=k/m$ as the free spring.

Your contrived distance-dependent friction force,

$$ ma=-m(\mu+kx)g $$

has exactly the constant-plus-linear form of a dangling spring, so you can find the stopping position by treating the stopping process as a partial oscillation, from $x=0$ (with nonzero initial velocity) to the oscillation’s turning point.


You are correct that this problem can be treated as an oscillator:

$$ x'' + \beta ^2 x + C $$ where $\beta^2 = gk $ and $C = \mu_0 g $

The equation is for a SHM plus a constant. This means the solution will be the sum of the SHM solution and the particular solution for this equation. When the math is done, the solution takes the general form: $$ x = Asin(\beta t) + Bcos(\beta t) - \frac C {\beta^2}$$ and $$ x' = A\beta cos(\beta t) - B\beta sin(\beta t)$$

With boundary conditions, these can be used to solve your problem. The point is, the SHM solution alone will not give the correct answer. The extra constant affects your values for A and B. The differential equation must be solved.


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