A brick sliding in an horizontal plane after an initial push (under Coulomb's dry friction) - part 2

Question

A brick sliding in an horizontal plane after an initial push (under Coulomb's dry friction) - part 2

Intro

This is a follow up of a previous question.

Main Body

From these Wikipedia sites: Contact dynamics | Non-smooth approach and Coulomb damping | Illustration I know the differential equation that model the Newton's 2nd law and rules this system is given by: $$ m\cdot x'' = -F - \left(\text{sgn}(x')\right)\cdot \mu_k\cdot m\cdot g \tag{Eq. 1}\label{Eq. 1}$$

Now, assuming that the initial push have already happened, the brick is sliding with an initial velocity $x'(0)$, so the force that have produce this push could be already being considered as $F = 0$ (is absent now from the point of view of the brick), so the only force present is the friction that slows down the brick until it stop moving. If I am not mistaken, dividing both side with $m$ and matching $k \equiv \mu_k >0$, I will have that the product $k\cdot g >0$ with $g = 9.8\,\frac{m}{s^2}$ the Earth's gravity acceleration constant recovering the differential equation: $$x'' = -k\cdot g\cdot \text{sgn}(x') \tag{Eq. 2}\label{Eq. 2}$$

I found here that the solution could be described as: $$x(t) =x(0)+ \frac{kg\cdot\text{sgn}(x'(0))}{2}\cdot\left[\left(\frac{|x'(0)|}{kg}\right)^2-\left(\frac{|x'(0)|}{kg}-t\right)^2\cdot\theta\!\left(\frac{|x'(0)|}{kg}-t\right)\right] \tag{Eq. 3}\label{Eq. 3}$$

which stop moving at the finite extinction time $$0<T = \frac{|x'(0)|}{kg}<\infty \tag{Eq. 4}\label{Eq. 4}$$

and moves a distance given by: $$\Delta x = |x(T)-x(0)| = \frac{|x'(0)|^2}{2kg}\tag{Eq. 5}\label{Eq. 5}$$

so the solution kind of matches the quadratic physics of free fall: $$\implies T=\sqrt{\frac{2\Delta x}{kg}}\tag{Eq. 6}\label{Eq. 6}$$

but is kind of going backwards, which can be tested in Desmos.

Question

What make me doubt about \eqref{Eq. 3} solution validity, is that the braking path/trajectory depends only on the initial speed, but not in the mass of the object, this makes that everything stop moving at the same rate $kg$, both things sounds counter-intuitive to me. I don't know if I messed up on the description, or if instead, it is right and just my intuition is wrong.

PS: I tried at home the following experiment: I pushed with a ruler in vertical position, at the same time (so I am assuming that with the same force - correct me if I am wrong please, but surely both share the same initial speed), 2 different objects over a wooden table, each object with different sizes: an USB flash memory (small/light), and a home stapler (big/heavy), and for slight pushes they kind of stop at the same time and travel the same distance, but for strong pushes the stapler runs way longer and travel further, I imagine because have much inertia since is way heavier, this somehow in line with my intuition that something is wrong.

Further analysis in the original post (see edit section): images, citations, and ideas where removed due the question got closed due its extension.

Answer 1

The the fact that the time only depend on the initial speed is correct, if you make the approximation $f_{\text{friction}}=\mu m g$.

The computation can be actually done in a simpler way, after the force is removed, we assume that the speed now is $v_0$ (which can be also written as $x'(0)$ in your question).

Applying Newton's second law, because the initial force is now absent, take the direction of the speed of the object to be the direction of positive x axis, we have $$ma=-\mu m g$$ here a is the acceleration of the object, also known as $x'(0)$. and we know that $a=- \mu g$.

Since the acceleration is constant ($x''(t)=constant$), we can do an integration and find that the acceleration declines linearly, so $$t=\frac{v}{a}=\frac{x'(0)}{\mu g}$$and$$v(t)=v_0-\mu g t$$then do the integration again and you find that $$x=v_0t-\frac{1}{2} \mu g t^2$$ it matches your result.

The reason why it seems anti-intuitive is that you make a approximation $f_{\text{friction}}=\mu m g$. This assumption is valid in a lot of physics system we study. However, in reality you see a lot of scenario where this assumption does not hold. When there's an extra term relevant to the velocity(which often happens in reality), the t you are calculating will depend on the mass.