Please provide a solution for this question (in English, preferably). Link: https://mathus.ru/olymp/vseros1998f11fin.pdf
Let’s denote x(t) and v(t) as coordinate and velocity of a cube relative to table and X and V as for the plate. The motion of this system is quite complicated because the directon of friction force depends on the sign of relative speed, but let’s first assume that v(t)>V(t), so
m\ddot x = -\mu mg, \qquad M\ddot X = -kX + \mu mg.
We have x(0)=X(0)=V(0)=0 and v(0)=v_0, thus we get
v(t)=v_0-\mu gt, \quad x(t) =v_0 t-\frac{\mu gt^2}{2}, \\ X(t) = \frac{\mu mg}{k}\left(1-\cos\left(\sqrt{\frac{k}{M}}t\right)\right), \quad V(t) = \frac{\mu mg}{\sqrt{kM}}\sin\left(\sqrt{\frac{k}{M}}t\right).
How to maximize the released heat? We need the system to stop at some specific moment so all of the X, v, V parameters would be equal to zero. In this case, the further motion is impossible, and Q_{\text{max}}=mv_0^2/2.
If we put time to be equal
t_n = 2\pi n\sqrt{\frac{M}{k}}, \qquad n\in N,
we get X(t_n)=V(t_n) = 0. Let’s require a condition v(t_n) = 0. In this case we get t_n = \displaystyle\frac{v_0}{\mu g}, or
\mu = \frac{v_0}{2\pi g}\sqrt{\frac{k}{M}}\frac{1}{n},
and also in this moment
x(t_n)=\frac{v_0^2}{2\mu g} \leq L.
If we put n\geq4, we get x(t_n)>L, which should not happen, so we get three values of \mu which are given below.
The main assumption I did is that v(t)>V(t) throughout the problem. The velocity of plate could be higher around the t\approx t_n because
\lim_{t\rightarrow t_n} v(t) \rightarrow 0, \qquad \lim_{t\rightarrow t_n} V(t) \rightarrow 0.
In this interval I can describe velocities in terms of acceleration, i.e. v(t)\approx \ddot x(t_n) (t - t_n) and V(t) \approx \ddot X(t_n) (t - t_n), so in order to meet our conditions we should check for
\ddot x(t_n)<\ddot X(t_n),
(I put “<” instead of “>” because t-t_n<0) or
-\mu g < \frac{\mu mg}{M}\cos{2\pi n}.
Also you may prove the second part of the question graphically, which is quite easier than an analytical method.
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Thanks, I’ll take a look at the second part.