Static and kinetic friction
 Question: A box is lying still on an inclined plane. The inclination angle (from the horizontal) of the plane is $\alpha=30^\circ$ .  At a certain moment, the box is being punched. This punch sets the box in motion up the inclined plane. The box climbs until it comes to a stop. The scenario is depicted in the clip. The coefficient of kinetic friction between the box and the plane is  $\mu_k=0.5$   What is the lower limit that can be placed on the value of the static friction coefficient, $\large \mu_s$?
3 People tried to answer this question

 It is always the case that $0<\mu_s<1$
 $\mu_s>0.58$ because in the case described here it must be that $tan(\alpha)<\mu_s$
 $\mu_s>0.5$ since static friction coefficient is always larger than the kinetic friction coefficient.
 The scenario is not realistic: What comes up must come down. No limit can be placed on $\mu_s$

True. If the box can lie still, it must be that the the static friction force, $f_s$, and the component of the weight  balance, so $f_s=mg\rm{sin}(\alpha)$. In addition, $f_s\leq \mu_s N$. The normal force balace the prependicular component of the weight, therefore $N=mg\rm{cos}(\alpha)$. combining these three relation we get $\rm{tan}(\alpha)<\mu_s$

Not true. The box will come to a stop and then remain at rest due to static friction force.

Wrong. It might be that the static friction coeffcient is larger that 0.5 but a stronger lower limit can be placed.

Wrong for many reasons. 1. This is not a lower limit. 2. This is not true. it is possible to have $\mu>1$

 Points on map: