A block of mass m moves on a horizontal surface with coefficient of friction mu , subject to a constant force F_0 acting vertically downward towards the surface. If the block has initial velocity v , how far does it travel before it stops?

There are 4 forces acting on the block:gravitational force mg downwards,the force F_0 downwards,the reaction force N upwards,the friction force F_f horizontally.
The net force F_(n et), which is their vector sum, gives the block some acceleration a. By Newton's Second law F_(n et) = ma. Because the block moves horizontally, its speed and acceleration are also horizontal. Thus the net force is also horizontal that means all vertical forces are balanced, N = mg + F_0, and F_(n et) = F_f.
It is known that F_f = mu N = mu(mg + F_0). Therefore, the acceleration magnitude is a =mu(mg + F_0)/m =mu(g + F_0/m). Also it is known that friction force is opposite to the direction of movement.
Therefore, the speed of the block is V(t) = v - at and the block stops at time t_1 = v/a (when the speed becomes zero). The distance as a function of time is therefore  D(t) = vt - (at^2)/2, tlt=t_1,  and the farthest distance is
D(t_1) = vt_1 - (at_1^2)/2 = vt_1 - (vt_1)/2 = (vt_1)/2 =v^2/(2mu(g + F_0/m)).
This is the answer.
https://www.physicsclassroom.com/class/newtlaws/Lesson-3/Newton-s-Second-Law