4. To maintain the stone’s motion, a horizontal force (in the +xdirection) is needed that cancels the
retarding effect due to kinetic friction. Applying Newtons’ second to the xand yaxes, we obtain
F−fk=ma
N−mg =0
respectively. The second equation yields the normal force N=mg , so that (using Eq. 6-2) the kinetic
friction becomes fk=µkmg. Thus, the first equation becomes
F−µkmg =ma =0
where we have set a= 0 to be consistent with the idea that the horizontal velocity of the stone should
remain constant. With m=20kgandµk=0.80, we find F=1.6×102N.