1. The problem statement, all variables and given/known data A 104.0-kg block is on a horizontal frictionless surface, on top of it is a second smaller block of 68.0-kg. A horizontal force F of 613.4 N is applied to the upper block which accelerates with 5.84 m/s2. 1. Find the coefficient of kinetic friction between the two blocks. 2. Find the acceleration of the lower block while the two blocks are in contact. 2. Relevant equations F=ma, fk=(coeff kinetic)* F_Normal 3. The attempt at a solution I found the coefficient of kinetic friction by taking the horizontal force (613.4N) and subtracting from it the force needed to move the block which is F=ma (68.0*5.84). That gave me a fk of 216.28N which I then divided by the normal force (mg= 666.4) to get the coefficient. I am having trouble finding the acceleration of the bottom block, I thought doing F=ma (613.3-216.28)=(104+68)a would give me the acceleration but it doesn't. Can someone give me a hint as in what to do because I can't figure it out.