A 5kg block is placed on top of a 15kg block that rests on a frictionless table. The surface between the top and bottom blocks is roughened so that there is no slipping between the blocks. A 40N horizontal force is applied to the top block. What is the minimum coefficient of static friction necessary to keep the top block from slipping on the bottom block? F=ma f=mew FN Ok, in my first attempt I figured that if the blocks don't slide apart, then the entire sysstem would be accelerating. I said 40N = (20)a, and found that a=2m/s^2 Using this I said (5kg)(2m/s^2) = mew FN 10 / (9.8 X 5) = .20 However the answer says that it is actually .6 Then I though that maybe the system could not accelerate at all because the opposing frictional force will stop it from accelerating. Im not even sure if im on the right track though, and I cannot figure this out! Thank you!