Here's the problem:
"Imagine there are two blocks laid on top of each other. One has a mass of 4kg and a string attached to it exerting force F and underneath it is a block of mass 3kg with no string attached to it.
The coefficient of static friction is 0.60 between the two blocks described The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F⃗ causes both blocks to cross a distance of 5.0 m , starting from rest.
What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?"
I'm not sure I fully grasp the understanding since my answer doesn't look correct.
Here's my attempt:
On the 4 kg block, we know that the forces in the x direction are:
$ M_4*a_{4x}=F_{pull}-f_{static 3 on 4}\\ =F_{pull}-\mu _{s}*N_{3 on 4}\\ ->4*a_{4x}=F_{pull}-0.6(4*9.8) $
On the 4 kg block we know that the forces in the y direction are:
$ N_{3 on 4}-4*g=0\\ N_{3 on 4}=4*9.8\\ $
On the 3 kg block the forces in the X direction are:
$ f_{static 4 on 3}-f_k =m_3*a_{3x}\\ 0.6(4*9.8)-\mu_{k}*68.8=3*a_{3x}\\ 3a_{3x}=0.6(4*9.8)-0.2(68.6)\\ a_{3x}=3.267\\ $
On the 3 kg block the forces in the Y direction are: $ N_{3}-N{4 on 3}-m_{3}g=0\\ N_{4 on 3}=4*9.8\\ N_{3}=4*9.8+3*9.8=68.8\\ $
Now since the two blocks move together their accelerations have to be the same. Namely, $a_{3x}=a_{4x}$. This means from above: $ (F_{pull}-0.6(4*9.8))/4=3.267\\ F_{pull}=36.688 N\\ $
The normal kinematic equations here show that; $ 5=x_{f}-x_{i}=0.5(3.267)t^{2}\\ -->5=1.75s. $
Can you possibly comment where I might have gone wrong and whether I'm not understanding anything correctly? Any advice would be extremely helpful.