I have an alternate scenario to offer that looks at the problem from a different perspective. I am going to assume that m2 is not rigid, but deformable, and is characterized by an elastic shear modulus G. I know how to do problems like this because of my prior experience with solid mechanics. What I'm going to show in this scenario is that, for a deformable m2, it is possible for slip to occur at the upper interface with m1, while, at the same time, the lower surface of m2 does not move.
Let A be the area of m2 in contact with m1 and with the table, and let h be the thickness of m2. The displacement to the right at the bottom of m2 is going to be zero. The displacement to the right at the center plane of m2 is going to be δ1 and the displacement to the right at the top surface of m2 is going to be δ2. If we did not allow a displacement at the top surface (i.e., temporary slip with kinetic friction, followed by re-establishment of stick) the frictional force at the top surface would be 25N, which would exceed the critical force for release of static friction. Therefore, the top surface of m2 must slip at the kinetic friction force of 12.3 N (temporarily, until static friction is re-established). Once static friction is re-established, the force at the top of m2 will remain at 12.3 N. That is where we are going to examine what prevails.
Now for the analysis:
Shear strain in the lower half of m2 = δ1/(h/2) = 2δ1/h
Shear strain in upper half of m2 = 2(δ1-δ2)/h
Shear stress in lower half of m2 = 2Gδ1/h
Shear stress in upper half of m2 = 2G(δ1-δ2)/h
Frictional force on top surface = 12.3 N = 2GA(δ1-δ2)/h
Frictional force on bottom surface of m2 = 50.0-12.3= 37.7 N = 2GAδ1/h
If we solve for δ1 and δ2, we obtain:
##δ_1=\frac{37.7h}{2AG}##
##δ_2=\frac{25.4h}{2AG}##
So the bottom of m2 will not be displaced to the right at all, and the center plane of m2 will be displaced to the right about 50% more than the top surface.
In the end, m2 would not be accelerating.
Chet