The same question was posted on Physics Stack Exchange : Blocks spring system. It contains a good explanation.

The key point is that the force which causes $m_2$ to move is **not** $F$, it is $F_2=\mu m_2 g$ just as you expect. $F_2$ equals $F$ plus the rate of change of momentum of $m_1$.

You are thinking of the situation in which the two blocks are initially in contact. In that case $m_1$ cannot move until $m_2$ also moves, and the force needed to move both is $\mu (m_1+m_2)g$. In the present case $m_1$ has accelerated over some distance, building up momentum. It then **collides** with $m_2$ via the spring. The force between the two blocks during the collision equals $F$ plus the rate of change of momentum of $m_1$.

Another analogy is to think of a policeman trying to force open a locked door, which requires a force $F_2$ to break the lock. If he leans against the door and pushes against it the maximum force he can apply is the static friction force $F_1$ between his shoes and the ground. However, if he takes a run up to the door and collides with it he can apply more force. The force which accelerates the policeman is still the friction force $F_1$, but when this is applied over a distance it builds up and stores momentum $P$ in the policeman. If the policeman bounces off the door elastically in time $\Delta t$ the force actually applied to the door is then $F+\frac{2P}{\Delta t}$. The 2nd term is the rate of change of momentum of the policeman, and can be much larger than $F$.

edited Mar 21, 2017 by sammy gerbil