newtonian mechanics - Why is the tension between two masses connected by a rope and undergoing a force along the direction of the rope less than that force?

Two different known masses at rest on a frictionless surface are connected by a rope. A known force F is applied horizontally to mass 2 so that the boxes begin to accelerate in that direction. The acceleration of both masses together is given by a=F/(m1+m2) (I think they can be considered the same object because of the rope and the lack of friction). The only horizontal force acting on mass 1 is the pull coming from the tension in the rope. Therefore the tension T is given by T=m1*F/(m1+m2) (again, the acceleration is the same because of the rope and the lack of friction, I think).

Intuitively, this makes absolutely no sense to me, because the tension in the rope ends up being less than the force acting on mass 2. What's going on? Are you supposed to not count both masses in the calculation of acceleration? If so, why?

Answer

It is best to draw free body diagrams for the two masses.

$F$ is the applied force and $T$ the tension in the massless and inextensible rope joining the two masses.
There is no friction and both masses have the same acceleration $a$.

Applying Newton's second law for each of the masses:

$T = m_1\;a$ and $F-T= m_2\; a \Rightarrow F = (m_1+m_2)\;a$ so $F>T$

You can think of it as the force $F$ is accelerating both masses whereas the force $T$ only has to accelerate mass $m_2$.