How is this seemingly counterintuitive result in a rotation problem explained? In the classic spool problem, where a spool of string of mass M is unrolled with a force F, like in the diagram, after working through the equations with Newton's laws of rotation and translation, we get a acceleration of 4F/3M, more than F/M, with a process like.

Standard solution(f for static friction force):
F+f=MA
(F-f)R = Ia = 1/2 MR^2 * A/R --> F-f= 1/2 MA
or(setting it up with friction pointing other way)
F-f=MA
(F+f)R = Ia = 1/2 MR^2 * A/R --> F-f= 1/2 MA
2F=3/2 MA
F=3/4 MA --> A= 4M / 3F
The textbook explanation is that there is a force of static friction F/3 pointing in the same direction as the applied force that causes this, but if the linear acceleration was 4/3 (F/M) as well as the spool accelerating rotationally, wouldn't this violate conservation of energy as only F force is applied to the spool? Where does this "extra" force come from or how can we explain it? I'm not sure how the spool can accelerate faster and gain rotational energy compared to a sliding frictionless block of the same mass with the same force applied to it. Are there problems with the underlying assumptions in the calculations?
Textbook solution is this:

 A: Your issue is that a force $F$ applied through a distance $\Delta x$ results in more kinetic energy here than if the same force was applied through the same distance on a sliding frictionless block. 
The problem is the equation $W = F \Delta x$ only applies to point particles, or objects that move totally uniformly like point particles. When you have a rotating and moving object, $\Delta x$ is ambiguous; it should be the $\Delta x$ of the point where the force is applied. In this case, the top of a rolling ball moves twice as fast as its center, so $\Delta x$ is actually twice what it appears to be.
Then the final energy should be at most $2W$, not $W$. You can check that it's indeed less than $2W$.
(Another way to see that $\Delta x$ is actually double is to consider what is actually doing the work. Suppose there's a machine pulling in the rope far off to the right. In the rolling situation, the machine pulls in twice as much rope length as in the sliding situation. From its perspective, it might be just lifting a weight twice as far, so the energy output has to be double.)
