Answer
$ a_x=\dfrac{2F}{m}-\mu_r g $
Work Step by Step
First, we need to draw a force diagram of Jorge.
As we see in the right figure of the pulley:
Since the rope is massless and the pulley is frictionless, so the net force is zero.
So that the force applied by Jorge is equal to the tension on the wire.
$$F=T$$
And hence, the reaction force from the rope is still $T$ on Jorge.
Now let's work on the left figure of Jorge's force diagram to find his acceleration.
$$\sum F_x=2T-F_r=ma_x$$
$$2T-F_r=ma_x $$
And we know that the rolling friction is given by $\mu_rF_n$;
$$2T-\mu_rF_n=ma_x \tag 1$$
Now we need to find the normal force $F_n$ exerted on Jorge.
$$\sum F_y=F_n-mg=ma_y=m(0)=0$$
Thus,
$$F_n=mg$$
Plugging into (1);
$$2T-\mu_r mg=ma_x $$
Thus, his acceleration is given by
$$a_x=\dfrac{2T-\mu_r mg}{m}$$
Recalling that $F=T$, as we found above. So,
$$a_x=\dfrac{2F-\mu_r mg}{m}$$
$$\boxed{a_x=\dfrac{2F}{m}-\mu_r g}$$
