I'm calculating more energy out than I put in

[jbriggs444]

I think I'll leave it the way it is.

OK. I will try to follow your analysis and see where it goes wrong.

The force being constant produces a constant translational acceleration a = F/M and a constant rotational acceleration α. = L / I where L (torque) =F⋅R and I is the moment of inertia. The work required to do this has two parts one rotating it and one moving the center of mass horizontally. The force is applied for a distance of its length x causing the sphere to rotate θ radians where X=R⋅θ.

You appear to be assuming that although there is rolling without slipping that the frictional force of the sphere on the platform imparts no momentum, provides no torque and does no work. That is, that the frictional force turns out to be zero.

I disagree with this, but will accept it for the purposes of argument.

Let us go ahead and apply the force $F$. The claim is that the resulting acceleration is $a = F/M$.

We can go ahead and apply the torque as well. The claim is that the resulting angular acceleration $\alpha = \frac{L}{I} = \frac{F \cdot R}{I}$. The moment of inertia of a sphere is $\frac{2}{5}MR^2$. So we end up with $\alpha = \frac{5F}{2MR}$. If we multiply by $R$ to convert this angular acceleration rate to a rate of horizontal acceleration, we get $a = \frac{5}{2}F/M$.

The angular acceleration is 2.5 times too fast. We would not actually maintain rolling without slipping under these conditions.

Non-zero friction with the platform has to provide enough retarding torque and advancing horizontal force so that the rotational acceleration and the horizontal acceleration match.

 

[A.T.]

The sphere moves without slipping horizontally on a flat surface as shown below.

Note that the OP doesn't specify this. And as @jbriggs444 points out, you cannot assume zero friction for accelerated rolling without slipping in general.