This post is a revised version of my previous post 66 due to suggestions of
@jbriggs444 and other posts that changed my thinking.
A sphere or radius R and mass M is pushed tangentially with constant force ##F## for ##X## meters as with a flat bar of length ##X## moving across the top of the sphere. The sphere moves without slipping due to a frictional force ##f## horizontally on a flat surface as shown below. The frictional force contributes to the net force on the sphere as well as the rotation.
The force being constant produces a constant translational acceleration ##a = F/M-f/M## and a constant rotational acceleration ##\alpha = L / I ## where## L =(F+f)⋅R## and ##I## is the moment of inertia. I will assume for this discussion that ##I=kmR^2##. 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 twice the length ##X## of the bar causing the sphere to move only a distance##X## and rotates the sphere θ radians where ## \theta =X/R##.
Using the above relationships we determine that $$a=2F/M(1+k)$$ and $$\alpha= 2kRF/I(1+k)$$
The translational work is## W_{T}=F⋅X## since f is not applied over a distance along the surface. The rotational work is ##W_{R} = (F+f)R \theta##. The total work ##W_{total} = 2FX##
After the force has been applied the sphere moves with constant velocity ##V## and rotates with an angular velocity ##\omega##. The two which are related by $$V = R\omega$$.
This results in the kinetic energy of translation ##KE_T=\frac{1}{2}⋅mV^{2}## and the kinetic energy of rotation ##KE_R=\frac{1}{2}⋅I\omega^{2}##.
Because of the constant acceleration, the translational and rotational velocities are related to their respective distances of the applied force by similar equations.
$$V ^{2} = 2⋅ a ⋅x $$
and
$$\omega ^{2} = 2⋅ \alpha \theta$$.
Using these relationships avoids the consideration of the time the force is applied.
Using the above relationships we find
$$V^2= 4FX/(M(1+k))$$
and
$$\omega^2=4kFX/(I(1+k))$$
The total kinetic energy is
$$KE_{final}= \frac{1}{2}⋅I\omega^{2}+\frac{1}{2}⋅mV^{2}$$
So $$KE_{final}= \frac{1}{2}⋅I\omega^2=4kFX(I(1+k))+ \frac{1}{2}⋅M4FX/(M(1+k)$$ thus
$$KE_{final}= 2kFX/(1+k)+ 2FX/(1+k)$$ which is ##2F## the work done by ##F##
If the frictional force is eliminated and the sphere slides, it will rotate due to ##FR## translate due to ##F##. We find interestingly that ##KE_{T} =FX## and ##KE_{R} =FX##. The work is shared equally between translation and rotation.