homework and exercises - Velocity of a body in a pulley problem involving a spring fixed to the ground

So the picture shows the whole situation. My question is, how do you find the velocity of the object after it had descended 10 cm (0.1 m) ? Pulley's radius 0.2 m, moment of inertia 0.2 kg/m

I have a problem because the pulley is supposed to have friction if it's rotating,right? Because the torques on either side shouldn't be equal. My teacher used $$ \frac {Iw^2}{2} + \frac {mv^2}{2} + \frac {kx^2}{2} = mgh$$ Where $$ v=rw $$ and $$h=0.1m $$ as given. So the answer came as 0.5 m/s. My problem is, if there is friction why aren't we taking the energy loss due to that into account?

$mgh $ would have to be spent on that as well. Plus where do the individual tensions on each side come into play? I understand that $mgh $ kind of accounts for those tensions, but not exactly how.

If mgh is doing work on the pulley, does that include work done by/against tension as well as the energy loss due to friction?

Plus, tension on one side is stretching the spring, right? So why shouldn't we consider that separately?

If somebody could clarify I'd be grateful. Thanks so much. No working out needed, I just want to know exactly what happens to the energy