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I'm afraid that I really don't understand the answer given below!
It seems to me that the problem is simply one of conservation of energy.
A ball that "Rolls without Slipping" has two parts to its total
kinetic energy when moving along a horizontal surface:
(1/2) m v^2 <=> translational kinetic energy
(1/2) I w^2 <=> rotational kinetic energy ("w" is omega)
Since the ball "Rolls without Slipping" there is NO ENERGY LOST TO
FRICTION because there is ZERO MOTION of the instantaneous point of
contact in the direction of the frictional force (leading to a zero work
integral).
When the ball encounters an incline and assuming it still "Rolls without
Slipping", it goes up the incline with NO LOSS TO FRICTION, transferring
ALL of both its translational and rotational kinetic energies directly