newtonian gravity - Potential energy and work-energy theorem

Let us suppose that a ball is present on earth's surface, gravity acts on it. Now if a force is applied on the ball in the opposite direction of gravity such that the applied force counters gravity and the ball starts moving upwards with constant velocity 'v'. At height 'h' lets says that it has energy $mgh + \frac{mv^2}{2}$ and since it is moving with constant velocity so the change in kinetic energy is zero. Therefore by Work-Energy theorem the net work done by all the forces would be zero(Since change in K.E. is zero) so at some other height say s such that s > h the total energy of the ball would be $mgh + mgs + \frac{mv^2}{2}$. My question is since the net work done by all forces is zero where does the extra P.E.(mgs) come from?

Any help is highly appreciated.