# 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.

• Answered here: physics.stackexchange.com/a/396198/45164 Aug 20 '18 at 6:58
• @MarkH The OP is confused because it is not realised that the ball alone cannot store gravitational potential energy and also the system under consideration is not defined. Aug 20 '18 at 9:08

What have missed out in your question is a clear statement of what is contained in the system that you are considering.
If the system is the ball alone then it cannot be a store of gravitational potential energy that is stored in the ball and Earth system.

If you consider the ball and Earth system and apply external forces to separate the ball and the Earth then the work done by the external forces increases the gravitational potential energy of the system.

If you consider the ball alone as the system then the system is acted on by two external forces.
The downward gravitational attraction of the Earth (ball's weight) and the upward force that you are applying on the ball which is of the same magnitude as the weight of the ball.
Thus the net force on the ball is zero,; no net work is done on the ball; the kinetic energy of the ball stays constant.