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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.

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    $\begingroup$ Answered here: physics.stackexchange.com/a/396198/45164 $\endgroup$
    – Mark H
    Commented Aug 20, 2018 at 6:58
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    $\begingroup$ @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. $\endgroup$
    – Farcher
    Commented Aug 20, 2018 at 9:08

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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.

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  • $\begingroup$ So if changing the choice of the composition of system changes net work done, cant I also say that energy will differ when viewe differently? $\endgroup$
    – Aveer
    Commented Jun 12, 2022 at 14:53
  • $\begingroup$ It is the net work done on a system by external (to the system) forces which needs to be considered. $\endgroup$
    – Farcher
    Commented Jun 12, 2022 at 16:00
  • $\begingroup$ Yes, so by changing what is included in the system, I can chnage the energy levels? (as external forces always change the ME if unbalanced) $\endgroup$
    – Aveer
    Commented Jun 13, 2022 at 7:47
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Maybe I'm misunderstanding your question.
If the ball is moving "upward" means the "upward" force is slightly greater than gravity, so, total force on the ball is not zero and by Newton's first law, the ball can't be moving with constant velocity.

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  • $\begingroup$ No it is moving with constant velocity so applied force is equal to gravity. $\endgroup$ Commented Aug 20, 2018 at 6:48
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The work-energy theorem says that work (net forces times distance) turns into kinetic energy, not total energy. Maybe you misread this theorem? Just google it.

So with that, the work-energy theorem works out right: no net forces, no change in K.

The change in potential energy comes from non-gravitational forces moving the ball against the gravitational field.

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  • $\begingroup$ Then where does the potential energy come from. $\endgroup$ Commented Aug 20, 2018 at 6:58
  • $\begingroup$ That's the third paragraph: from the non-gravitational forces pushing the ball up. $\endgroup$ Commented Aug 20, 2018 at 7:13
  • $\begingroup$ The ball alone cannot store gravitational potential energy. The system that you are considering needs to be defined to answer such a question. $\endgroup$
    – Farcher
    Commented Aug 20, 2018 at 7:35

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