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If we lift and object up, the net work is clearly zero because the kinetic energy after the lift and before the lift are the same (0). However, the object still seems to gain gravitational potential energy. How did it gain that energy?

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    $\begingroup$ By the external force you applied maybe? $\endgroup$ – MMa Nov 22 '17 at 5:28
  • $\begingroup$ @Mohammed I was under the impression that the gravitational force cancelled out my external force. Thus, work done by gravity cancelled out work done by me. Thus, energy provided by me was cancelled out by energy provided by gravity. $\endgroup$ – Typical Highschooler Nov 22 '17 at 5:45
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The kinetic energy is 0 even after you did work on it, is because gravitational force did equal and opposite work on it I.e. negative work. But gravity is a conservative force. The negative work done by such forces gets stored as potential energy. It's a property of conservative forces.

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You are quite right in saying that the net work done on the object by the force that you exerted on the object and the gravitational force that the Earth exerted on the object is zero and hence the change in kinetic energy of the object is zero.
The net work done is zero because the force that you exerted on the object and the gravitational force that the Earth exerted on the object are equal in magnitude and opposite in direction.
So far the object alone has been treated as a system and that object alone cannot have gravitational potential energy.

However when both the object and the Earth are treated as a system the system can have gravitational potential energy.
When you exert a force which is external to the object & Earth system and do work separating the object and the Earth the object & Earth system gains gravitational potential energy.

Note that you must actually exert two equal magnitude and opposite direction external forces to increase the separation between the object and the Earth. One of the forces is exerted by you on the Earth and the other force is exerted by you on the object.

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  • $\begingroup$ I found your answer quite interesting, but could not understand what you meant in the last paragraph. Why are we required to exert equal and opposite forces on the system? Can't we exert any force on the system ? $\endgroup$ – Abhinav Dhawan Nov 22 '17 at 13:35
  • $\begingroup$ @AbhinavDhawan If you apply a force on the object alone then the object will move and so will the centre of mass of the object & Earth system. So the system will gain kinetic energy. $\endgroup$ – Farcher Nov 22 '17 at 17:25
  • $\begingroup$ Is this also happening when we are lifting a rock or any other thing. (We may not be in contact with earth...) ? $\endgroup$ – Abhinav Dhawan Nov 22 '17 at 17:28
  • $\begingroup$ @AbhinavDhawan If you are in mid air and exert an upward force on a rock the rock will move anf the Earth will not move due to the force that you exerted on the rock and so the centre of mass of the rock & Earth system will change. $\endgroup$ – Farcher Nov 22 '17 at 19:30
  • $\begingroup$ Then won't the Earth-rock system gain any potential energy ? $\endgroup$ – Abhinav Dhawan Nov 23 '17 at 3:06
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The total work done on the object is related to its gain or loss of kinetic energy by $W = \Delta K$. The total work, in your situation, is due to the force you exert when lifting the object ($W_\text{lift}$) and the force gravity does against this lift ($W_\text{gravity}$). The overall relation is $W_\text{lift}+W_\text{gravity}=\Delta K$.

Gravity is a conservative force, meaning the energy it "spends" doing work comes from variations of the gravitational potential energy stored in the system (energy stored in the gravitational interaction among parts of the system), $W_\text{gravity}=-\Delta U$.

If the object goes from rest at some height to rest at a higher height ($\Delta K=0$), that only means in the end all energy you "spent" lifting it ($W_\text{lift}$) was converted into potential energy ($\Delta U$), $\Delta U = - W_\text{gravity}= W_\text{lift}$. Energy was not cancelled, as you may have thought, but converted. Indeed, if you suddenly release the body, this potential energy will again be converted, but this time into kinetic energy (the object will gain velocity as it falls).

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Using an example, a 10 kg object at rest on the ground (r=0) has zero Kinetic and Potential Energy (simple model).

Lifting that 10 kg object up 3 meters requires about 294 joules - there is kinetic energy required for the lifting that is converted to potential energy when the lifting stops and requires no further work - or does it? The stored potential energy can be converted back to kinetic energy (for example about 30 watts for 10 sec) by dropping the object from height and reduce to zero Kinetic and Potential energy state at rest on the ground.

I would like to better understand how physics explains the continuous "work" required to maintain the potential (gravitational) energy at a stationary height. Every backpacker knows that the "work" isn't done after lifting a heavy backup up on their shoulders. Potential energy stored in heavy loads sitting on a table continue to exert pressure from tabletop to ground with that continuous "work" being spread throughout the physical support structure and may eventually even cause the table to fail!

Where is that continuous energy requirement being accounted for? Molecular and environmental heat, relativistic mass gain and loss? I know that I wouldn't be able to hold 10 kg above my head for long even though the standard calculation would say that, theoretically, no further work is required once height is reached.

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If you lift an object up, the net work is not zero. The work is equal to the force required to lift the object times the distance you raised it. By lifting it, the object gains potential energy (not kinetic energy). You can get the energy back by letting it fall, which converts the potential energy back to kinetic energy.

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  • $\begingroup$ Why is it not zero when final KE = initial KE = 0? $\endgroup$ – Typical Highschooler Nov 23 '17 at 2:50
  • $\begingroup$ Kinetic energy is equal to one-half mass times velocity squared. Since the velocity of the object is zero both before and after raising the object, you are correct that the kinetic energy is zero in both cases. That doesn't mean the net work is zero. Work (the energy expended) is equal to force times distance. So it takes work (energy) to raise the object. Since energy can't be created or destroyed, that energy has to go somewhere. In this case, it is stored in the form of potential energy (not kinetic energy), due to the increase in the object's height. $\endgroup$ – David Rose Nov 27 '17 at 17:05
  • $\begingroup$ You have covered what the lifter's work does. This does not cover the work that gravity does. Could you please explain that? $\endgroup$ – Typical Highschooler Nov 28 '17 at 23:32
  • $\begingroup$ Imagine you are on an asteroid with negligible gravity, and that you "lift" an object up off the surface of the asteroid against the force of a spring, instead of against gravity. The work you do (force times distance) is stored in the stretched molecules of the spring. Similarly, when you lift an object on the Earth, the energy is strictly stored in the gravitational field. But in Newtonian physics we generally simplify by assuming the energy is a property of the lifted mass itself, and don't bother with the intermediate step of gravitational field energy. $\endgroup$ – David Rose Dec 4 '17 at 18:30

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