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If I lift an object with a specific force, the force does work on the object and energy gets stored in the object in the form of potential energy (m * g * h (h = height)). But there is also a force downwards - the gravitational force which also does work, doesn't the energy equal the gravitational work minus the work I do to lift it up, but that is very close to 0 which seems weird compared to excercises i've done on work.

Because when a car drives, it has a force on it which pushes it forward therefore the force does work on the car, but there is also a force of friction on the car, which also does work which brings to my first question: do you subtract the friction work with the work that drives it forward to ultimately get the energy or the kinetic energy in this case of the car?

second question: if the work is net zero when it's in certain height how does this work of net zero become kinetic energy if there isn't any energy when i drop it from that certain height?

I hope you somewhat understand my problem, if there's any grammatical mistakes or such just tell me and I'll fix

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    $\begingroup$ Possible duplicate of Net work done on the body when we lift it and put it on the table is zero? $\endgroup$ – sammy gerbil Jun 7 '17 at 16:15
  • $\begingroup$ What exactly is the question? The text needs some line-breaking and periods. $\endgroup$ – Steeven Jun 7 '17 at 16:19
  • $\begingroup$ @sammygerbil, the answer to the question is kind of my problem , if the work is net zero when it's in certain height how does this work of net zero become kinetic energy if there isn't any energy when i drop it from that certain height? $\endgroup$ – Iram Haque Jun 7 '17 at 16:19
  • $\begingroup$ Fixed the questions @Steeven $\endgroup$ – Iram Haque Jun 7 '17 at 16:22
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    $\begingroup$ As the answer explains : you did +ve work, gravity did -ve work. That -ve work is stored as PE in the gravitational field. It came from the chemical energy in your body. When the object is dropped, its KE comes from the stored gravitational PE, not from the net work done on the object. $\endgroup$ – sammy gerbil Jun 7 '17 at 16:38
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There is a net work done when you consider the entire process of picking the object up from rest, it's just not something you'd use in calculations so you usually don't talk about it and people will find ways to work around it.

But the fact is that even if you pick up the object at "constant speed" the whole time, there was still an initial impulse force you had to apply to it to make it accelerate from zero up to the desired constant velocity. Afterward, because you are lifting it at constant velocity you can see that that means the acceleration is zero again. That is a short lived net force greater than gravity, and accounts for the change in kinetic energy.

However, to understand why the object will continue to gain potential energy despite the fact that as you lift it at constant velocity you are doing zero net work, is that you don't use the net work done here. When trying to determine the change in potential energy of an object submersed in a "field" (e.g. the gravitational field) you consider only the work done against that field, not by that field. So when calculating the change in gravitational potential energy you consider the work done by all forces except gravity. However, the net work (just during the lifting process, not considering any initial acceleration caused by you lifting the object from rest) is still zero.

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  • $\begingroup$ So you essentially ignore the gravitational work during the whole process when for example calculating the potential energy? Is this also true when my example of the car, which also has a frictional work, do you not take into account the frictional work and just ignore it and continue with the work which pushes the car? $\endgroup$ – Iram Haque Jun 7 '17 at 17:51
  • $\begingroup$ Yes, but remember that's only for finding "potential energy" due to something, for net work you'd still include ALL the forces. For the car example you would only ignore friction if you were asked to find the "potential energy of friction", which I'm not sure actually makes any conceptual sense. For a car driving, I don't think there is any real meaningful "potential energy". The net work for the car would be the work done by the engine minus the work done by friction. $\endgroup$ – ison Jun 7 '17 at 18:01
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It is not strictly true that when an object is at a height its total energy is zero, at that point it has a higher gravitational potential energy then it being a certain height below. To clarify:

Gravitational potential energy is given by -GMm/r (M is mass of earth and m is mass of object) where r is distance from center of earth. So when r is larger then GPE is less negative(higher in value) than when r is smaller( closer to earths surface ) so in the end the object's potential energy DOES decrease and thus it's kinetic energy increases. The potential energy lost is converted to the kinetic energy. In our school we often simplify problems by taking GPE zero at the earths surface hence confusion arises. Hope this helps.

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