- #1

DrBanana

- 35

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Suppose somehow an object is moving upwards with a speed ##v##, at this point I start applying a force ##F## that is equal to its weight, so the net force on the object is zero. So it will continue moving upwards with its initial speed. Suppose after the height difference is ##h##, I stop applying the force. At this moment, the object has its initial kinetic energy, but now it also has gained some potential energy. If the net force on the object was zero, the net work must also have been zero. But that should have meant no energy flowed to the object, so how did it gain potential energy? I know this question has been asked a lot of times.

The usual explanation goes like this (Work applied by me, potential energy, kinetic energy, work done by field):

##W_{applied}=\Delta U + \Delta K##

##F*h=-W_{field}+ \Delta K##

##F*h=-(-F*h) + \Delta K##

##\Delta K = 0##

Therefore ##W_{applied}= \Delta U##

I know my confusion arises from the fact that I've only been considering the box, but if potential energy is concerned, I should consider the box-earth system. In that case, the work done by the force applied by me comes externally, and gets stored in the system as potential energy. In this case, equality of energy is achieved (that is, energy in = energy supplied). However, I'd like to know the 'internals' of this, as I'm still having a hard time digesting it. I apply a force on the box, but the work done by that force gets treated as force done on the system, why? Does gravity do work on the system, or on the box? If the work done by my force on the system gets stored as potential energy, where does the work done by gravity 'go'?

Edit: I think I've boiled my problem down to this: Is the potential energy gained by the system due to the work done by my force, or due to the negative work done by the gravitational field? How can both be true?

The usual explanation goes like this (Work applied by me, potential energy, kinetic energy, work done by field):

##W_{applied}=\Delta U + \Delta K##

##F*h=-W_{field}+ \Delta K##

##F*h=-(-F*h) + \Delta K##

##\Delta K = 0##

Therefore ##W_{applied}= \Delta U##

I know my confusion arises from the fact that I've only been considering the box, but if potential energy is concerned, I should consider the box-earth system. In that case, the work done by the force applied by me comes externally, and gets stored in the system as potential energy. In this case, equality of energy is achieved (that is, energy in = energy supplied). However, I'd like to know the 'internals' of this, as I'm still having a hard time digesting it. I apply a force on the box, but the work done by that force gets treated as force done on the system, why? Does gravity do work on the system, or on the box? If the work done by my force on the system gets stored as potential energy, where does the work done by gravity 'go'?

Edit: I think I've boiled my problem down to this: Is the potential energy gained by the system due to the work done by my force, or due to the negative work done by the gravitational field? How can both be true?

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