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Non-zero potential energy

  1. Apr 20, 2015 #1
    I have a question about the conservation of energy. Imagine that there is an object on a level plane with zero kinetic energy and zero gravitational potential energy. Now imagine moving this object at some constant velocity in the y-direction for a few meters. Since it was moved at a constant velocity, there was no chance in kinetic energy, and thus no work was done. However, at the new height, there is non-zero potential energy mgh. If no work was done, how did the object go from having 0 kinetic energy and 0 potential energy to having 0 kinetic energy and mgh potential energy? Where am I going wrong?
  2. jcsd
  3. Apr 20, 2015 #2
    Okay, here's my two cents...
    The object starts on the plane with zero kinetic energy and we define the plane to be at a point where the graviational potential energy is zero. When we move the object up to a distance height we must do work against the weight of the object (gravitaitional force) Work done is defined as Force (mg) multiplied by distance moved in the direction of the force (h). Therefore once we raise the object a height h we have done mgh work against gravity and potential energy = -work done for a force field such as that of the earth's gravitational field.

    Not sure if that was a particularly clear or precise answer but i hope it helps
  4. Apr 20, 2015 #3
    I still don't understand. You do positive work, gravity does work that is equal and opposite to the work done by you, so the net work done on the body is zero. Therefore, if there was no net work, no change of energy, then why is there mgh Joules of potential energy at the end?
  5. Apr 20, 2015 #4
    The change in potential energy is not equal to the net work but just the work done by gravity force. (with minus sign).
    The net work is equal to the change in kinetic energy, which is zero if the object ends at rest.
    You should not confuse the work-energy theorem with the definition of potential energy. In the first one is the net work, in the second one just the work of the force whose PE you calculate.
  6. Apr 20, 2015 #5
    That actually helps clear things up. However, I don't completely understand. According to the work-energy theorem, the net work done on an object when moving it vertically is zero. But at the same time, the object has potential energy that it it did not have before... I can't figure out where this potential energy comes from...
  7. Apr 20, 2015 #6
    Potential energy is a property associated with the gravitational field. Yes it is correct that we say the ball now has certain amount of potential energy but that is there only because of it's interaction with the gravitational field. Potential change between two points in space is defined as [tex] \phi = -∫E.dr , r1,r2 [/tex] where [itex] E [/itex] is the field (in this case gravitational) whose potential we wish to calculate. I prefer interpreting this potential as a property of the field. Now for an object of mass m, its potential energy change between these two points will be [itex] m\phi [/itex].

    See the net work and "gravitational" potential energy are two separate entities. Had the ball at height h had a velocity say v in horizontal direction, (in your example) the net work done on it would have been ## \frac{1}{2} mv^2 ## and the increase in potential energy would have been the same as ## mgh ## (which already mentioned is negative of work done by gravity).

    Here is a simple example which might help your understanding. If the ball is at height h from ground level at rest and is dropped from there in free fall then there are two ways to analyse it.
    • Go for energy conservation and use final energy equal to initial energy, that is $$ -mgh + \frac{1}{2} mv^2=0 $$ where the reference for potential energy is taken at height h from the ground.
    • Use work-energy theorem and use $$ mgh = \frac{1}{2} mv^2 $$
    Further I would like to add that the work-energy theorem is merely a consequence of the energy conservation law we know. It is one and the same thing. Which one you use depends on you.
  8. Apr 20, 2015 #7
    I don't see the contradiction between the two descriptions. Maybe the things are not clear yet.:)

    When you use the work-energy, all forces are accounted for by their work. There is no potential energy in it.
    If you use potential energy for some forces, then the work of the other forces, with no potential associated, will equal the change in kinetic and potential energy.
    In this case, as the change in KE is zero, the work done by the other forces (in this case just the force exerted by the hand on the stone) will be equal to the change in PE.
    So you can say that the PE "comes" from the work done by hand. Or by gravity, up to as sign. It does not really matter.
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