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Homework Help: Conservative Forces and Fg Questions

  1. Nov 17, 2012 #1
    Im SOOOO Confused.

    Im studying Work, Energy and Society and I'm really darn confused.

    I have the highest mark in the course, and If I'm confused, I'm wondering how others are feeling.

    Whenever I have multiple questions, and they may sound stupid but please help me out:

    1. Does Fg do work? I never understand when Fg does work when it doesn't. Please help me understand if Fg does work, and if it does, when.

    2. I don't understand what my teacher meant when he said "For each conservative force there is a form of energy associated with it. Therefore we never calculate energy done by conservative forces"

    3. When I solve problems, I'm never sure when I need to set directions as positive or negative, or when I need to rely on Cosθ (Ex: Work=FxDistancexCosθ) but when you work with energy, if something moves up its + when something moves down its -?

    4. Is gravitational potential energy is equal to the work done by the force that would lift hte object to that height? Yes or no?

    5. Is Gravitational potential energy the work done by Fg? I thought Fg does no work :/

    PLEASE Help me.

  2. jcsd
  3. Nov 17, 2012 #2


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    If it is a force being displaced in the direction of or opposite the force, does it do work?
    Forces don't 'do' energy, but they can do work and change or transform the energy of a system. This statement is not clear.
    The direction of an objects motion does not determine the sign of its energy. What is the Kinetic Energy of an object of mass m moving down at a speed v? What is the Kinetic Energy of an object of mass m moving up at a speed v?
    not always.
    Work results in energy changes. Work and energy are not the same.
    Why did you think that?
  4. Nov 17, 2012 #3

    Simon Bridge

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    If an object is displaced (changes position) when an unbalanced force acts on it, then the unbalanced force does work if the displacement is in the same direction as the force.
    You are right - it does not make sense - we frequently calculate changes in energy in conservative force fields. I think your teacher is trying to say something else and it didn't come out right - you should ask about it: ask for an example.

    Perhaps your teacher means that when you move in a closed path in a conservative field, the total change in energy is zero.
    You should do your problems the other way - decide on which direction is positive first and then change the equations (like those trig functions) to suit the resulting relationships. Use the math like a language to describe the setup. When you use an example from class to help you solve a problem, use the method the example illustrates, not the equations the example ends up with: the method is general, the equations are only for that exact kind of problem.

    If you post an example (as a new question) I can show you what I mean.
    gravitational potential energy is the work done to lift an object through that height.


    The work done lifting an object against gravity is not done by gravity - it is done by whatever is doing the lifting - as you can easily verify by lifting something heavy.

    When an object falls from a height, then the displacement is in the same direction as the force, and gravity is doing the work.

    When we lift an object we are storing energy in the gravitational field and when it falls, that energy is being released.
  5. Nov 18, 2012 #4
    The thing I think Im having trouble with is this:

    1. When energy is being conserved the work being done on the object is 0.

    2. Because Fg is a conservative force, If there are no non conservative forces on it, then the total work done on the object is 0.

    Therefore the mechanical energy at moment 1 is the same as the mechanical energy at moment 2.

    But what's confusing me is that when you lift an object over your head, gravity now does do work.

    That's what is confusing me.

    So what I can conclude is that gravity does work, but when it is the only force doing work, there is no change in energy therefore the energy in conserved? But that kind-of contradicts itself, because work is the change in energy. Please help me figure out where I'm slipping up.

    Please Help,

    Last edited: Nov 18, 2012
  6. Nov 18, 2012 #5
    What I didn't understand is when you have an object at a height of 10M above a reference point per-say, would you need to set its height as positive or negative? Or is the gravitational potential energy positive or negative depending on whether the object is above or below the reference point?
  7. Nov 18, 2012 #6

    Simon Bridge

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    This is not correct - energy can be transferred from one source to another. For instance, when lift a heavy object you get tired. You have lost energy - so you did work - the object gains energy - so it had work done on it. Energy is still conserved.

    The statement you are thinking of is that the total work in moving about a closed path in a conservative force field is zero.

    As you see above - this is an error. An object moving along a closed path in gravity has zero net work done on it. When an object goes up, it gains energy, which it loses when it comes back down again - zero net change in energy, zero net work. However, work is being done on both legs of the trip.

    Work is happening from moment to moment.

    It doesn't matter - you can make your coordinate system any way you like. However, you have to make the physics guide the equations you write.

    i.e. if I put the reference level for gravity at the level of the table-top, and all the motion is above the table, then it is convenient to put the +y axis pointing upwards. In this coordinate system, the gravity force is pointing in the -y direction. Therefore Fg=-mg and lifting objects in the +y direction requires positive work so P=+mgy.

    OTOH: I may be dropping objects from the table (as in a ballistics experiment). In that case, all motion is below the tabletop and it becomes convenient to make +y direction downwards. In that case, Fg=mg and U=-mgy because I have to move in the -y direction to do positive work.

    But I am free to make the axis point in any direction I like, so long as the physics stays the same. The axis is just the map - the physics is the territory - it is important not to get them confused.
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