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Having trouble thinking about conservative forces

  1. May 22, 2012 #1
    I'm having trouble with the concept of conservative forces. I've been studying things of that nature for awhile now and can make most calculations on my own, but there are a few specific details that confuse me. It's easiest to explain my confusion with two simple examples.

    First Example:
    Let's just consider a particle moving in the Earth's gravitational field. Let the potential energy be zero at a height of zero. So, the total potential energy is just mgh. It makes perfect sense that when this particle is moving in the field that energy will be conserved (i.e. an increase in height leads to an increase in potential energy and an equal decrease in kinetic energy). The part that bothers me is thinking about a person lifting an object of mass m a height h. Now, the work that person must do to lift the object is simply mgh. Also, the work done by the gravitational field on that object during the lifting process is -mgh. There is a net work of zero done on the particle (which is good since the kinetic energy didn't change, and there is no violation of the work-energy theorem). The problem is it appears to me that no energy is put into the particle. Whatever energy the person puts into the object, the gravitational field takes out of it (since mgh + -mgh = 0). However, as we all know the potential energy did increase by mgh. Where is this energy coming from?

    Second Example:
    Consider a particle of mass m attached to a horizontal spring at it's equilibrium position. In order to determine the potential energy as a function of displacement x, we strech the string and determine the amount of work we had to do on the particle. However, I have the same issue again as above. I did 1/2 kx^2 amount of work on the particle while the sping did -1/2 kx^2 worth of work on the particle (i.e. no net work). But, it's potential energy increased by 1/2 kx^2. Where is this energy coming from?

    The only answer I can seem to come up with, is the fact that you can only define a potential for conservative forces. Therefore we only set the change in potential energy equal to the work done by the conservative force (i.e. the gravitational force, not the man's force during lifting). The net work done on the object is still zero, so there is no change in kinetic energy. But, the total work done by conservative forces (for the gravitational example) is -mgh (resulting in an increase of mgh in it's potential energy). This whole idea is still slightly confusing. Does energy conservation work out in this case? Can someone explain what is really going on here? It would be great if you could point me to a source to read up on conservative forces that discuss my troubles.

    Thank you to anyone who replies.
     
  2. jcsd
  3. May 23, 2012 #2
    silmaril89
    The way to visualize this is that we are transferring energy between two entities. Consider transferring energy from A to B (with no losses). "A" does positive work on "B". "B" does equal negative work on "A". Consider the elastic collision between a moving pool ball (A) and an identical stationary pool ball (B). After the collision, all of A's energy has been transferred to B. "A" did positive work on "B", "B" did equal negative work on "A". The fact that they sum to zero is just saying that total energy between the two was conserved.
    In the gravitational potential energy case the energy has been transferred from the human body of the lifter (muscle energy, chemical energy, food calories consumed), to gravitational potential energy of the lifted weight.

    The pool ball case was reversible, ball "B" could go up a ramp, back down, and recollide with "A" and return all of the energy back to ball "A". The person lifting case is probably not reversible. It is unlikely we can find a way to have the weight fall fall back down and have that energy restored to the persons muscles.

    We can create a reversible example with gravitation potential energy: a compressed vertical spring contains energy. The spring is released lifting a weight into the air. When the weight reaches its highest point and the spring is relaxed, we have transferred all of the spring energy to gravitational potential energy. The spring did + work on the weight (and lost energy in the process), and gravity did (-) work on the weight (and gained energy in the process). Next the ball falls, reversing the process, and gravity will transfer the energy back to the spring.

    Hope this helps
     
    Last edited: May 23, 2012
  4. May 23, 2012 #3

    Andrew Mason

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    The energy comes from a loss of kinetic energy of the body. If the body moves up at constant speed through application of a force, you can think of it still as a loss of kinetic energy. It is just that the kinetic energy is being constantly replaced by the applied force as gravity takes it away. Potential energy is a kind of bookkeeping tool. When gravity takes away kinetic energy, we use potential energy to keep track of how much kinetic energy has been lost.
    This is essentially the same problem. Potential energy can be thought of as a means of keeping track of the loss of kinetic energy. If you stretch the spring and hold m still, it has 0 KE. When you let it go, the spring does positive work on the body (increasing its KE) until it passes the equilibrium position. Then the body uses its KE to do work on the spring.

    AM
     
  5. May 23, 2012 #4
    In your first example, if you only look at the object-earth system, then yes, it gains mgh energy in the form of potential energy, and the total energy is not conserved. However, if you include the person lifting the object in the system, then the object still gains mgh energy, but the person loses mgh energy in the form of work done on the object, so the total energy is conserved. The energy that the object gained came from the person lifing the object, so when you take both into account, the net energy change is zero. The negative work done by gravity is also the work done by lifting, just in a different perspective, so trying to take both into account, you are essentially "double counting" the same work.

    The same argument applies for your example two, just replace mgh with 1/2 kx^2 and replace lifting with stretching.

    And conservative forces are ones where the amount of work done to move an object does not depend on the path that the object takes. Gravity is a conservative force, since only vertical displacement counts(a curvy path is the same amount of work as a straight line), and friction is a nonconservative force since the amount of work done depends on the total disance travelled (a curvy path would take more work than a straight line).
     
    Last edited: May 23, 2012
  6. May 23, 2012 #5
    @the_emi_guy:

    What you said, certainly does help me move in the right direction when thinking about this.

    However, let "A" be the object being lifted and "B" be the man lifting the object. In my example, I have "A" did positive work on "B", and the gravitational field did negative work on "B". This is different then what you discussed.

    @Andrew Mason:

    I think this is the most helpful way for me to think of it. I like to think of potential energy as a "bookkeeping tool". It also helps to still think of the fact that the object is losing kinetic energy to the gravitational field which is being replaced by the work done by the person.

    @Jasso

    Thank you, this was helpful as well.


    I think I actually do understand it now. Thank you to all of you.
     
  7. May 23, 2012 #6

    Andrew Mason

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    Instead of lifting an object through a height h you could propel it with exactly the amount of kinetic energy needed to just reach height h (at which point it is held there). There should be no difference in your analysis of work and energy.

    It would also help to keep in mind that there are two forces acting on two bodies so the work is not done on the same body.

    There is m the object, and M the earth. Both exert forces on each other and both move relative to the centre of mass, which remains fixed in space. However, the work done by m on M is not so easy to see because M moves a very small distance (lets call it s).

    As m moves from the surface to height h above the surface, m does work on M equal to M(-g)s = ΔKE. At the same time M does work on m: mgh = -ΔKE.

    AM
     
  8. May 23, 2012 #7
    You are not losing kinetic energy if it is moving at constant speed.
     
    Last edited: May 23, 2012
  9. May 23, 2012 #8

    jtbell

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    When you're doing an energy analysis of a situation that has a conservative force like gravity, you include either the work done by that conservative force or the potential energy associated with that force; never both at the same time.

    Consider the work-energy theorem, and split the total work done on an object into the work done by conservative forces (e.g. gravity) and non-conservative forces (e.g. you):

    $$\Delta K = W_{net}$$
    $$\Delta K = W_{c} + W_{nc}$$

    Rearrange it a bit:

    $$\Delta K - W_{c} = W_{nc}$$

    The change in potential energy is the negative of the work done by conservative forces, so:

    $$\Delta K + \Delta U = W_{nc}$$
     
  10. May 23, 2012 #9
    @jtbell

    Thanks, that again makes sense. The use of equations makes it quite easy to see what is happening. Is it possible to show me with equations what it was that I was doing? You mention that you use either the work done by a conservative force, or the potential energy associated with that force. I was using both? If so, can you show me mathematically what I was doing?
     
  11. May 23, 2012 #10

    jtbell

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    In your first example, you wrote:

    Your mistake here was talking about "what the gravitational field (force) takes out of it". Your description includes the gravitational potential energy, so you must omit the gravitational force (and the work done by it) from your description. In terms of my final equation:

    $$\Delta K + \Delta U = W_{nc}$$
    $$0 + \Delta U = W_{nc}$$
    $$\Delta U = W_{nc}$$

    The potential energy increases by an amount equal to the work that you do on the object while lifting it, because the final change in kinetic energy is zero (starting at rest at height 0 and ending at rest at height h).
     
  12. May 24, 2012 #11

    Andrew Mason

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    That is true. But if it is moving at constant speed, it will not stop when you remove the applied force - it will keep going up until its kinetic energy is exhausted by gravity i.e. converted into gravitational potential energy.

    If you apply an upward force of mg to a stationary body on the surface of the earth, it does not move. So you give it an initial upward push of force F>|mg| through a distance Δh1 and it ends up with kinetic energy KE = (F-mg)Δh1. You then stop the applied force and it goes up an additional height Δh2 = KE/mg = (F-mg)Δh1/mg before stopping.

    From this we see that: FΔh1 = mg(Δh1 + Δh2)

    AM
     
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