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Question about work

  1. Nov 15, 2009 #1
    Hi guys, I'm fairly new to the forum, so I don't know whether a similar question has been posted or not, but here's my question.

    If a box is being lowered at a constant velocity, is there work done on the box.

    My understanding tells me that since W=ΔEk+ΔEp, work done on the box is equal to mgΔh.
    If this is correct, then how can one explain the cause of this work? It seems a bit confusing because there is no net force acting on the object, and yet there's still work done.

    If possible, can you guys back up your reponse with reliable sources. thanks.:smile:
     
  2. jcsd
  3. Nov 15, 2009 #2
    [tex]\Delta[/tex] kinetic energy + [tex]\Delta[/tex] potential energy = [tex]\Delta[/tex] total energy in system

    If all the forces involved in the system are conservative forces the total energy in the system is zero. If this holds then:

    [tex]\Delta[/tex] kinetic energy = - [tex]\Delta[/tex] potential energy

    therefore, your expression mg[tex]\Delta[/tex]h = [tex]\Delta[/tex].5mv^2 = 0 because the velocity doesn't change. This doesn't mean that gravity doesn't do work, it means that the TOTAL work done is zero
     
  4. Nov 15, 2009 #3
    First of all, I don't think that the applied force is a conservative force, and also change in height is not zero. This question is such a paradox, because the people that I've asked have different answers.
     
  5. Nov 15, 2009 #4
    example:

    If I am sliding a box across the floor with friction at a constant velocity, I apply a force on the box over a distance so I do work. Friction also applies a force (equal and opposite) over a distance but the box does negative work because the friction force is in the opposite direction of the displacement.

    All this happens and we conclude:

    1) I do work
    2) friction does work
    3) the velocity doesn't change, therefore the net work = 0
    4) the box doesn't accelerate because the net force is zero

    perhaps a specific problem will help?
     
  6. Nov 15, 2009 #5
    I have trouble following your response, because potential energy is involved in this problem. For there to have no work to be done on the box, the applied force has to be neglected, or else it wouldn't satisfy the work energy theorem.

    If no work is done, that means:

    Ek+Eg=Ek'+Eg'

    but there is no potential gravitational energy when the box is lowered to the ground.

    Ek-Ek'+Eg-Eg'=0
    ΔEk+ΔEg=0
    ΔEg=0

    which does not hold true since there is a change in potential gravitational energy.
     
  7. Nov 15, 2009 #6
    If we are lowering an object towards the ground at a constant velocity the force we apply is in the positive direction, opposite the displacement. Therefore, we are doing negative work on the object. Gravity is exerting a downward force in the direction of the displacement, therefore gravity does positive work. The potential energy changes but does not convert to kinetic energy.

    Without a specific problem I believe this will become repeatable.
     
  8. Nov 15, 2009 #7
    There are two ways of answering the question, I believe.

    If your system is the block and only the block, then the force of gravity is an external force, along with whatever force is acting on the block to keep it from accelerating.

    If you system includes both the block and the earth, then the system has potential energy and therefore negative work is done on the system.

    Another way to think of this is that if you have a block attached to a spring, and then define your system as just the block, you don't really care what type of force the spring exerts on the block, merely that it is an external force doing work on the block. The spring itself can be said to store potential energy; however, in the case of gravity there is no spring so you have to include both objects to truly have potential energy.
     
  9. Nov 15, 2009 #8
    Well, I guess it's more about theory than the question itself. I understand that both of the forces are doing work, and one is positive and the other is negative. And if I understand correctly, you're saying that potential energy in this case cannot be converted into kinetic energy? But wouldn't it still be stored in the object? Since when the box is let go, all of its potential gravitational energy will be converted into kinetic energy?
     
  10. Nov 15, 2009 #9
    some of the potential is still stored in the object (hence why if you were to let the box go all its potential would convert to kinetic), BUT as we are lowering the box the energy that would have produced kinetic energy is being dissipated through the work done by the force in the positive direction (our hand).
     
  11. Nov 15, 2009 #10
    The "fun" part about this is that springs are also a type of conserved forces...meaning they too can have potential energy. That'd mean that if the applied force is actually a conservative force(which i have no clue about whether it is one or not), work would be zero since the distance between the box and the starting position has increased while the distance between the ground and the object has decreased. So then is the applied force a conservative force?
     
    Last edited: Nov 15, 2009
  12. Nov 15, 2009 #11
    But I think the work done by our hand is negative(since the force is in the opposite direction as the displacement), so energy cannot be transferred from potential energy to kinetic energy.
     
  13. Nov 15, 2009 #12
    that is exactly what I am saying ^^
     
  14. Nov 15, 2009 #13
    that is why, the potential changes and the kinetic stays the same
     
  15. Nov 15, 2009 #14
    wow, that's deep. Can you fully explain it?
     
  16. Nov 15, 2009 #15
    :cool:
     
  17. Nov 15, 2009 #16
    @#$%^&* I'm so confused now :(
    Do you have resources that are about this kind of problem?
     
  18. Nov 15, 2009 #17
    Unfortunately I believe that it isn't that deep.

    Because the force your hand exerts of the system isn't a conservative force, the work energy theorem doesn't hold.
     
  19. Nov 15, 2009 #18
    not that it doesn't hold, but the mechanical energy isn't conserved
     
  20. Nov 15, 2009 #19
    my resource is University Physics by young & freeman
     
  21. Nov 15, 2009 #20
    Well then what energy is transferred in or out of the system?
     
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