Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work-Energy Theorem Question

  1. Dec 7, 2015 #1
    Assuming you are lifting a block up 1 meter from rest to rest with constant work. You know that the work is -deltaU or 10. However, you also know W=deltaKE which is 0. You finally know that W=Fx=10*F. How do you explain why the numbers are different? Thanks!
     
  2. jcsd
  3. Dec 7, 2015 #2

    Ken G

    User Avatar
    Gold Member

    It's the difference between work by a single force, versus net work. The work-energy theorem only applies to net work, which is zero. There is positive work by the lifting force, and negative work by gravity, so the net work is zero.
     
  4. Dec 7, 2015 #3
    Thanks for your reply! I still have questions however. If this is true, when does W=deltaEmech, W=deltaKE, work=-deltaU? Also, since W=deltaEmech, shouldn't the total be deltaKE+deltaU=deltaU=-10 not 0?
     
  5. Dec 7, 2015 #4

    Ken G

    User Avatar
    Gold Member

    W doesn't equal delta E mech, it equals delta KE. But you need to use the net work-- the sum of all the work done by all the forces on the object. There would be no point giving delta E mech its own name, it is always zero unless there is heat generated (and when that happens, life can get complicated, because you have friction and inelastic collisions and all that jazz). So if you would give delta E mech a name, it would just be the negative of the heat dissipated.
     
  6. Dec 7, 2015 #5
    Thanks again! However, my textbook states that work done on a system, assuming no friction is deltaEmech. Also, if net W=deltaKE=-deltaU, why is deltaKE 0 and why is -deltaU 10?
     
  7. Dec 7, 2015 #6
    Assuming no friction (or other non-conservative forces) the mechanical energy is conserved so the change is zero.
    However the work doesn't have to be zero. Maybe you misunderstand what the book actually says.
    It seems this is a common matter of confusion. It was discussed here several times.
    What book is that? Can you show an image?
     
  8. Dec 7, 2015 #7
    I am using Halliday Fundamentals of physics 6th edition. nasu or someone else, can you explain why I am getting different numbers though for the problem? And, yes, this concept is confusing me! A more specific question, I guess, is when is W=deltaEmech, when is W=deltaKE, and when is work=-deltaU? Thanks!
     
  9. Dec 7, 2015 #8

    Ken G

    User Avatar
    Gold Member

    It would be quite unusual to say that work is delta E mech, unless they are not counting work done by the forces associated with potential energy functions (called "conservative forces") as work! But it normally would be called work, even the conservative forces. If you don't count the work done by conservative forces in the work-energy theorem, then the work done equals the change in mechanical energy (if there's no heat loss). I can't recommend that conceptual approach however, work is too valuable a concept to waste on purely nonconservative forces. Just count all forces as able to do work, and then if there is no heat loss, work is always equal to the change in kinetic energy, you can derive that from F=mA.

    So in summary:
    work by a single conservative force = - delta U
    is only for work done by conservative forces (forces with a potential energy associated with them). This isn't a statement about reality, it is just the definition of potential energy-- it's bean counting.

    net work = delta KE
    is the work-energy theorem, equivalent to F=mA. This is a statement about how reality works, given the more general definition that net work is the sum of all the work by all the forces (including friction).

    work = delta mechanical energy
    is only true if you don't count work done by conservative forces as work on the left-hand-side (their - delta U is pulled to the other side of the equation, and absorbed into mechanical energy, just a different way to count those beans). So the meaning of "work" here is external forces applied manually (like a person pulls on a rope or something), or frictional forces. If you include work done by frictional forces, you are not including heat that is dissipated in the mechanical energy, that's just KE+U.

    One final note-- when you do inelastic collisions, like objects that stick together, it is generally hard to include the work done there, so you usually drop the attempt to bean count using work, unless you are told the heat dissipated in the elastic collision. Usually you just don't use conservation of energy at all for inelastic collisions, but you can if there is friction, if you can track how much work the friction does.
     
    Last edited: Dec 7, 2015
  10. Dec 7, 2015 #9
    Actually, the textbook says this formula is true for nonconservative forces. It says the work done by an external force for conservative forces is W=deltaEmech and for nonconservative, it is W=deltaEmech+deltaEtherm

    EDIT Usign another textbook (college Physics by Serway), it seems that it is a little clearer. It explicitly state that non-conservative work is deltaEtherm. Now, can some one help me with the individual problem? Why do I get different numbers by doing W=deltaKE=0 and W=-deltaU=10? Thanks!
     
    Last edited: Dec 7, 2015
  11. Dec 7, 2015 #10

    Ken G

    User Avatar
    Gold Member

    It's because the first answer uses net work, which is zero, and the second answer only uses the work done by the person lifting the object, which is positive. There is negative work done by gravity, which makes the net work zero, but that's a conservative force rather than an "external" force, so one of the books is counting that work separately in its bean counting. I don't recommend that at all.

    What you can take to the bank is the work-energy theorem, which says, count all the work done by all the forces (including friction), add it up, and it equals change in KE. If any of the works that are done cannot be tracked, you will need to be just told what they are, or not use conservation of energy at all.
     
  12. Dec 7, 2015 #11
    Wait, so W=deltaKE gives total work and work=-deltaU gives work by a specific force?
     
  13. Dec 7, 2015 #12

    Ken G

    User Avatar
    Gold Member

    Yes, if U refers to a given potential energy term (that's the single force). If U adds them all, then it refers to all their work added up.
     
  14. Dec 8, 2015 #13
    Oh... since Ui would be 0 and Uf =Ug+UF=0. Ok! So for W=KE, if we knew the individual deltaKE for the force, it would be 10. I have one last question. W=Fx also. Thus, assuming F is constant, W done by the force would be W=F*1=F. Thus, work depends on F. However, W=-deltaU=10 gives us that it is always 10. Why are these yielding different results?
     
  15. Dec 8, 2015 #14

    Ken G

    User Avatar
    Gold Member

    They aren't-- you are forgetting that if you would lift something against gravity, you don't get to use whatever F you like, it must balance gravity. If you use less F, it will drop instead of lift, and if you use more F, you get some excess KE by the end.
     
  16. Dec 8, 2015 #15
    Yes. However, what if the final velocity is not 0. Then the extra KE would just cause a bigger final velocity which won't influence the PE. Thus, even if you have extra KE, the work still should be 10.
     
  17. Dec 8, 2015 #16

    Ken G

    User Avatar
    Gold Member

    The work is only 10 if the force is 1, and that barely balances gravity, so there'll be no excess KE. The principles work.
     
  18. Dec 8, 2015 #17
    Firstly, since the work is 10, W=F(gravity)*x+F(force)*x=>10=-10+F, so isn't F=20.
    If this is so, isn't it possible to choose the magnitude of the force you apply to the object? Why does the force HAVE to be g+10?
    Sorry if I am being rally naive about this topic.
     
  19. Dec 8, 2015 #18

    Ken G

    User Avatar
    Gold Member

    I should have said the force is 10, I forgot it is the distance that is 1. Still, there's no force that is 20 here, if you lift something up against gravity, such that it acquires no significant velocity but rises up a distance of 1 (let's not bother with units, it's metric), the work looks like this:
    work by external force = 10*1 = 10
    work by gravity = -10*1 = -10
    net work = 0
    change in kinetic energy = 0
    which all checks with no significant velocity, verifying that we did indeed need an external force of +10 to get this done.
     
  20. Dec 8, 2015 #19
    You are mixing the work done by different forces. And using units won't hurt either.:)

    If you lift a 1kg block (you never mentioned the mass of the block, I think) to a height of 1 m "from rest to rest" the net work is zero. The change in KE is zero. This is the work-energy theorem.
    The work done by gravity is -10J (assuming g=10 m/s2). The force of gravity is opposite to the displacement so the work is negative.
    You can associate this work with a change in gravitational potential energy, ΔPE=-W=10J . The gravitational PE has increased by lifting the body. The KE did not change. So the change in mechanical (PE+KE) is +10J.

    Because the net work is zero, the work done by the force lifting the body should be +10J, to cancel the -10J done by gravity.
    So the work done by this force is equal to the change in mechanical energy.

    As long as you use the definition consistently there is no contradiction.
     
  21. Dec 8, 2015 #20
    Oh wait... Yeah! I forgot that potential is for a specific force. Can you explain to me specifically in what cases -deltaU and deltaKE gives you the work of a specific force and when it gives you total work? Also, why must F be 10. Can't you just apply any force to the object you are lifting? Sorry if I am bothering you too much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Work-Energy Theorem Question
  1. Work-Energy Theorem (Replies: 2)

Loading...