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Work done by force acting on a spring

  1. Sep 4, 2012 #1
    Let say we have a horizontal mass-spring system oscillates without friction on a smooth surface.
    One side of the spring is attached / fixed to a wall. The other is where the mass is attached to.
    Let say we choose our system to be the mass and the spring (negligible mass), so our free body diagram for the system that we have chosen consists of the following forces:

    Normal force on the mass. (A)
    Weight on the mass. (B)
    Spring force on the mass (C). Same force (D) will be acting on the spring according to Newton's third law.
    Another force (E) is exerted on the spring by the wall.

    Question: What is the work done by force E on the spring? Explain.
     
  2. jcsd
  3. Sep 4, 2012 #2

    AUM

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    You say the spring is attached to the wall, so is the force pulling outward, or downward?
     
  4. Sep 4, 2012 #3
    When the spring is compressed, that force is a push. When spring is stretched, that force is a pull.
     
  5. Sep 4, 2012 #4

    Doc Al

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    Staff: Mentor

    What do you think? What's the displacement of the wall?
     
  6. Sep 4, 2012 #5
    The wall is not moving. Its displacement is zero. Force E is exerted on the spring by the wall.
     
  7. Sep 4, 2012 #6

    Doc Al

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    Right. So what does that imply about the work done by the wall?
     
  8. Sep 4, 2012 #7
    Is this homework?
     
  9. Sep 4, 2012 #8
    The net work done on the wall for sure is zero. Is there any relationship between the work done on the wall with the work done on the spring by the wall? I do not know anything about that.
     
  10. Sep 4, 2012 #9
    No. I just wondered when I read a book and did some exercises.
     
  11. Sep 4, 2012 #10

    Doc Al

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    The work done on the wall is zero. Similarly, the work done on the spring by the wall is zero, as the point of application of the force does not move.
     
  12. Sep 4, 2012 #11
    I suggest you draw yourself a diagram.

    There are two forces acting in (on) the spring.

    One of them does not move its point of application.
    One of them moves its point of application.

    Which one do you think does the work involved?
     
  13. Sep 4, 2012 #12
    Then force D will do some work on the spring since the point of application is moving because this end of the spring is oscillating together with the mass. So, there will be a net work done on the spring.
    Force E does zero work on the spring while force D does some work on the spring. So, there is indeed a non-zero net work done on the spring. But since the mass is negligible, according to Newton's second law, the net force acting on the spring will be zero which means D and E have the same magnitude in opposite direction. And if the net force acting on the spring is zero, the net work done on the spring must also be zero. Thus, there is a contradiction here. Therefore, can we say force E does zero work on the spring because the spring is at rest and not moving, so it doesn't have any displacement?
     
  14. Sep 4, 2012 #13

    Doc Al

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    Why do you think that?

    That doesn't follow at all. If you pull on the ends of a spring, say, the net force is zero yet you are doing work to expand the spring.
     
  15. Sep 4, 2012 #14
    I think you are confusing yourself by referring to 'net' force and 'net' work.

    Just consider plain old fashioned force and work.

    How's the diagram coming along?
     
  16. Sep 4, 2012 #15
    I know that the net work done on a body equals to the sum of the work done by each of the forces acting on the body which also equals to the work done by the net force acting on the body.

    I can't see the whole picture. I don't get the same thing when I view it from different angles. Something is missing here and I don't know what.
     
  17. Sep 4, 2012 #16

    Doc Al

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    Right.
    Wrong.

    The work done by a force equals the force times the displacement of the point of application of that force (in the direction of the force).
     
  18. Sep 4, 2012 #17
    Net work done on a body = (F1 dot s) + (F2 dot s) + (....) = (F1 + F2 + ...) dot s
    = (Net force) dot s
    = Work done by net force

    For simplicity, I assume all the forces are constant so that we can use (F dot s).

    I am going to bed. Thanks for spending your time, both of you.
     
  19. Sep 4, 2012 #18

    Doc Al

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    You are also assuming that each force has the same displacement 's', which is not the case.

    What you're probably confusing this with is this:
    F1*scm + F2*scm + F3*scm + ... = Fnet*scm = ΔKEcm

    Where scm is the displacement of the center of mass.
     
  20. Sep 4, 2012 #19
    Good night, Leong.

    You are making this wa.a.ay to complicated.

    The word 'net' imples a sum of some negative and some positive quantities.

    Work is always positive (reckoned in the same system).

    Forces can be positive or negative and so the net force idea has meaning, although I prefer the word 'resultant' since direction is also involved.
    Thus a zero resultant is entirely possible as with the zero vertical resultant in your example.

    If work is done, on the other hand, you cannot 'balance it out' as with forces. Work represent the quantity of energy passed from one system to another.

    If system A does work on system B it is not true that system B does a corresponding (or any other) amount of work on system A.

    So the force extending the spring sideways does work on the spring, and in so doing passes an amount of energy equal to this work to the spring.

    But the spring does no work on the source of that force.

    A good expression is total work, if you must add up separate amounts.
     
  21. Sep 4, 2012 #20

    Doc Al

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    Work can certainly be negative.

    OK.

    Sure it does. (But here we are talking about the work done on the spring.)
     
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