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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.)
     
  22. Sep 4, 2012 #21
    I know that s is the displacement of the body, not the force. Whether there are 2, 3 or 10 forces acting on the body, the body will only move in one direction and therefore one displacement. Thus, when we find the work done by each of the force, we use the same s, the displacement of the body.
     
  23. Sep 5, 2012 #22

    Doc Al

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    That's obviously not true for a non-rigid body like a spring.
    That's not the definition of work. Again, the work done by a force is found by using the displacement of the point of application of that force. What if you pulled a spring apart so that its center never moved? Would you say that no work was done, since the displacement was zero? Do you see the problem?

    As I said earlier, using the displacement of the center of mass times the force can be quite useful, but it is not the same thing as the work done by each force.
     
  24. Sep 5, 2012 #23
    Thank you for your input.
     
  25. Sep 5, 2012 #24
    This subject is clearly still active and merits further discussion.

    All 3 very good points.

    However

    This deserves closer examination.

    I was brought up with the tighter distinction work is done when a force moves its point of application.

    This is because the point of application may be moving for other reasons so the spring may be mounted on a concrete wall of a building at rest or the bulkhead of a train moving at 60 mph. Either way I will perform the same amount of work if I extend the spring, although in the second case the point of application is moving at 60 mph in addition to the extension speed.

    We can see an example of this in Leong's problem.

    Since no diagram has been forthcoming I have sketched some out in my ususal scruffy manner.

    I have separated the elements to better see the forces acting and the work done and used Leong's notation, including s for distance.

    Sketch 1 shows the spring and mass in equilibrium at rest on the surface. Since the spring is not extended only two forces are acting. The mass is at s0
    No work is being done.

    Initially the spring must be pulled out by a force F to say position s1.
    This immediately activates the full system of forces described by Leong.
    I have added the missing force D at the wall end of the spring as already discussed .

    Forces C, D and E are not the cause of the motion so, as with the train, do no work.
    The only force doing any work in this situation is F
    I am going to use the slowly applied criterion so the work done = 1/2 F (s1-s0)

    This is the energy input to the system by the initial pull F.

    At this time the system is in equilibrium as in sketch 2. F is required to maintain this equilibrium.

    At some point F is removed and the system begins its oscillation as in sketch 3.
    The system is no longer in equilibrium since it is now in motion.

    C is the pull of the spring on mass m and is the force that does the work on the mass.
    That is the spring does positive work on the mass, accelerating it to some velocity v.

    So energy from the spring is transferred from the spring to the mass.

    It is an interesting exercise to account for the values of C, D, E, and F in the light of Newton’s laws N1, N2 and N3 for each of the 3 sketched situations.
     

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  26. Sep 5, 2012 #25

    Doc Al

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    I agree that nothing much of interest would be added by viewing this problem from a frame in which the wall is moving. But I don't see that point as relevant to this problem.

    Nothing's moving, so no work is being done.

    OK, though I would have distinguished the forces at each end of the spring. As D1 and D2, perhaps.

    That really depends on what system you choose to study. If you want to look the mass by itself, then you need to consider both F and C as doing work. (The net work done is zero, of course, since the mass doesn't change KE.) And if you wanted to look at the spring alone, then you'd need to consider D2 as doing work on the spring.
    OK.

    OK.

    Interesting that you consider C as doing work now, but didn't before. :wink:

    Right.
     
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