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Work done in vertical springs & External Agent

  1. Oct 17, 2015 #1
    1. The problem statement, all variables and given/known data
    I've encountered an answered problem in this book titled, "Physics for Scientists and Engineers with Modern Physics, 9th Ed" by Serway at page 187.

    So there is a spring which is hung vertically.
    1. The first condition was the spring at an equilibrium state (ΔX=0)
    2. The second condition, a block with a mass of 0.55 kg was attached at the lower end of the vertical spring and the spring stretched 2.0 cm

    The work done by the spring was around -0.054 J and the work done by gravity on the object was around 0.11 J

    So heres my question :
    1. Why is the work done by the spring and the work done by gravity are not the same amount ? If the work done by the spring is negative and the work done by gravity is positive, should it not counter each other to achieve an equilibrium state or not moving at all ?
    2. What does the negative and positive meant on the works ? Does it refer to direction (up or down) ?
    3. I also found a formula for work by external agent. Is gravity also included as an external agent ? If not, why ? Isn't gravity came from outside of the block ?
    2. Relevant equations
    Not needed

    3. The attempt at a solution
    -
     
  2. jcsd
  3. Oct 17, 2015 #2
    Did you check to see if the mass is at equilibrium at the location 2 cm below the unstretched length of the spring, or do you still have to support the mass with your hand when the mass is at this location?
     
  4. Oct 17, 2015 #3

    JBA

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    I need to see what equation you used to calculate the PE change in the spring.
     
  5. Oct 17, 2015 #4

    haruspex

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    I assume that in the second condition the mass is hanging at equilibrium and stationary.
    If you were to attach the mass at the unstretched length and let go, what would happen? When the mass reaches the equlibrium position (a stretch of 2cm) would it be at rest?
     
  6. Oct 17, 2015 #5
    The equation I used for the work done by spring is
    Ws= 1/2.k.(x initial2-x final2)
    and for the work done by gravitational force is
    W= m.g.ΔX

    Umm, well the problem doesnt said anything about reaching a condition of rest when it strectched 2cm, so I just assume it was in a condition of rest.

    But if it were to reach to reach a state of equilibrium at somewhere x cm, would the work done by the spring equals the work done by the gravity ? And what does the positive and the negative in work means ?
    And why not use the formula for work done by external agent (W ext=1/2.k.(x final2-x initial2) for gravitational force ?
     
  7. Oct 17, 2015 #6

    JBA

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    I have verified your PE calculations and have worked with springs for many years. The initial deflection is irrelevant to the PE calculation on the spring. The spring rate for a spring determined by the force and deflection values determined at any point in the spring travel regardless of it prior loading or deflection. The problem expression clearly implies that the spring is once more in equilibrium again at the 2 cm deflection after the .55 kg load is applied.

    in line with Haruspec's above post, with the given calculated spring rate, the .55 kg mass would need to travel exactly twice the 2 cm given distance to equal the gravity PE. Is that fact trying to tell us something related to the velocity of the mass as it drops to, or through, its final resting equilibrium point?
     
  8. Oct 18, 2015 #7

    haruspex

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    You didn't really answer my questions. Let me try again.
    You have a spring hanging at some unstretched length. You do two experiments.
    In the first experiment, you attach a mass and gradually let the mass descend until it hangs without moving. The extension of the spring is 2cm.
    In the second experiment, you start with the unstretched spring, attach the same mass, and let go. Will the mass come to a stop as soon as it has descended 2cm? If not, what is going on? It has lost the same GPE as in the first experiment, and the spring has gained the same PE as in the first experiment.
     
  9. Oct 18, 2015 #8
    Oh I'm very sorry. Well, I suppose in the second experiment the spring will stop as it has descended 2cm, because the spring's condition in the first and in the second are the same. If in the first experiment, it stops at 2cm with a mass of m kg, the in the second experiment it should also stop at 2cm with the same mass attached to it.
    I'm sorry if it's not the answer you hope. If I'm wrong, please correct me, thank you so much in advance.
     
  10. Oct 18, 2015 #9

    haruspex

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    No, that's the point -it won't stop. This is the diference betwen letting the mass down gently (during which the weight of the mass exceeds the spring tension, and therefore does work on your hand) and letting the mass fall. If you let it fall, all the while the extension is less than 2cm the weight of the mass exceeds the tension, so the mass continues to accelerate. At the 2cm mark it will have KE, which will carry it on through the equilibrium position (for anothr 2cm). This explains your energy discrepancy.
     
  11. Oct 18, 2015 #10
    Oh I understand now. Thank you so much sir
     
  12. Oct 18, 2015 #11

    JBA

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    Today I found a free pdf downloadable copy of the referenced "Physics for Scientists and Engineers with Modern Physics, 9th Ed" by Serway" textbook with the problem and solution, including the given different energy values.
    (Anyone interested in this textbook only needs to Google the above textbook name and select the https://faculty.psau.edu.sa/filedownload/doc-3-pdf-67a5de9fa89738da0c6835ef457b5878-original.pdf [Broken] option to download the entire textbook.)
    Michael, however failed to include in his thread statement that, in that book, at the end of the problem solution, the reader is referred to the following book Section on Kinetic Energy for the explanation and method for determining the missing energy that completes the energy solution.

    The solution, as implied above and suggested by haruspex lies in finishing the analysis by calculating the KE involved.
    From the referenced KE Section of the textbook, we learn the completion of the solution is by calculating the velocity of the block as it falls v = sqrt(g*h) =.443 m/sec; then, the resulting block KE = 1/2 m*v^2 = .054 j; we have the required missing energy in the original calculation.

    Michael, if you missed that reference note, then your confusion is entirely on you for lack of diligence. On the other hand, if you saw that reference but did not understand the explanation and did not note that fact in your initial thread statement, then "shame on you" because this issue might have been resolved much sooner otherwise.
     
    Last edited by a moderator: May 7, 2017
  13. Oct 18, 2015 #12
    Oh I'm sorry I've missed the reference note, sorry for my lack of diligence sir.I'll try to be more diligent next time. Also thank you for everyone trying to help
     
    Last edited: Oct 18, 2015
  14. Oct 18, 2015 #13

    JBA

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    That's OK Michael, if you had not started this thread I would have missed a good learning opportunity and access to an excellent physics reference textbook. At least you included the textbook reference that helped officially close the issue for everyone involved.

    Just to finish the discussion as to how energy can be balanced with KE = 0 at the new equilibrium point, see the below paragraph from the textbook that addresses that issue.

    "Physics for Scientists and Engineers with Modern Physics, 9th Ed" by Serway at page 190.
    "The only way to prevent the object from having a kinetic energy after moving
    through 2.0 cm is to slowly lower it with your hand. Then, however, there is a third
    force doing work on the object, the normal force from your hand. If this work is
    calculated and added to that done by the spring force and the gravitational force,
    the net work done on the object is zero, which is consistent because it is not moving
    at the 2.0-cm point."
     
  15. Oct 19, 2015 #14
    Yes, thank you very much sir for helping me also for the additional info. And sorry again for the inconvenience. I'll be more diligent next time.
     
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