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Conservation of energy

  1. Jun 27, 2009 #1

    ham

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    I am a tutor. I ran across this problem is a high school physics textbook. See attachment x162, my solution is on x163. I thought that mechancical energy is conserved and that loss of potential energy should be equal to gain in elastic (stored) energy. Ithink the spring constant must be wrong???
     

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  2. jcsd
  3. Jun 27, 2009 #2

    Doc Al

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    Staff: Mentor

    Note the phrase "allowed to come to rest". If the weight were just dropped, there would be both spring PE and kinetic energy. That extra kinetic has been dissipated somehow. (The mass may have been lowered gently to the equilibrium position, or just allowed to oscillate until the kinetic energy was "lost" to internal friction or air resistance.)
     
  4. Jun 27, 2009 #3

    ham

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    I think that although not specifically said, there is no wind or other resistance here. If the mass were dropped, it would oscillate as the potential energy in the spring would be transferred between kinetic energy and gravitional potential energy. At rest, there would be no kinetic energy so I would think all of the PE would be transwferred to stored elastic energy. I still think the book answer is wrong.
     
  5. Jun 27, 2009 #4

    Dale

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    This problem is not a conservation problem. It is just trying to teach the student how to calculate the elastic potential energy.
     
  6. Jun 27, 2009 #5

    ham

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    Dale, Regardless of the objective of the problem, I think there is an error here. Please check my calculations and tell me whee I am wrong.
     
  7. Jun 27, 2009 #6
    my bet is that the problem wants you to apply the following equation:

    [tex]U= \frac{1}{2}kx^{2}[/tex]

    since Force is related to Potential Energy by the following relation

    [tex] F= -\frac{dU}{dx} [/tex]

    and the force of a spring is [tex]F= -kx[/tex]

    the negative integral with respect to x would give

    [tex]U= \frac{1}{2}kx^{2}[/tex]

    and x is the change from the relaxed point with 0 mass(really just the spring mass)
     
  8. Jun 27, 2009 #7

    ham

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    Smash, I agree that that equation e = (1/2) kd^2 should be used. This is not the issue. I think the k value, or something else, is wrong in order to conserv energy.
     
  9. Jun 27, 2009 #8

    Dale

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    You just didn't need to calculate the mass of the weight nor the change in gravitational potential energy. Your calculations for the elastic potential energy are correct.

    The k value is fine. This is not an energy conservation problem.
     
  10. Jun 27, 2009 #9

    Doc Al

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    If dropped without any dissipative forces, it would oscillate. It would not come to rest at the equilibrium position, but (momentarily) at the bottom of the oscillation. The book's answer is correct, given the stipulation "allowed to come to rest". That stipulation implies that all kinetic energy has been dissipated. (Otherwise it just bounces up and down, with maximum kinetic energy at the equilibrium position.)

    Mechanical energy of the spring-mass system is not conserved here.
     
  11. Jun 27, 2009 #10
    my mistake i failed to see what you were really asking here.

    Doc Al got it though. Spring systems that come to rest are nonconservative. that is where your loss in energy is
     
  12. Jun 27, 2009 #11

    Dale

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    Wow, pretty impressive. Three replies in the same minute! My kids would say "Jinks, you owe me a coke"
     
  13. Jun 27, 2009 #12

    ham

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    Doc Al. Theoritically, energy in a sping is stored and not used, like energy in a capacitor and inductor. If the mass was dropped, it should oscillate indifinately. Lets asume we dont drop the mass but lower it slowly. The change in PE would be 6.52, and the stored energy, per calculation would be 3.26. Where did the PE go? I agree that at the equiliberium point of the spring, the KE would me at max. If we plotted the energy and displacements of this problem, we would have a sinusoid going forever (with no friction) ?

    Agree????
     
  14. Jun 27, 2009 #13

    Dale

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    But the problem specifies that the energy is taken out of the system (used) by the satement "allowed to come to rest".
     
  15. Jun 27, 2009 #14

    Doc Al

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    Right.
    The energy was dissipated in whatever mechanism lowered the mass. If you lower it with your hand, your internal energy will increase as you absorb that energy. (Think about walking downstairs. You start at rest at the top of the steps and end up at rest at the bottom. Where does the energy go?)
    Sure.
     
  16. Jun 27, 2009 #15

    ham

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    Dale, I see your point. However, given that this is a high school physics problem, I don't think the intent of the problem was to consider damping where I would agree some energy would go into the damping medium such as air friction. If we asume there is no friction, and we lowered the mass gradually so it would not oscillate but rather hang there, would you agree that there is a problem?

    I have to leave for 1 hour and will return. I hope you guys will still be online.
     
  17. Jun 27, 2009 #16
    after thinking for a minute, I came to an answer, but Dale beat me to it.

    Lets think about it:

    Originally, spring is at equilibrium. When you add the mass, the spring oscillates, the height changes. This change in gravitational energy is split up; half goes into Kinetic Energy, half goes to Elastic Energy in the spring. If you let the spring come to rest, the only way to do this is to take energy OUT of the system. The only energy you can take out here is the kinetic. So that's what happens to the energy you're missing
     
  18. Jun 27, 2009 #17

    Doc Al

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    No, there's no problem (as I have attempted to explain).

    To "answer the question" all one need do is plug into the formula for spring PE. But I applaud your efforts to understand why this doesn't fit with your assumption that mechanical energy is conserved. (The answer is that mechanical energy is not conserved. There are additional dissipative forces at work.)
     
  19. Jun 27, 2009 #18
    when you lower the mass gradually, what you are doing is applying an upward force so that the acceleration of the mass is very small, so that at the bottom there is no motion. Therefore, you are applying a force, over a distance, which is Work. Work= change in Kinetic Energy. Normally the Kinetic energy, if you allow the spring to oscillate, would be the same 3.26 value at the bottom point of oscillation. Therefore, for there to be no kinetic energy, you have to apply an equal amount of work to the system, which takes the energy out.

    Efinal= Einitial - Work
     
  20. Jun 27, 2009 #19

    RonL

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    If you compare the spring and weight, to the bounce of a basket ball, you have the same manor of energy dissipation, however the spring allows a smooth change of direction and a much longer oscillation period.
     
  21. Jun 27, 2009 #20

    Dale

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    Just go to your answer, use the first and the fourth line, delete the second and third lines. That is all that this question is asking. It is definitely a high-school level problem.

    It does not matter how it comes to rest at the stretched length, whether it is through friction or gradually lowering the mass. All that the question is asking for is what is the elastic potential energy at that stretched length. It is completely independent of any specific mechanism for removing the energy.

    The "allowed to come to rest" specifies that energy is removed but does not give details of how it is removed since the details are irrelevant for the problem. You can prescribe many different mechanisms for removing the energy, all will result in the conservation of energy (i.e. the work done on the mechanism will exactly equal the change in PE) but none are relevant for the problem since it is not a conservation problem.
     
    Last edited: Jun 27, 2009
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