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Mechanical properties of Solids Concept

  1. Sep 27, 2010 #1
    When a block of mass M is suspended by a long wire of length L, the elastic potential energy stored is given by 1/2 x Mg x l, where l is the elongation produced.
    The loss in gravitational potential energy of the mass-earth system is Mgl. I wonder where does the other Mgl/2 go?
     
  2. jcsd
  3. Sep 27, 2010 #2

    Doc Al

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    What do you think? Hint: How was the mass lowered from the original position to its final position?
     
  4. Sep 28, 2010 #3
    The mass lowers due to its weight (stress) on the wire, which is stored as elastic potential energy of the rod.
     
  5. Sep 28, 2010 #4

    Doc Al

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    Are you just dropping the mass or are you lowering it gently?
     
  6. Sep 28, 2010 #5
    Just dropping
     
  7. Sep 28, 2010 #6

    Doc Al

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    OK. The gravitational PE goes into both elastic PE and KE. Eventually, that KE will be 'lost' to internal energy.
     
  8. Sep 28, 2010 #7
    So the elastic PE includes 2 quantities- PE due to elongation and PE due to shrinking of wire?
    How is the KE lost to internal energy - You mean by the production of heat etc?
     
  9. Sep 28, 2010 #8

    Doc Al

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    :confused:
    Yes. If you just drop the load, the mass will oscillate about the equilibrium point. Due to internal friction, eventually it will come to rest.
     
  10. Sep 28, 2010 #9
    When the wire is stretched, there is some longitudinal elongation produced as well as decrease in cross sectional radius (shrinking in that sense).
    Here the wire in not completely elastic to execute oscillations.
     
  11. Sep 28, 2010 #10

    Doc Al

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    You only mentioned elastic PE in your original post, so I thought that's what you were asking about. The 'additional energy' can go into various other forms, including inelastic deformation.
     
  12. Sep 28, 2010 #11
    Thanks!
     
  13. Sep 28, 2010 #12
    Let me suggest a solution without ad hoc energy losses.
    The mass m is moving under the influence of the force f = g - kx (k is the spring constant of the wire). So you can´t apply
    Epot = mgh here the way you do it. If you work it out keeping this in mind, all will be ok.

    Edit:
    A different view of the situation: Energy has a sign (you can put energy into a system or you can take energy from it). Don´t forget you have two forces (elastic/gravity) here. Take a look at their directions.
     
    Last edited: Sep 28, 2010
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