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Homework Help: Spring Problem (Need to know if its right)

  1. Mar 13, 2006 #1
    A glider of mass 0.150 kg moves on a horizontal frictionless air track. It is permanently attached to one end of a massless horizontal spring, which has a force constant of 9.0 N/m both for extension and for compression. The other end of the spring is fixed. The glider is moved to compress the spring by 0.180 m and then released from rest.

    (a) Calculate the speed of the glider at the point where it has moved 0.180 m from its starting point, so that the spring is momentarily exerting no force.

    So potential = kinetic energy
    1/2kx^2=1/2mv^2
    v^2=kx^2/m
    v^2=(9)(.18^2)/.15
    v=1.39

    (b) Calculate the speed of the glider at the point where it has moved 0.250 m from its starting point.

    Here x=.07 because isnt x the distance from the equilibrium point, which is at .18? If so, use the same equation

    1/2kx^2=1/2mv^2
    v^2=(9)(.07^2)/.15
    v=.542 m/s

    Thanks for checking
     
  2. jcsd
  3. Mar 13, 2006 #2
    a - it is given the question itself that it is mommentarily at rest.
     
  4. Mar 13, 2006 #3

    Doc Al

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    Staff: Mentor

    This is correct (but don't leave off the units). Note that the total energy at the point in question is purely KE; the elastic PE is zero at that point. Initial total energy = final total energy.

    Careful! "PE = KE" is not true in general. What is true is "initial total energy = final total energy". Set the total initial energy (which is the same as in part (a)) equal to the sum of PE + KE at the point in question.
     
  5. Mar 13, 2006 #4
    Sorry - I am really frustrated with the misreadings I make.
     
  6. Mar 13, 2006 #5
    So in part b:
    Initial = Final energy
    Initial = 1/2mv^2 = .5 * .15 * 1.39^2 = .1449 J
    Final = KE + PE
    KE=1/2 * .015 * v^2
    PE=1/2kx^2=.5*9*.07^2

    Is that the right set up?
     
  7. Mar 13, 2006 #6
    Initially KE = 0 and there is only PE. The object is at rest initially.
     
  8. Mar 13, 2006 #7
    Oh so PE=1/2kx^2=.5*9*.18^2?
     
  9. Mar 13, 2006 #8

    Doc Al

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    Staff: Mentor

    Right. That's the initial PE (from the starting point point where it was released); since it was released from rest, that's also the total energy.
     
  10. Mar 13, 2006 #9
    Alright, so I did it out..

    Initial Energy=Final Energy
    Initial energy is stated above
    Final Energy=KE+PE=
    .5*9*.18^2 = (.5*.015*V^2)+(.5*9*.07^2)

    Solving for V gives V=4.06 m/s?
     
  11. Mar 13, 2006 #10

    Doc Al

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    Staff: Mentor

    Better check your arithmetic since the velocity you solved for in part (a) was the maximum possible (all of the energy was KE in that case). :wink:
     
  12. Mar 13, 2006 #11
    Ah whoops... .15 not .015...so 1.284 m/s...Thanks
     
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