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Homework Help: Other spring question

  1. Nov 1, 2015 #1
    1. The problem statement, all variables and given/known data

    A bloc of 0,2 kg is attached on a horizontal spring without any friction on the floor. The other side of the sprin gis attached to a wall. He does a movement of back and forth . Between the two extremities of the back and forth movement, the bloc has a displacement of 40 cm. The maximum speed of the bloc while he is oscillating is 0,5 m/s. What is the constant of the spring ?
    2. Relevant equations
    Kf=1/2*m*(vf)^2

    3. The attempt at a solution
    So we know that the maximum speed is 0,5 m/s. Then the kinetic energy is going to be kf= 1/2*0,2 kg* 0,5^2 = 0,025 J

    Other than that, I'm not sure where to go. I know that my speed at the two extremities are going to be 0 m/s...
     
  2. jcsd
  3. Nov 1, 2015 #2

    PeroK

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    If the speed is 0, where has all the kinetic energy gone?
     
  4. Nov 1, 2015 #3
    To the potential energy.
     
  5. Nov 1, 2015 #4

    PeroK

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    Correct. And what do you know about the potential energy of a spring?
     
  6. Nov 1, 2015 #5
    This and this :

    U=1/2*k*e^2

    Kf+Uf=Ki+Ui
     
  7. Nov 1, 2015 #6

    gneill

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    So, where might the KE have gone (where's it hiding) when the block is at one of the extremities?
     
  8. Nov 1, 2015 #7
    It became a potential energy. This :
    U=1/2*k*e^2
    Kf+Uf=Ki+Ui
     
  9. Nov 1, 2015 #8

    gneill

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    Right. Continue...
     
  10. Nov 1, 2015 #9
    So :

    Kf+Uf=Ki+Ui

    0,025 J + 1/2*k*e^2 = 0 J + 1/2*k*e^2
     
  11. Nov 1, 2015 #10

    gneill

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    Where is the block when its speed is maximum? Where is the block when its speed is zero?
     
  12. Nov 1, 2015 #11
    Well, when the speed is 0 its going to be at the two extremities. When its at the maximum its going to be somewhere in the middle.
     
  13. Nov 1, 2015 #12

    gneill

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    Right.
    Not just somewhere: There's a precise location. You should be able to deduce it from the energy equation. When KE is maximum, what is the PE?
     
  14. Nov 1, 2015 #13
    Oh, there is going to be a potential energy of 0 because it will have been transfered all to the kenetic one right ?
     
  15. Nov 1, 2015 #14

    gneill

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    Right. So what can you say about the values of maximum KE and maximum PE? What are the values of displacement associated with each?
     
  16. Nov 1, 2015 #15
    So the maximum is going to be 0,025 for the kinetic one and 0,025 for the potential one at the two extremities.
     
  17. Nov 1, 2015 #16

    gneill

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    Assuming that you meant Joules for the units, yes. And you should be able to locate the displacements involved from the given information. Then just write out your energy equation accordingly.
     
  18. Nov 1, 2015 #17
    I'll go try this and give you my thanks if it works !
     
  19. Nov 1, 2015 #18
    Ok so I obtained : 0,025 J + 0.5*k*e^2 = 0 J + 0,025 J
     
  20. Nov 1, 2015 #19

    gneill

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    First state what the conditions are for each side of the equation. Numbers alone aren't telling the whole tale. It doesn't help to have your known maximum numerical value on both sides of the equation at the same time, since they'll just cancel out.

    Remember that when KE is maximum the PE is zero, and when PE is maximum the KE is zero. Choose your two block positions so that you can use your maximum energy value to best advantage.
     
  21. Nov 1, 2015 #20
    This the equation ; Kinetic energy final + potential energy final = kinetic energy initial + potential energy initial

    I thought that I should do the following :

    Kinetic energy final : 0,025 J

    Potential energy final : 0 J

    Kinetic energy initial : 0 J

    Potential energy final : 0,025 J
     
  22. Nov 1, 2015 #21
    It's really the position which is bugging me... I dont knwo what to do with 40 cm...
     
  23. Nov 1, 2015 #22

    gneill

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    Presumably that last entry should be "Potential energy initial"?
    If the KE final is 0.025 J, that must mean that the final position is at the midpoint. And your initial position is at one of the maximum displacement points.

    So when you write your equation, don't use a number for the maximum displacement, use the expression that includes the displacement. What is the maximum displacement? .....

    Doesn't that tell you what the maximum displacement should be?
     
  24. Nov 1, 2015 #23
    I think Im going to cry. Why does the maximum happen to be at the middle point and not at the third, fourth, etc.? I don't even know where my initial position is (Without contracting or lenghting my spring)

    And yeah I meant initial its a mistake
     
  25. Nov 1, 2015 #24

    gneill

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    In an oscillator like a spring/mass system, energy is traded back and forth between kinetic energy of the mass and potential energy of the spring. The kinetic energy depends upon the speed of the mass, while the potential energy depends upon the stretch or compression (displacement) of the spring from its equilibrium position. In this case the equilibrium position corresponds to the location where the spring is relaxed: zero displacement.

    When kinetic energy is maximum the potential energy is minimum, i.e. zero. Zero potential energy means the displacement is zero, since PE = (1/2) k 02. That places the maximum KE at the midpoint where displacement from equilibrium is zero. The mass zips through the midpoint at its maximum speed, on its way to the next maximum displacement.

    When potential energy is maximum the kinetic energy is minimum, i.e. zero. Zero kinetic energy occurs when the spring is maximally stretched (the mass is at maximum displacement and has zero speed). Then the potential energy is PE = (1/2) k Δxmax2, where Δxmax is the maximum displacement value.

    You have been given the maximum speed and the total displacement. So you can calculate both the maximum KE (which you've done) and the maximum displacement (which you haven't done yet).
     
  26. Nov 1, 2015 #25
    Sorry, I guess Im tired today because its going through my head. I did this picture : http://imgur.com/gUvMPQz

    I know that my displacement is e=L-Lnatural

    Lnatural = 0 because this is the position where there isn't any force applied on it. Now, where is the 0 position on my picture ? I understand why my displacement is 0 because of the PE formula.
     
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