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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
     
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