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Rate of spring doing work.

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data
    A 0.18 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 680 N/m) whose other end is fixed. The ladle has a kinetic energy of 9.8 J as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed 0.019 m and the ladle is moving away from the equilibrium position?



    2. Relevant equations



    3. The attempt at a solution
    (a) I found that answer is zero. This is because spring has no work done at the particular moment.

    (b) I let 1/2mv^2 = 9.8J
    Solve for v and I get 10.43m/s.
    According to the question (b), it states that need to find the instantaneous power of the spring. Therefore, I let Fs = -Kd
    Fs= force from spring
    d= distance.
    Then, substitute Fs into power formula, which is P= fv . The answer I obtained, is 134.76W.
    But the answer is wrong. May I know the reason ? Could it be I mess up with the positive and negative sign ? Isn't "d" negative in this situation because it's being compressed ? Thank you.
     
  2. jcsd
  3. Nov 5, 2011 #2

    PhanthomJay

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    So tebes,
    1.) The speed of the ladle at x = 0.019 m is not the same as its speed in the 'equilibrium' position.
    2.) If the force and displacement are in the same direction, the work done is positive. If the force and the displacement are in opposite directions, the work done is negative.
     
  4. Nov 5, 2011 #3
    So, I need to find the velocity when the spring the is compressed 0.019m by letting
    1/2mv^2 equals to the Fs I obtained . Then, apply the power formula again . Am I right ?
     
  5. Nov 5, 2011 #4

    PhanthomJay

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    No-o. The kinetic energy change of the ladle as it moves from 0 to 0.019 must equal the potential energy change of the spring as it moves from 0 to 0.019. Solve for v', and then use power equation, being careful with your plus amd minus signs. don't let them sting you :cry:..
     
  6. Nov 5, 2011 #5
    Shame on me.
    The equation would be like this
    -1/2mv^2=1/2k(xi^2-xf^2)
    This is because kinetic energy is losing to the "spring".
    Then, solve for v and then substitute in power equation.
     
  7. Nov 5, 2011 #6
    For V, I obtained 3.2 m/s. Then, substituted into P=FV.
    P= Fs(3.2)= - 148.608 W
    but the answer is wrong.
     
  8. Nov 6, 2011 #7

    PhanthomJay

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    No-oh, it's the change in KE plus the change in PE = 0. So it should be 1/2mvf^2 - KEi = 1/2kxf^2 - PEi, where KEi = 9.8 and PEi = 0.
     
  9. Nov 6, 2011 #8
    Sorry, there is something still baffle me. May I know why there is no negative sign in front of 1/2kxf^2 ? according to formula, 1/2k(xi^2-xf^2), when xi=0, then formula would be - 1/2kxf^2.
     
  10. Nov 6, 2011 #9

    PhanthomJay

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    The change in potential energy is PEfinal - PEinitial. What you have written is not the change in PE, it is the work done by the spring (Ws = - Δ PE).
     
  11. Nov 6, 2011 #10
    Sorry, I didn't know that because PE is the material in next chapter. But I'll read the chapter in advance.
     
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