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Power problem Need help

  1. Apr 17, 2005 #1
    This problem is simple and i cant figure out what I am doing wrong. I am supposed to find the speed of a bucket that is being lifted out of a well with a power output of 118W. The mass of the bucket is 4.5 kg and the weight of the rope is ignore. Dividing the power by the force should give me the velocity but it doesnt. I used 118 over 4.5*9.81 and the answer is wrong. Can anyone tell me what i am doing wrong?
     
  2. jcsd
  3. Apr 17, 2005 #2
    Power = work/time
    work = fdcos(angle)
     
  4. Apr 17, 2005 #3
    I dont have the time so i am using P=fd/t=F(d/t)=Fv which is p/f=V. The force should be m*a and a should = gravity right?
     
  5. Apr 17, 2005 #4
    Yes, that looks right. :smile:
     
  6. Apr 17, 2005 #5
    but the answer is wrong when i submit it...
     
  7. Apr 17, 2005 #6
    Wait a minute...
    how did U get from P=W/t to P=fd/t??? :confused:
     
  8. Apr 17, 2005 #7
    You can also use
    [tex]Work = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/tex]
     
  9. Apr 17, 2005 #8

    OlderDan

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    Part of the work being done by the power source is increasing the potential energy of the bucket, and part is increasing the kinetic energy. You need to consider both.
     
  10. Apr 17, 2005 #9
    It's asking for the FINAL velocity, right? The velocity can't be constant.
    Use W=Fd and the change in kinetic energy formula.
     
  11. Apr 18, 2005 #10

    OlderDan

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    Alisa,

    Your original statement is correct, and so is your calculation if the system has achieved a state of constant velocity and constant force. What is the form of the answer you are supposed to get? Perhaps the problem is not so easy and you are supposed to be looking at something other than the constant velocity condition.
     
  12. Apr 18, 2005 #11
    I still havent figure it out and its due it 10 minutes
     
  13. Apr 18, 2005 #12

    OlderDan

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    Then believe in what you did and try to defend it until you understand what (if anything) you did wrong.
     
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