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Work / Power done to a particle

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A 1 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 6.6 + 12.6x - 1.8x2, where Fx is in Newtons and x is in meters.

    Find the work done by this force on the particle as the particle moves from x = 0 m to x = 1.5 m.

    Find the power delivered to the particle when it is at x = 1.5 m.

    2. Relevant equations

    Integrals
    V = (2K/m)1/2
    P = Fv

    3. The attempt at a solution

    I solved the first part by solving the integral and plugging in 1.5. The answer was 22.05.

    The second part I thought I had correct:

    V = (2K/m)1/2, where K = W
    V = (2(22.05)/1.5)1/2
    V = 119.56 W

    I tried variations on this number because I was certain I was correct, I even tried the negatives. Perhaps my equation for V is incorrect?
     
  2. jcsd
  3. Sep 27, 2009 #2

    Doc Al

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    Staff: Mentor

    m ≠ 1.5
     
  4. Sep 27, 2009 #3
    Wow, m equals mass?

    V = (2K/m)1/2, where K = W
    V = (2(22.05)/1)1/2
    V = 6.64 m/s

    W = 22.05 x 6.64

    W = 146.43
     
  5. Sep 27, 2009 #4

    Doc Al

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    Staff: Mentor

    Uh, yeah... I hope you're not just plugging into formulas without knowing what they mean. :wink:

    That formula for speed comes from the definition of kinetic energy, K = 1/2mv^2.

    OK.
    :confused: Why are you multiplying W by the speed? (And then calling it W!)
     
  6. Sep 28, 2009 #5
    I was just becoming very frustrated with this question because it seemed so easy but I guess I don't really understand the concepts yet.

    I am trying to solve for power which is P - Fv. I solved for v and I guess I thought that my previous answer was the force but it wasn't, it was just the average power.

    I believe that in order to solve this problem, I have to plug in 1.5 into the Fx equation given to solve for the force and multiply that by the velocity.

    Note - I already used up all of my tries for this problem so I would just like to understand the concept of the question now.
     
  7. Sep 28, 2009 #6

    Doc Al

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    Staff: Mentor

    That's right. (In the future, try to take it slower. Double check each step.)
     
  8. Sep 28, 2009 #7
    Thanks, will do.
     
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