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Elevator problem?

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data

    A skier of mass 69.7 kg is pulled up a slope by a motor-driven cable.
    (a) How much work is required to pull him a distance of 59.8 m up a 29.9° slope (assumed frictionless) at a constant speed of 1.90 m/s?
    (b) A motor of what power is required to perform this task?
    hp

    2. Relevant equations

    (a) w=f*change in distance*cos angle
    (b) power=work/change in time

    3. The attempt at a solution

    gravity is the only force in play. so the equation for work would be
    w=(9.8m/s^2)(59.8m)(cos29.9)=508.03

    take that value and divide it by change in time to get power

    *correct?
     
  2. jcsd
  3. Feb 20, 2007 #2

    Dick

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    Homework Helper

    "cos angle" refers to what angle? With respect to your given angle cos is the wrong trig function. You are also missing a mass factor which would have been obvious if you had checked the units on your answer.
     
  4. Feb 20, 2007 #3
    No. w=f.s=m.a.s
    You havent taken the mass of the skier into account. The acceleration will be along the incline (which youve taken incorrectly).
     
  5. Feb 20, 2007 #4
    Power can be calculated by p=fv
     
  6. Feb 20, 2007 #5
    Ok so the F in the problem is 9.8m/s^2. I would use cos0. However, the change in distance is unknown. Using the formula d=time * speed, I found it to be 8.75m. So, now would I use the work formula with the numbers (9.8m/s^2)(8.75m)(cos0). Which equals 85.57J. After I find that I would divide work by time (5secs) to find average power which is 17.15 W...just my thinking??
     
  7. Feb 20, 2007 #6
    F is not 9.8 m/s^2. That is a. You are forgetting to take into account mass of the skier. force = mass * acceleration
     
  8. Feb 20, 2007 #7
    Ok-so hopefully this is it:
    W=(mass of skier*acceleration)(change in distance)(cos29.9)
    W=(132.43)(59.8)(cos29.9)=6865.23.
    Then take that value and divide by the time which is 31.55 seconds.
    Which equals 217.6 hp
     
  9. Feb 20, 2007 #8
    The SI unit of power is watts. There are about 746 watts per h.p. but you better look up that number, i am pulling it out of my head.
     
  10. Feb 20, 2007 #9
    Is the way I worked the problem out correct? I just need to convert the 217.6 watts to hp? I have had trouble solving this problem. It would be nice to know if this thinking is correct.
     
  11. Feb 20, 2007 #10
    The reasoning seems okay to me.
     
  12. Feb 20, 2007 #11
    UGH! I'm still stuck. Any help would be greatly appreciated!!
     
  13. Feb 20, 2007 #12
    Where are you stuck?
     
  14. Feb 21, 2007 #13
    alright...i think i've got it. I have to run to class, but I will post later witht the answer I finally came up with! YAY!
     
  15. Feb 21, 2007 #14
    actually, nevermind that answer was also incorrect. sorry to get our hopes up!:frown: Still working...
     
  16. Feb 21, 2007 #15
    whoops, you shouldnt be looking at the cosine of theta. Sorry, just caught that.
     
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