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What was the minimum power required?

  1. Jul 30, 2007 #1
    1. The problem statement, all variables and given/known data
    A truck weighing 3.0*10^4 newtons was driven up a hill that is 1.6*10^3 meters long to a level area that is 800 meters above the starting point. If the trip took 480 seconds, what was the minimum power required?

    3. The attempt at a solution

    My problem here is to find out the total work. I know that the work done by vertical position and along the hill is the same, and then divided by the time, so to get power. But I tried to use W=PE+KE, but it didn't come out the same answer. I used: PE=mgh=3.0*10^4*800, KE=1/2mv^2=1/2*30000/9.81*1600/480. then i added them up and divided by 480. It just didn't come out right. I dont know why i can't use this method?
     
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  3. Jul 30, 2007 #2

    Dick

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    Conceptually, your problem is that W is not equal to KE+PE. It's equal to CHANGE in KE plus CHANGE in PE. If you take the truck driving up the hill at constant speed, there is no change in KE. And I don't get why there is no 9.81 in your numerical expression for PE, and I don't get at all how you came up with the numerical expression for KE.
     
  4. Jul 30, 2007 #3

    mgb_phys

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    Ignore KE, all you are looking for is change in PE in a given time = power.

    ps. There is no 9.8 in the pe because he is given the weight of the truck not it's mass.
     
  5. Jul 30, 2007 #4

    Dick

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    Good point. And that explains the 9.81 in the KE as well but not the failure to square the v. Aside from the fact KE shouldn't be there at all.
     
    Last edited: Jul 30, 2007
  6. Jul 31, 2007 #5
    Is that the formula for KE is different from the formula for change in KE, right? Former one is that simply speed, and the later one you need to find the initial and final velocity?
     
  7. Jul 31, 2007 #6
    The solution for the problem is PE/480=5.0*10^4? But i think my teacher said the answer was 1.0*10^5 because she found out the total work was 3.0*10^4*800*2. She said that the work on incline of the hill is the same as lifting the truck in the vertical direction. But why should times 2? We don't have to do work twice? We just need to choose one way to let the truck go up?
     
  8. Jul 31, 2007 #7

    PhanthomJay

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    It is assumed that there is no change in speed as the truck goes uphill, that is, it is moving at constant velocity, so that there is no change in kinetic energy. In which case, the total work done by all forces acting on the truck is zero. So the book answer for power delivered by the truck appears correct. The work done by gravity is the negative of the work done by the truck. Your teacher probably added the two to get a total work of twice the work done by the truck, when instead she should of subtracted the two to get a result of zero total work. Bit in any case, the problem does not ask for total work, just the truck piece. BTW, The 'v' in the kinetic energy equation is not a velocity, is a speed. KE is a scalar quantity.
     
  9. Aug 2, 2007 #8
    You mean the total work is 0?
     
  10. Aug 2, 2007 #9

    PhanthomJay

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    Yes. Some would call it the net work. By 'total' work or 'net' work, we mean the work done by ALL the forces acting on the object. Since there is no change in the kinetic energy of the system, the net, or total, work, is zero. This follows from the work-energy theorem:[tex]W_{tot} = W_{net} = \Delta KE[/tex]

    Just don't confuse total work with the work done by the individual forces. Total work includes work done by gravity.
     
    Last edited: Aug 2, 2007
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