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Work, Energy & Power Problem

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data

    A Tank of mass 80 metric tons is travvelling at a uniform speed of 54Km/hr on a level terrian. It then starts travelling uphill on an incline of 1 in 10 (sine slope). Claculate the extra power required from the engine in megawatts to maintain the smae speed on the incline.

    m = 80,000 kg
    v = 15 m/s

    2. Relevant equations

    Kinetic Energy = [tex] \frac{1}{2}mv^2 [/tex]

    Potential Energy = [tex] mgh [/tex]

    3. The attempt at a solution
    You see, this is my problem. I'm at a loss on how to start I think I might be reading into this to0 much.

    Is the extra power in MW required impossible to work out unless a distance travelled uphill is given in which case the following would be used:

    Work = [tex] Fs\cos \phi [/tex]

    In which case how can I work the Force propelling hte tank in a horizontal direction?

    The next part of the question is When it has travelled 100m up the incline the driver stops and has a drinks and so on. . . . . . . (I'm happy with this section of the question). Is this just the authors question writing or am I just reading too much into the original problem?

    Thanks in advance Jonno.
     
    Last edited: Sep 19, 2007
  2. jcsd
  3. Sep 19, 2007 #2

    Doc Al

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    Staff: Mentor

    You don't need the distance you need the rate at which height is gained, which you can figure out given the speed and the angle of the incline. All you need to figure is the additional power needed to go uphill (against gravity).
     
  4. Sep 19, 2007 #3
    Thanks Doc Al,

    I see where you're coming from.

    So if the tank gains 1 metre in height for every 10m travelled in a second gains height at a rate of 1.5m/s

    so using E = mgh I can work out the extra energy in joules and convert to Power (J/s) From then?

    Thanks again.
     
  5. Sep 20, 2007 #4
    if you substitute h [m] by hdot [m/s] (with the dot on top, meaning it's the time-derivative) then E also becomes a time derivative. That's what you want. E [J] becomes P [W] (J/s=W)
    so hdot is 1.5 m/s yes yes!!
     
  6. Sep 20, 2007 #5

    Astronuc

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    Staff: Mentor

    Going back to [tex] Fs\cos\phi [/tex] - change s to v, and use sin for changing elevation, as opposed to cos for horizontal displacement.
     
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