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Homework Help: Another question that is troubling me

  1. Sep 3, 2007 #1
    extra power required to maintain the same speed on an incline

    1. The problem statement, all variables and given/known data

    a tank of 80 metric tonnes is travelling at a uniform speed of 54km/r on level terrain, it the starts travelling uphill on an incline of 1 in 10 sine slope)
    Calculate the extra power required from the engine in Megawatts to maintain the same speed on the incline.
    2. Relevant equations

    F=(% grade of hill)*M*G

    3. The attempt at a solution

    I calculated the power required to go uphill as
    F=1/10 * 80000*9.8 = 78400N
    therefore power = F*V = 78400*15m/s = 1.176MW

    the problem I have is that I cant seam to find an equation to work out the force on the flat which does not have the acceleration variable, do I just put this to 1 and solve that way? which would give a force of 784000N, P=F*V = 784000*15 = 11760000.
    take the first answer away from the second which leave a difference in power of 10.584MW.
    Really not too sure.
    I know of other equations such as Ek=1/2Mv^2 for power to accelerate a vehicle but dont think this is the equation to use,
    Last edited: Sep 3, 2007
  2. jcsd
  3. Sep 3, 2007 #2


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    Oops... sorry I posted wrongly.. I think you need to calculate friction.
    Last edited: Sep 3, 2007
  4. Sep 3, 2007 #3
    can friction be calculated without knowing the proopertys of the road surface?
  5. Sep 3, 2007 #4


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    I thought at first, maybe you'd need to consider how the friction changes when it goes up the incline...


    [tex](F_{engine1} - \mu{m}g)v = 0[/tex]

    Then up the incline

    [tex][F_{engine2} - \mu{m}gcos(\theta) - mgsin(\theta)]*v = 0[/tex]

    So then subtracting these two:

    [tex]F_{engine2}*v - F_{engine1}*v = [(cos\theta-1)\mu{m}g + mgsin(\theta)]*v[/tex]

    I think the [tex](cos\theta-1)\mu{m}g[/tex] is small enough to be ignored (since they didn't give any [tex]\mu[/tex] for the problem) Also [tex]|cos(\theta) - 1|<<<|sin(\theta)|[/tex] (we have sintheta is 0.1... and I'm getting costheta-1 as -0.005) and [tex]\mu < 1[/tex] so the first term on the right side is much less than the second...

    So I think it's safe to say:

    [tex]F_{engine2}*v - F_{engine1}*v = mgsin(\theta)*v[/tex]

    So I'd say 1.176MW is correct...
  6. Sep 3, 2007 #5


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    You can't "calculate friction" because no information is given about friction that would allow you to do that.

    I think I would do this in terms of "conservation of energy". Assuming that velocity is constant, the only change in energy would be the increase in potential energy: mgh. How fast is that changing?
  7. Sep 3, 2007 #6
    to be honest, now I am not sure what is correct now,

    I think my answer is correct for the power required to go up the hill (1.176MW), however I dont understand how no power is used to keep the tank moving at a constant speed.

    If the answer is 1.176MW could someone please explain in laymans terms.

  8. Sep 3, 2007 #7


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    Suppose the terrain is completely frictionless... the truck is sliding along the flat terrain at a constant speed of 54km/h... how much power is the engine giving the truck?
  9. Sep 3, 2007 #8
    Ok, so in my answer I can state that, because there is no friction given it is assumed there is no friction when the tank is on the flat, hence the extra power required is the total power required to get the tank up the hill at 54km/hr.

    the thing I am a bit dubious about is that there is 10 marks for this
  10. Sep 3, 2007 #9


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    Yeah like that... or say that we must assume the frictional force is the same on the incline as the flat part(or that the difference negligible) since mu is not given...

    Basically the idea is that since the truck isn't accelerating, the net force on the truck is 0. Assuming the friction is the same:

    Ftruck1 = friction (flat part)
    Ftruck2 = friction + mgsintheta (inclined part)

    So this way also you get the difference in power as 15*mgsin(theta)
  11. Sep 3, 2007 #10
    thankyou very much
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