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Automobile Power

  1. Feb 19, 2008 #1
    [SOLVED] Automobile Power

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

    A truck engine transmits 28.0kW(37.5hp ) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0km/h ( 37.3mi/h) on a level road.

    a) resistance force? P = Fv, P/v = F = 1680N

    this is where I have trouble

    b) Assume that 65% of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 ? Give your answer in kilowatts .

    2. Relevant equations

    P = Fv


    3. The attempt at a solution

    so it says the forces acting are 65% of the resisting force (.65*1680 = 1092) and the air resistance is the square of the speed, so I used (with speed in 8.3m/s):

    P = Fv = 1092(8.3^3) = 624,391 W = 624kw wrong

    also tried:

    P = Fv = (1092+8.2^2)8.3 = 9,621 W = 9kW wrong

    I think I'm using the wrong force, any help is appreciated.
     
  2. jcsd
  3. Feb 19, 2008 #2

    cepheid

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    65% is due to FRICTION. You need to multiply F by 0.35...that's how much force is due to air resistance.
     
  4. Feb 19, 2008 #3

    cepheid

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    Also, once you calculate how much power is lost due to air resistance at 30 km/h, that's not your final answer is it? What else do you have to do? (Friction is still there, after all).
     
  5. Feb 19, 2008 #4
    that's pretty much it, the next answers are just to give the power in hp and the same for another velocity, which I'll be able to find if I can find how to do this problem.

    you say I need to multiply 1680 by .35 to find air resistance, but it came out like this:

    ((F*.65)+(F*.35))*8.3 = 13,944 W = 14kw which is the wrong answer

    or should I do: ((F*.65)-(F*.35))*8.3 = 4183W = 4kw?
     
  6. Feb 19, 2008 #5

    cepheid

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    I think you need to think this through a little better. What do we know?

    [tex] P = F_\textrm{total} v [/tex]

    [tex] F_\textrm{total} = F_\textrm{fric} + F_\textrm{air} [/tex]

    [tex] F_\textrm{fric} = 0.65F_\textrm{total} [/tex]

    AT 60 km/h,

    [tex] F_\textrm{air} = 0.35F_\textrm{total} [/tex]

    BUT AT 30 km/h

    [tex] F_\textrm{air} \not= 0.35F_\textrm{total} [/tex]

    BECAUSE F_air changes with velocity. So how can we figure out the new F_air?

    We know that:

    [tex] F_\textrm{air} \propto v^2 [/tex]

    and v has halved.
     
  7. Feb 19, 2008 #6
    it did say that the force of air resistance was proportional to the speed squared, might it be:

    ((F*.65)+(8.3)^2)*8.3 = 8491 W = 8kW?

    edit: oh speed is halved, so maybe:

    ((F*.65)+(4.2)^2)*4.2 = still 4kw.
     
  8. Feb 19, 2008 #7

    cepheid

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    Just work on solving for the new F_air at 30 km/h first, which you still haven't done properly. Don't worry about the rest of it just yet.

    EDIT: here's a further hint. F_air is a function of v:

    [tex] F_\textrm{air}(v) \propto v^2 [/tex]

    At 60 km/h,

    [tex] F_\textrm{air}(60) = 0.35 F_\textrm{total} [/tex]

    But at 30 km/h

    [tex] F_\textrm{air}(30) = \textrm{unknown}[/tex]

    But we can solve for it, because using the information given about how F_air varies with v, we can determine what this ratio is:

    [tex] \frac{F_\textrm{air}(60)}{F_\textrm{air}(30)} [/tex]
     
    Last edited: Feb 19, 2008
  9. Feb 19, 2008 #8

    edit, so Fair(60)/Fair(30)

    might it be:

    .35(1680) = 588/8.3 = 70?

    edit: I have to solve for a new force at 30km/h don't I?
     
    Last edited: Feb 19, 2008
  10. Feb 19, 2008 #9

    cepheid

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    It doesn't say that. It says that F_air is PROPORTIONAL TO v^2 (which is what the symbol I used means). F_air can't be equal to a velocity squared, because the two things don't even have the same dimensions. That would be as nonsensical as saying:

    2 metres = 3 seconds

    Just as you can't equate a distance to a time interval, you can't equate a force to something that is not a force.

    When we say that F is proportional to v^2, what we mean is that:

    [tex] F = Cv^2 [/tex]

    Where C is some constant of proportionality with units of N/(m/s) that takes care of the fact that the two quantities don't have the same units. But this is all irrelevant. You don't need to know what C is to solve this problem. The fact that if v changes by some factor, F changes by that factor squared is all you really need to know.
     
  11. Feb 19, 2008 #10

    cepheid

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    Yes, that is the whole point of what we are trying to do here
     
  12. Feb 19, 2008 #11
    sorry if I'm being a little difficult, I'm really not getting the concepts for this problem, but what I'm getting from your explanation is that since the velocity is halved the force will be halved, now F(30) = 840N.

    so in that case Fair(30) = Fair(60)/2 = (.35Ftotal/2) = 294N?
     
  13. Feb 19, 2008 #12

    cepheid

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    ALMOST! That would be true if the force were just proportional to velocity. But the force is proportional to velocity SQUARED, which means that if the velocity is halved, the force is quartered. Think about it this way:

    [tex] F_\textrm{air} \propto v^2 [/tex]

    Therefore:

    [tex] \frac{F_\textrm{air}(30)}{F_\textrm{air}(60)} = \left(\frac{30}{60}\right)^2 [/tex]

    [tex] = \left(\frac{1}{2}\right)^2 = \frac{1}{4} [/tex]
     
  14. Feb 19, 2008 #13
    than (.35(1/4(1680)) = 147N?
     
  15. Feb 19, 2008 #14

    cepheid

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    Right! Finish it off! =)
     
  16. Feb 19, 2008 #15
    alrighty then.

    Ftotal = ((.65Ftotal)+147N) = 1239N

    P = 1239N(8.3m/s) = 10284 W = 10kW

    right? (crosses fingers)
     
    Last edited: Feb 19, 2008
  17. Feb 19, 2008 #16
    I think it's actually:

    P = ((.65Ftot) - 147)8.3 = 7843.5 W = 7.8 kW

    edit I feel really stupid:

    ((.65Ftot)-147)*(8.3^2) = 65101 W = 65 W
     
    Last edited: Feb 19, 2008
  18. Feb 19, 2008 #17

    cepheid

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    It looks ok to me. I though you had the answer already.
     
  19. Feb 19, 2008 #18
    I'm not getting the right answer...

    I tried 65 kW and still got it wrong, then I tried ADDING the 147N to the (.65Ftot)8.3^2 and got 85kW and still got it wrong, do I need the new force I got instead of the original resisting force?
     
  20. Feb 19, 2008 #19

    cepheid

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    Dude! When I said it looked alright to me, I meant what you wrote in post #15! What's up with all the other stuff? Your subsequent equations make no sense!
     
  21. Feb 19, 2008 #20
    oh shoot! yep got the answer, well I got 10kw but the stupid mastering physics wanted 10.3 kw so technically got the answer wrong, oh well, really THANKS SO MUCH FOR ALL YOUR HELP!
     
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