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Speed of car down hill with same power as up.

  1. Jun 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A snowmobile climbs a hill at 6.526m/s. The slope of the hill is a 1 ft rise for every 40 ft of distance. The resistive force of the snow is equal to 3.3 percent of the weight of the snowmobile. How fast will the snowmobile move downhill, assuming the engine delivers the same power?


    2. Relevant equations

    Would the resistive force be .033*mg, or .033*mg*cos[tex]\theta[/tex]

    3. The attempt at a solution

    Fd = Driving force, P=F*v -> F=P/v
    Fr = Resistive force

    Up the hill: ma=0=Fd-mgcos[tex]\theta[/tex] -Fr

    Fd= mgcos[tex]\theta[/tex] + Fr

    Down: ma= Fd + mgcos[tex]\theta[/tex] - Fr

    -I'm confused if they want the final velocity (due to acceleration) or if madown = 0?
    if it's the final V, then I need vf = vo +at,
    where I can find t with the quadratic equation by: X= Vo*T + [tex]\frac{1}{2}[/tex]a*T2

    I'll also try by means of energy, but I still need to know about the resistive forge :\

    Thanks for your help in advance.
     
    Last edited: Jun 11, 2009
  2. jcsd
  3. Jun 11, 2009 #2

    cepheid

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    The expression 0.033*mg seems to me to be the only sensible way of interpreting the statement, "The resistive force of the snow is 3.3.% of the weight of the vehicle.

    Isn't the "down the incline" component of the weight given by mgsinθ, not cosθ?

    Based on your calculations, is the net force on the downhill side non-zero (intuitively, it would have to be...)? If so, the question is ambiguous. -- the best you could do would be to say that its acceleration is a and that it.s velocity therefore varies linearly from v_initial to v_final over a distance (or time interval) of [whatever you want].
     
    Last edited: Jun 11, 2009
  4. Jun 11, 2009 #3
    Good call, I'll just treat it as friction then.

    Yeah, I would think it would have to speed up with the same power going down.
    But, is the Driving force now Po/Vo or Po/Vf? I keep getting snagged...
     
  5. Jun 11, 2009 #4
    I misinterpreted what you said. I think I was using the normal as the Down-weight. So I'll keep the resistive force 0.033*mg.
     
  6. Jun 11, 2009 #5

    cepheid

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    Nevermind. The short answer to your question is that it is Po/Vf, and Vf is indeed constant. Po is constant as stated in the problem.

    Explanation:

    I was being silly. It's interesting to note that according to my calculations, the frictional force along the incline is larger in magnitude than the component of the weight along the incline. This means that, if *sitting* on a downward slope, this vehicle will actually NOT move unless a driving force greater than 0.033mg - mgsinθ, is supplied. I say greater than because presumably the coefficient of static friction would be greater than 0.033. However, in our case, we're already moving when we reach the top, so we'll just use kinetic friction (which is given as mu = 0.033). The bottom line is that is IS possible to move at a constant speed down the hill with the engine still driving forward (because the net force is NOT downward in the absence of the engine). So you can deduce the final speed as follows:

    1. Calculate Fd (uphill) from a force balance (as you've done)
    2. Deduce Po from Fd (uphill) and v_uphill (which is given)
    3. Calculate Fd (downhill) using a force balance, which works because the component of weight downhill is smaller than the friction.
    4. Decude v_downhill from Fd (downhill) and Po (which is the same)
     
  7. Jun 11, 2009 #6
    Gah damn this is actually the first was I did it, except using cos[tex]\theta[/tex].. but that's right! THANKS!

    My answer was 47.29m/s which seems like a huge increase for such a small slope. I'm still having an intuitive block, which is why I deterred from continuing with that method; why isn't it accelerating down the hill?
     
  8. Jun 11, 2009 #7

    cepheid

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    I get 47.2537562 m / s, which is close enough. Let's talk about this conceptually:

    Q. Why is it such a huge increase?

    A. Let's break up the question into two parts:

    Q1: Why does the velocity increase at all?

    A1: Because on the way up, your weight is hindering you, as is friction. This means you need to supply a larger force to counteract BOTH of them. Given that you have a constant available power, a larger force means a lower velocity. On the way down, your weight helps you instead of hindering you. So a smaller force is required, to counteract friction only. Again, given a constant available power, a smaller force means a larger velocity than in the uphill case.

    Q2: Why does it increase by SO MUCH?

    A2: If you calculate the component of the weight that is along the incline (|| to the incline), you will notice that it is less than, but very close to the friction force. If the friction force is x, then the || component of weight is close to x (slightly less). This means that in the uphill case, where the force hindering you is the SUM of the two, this total force to be overcome will be close to 2x, requiring a correspondingly large driving force. On the way down, the force hindering you is the DIFFERENCE between the two, which will be close to zero, requiring a MUCH smaller driving force.


    Q. Why is the snowmobile NOT accelerating down the hill?

    A. Because the NET force is zero. The force being supplied by the engine PLUS the || component of weight are together just enough to counteract the frictional force on the incline (which still hinders your downhill motion).
     
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