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Work and Energy (snowmobile up- vs. downhill)

  1. Apr 22, 2012 #1
    Taken from An introduction to mechanics - Kleppner, Kolenkow.
    Problem 4.17

    1. The problem statement, all variables and given/known data
    A snowmobile climbs a hill at 15 mi/hr. The hill has a grade of 1 ft rise for every 40 ft. The resistive force due to the snow is 5 percent of the vehicle's weight. How fast will the snowmobile move downhill, assuming its engine delivers the same power?


    2. Relevant equations
    frictional force f=μFn

    P=dW/dt = F dx/dt = Fv

    3. The attempt at a solution

    angle of hill , tan θ = 1/40 , θ ≈ 0.025
    Uphill forces : frictional force f = 0.05mg cos(θ)
    and also mg sin(θ)

    P = mg(0.05cos(θ) + sin(θ))v

    Downhill forces : same frictional force, but now subtract for the gravitational pull in other direction. P is the same so,
    vdown = P/F = \frac{mg(0.05cos(θ) + sin(θ))v}{mg(0.05cos(θ) - sin(θ)} ≈ 31 mi/hr

    Answer says 45 mi/h in the book. So what have I missed, .. thanks for any suggestions.
     
    Last edited: Apr 22, 2012
  2. jcsd
  3. Apr 22, 2012 #2
    First thing, answer in book is correct.

    "hill forces : frictional force f = 0.05mg cos(θ)"

    above is not correct.

    Hint: Determine the power consumed going up the hill in terms of weight. You have the speed, you have the rate of ascent, and you have the power consumed due to friction if you correct the above expression.

    The only impediment going down hill is the friction. You have 'extra' power assisting you because of the rate of change of potential energy which can be expressed in terms of velocity and the 1 ft drop for every 40 ft of travel down the hill.
     
  4. Apr 23, 2012 #3
    Oh ok, thanks for your reply.
    An example of how I misinterpreted the given information. Thought the 5 percent of its weight was somehow 5 percent of its weight against the normal to the hill.
     
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