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Work and Engery Problem

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data

    A 1400kg car ascends a mountain road at a steady 60km/h. The force of air resistance on the car is 450N. If the car's engine supplies enough energy to the drive wheels at the rate of 38kW, what is the slope angle of the road?

    m = 1400kg
    F(air) = 450N
    V=60km/h = 60000km/3600s = 16.667ms/
    W(total) = 38000W
    theta = ?

    2. Relevant equations

    W(total) = K(final) - K(initial)

    K = 0.5mv^2


    3. The attempt at a solution

    I am not sure how to start this problem, but here is my first attempt:

    38000 = 0.5(1400)(16.6667^2) - 450cos(theta)

    The number was not in the domain of cos. I'm stuck. Any ideas? Thanks!
     
  2. jcsd
  3. Oct 12, 2009 #2

    rl.bhat

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    To move the car with a steady speed, it must overcome the component of weight along the inclined plane + the air resistance.
    And the power = F*v.
    What is the component of the weight along the inclined plane?
     
  4. Oct 12, 2009 #3

    kuruman

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    Gold Member

    Where did you get -450cos(theta)? What does it represent? You want to say that power produced by the engine is equal to the power dissipated by air resistance and rate of work done against gravity to raise the car.
     
  5. Oct 12, 2009 #4
    Hi everyone! Thanks for the responses.


    Ah ok, I wasnt sure where to put gravity in the equation. As for the -450cos(theta), its the work done on the car. I assumed that it was a horizontal force and broke it down into the components.
     
  6. Oct 12, 2009 #5
    the way I like thinking about it for gravity is like so;

    you say that it's travelling at 16 2/3 m/s, so what distance is it travelling in the vertical distance for? by resolving, (with 16 + 2/3) as the hypotenuse, and finding the distance it travels upwards, all you need to do to find out the work done against gravity is multiply that distance against the force of gravity, which I imagine will look something like;

    (for work done against gravity)

    [tex]16\frac{2}{3} sin(\theta)1400 g[/tex]

    do you see why this works?

    Work done = force * distance moved in direction of that force because gravity is acting downwards, you need to find the direction travelled upwards!
     
  7. Oct 12, 2009 #6
    Ah! So now all I have to do is find the direction of the air resistnace and put it all together in the work-energy theorem.
     
  8. Oct 13, 2009 #7
    I set the equation 16.667sin(theta)1400(0.81) = 0.5(1400)(16.667^2) and got a theta of 9.8 degrees. Which is close to the answer.
     
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