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Work problem help!

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data

    A highway goes up a hill, rising at a constant rate of 1.00 m for every 40 m along the road. A truck climbs this hill at constant speed v_up = 18 m/s, against a resisting force (friction) f equal to 1/20 of the weight of the truck. Now the truck comes down the same hill, using the same power as it did going up. Find v_down, the constant speed with which the truck comes down the hill.

    ASSUME: the resisting force (friction) has the same magnitude going up as going down.

    2. Relevant equations


    →W = mgd
    →KE = W = ½ * mv²

    3. The attempt at a solution

    Tried to use this form:

    ½ * m * v_up² - ½ * m * v_down²

    BUT doesn't seem to find the correct path.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 6, 2012 #2
    Didn't calculate in the form of work done, resolved the forces.

    First, The equation when the truck climbs the hill. (take power=P)

    P/v_up=(1/20)m+mg sin(tan-1(1/40))

    Then, the equation when the truck is going down.

    P/v_down=(1/20)m-mg sin(tan-1(1/40))

    divide the first equation by the second equation to eliminate the unknowns.
    You'll get the answer :)

    Tell me if my equations are unclear
  4. Oct 6, 2012 #3
    The equations seem to be incorrect. I enter in the solution, but not right.
  5. Oct 6, 2012 #4
    How did you get mgsin(arctan(1/40)) by the way? I see that you use the kinetic formula, but it's not right.
  6. Oct 6, 2012 #5
    eh? should be correct, I can't see where I did wrong >.<
    P/v_up is the constant force applied by the truck.
    When it is going up, (1/20)m is the friction and mg sin(tan-1(1/40)) is the resultant gravitational force acting on the truck.
    When going down, direction of resultant gravitational force is the same as the constant force. So I put a negative sign there.
    o_O I can't see what I did wrong.
  7. Oct 6, 2012 #6
    Sory I just noticed my careless mistake :P .
    The resistant should be (1/20)mg instead of (1/20)m.
    try it, should be correct this time.
  8. Oct 7, 2012 #7
    What about v²? Does that take in account?
  9. Oct 7, 2012 #8
    v2? you mean kinetic energy right? I'm resolving everything in forces, not energy form, so it doesn't take into account
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