1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook