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Using Energy to Solve Work Problem

  1. Oct 20, 2012 #1
    1. The problem statement, all variables and given/known data

    A truck is traveling at 11.1 m/s down a hill when the brakes on all four wheels lock. The hill makes an angle of 15.0° with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is 0.750. How far does the truck travel before coming to a stop? Use energy to solve this problem.



    2. Relevant equations

    fk = Ef - Ei
    fk = μk*FN
    FN=mgcos15°


    3. The attempt at a solution

    First, I found the work done by the friction force: Wfk = S(μkFN)cos180°
    Wfk = μk(mgcos15°)(-1)

    Then I solved for the work done by friction using the energy equation:
    W = 1/2(mv2) +mgy - 1/2(mvi2) -mgyi
    W = 0 + mgy - 1/2(mvi2) - 0

    then I set the two equations equal to one another:
    μk(mgcos15°)(-1) = mgy - 1/2(mvi2)

    the m's cancelled out and I solved for y:

    y = μk(gcos15°)(-1) + 1/2(vi2) / g

    and plugged in values:

    y = (0.750)(9.8m/s2)(cos15°)(-1) + 1/2(11.0)2 / 9.8m/s2

    and got: y = 6.90 m

    knowing that the truck was traveling at 15°, sin15° = y/s
    s = 6.90/sin15°
    s = 26.7m

    the correct answer should be 13.5m. Any help/advice would be much appreciated.
     
  2. jcsd
  3. Oct 20, 2012 #2

    PeterO

    User Avatar
    Homework Helper

    Should that 180° angle you used have actually been 15° ?
     
  4. Oct 20, 2012 #3
    I thought that the in the formula W = sfcos∅ , the angle ∅ should represent the angle between the direction of s and the corresponding force, which in this case would be friction...and those two directions are 180° apart, right?
     
  5. Oct 20, 2012 #4
    Where did s go in your Wfk equation? Also, what is the first equation in the relevant equations section? Don't use y in your potential energy equation, write in terms of s.
     
  6. Oct 21, 2012 #5
    Oops, looks like I dropped the s. Looks like that would make

    y = sμk(gcos15°)(-1) + 1/2(11.0)2 / 9.8 m/s2

    could I then relate s and y through sin? Am I even on the right track?

    The first equation is Wother = ΔE , other in this case meaning friction. Was given to me by my professor. But s and y are representing two different things, and y is a reflection of the incorporation of gravitational potential energy into the problem so I don't see how I can get around using it?
     
  7. Oct 21, 2012 #6
    Go back a step or two and write out the Wfk=ΔKE+ΔPE equation. Don't use y, use s and θ. Also, keep a close eye on the signs. s is downwards, but both ΔEs should be negative.
     
  8. Oct 21, 2012 #7
    Finally got the correct answer, thank you very very much :smile:
     
  9. Oct 21, 2012 #8
    You're welcome.
     
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