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Kinetic energy- pushing an object up an incline.

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data
    An object is given an initial velocity of 3.0 m/s up a 20 degree incline. Mass= 22.3 kg and coefficient of kinetic friction is .93.


    2. Relevant equations

    W = (1/2)mv1^2-1/2mv0^2

    W = Fdcos(u)


    3. The attempt at a solution

    So I added the forces together, mgsin(20) + .93mgcos(20)

    and this times cos(20) should be the total work, right?

    so then since the change in kinetic energy equals work i set -(1/2)mv^2= -Fdcos(20)

    plugging in the mass for m, initial velocity for v, the forces due to friction and gravity for F, and solving for d, i got the wrong answer. where'd i go wrong? I also tried this taking out the cos(20) in the second work equation and still got it wrong.


    Thanks, this is my first time using physics forums and i really hope i can get to understand this stuff...
     
  2. jcsd
  3. Oct 11, 2008 #2
    What is the question asking for?
     
  4. Oct 11, 2008 #3
    sorry. the distance it's pushed up the hill.
     
  5. Oct 11, 2008 #4
    Well I say you've done it fine by adding up the two forces.
    What I would have done then would be to tilt the axes 20 degrees such that it's parallel with the incline. We'll call this axis the x.

    We know that the magnitude of net force acting on the x axis is mgsin(20) + 0.93cos(20).

    Therefore just use [tex] \frac{1}{2}mv^2 = F\Delta d [/tex]
     
  6. Oct 11, 2008 #5

    that's what i did :-\

    it came out to .3979 ish and that wasn't right...
     
  7. Oct 11, 2008 #6
    ugh. my calculator was in radians.

    *facepalm*
     
  8. Oct 11, 2008 #7
    How do i mark a thread as solved?
     
  9. Oct 11, 2008 #8
    You can just let it die away
     
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