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Work and kinetic energy

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data

    a 300 kg crate is on a rough surface inclined at 30 degrees. A constant force P=2400 N is applied horizontally to the crate. The force pushes the crate a distance of 3.0 m up the incline, in a time interval of 9.0 s, and the velocity changes from v1=0.7 m/s and v2=2.9 m/s. What is the work done by the weight?


    2. Relevant equations

    WD=F.d

    3. The attempt at a solution
    i have found the parallel components:
    Fnet=Psin(theta) -mgsin(theta)
    Psin(theta)-mgsin(theta)=ma
    and the perpendicular components:
    Fnet=Fn-mg
    Fn=mg
    I'm not 100% sure any of this is correct, and now I don't know where to go with it
     
  2. jcsd
  3. Oct 27, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi shaunanana! Welcome to PF! :smile:

    (have a theta: θ and try using the X2 tag just above the Reply box :wink:)

    You're only asked for the work done by the weight (the mg force), so most of the information in the question is unnecessary.

    And work done = force "dot" distance.

    So work done by the weight = … ? :smile:
     
  4. Oct 27, 2009 #3
    Oh wow, that's so simple...maybe I should stop trying to do my homework so late at night!

    Thank you!!! :)
     
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