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Work Done by Weight

  1. Apr 7, 2013 #1
    1. The problem statement, all variables and given/known data

    A 700 kg crate is on a rough surface inclined at 30°. A constant external force P = 5600 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 7.3 s, and the velocity changes from v1 = 1.4 m/s to v2 = 2.5 m/s. What is the work done by weight?

    75512f21-629f-400f-914b-f9f928f9ba49.png

    2. Relevant equations

    W = Fd, KE = 0.5mv^2, PE = mgh

    3. The attempt at a solution

    I drew the free-body diagram, but I am stumped on how to go about the problem. What is the work done by weight, and how can I find it? I don't exactly know what it is.
     
    Last edited: Apr 7, 2013
  2. jcsd
  3. Apr 7, 2013 #2

    haruspex

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    I don't know what they mean by 'work done by weight'. My guess is they meant the work done by the force P. (Cut and paste error from another question?)
     
  4. Apr 7, 2013 #3
    I figured out the problem, and it is the work done by gravity on the box (ΔPE).
     
  5. Apr 7, 2013 #4

    haruspex

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    Really? So most of the information is irrelevant? (Or maybe there are more parts to the question?)
     
  6. Apr 7, 2013 #5
    Yes, there is another part of the question I am stuck on. It is now asking me to find the work done by the frictional force, and my attempts at the problem have been futile. The only things changed in the info in the original post is that now V2 = 2.3 m/s, and time = 8.3 s. Could you possibly help?
     
    Last edited: Apr 7, 2013
  7. Apr 7, 2013 #6
    You'll have to use an equation that relates conservation of energy and the energy lost by friction: [itex]KE_{i}+PE_{i}+WE_{i}=KE_{f}+PE_{f}+WE_{f}[/itex]
     
  8. Apr 7, 2013 #7

    haruspex

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    You know the distance, so it remains to calculate the frictional force. What equations do you get from your free body diagram?
     
  9. Apr 8, 2013 #8

    PhanthomJay

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    see important correction in red.
     
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