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Multistep stumper

  1. Oct 31, 2006 #1
    A 5.4 kg block is pushed 3.0 m up a rough 37° inclined plane by a horizontal force of 75 N. If the initial speed of the block is 2.2 m/s up the plane and a constant kinetic friction force of 25 N opposes the motion, *calculate the following.
    (a) the initial kinetic energy of the block
    J
    (b) the work done by the 75 N force
    J
    (c) the work done by the friction force
    J
    (d) the work done by gravity
    J
    (e) the work done by the normal force
    J
    (f) the final kinetic energy of the block
    J
    HELP!!, please
     
  2. jcsd
  3. Oct 31, 2006 #2
    what have u done to answer them yourself?
     
  4. Oct 31, 2006 #3
    [tex]W = |F||D|cos\theta[/tex] theta being the angle between force and displacement vectors. That should take care of the first 5. The last one looks like [tex] W_{net} = \Delta KE [/tex]
     
    Last edited: Oct 31, 2006
  5. Oct 31, 2006 #4
    thank you
     
  6. Oct 31, 2006 #5
    I tried to use the formula but I cant get the right answer.
     
  7. Oct 31, 2006 #6
    how would you do part be on this question?
     
  8. Oct 31, 2006 #7
    In part B the force is 75N the distance is 3m and the angle between them is 37 degrees, so using the formula the work should be 180 N*m
     
  9. Nov 4, 2006 #8
    Thanks, now i'm trying to find the work done by friction. so I did W=25N*5.2(distance)*cos(theta but it didn't work
     
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