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Homework Help: 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
    (b) the work done by the 75 N force
    (c) the work done by the friction force
    (d) the work done by gravity
    (e) the work done by the normal force
    (f) the final kinetic energy of the block
    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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