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Frictional force and work problem

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A slide similar to der stuka is 35.0 meters high but is a straight slope,inclined at 45.0 degrees with respect to the horizontal. If a 60kg woman has the speed of 20.0m/s at the bottom and the work done by frictional force on the woman is -8.58 x 1000J, find the magnitude of the force of friction.


    2. Relevant equations

    Work done by frictional force = Frictional force cos 45 x horizontal distance



    3. The attempt at a solution

    using the equation:
    -8580 = F cos 45 x 35

    i am getting double the correct answer...can u tell me what am i doing wrong
     
  2. jcsd
  3. Oct 15, 2009 #2

    Andrew Mason

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    You have to divide the height by cos 45

    AM
     
  4. Oct 15, 2009 #3

    tiny-tim

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    Hi billu77! :wink:
    yes … work done = force times distance, the height is 35, so the distance along the slide is … ? :smile:
     
  5. Oct 15, 2009 #4
    Your trig function is wrong. W=Fdcos(0), since the work the friction done is 8580 J, then W=8580. The distance the friction gone through is Hypotenuse, which is 35/sin(45) rather than 35*cos(45). F=8580/(35/sin(45))=8580*sin(45)/35=173. I am not sure what the initial velocity is doing though.
     
  6. Oct 15, 2009 #5
    thanks....i did not pay attention to the fact that frictional force is just acting along the slide and not at an angle.....appreciate ur help
     
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