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Work done by friction Help!

  1. Apr 5, 2005 #1
    A 16# block on a 15* incline is released w/a F=30# pulling up at 30* to the Horiz. it's Vo=3ft./s and V=9ft./s after moving 2ft. up the incline. Determine the Work done by Friction.

    I have started by trying to find the Accel. but I do not know that it is going to help me. Anyone have any insight on this problem?
     
  2. jcsd
  3. Apr 5, 2005 #2

    dextercioby

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    What is "#" and "*" ?I mean,what do they stand for...?

    As a general idea,i would reccomend using the theorem of variation of KE...

    Daniel.
     
  4. Apr 5, 2005 #3
    # is lbs.
    * is degrees

    i guess is should say i have found Accel to be 18 ft./s^2 , But then What?
     
  5. Apr 5, 2005 #4

    dextercioby

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    It's not right,the force & the mass have the same unit.If the mass is measured in "lbs",then the force should be measured in "lbsf"...

    Anyway,u don't need the acceleration..What are the 3 forces that do work on the body?

    Daniel.
     
  6. Apr 5, 2005 #5
    Sorry,

    The F is created by a hanging 30 lb. weight wraped arround a pulley.
    Then there is of course Friction and Normal forces however (coefficient of friction) is not given.
     
  7. Apr 5, 2005 #6

    dextercioby

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    Yea,the normal force from the incline does no work,however,so it should be kept outta the KE equation.Good then,so the theorem says that the variation of KE is due (and is numerically equal) to the work done by the forces acting on the body.

    I'll give u one more hint:those forces that do matter (i.e.enter the equation) are:tangent component of gravity,the 30lbf tension in the rope and kinetic friction force.

    Daniel.
     
  8. Apr 5, 2005 #7
    Here is a pic of the problem.

    KE=(1/2)mv^2
    KE1+delta.W=KE2
    W=Fd
    Right?
    How do you find F?
    And
    V^2= what V or Vo?


    I don't know, i have been staring at this problem for too long!
     
  9. Apr 5, 2005 #8
    Is The sum F=15lbs.cos(15)+usin(15)16lbs. ?????

    the 15lbs. coming from the pulley setup.
     
  10. Apr 5, 2005 #9

    dextercioby

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    Why [itex] 15 lbsf [/itex]...?What's the tension in the chord...?

    Daniel.
     
  11. Apr 5, 2005 #10
    Look at the attachment above. I may be thinking about this wrong. MY BRAIN IS FRIED!
     
  12. Apr 5, 2005 #11

    dextercioby

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    So you got 2 pulleys and one chord.One of the pulleys is free and the other is fixed.U need to find the tension in the chord.By the looks of it,neglecting the moment of inertia of the mobile pulley,i'd say u need to know the pulley's acceleration...

    Daniel.
     
  13. Apr 5, 2005 #12
    I can find the accel. of the block by way of
    V^2=Vo^2+2a(delta)x
    but it isnt going to help.
     
  14. Apr 6, 2005 #13

    mpm

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    This is just off the top of my head and looking at a few of the responses.

    To find work, you use the Equation W= Fd.

    I see you already know this.

    An earlier post said that the normal force was neglible to this problem although. That sounds kinda funny to me but I will go along with it for now.

    Anyway, you need to find the tension in the rope pulling toward the right.

    It looks like your system is in equilibrium right now, meaning that the Force pulling to the left is equal to the tension (force) in the rope pulling to the right.

    If this is the case, just find the tension in the rope and that should be your force to find the work done by friction.

    Again, I would think the normal force would play a part in this but maybe not.

    This is a typical physics question so your question may be better answered in the physics area.

    Hope I helped some.
     
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