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Work and Energy Help

  1. Oct 12, 2007 #1
    1. The problem statement, all variables and given/known data

    A 300 kg piano slides 4.0 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-36). The effective coefficient of kinetic friction is 0.40.

    Picture attached

    (a) Calculate the force exerted by the man.
    N
    (b) Calculate the work done by the man on the piano.
    J
    (c) Calculate the work done by the friction force.
    J
    (d) What is the work done by the force of gravity?
    J
    (e) What is the net work done on the piano?
    J

    2. Relevant equations

    Fg = Fn
    Fgx = Fgsin(30)
    Fgy = Fgcos(30)

    W = F(displacement)cos@


    3. The attempt at a solution

    Fg = mg = (300 kg)(9.8 m/s2) = 2940 N
    Fn = 2940 N

    Fgx = 2940sin(30) = 1470 N
    Fgy = 2940cos(30) = 2546.11 N

    Ff = uFn = (0.4)(2940 N) = 1176 N

    Fx = Fgx - Ff
    = 1470 - 1176
    = 294 N

    Can't get any of it =/
     

    Attached Files:

  2. jcsd
  3. Oct 12, 2007 #2

    Doc Al

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    Staff: Mentor

    Gravity acts vertically, while the normal force acts perpendicular to the incline surface. They are not equal. (If the surface was horizontal, then they would be equal.)
    Good. Which of those is equal to the normal force?

    The wrong value for the normal force is messing you up.
     
  4. Oct 12, 2007 #3
    ahhh right thanks :P.

    i still can't seem to get the work done by the man on the piano though..

    I do

    W = (451.56)(4.0)cos(30)
     
    Last edited: Oct 12, 2007
  5. Oct 13, 2007 #4
    Gah.. Still can't get it.. I'm on my last attempt.. If I screw it up I can't input anymore..
     
  6. Oct 13, 2007 #5

    learningphysics

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    Homework Helper

    What force did you get in part a) ?
     
  7. Oct 13, 2007 #6
    In part A I got 451.56 N

    EDIT: Actually I inputted 451.96 but theres a 1% margin for error. I got that part right.
     
  8. Oct 13, 2007 #7

    Doc Al

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    Staff: Mentor

    The force he exerts is parallel to the incline, so the angle (between force and displacement) is 0 degrees not 30.
     
  9. Oct 14, 2007 #8
    I'm still getting it wrong using cos(0). On my last try.
     
  10. Oct 14, 2007 #9

    Doc Al

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    Staff: Mentor

    Are inputing the correct sign? Since the man pushes up while the piano moves down, the work he does will be negative.

    More accurately, I should have said that the angle between force and displacement is 180 degrees, not really zero. :redface: (Sorry if that threw you off.)

    (I usually separate the calculation of the magnitude of the work and the figuring out of its sign.)
     
  11. Oct 19, 2007 #10
    What equation did you use to get 451 N in part A?
     
  12. Oct 19, 2007 #11

    Doc Al

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    Staff: Mentor

    Since the piano is not accelerating, the net force parallel to the incline must be zero. There are three forces to consider.
     
  13. Oct 19, 2007 #12
    Oh wait, I just had a calculation error, thanks though!
     
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