Work to push a sled up a hill: Friction, inclines,force,work

  1. 1. The problem statement, all variables and given/known data
    We are pushing a sled up a hill. Our push is horizontal. The hill is inclined at 15 degrees. The sled masses 35 kg. The coefficient of kinetic friction is .2. The hills height is 3.6 m. We push at a constant velocity. How much work do we do in pushing?


    2. Relevant equations
    Mass*Gravity Accel*cos(15) = Normal force/Gravity into surface
    Mass*Gravity Accel*sin(15) = Gravity slideways
    Normal*Coef of friction = friction force
    sin(x) = Opp/hyp or hyp = Opp/sin(x)
    Work = Force*Distance

    3. The attempt at a solution
    Normal force = 35*9.8*cos(15) = 331.313N
    Force down hill = 35*9.8*sin(15) = 88.77N
    Friction force = 331.313*.2 = 66.262N
    Length of incline = 3.6*sin(15) = 13.909M
    With constant velocity, forces will equal out
    gravity and friction = uphill push force = 66.262+88.77 = 155.038N
    Work = 155.038*13.909 = 2156J ?

    I thought this would be it assuming the force and distance must be in the same direction.

    forward push*cos(15) = uphill push
    forward push = Uphill push/cos(15) = 155.038/cos(15) = 160.507
    Work = 160.507*13.909 = 2232J ?

    The back of book answer is 2300J (2 signifigant digits)
    Where is the error?
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,041
    Science Advisor
    Homework Helper

    Hi krausr79! :smile:

    No, the push is horizontal, so that will add to the usual normal force. :wink:
     
  4. Thank you.
     
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