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Can somebody check my work see if I did the steps right?

  1. Oct 13, 2009 #1
    A boy pulls a 14.6kg sled 12.5m at a constant speed across a horizontal floor. if the rope makes a 32.4degrees?
    angle with the horizontal and the coefficient of friction is .489, what is the work done by each of the forces acting on the sled?
    2 days ago - 2 days left to answer.
    Additional Details
    my work,
    from the free body diagram I get

    EFy= n+Fsin@-mg=0 "@ = theta"
    EFx= Fcos@-f=0 "constant speed"

    solve Fy= n=mg-Fsin@
    plug it into Fx

    friction=un
    Fcos@-u(mg-Fsin@)=0
    Fcos@-umg+uFsin@=0 >>
    Fcos@+uFsin@=umg >>
    F(cos@+usin@)=umg>>
    F=(umg)/(cos@+usin@)
    F=63.3N >>

    work of the force
    WF=63.3N(12.5)cos0>> WF= 791J

    no solve by friction
    Fcos@-f=0 >>> f=Fcos@
    f=(63.3N)cos(32.5) >> f=53.4N

    work of friction
    Wf=(53.4N)(12.5M)cos180 >> Wf= -668J

    I ignore the N force and the gravety and Fsin@ because they are perpendicular.

    is my physic process good?
    is my answer right?
    thanks
     
  2. jcsd
  3. Oct 13, 2009 #2
    All the forces you calculated are OK. The total work done on the sled isn't 0. This isn't possible since the sled gains no kinetic or potential energy. Are you sure that both of the forces involved are pulling in the same direction as the motion of the sled?
     
  4. Oct 13, 2009 #3
    so the total work is W=WF+Wf

    W=791J-668J
    W=123J
    the total work done on the sled is 123J
     
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