Physics

A 24.5-kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 77.0 N and is directed at an angle of 30.0° above the horizontal. Determine the coefficient of kinetic friction.

asked by Donna
  1. ok, first find the vertical and horizontal components.

    for the vertical part, it reduces weight.

    frictionforce= (weight-upwardforce)*mu

    upward force= 77Sin30
    weight=24.5g

    Now for friction force, that with no acceleration, horizontal component=24.5*cos30=fricton force

    put those in the equation above, solve for mu.

    posted by bobpursley
  2. ok according to you the answer must be 1.5 but that is wrong :(

    posted by Donna
  3. x: Fcosα-μN =0,
    y: mg-N-Fsinα=0 =>N= mg-Fsinα,

    μ= Fcosα/N = Fcosα/(mg-Fsinα)=
    =77•cos30/(24.5•9.8-77•sin30)=
    =0.33

    posted by Elena

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