Physics
posted by Donna
A 24.5kg 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.

bobpursley
ok, first find the vertical and horizontal components.
for the vertical part, it reduces weight.
frictionforce= (weightupwardforce)*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. 
Donna
ok according to you the answer must be 1.5 but that is wrong :(

Elena
x: FcosαμN =0,
y: mgNFsinα=0 =>N= mgFsinα,
μ= Fcosα/N = Fcosα/(mgFsinα)=
=77•cos30/(24.5•9.877•sin30)=
=0.33
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