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.

Answer 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.

Answer 2

ok according to you the answer must be 1.5 but that is wrong :(

Answer 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