- #1
StephenDoty
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If a block of mass m is pulled in the direction of (x-component) Tcos(theta) and (y-component) Tsin(theta) across a rough surface with a constant velocity. What is the magnitude of the frictional force?
A) μkmg B) μkT cos θ C) μk(T – mg) D) μkT sin θ E) μk(mg – T sin θ)
Since fk is in the opposite direction as the motion and since there is no acceleration F=0=Tcos(theta) - fk
So
fk= Tcos(theta)
So is the answer B?
A) μkmg B) μkT cos θ C) μk(T – mg) D) μkT sin θ E) μk(mg – T sin θ)
Since fk is in the opposite direction as the motion and since there is no acceleration F=0=Tcos(theta) - fk
So
fk= Tcos(theta)
So is the answer B?