A rope pulls a 82.5 kg skier at a constant speed up a 18.7° slope with μk = 0.150. How much force does the rope exert?
1 answer:
Answer:
374 N
Explanation:
N = normal force acting on the skier
m = mass of the skier = 82.5
From the force diagram, force equation perpendicular to the slope is given as
N = mg Cos18.7
μ = Coefficient of friction = 0.150
frictional force is given as
f = μN
f = μmg Cos18.7
F = force applied by the rope
Force equation parallel to the slope is given as
F - f - mg Sin18.7 = 0
F - μmg Cos18.7 - mg Sin18.7 = 0
F = μmg Cos18.7 + mg Sin18.7
F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7
F = 374 N
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