Arope pulls a 82.5 kg skier at a constant speed up a 18.7â° slope with î¼k = 0.150. how much force does the rope exert?

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