A 150-N sled is pulled up a 28° slope at a constant speed by a force of 120 N.What is the coefficient of kinetic friction between sled and slope?
A) 0.53
B) 0.22
C) 0.37
D) 0.33

Question

Quizplus · Accepted Answer

The force acting parallel (up the slope) is 120 N and the perpendicular force is mg (where g is the acceleration due to gravity and m is the mass). The perpendicular force is m*g*cos(28°) and therefore, the mass is (m*g*cos(28°))/g = m*cos(28°). 
Since the sled is moving at a constant speed, friction must be equal to the parallel force (120 N), and the frictional force is given as f = μ*mg*cos(28°). 
Equating f to 120 N, we get μ = 120/(m*g*cos(28°)). Substituting the value of m*cos(28°) in terms of m, we get μ = 120/(m*g*cos(28°)) = 120/(m*g*cos(28°)) = 120/(150*cos(28°)) = 0.374. 
Therefore, the coefficient of kinetic friction between sled and slope is 0.37 (option C).

A 150-N Sled Is Pulled Up a 28° Slope at a Constant