A 50 kg skier is pulled up a 28° slope at a constant speed by a friend applying a force of 300 N. What is the coefficient of kinetic friction between the snow and the skier?
A)0.18
B)0.30
C)0.36
D)0.63

Question

Quizplus · Accepted Answer

The force applied by the friend is used to overcome both the component of gravitational force along the slope and the force of friction. Since the skier is moving at a constant speed, the net force acting on the skier is zero. This means the applied force is equal to the sum of the frictional force and the component of the gravitational force parallel to the slope.The gravitational force acting on the skier is $mg$, where $m = 50\,kg$ and $g = 9.8\,m/s^2$. The component of this force along the slope is $mg\sin(\theta)$, where $\theta = 28^\circ$.$$mg\sin(\theta) = 50 \times 9.8 \times \sin(28^\circ) \approx 229.8\,N$$The applied force is 300 N, so the force of friction $f_f$ is the difference between the applied force and the component of the gravitational force along the slope.$$f_f = 300\,N - 229.8\,N = 70.2\,N$$The normal force $N$ is the component of the gravitational force perpendicular to the slope, which is $mg\cos(\theta)$.$$N = mg\cos(\theta) = 50 \times 9.8 \times \cos(28^\circ) \approx 431.8\,N$$The coefficient of kinetic friction $\mu_k$ is the ratio of the frictional force to the normal force.$$\mu_k = \frac{f_f}{N} = \frac{70.2}{431.8} \approx 0.162$$Rounding to two decimal places, the closest answer is 0.18, which is option A.

A 50 Kg Skier Is Pulled Up a 28° Slope