The 60.0 kg skier shown below is skiing down a 35.0 degree incline where the magnitude of the friction force is 38.5N

Question

a. what is the acceleration of the skier?
b. what is the normal force on the skier?

Answer 1

Normal force = m g cos 35 {part B}
force down slope = m g sin 35 - 38.5
so
m a = m g sin 35 - 38.5
a = 9.81 sin 35 - (38.5/60) {part A}