Physics Dynamic Friction

Question

Include Free Body Diagram

11. A 70-kg skier slides down a hill inclines at an angle of 12 ̊ with the horizontal...

a. Calculate the skier’s acceleration if opposed by a 100-N friction force.

b. what is the coefficient of kinetic friction.

12. A 20-kg crate is lowered down a plane inclined at an angle of 37 ̊ with the horizontal by a rope that exerts on the crate a 50-N force directed parallel to the plane. A 40-N friction force is also parallel to the plane. (Opposes the downward motion)

a. What is the acceleration of the crate?

b. What is the coefficient of kinetic friction?

Answer 1

#11. a. force down slope = mgsin12 - friction 100N up the slope

net force = 70 x 9.81 x sin12 - 100 = 42.77N = 70 x a, so acceleration = 42.77/70 = 0.61m/s²

b. μ_k = F_f/F_R = 100/70•9.81•cos12 = 0.15

#12. total upward force = 50 + 40 = 90 N along incline

downward along incline = component of weight = 20 x 9.81 x sin37 = 118.07N

Net force = 118.07-90 = 28.07N = m•a therefore a = 28.07/20 = 1.4m/s²

b. μ_k = F_f/F_R = 40/20•9.81•cos37 = 0.26

Nancy L, if you need further clarification on this or any other problem, let me know.

Mr. A