Question

A 64 kg skier is on a 18 degree slope. The coefficient of kinetic friction between the skis and snow is .04. A) What is the weight of the skier? B) what is the normal force of the slope on the skier? C) why is the “parallel” force on the skier? D) what is the frictional force on the skis? E) what is the acceleration of the skier? F) how fast will the skier be going after 3.2 seconds?

Answer 1

a. mg
b. mgCosTheta
c. why? Because of gravity
d. mg*mu*cosTheta
e. in the parallel direction:
net force=ma
mg*sinTheta-mg*mu*CosTheta=m*a solve for acceleration a
f. vf=a*t^2

Answer 2

weight = m g = 64*9.81 = 628 Newtons

normal force = 628 cos 18 = 597 Newtons

parallel force = 628 sin 18 = 194 Newtons

mu * normal force = .04*597 = 24 Newtons

total force down slope=194-24= 170 Newtons
so a = F/m = 170/64 = 2.65 m/s^2
(check that is about .27 g which is a bit less than 9.81 sin 18 because of the friction)

v = a t if initial speed is zero
v = 2.65 * 3.2

Answer 3

vf = a t

Answer 4

Prof Damon is right on F, the world series kept me up late last night.