A 60kg skier rides down a ski hill with 15 degrees slope. At the bottom of the slope is a mogul (hump) witha radius of curvature of 30.0 meters.
What is the max height of the ski hill if the skier is to just barefly remain in contact with the ground while skiing over the hump?
If the coefficient of friction were 0.15 between the skis and the snow, what min height would be needed?
given
theta = 15 degrees
r = 30 m
minimum speed of the skier barefly remain in contact with the ground while skiing over the hump,
v = sqrt(g*r)
= sqrt(9.8*30)
= 17.1 m/s
let h is the max height of the ski.
Apply conservation of energy
m*g*h = (1/2)*m*v^2
h = v^2/(2*g)
= 17.1^2/(2*9.8)
= 14.9 m <<<<<<<<<<----------------------Answer
Workdone by friction = change in mechanical energy
fk*d*cos(180) = (1/2)*m*v^2 - m*g*h
mue_k*m*g*cos(15)*(h/sin(15))*(-1) = (1/2)*m*v^2 - m*g*h
-mue_k*m*g*h*cot(15) = (1/2)*m*v^2 - m*g*h
-mue_k*g*h*cot(15) = (1/2)*v^2 - g*h
-0.15*9.8*h*cot(15) = (1/2)*17.1^2 - 9.8*h
==> h = 33.9 m <<<<<<<<<<----------------------Answer
A 60kg skier rides down a ski hill with 15 degrees slope. At the bottom of...