Ted, who's ,mass is 75 kg, takes off down a 50 m high, 10 degrees slope on his jet-powered skies. The skis have a thrust of 200 N. Ted's speed at the bottom is 40 m/s. What is the coefficient of kinetic friction of his skis on snow?

Answer 1

friction force up slope = mu m g cos 10 = mu *75 * 9.81 * .985
= 725 mu Newtons up slope
gravity down slope = m g sin 10 = 75*9.81 * .1736 = 128 Newtons
Thrust down slope = 200 Newtons
so
total force down slope = 200 + 128 - 725 mu = 328 -725 mu
so
a = (328 - 725 mu)/75 = 4.37 - 9.67 mu
now kinematics
distance = 50/sin 10 = 288 meters
average speed = 40/2 = 20 m/s because a is constant and v = at start
so
t = 288/20 = 14.4 seconds for the trip
v = (1/2) a t^2
40 = (1/2)(4.37 - 9.67 mu) (14.4)^2
80/207 = 4.37 - 9.67 mu
9.67 mu = about 4
mu = about 0.4