Question

A ski jumper glides down a 37° for 15 m before taking off from a negligibly short horizontal takeoff. If her speed at take off is 12 m/s, find the coefficient of kinetic friction between the skis and slope.

Answer 1

given
theta = 37 degrees
d = 15 m
vi = 0 m/s
vf = 12 m/s

let mue_k is the coefficient of kinetic friction.

let a is the acceleration of the skier while glides down

now use, a = (vf^2 - vi^2)/(2*d)

= (12^2 - 0^2)/(2*15)

= 4.8 m/s^2

let m is the mass of the skier.

Net force acting on skier, Fnet = m*g*sin(theta) - fk

m*a = m*g*sin(theta) - mue_k*N

m*a = m*g*sin(theta) - mue_k*m*g*cos(theta)

a = g*sin(theta) - mue_k*g*cos(theta)

mue_k*g*cos(theta) = g*sin(theta) - a

mue_k = (g*sin(theta) - a )/(g*cos(theta))

= tan(theta) - a/(g*cos(theta))

= tan(37) - 4.8/(9.8*cos(37))

= 0.140 <<<<<<<<------------------------Answer