Question

A skier, starting from rest, coasts down a slope that makes a 25.0^o angle with the horizontal. The coefficient of kinetic friction between the snow and the skis is 0.200. The skier travels along the slope a distance of 11.4 m before skiing off the edge of a cliff. The skier falls a vertical distance of 3.80 m before landing on another downhill slope. Determine the speed with which the skier lands.

Answer 1

here,

theta = 25 degree

s = 11.4 m

the coefficient of kinetic friction , uk = 0.2

h = 3.8 m

let the speed at which the skier lands be v

using Work-energy theorm

the work done by friction = 0.5 * m * v^2 - (m * g * (s * sin(theta) + h) )

-uk * m * g * s * cos(theta) = 0.5 * m * v^2 - (m * g * (s * sin(theta) + h) )

- 0.2 * 9.81 * 11.4 * cos(25) = 0.5 * v^2 - (9.81 * (11.4 * sin(25) + 3.8) )

solving for v

v = 11.34 m/s

the speed at which the skier lands is 11.34 m/s