Question

while out skiing, a skier is on a flat terrain that has a coefficient of kinetic friction of 0.62 with his skies. In front of him is a long slope, which is angled 12 above the horizontal and is icy enough to be considered frictionless. At a distance of 8.7m in front of the start of the slope, the 80 kg skier is traveling at 13 m/s. He traverses the level ground and begins up the slope, allowing friction and gravity to stop him. How far along the slope has he traveled when he comes to a stop?

Answer 1

Journey from the start to the foot of the slope

u=13 m/s; =0.62; s=8.7 m; m=80 kg;

frictional force, f= mg

acceleration, a= -f/m =- g = -0.62*9.8= -6.076 m/s²

negative sign indicates that the force acts opposite to the direction of motion,

using, v² - u² = 2as

v = sqrt(u² + 2as) = sqrt(13² - 2*6.076*8.7)

v=7.95 m/s

Journey from the foot of the slope to the end

u= 7.95 m/s; v=0;

force which is responsible to stop the body from ascending, F = mgsin()

a = -F/m = -g*sin() = -9.8*sin(12) = -2.03 m/s²

again using,

v² - 7.95² = -2*2.03*s

s=15.56 m

0² - u² = 2as