A Skier starts from rest at the top of a hill that is inclined at 9.8degree with

Question

A Skier starts from rest at the top of a hill that is inclined at 9.8degree with respect to the horizontal. The hillside is 210 m long, and the coefficient of friction between snow and skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully

Explanation / Answer

Friction u = 0.075
Initial Velocity = 0
Acceleration a = g*sin(9.8) - uk * g cos(9.8)
Distance s = 210 m

Velocity at the bottom of hill V =
v^2 = u^2 + 2*a*s
v= sqrt(0 + 2*(g*sin(9.8) - uk * g cos(9.8))*210)
v = sqrt(0 + 2*(9.8*sin(9.8) - 0.075 * 9.8* cos(9.8))*210)
v = 19.9 m/s

Now Distance travelled in horiontal =
v^2 = u^2 - 2*a*s
s = 19.9^2/ (2*a)

a = uk * g
a = 0.075*9.8

s = 19.9^2/ (2*0.075*9.8)
s = 269.4 m

Distane travelled by skier , s = 269.4 m

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