Question

A skier starts from rest at the top of a hill that is inclined at 10.9° with respect to the horizontal. The hillside is 190 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?

Answer 1

Using energy conservation for this whole process:

KEi + PEi + Wf = KEf + PEf

KEi = 0, since skier starts from rest

KEf = 0, since finally skier is at rest

PEf = 0, since at final position height of skier is 0 m

PEi = m*g*hi

hi = Initial height of skier = L*sin 10.9 deg = 190*sin 10.9 deg

hi = 35.928 m

Wf = Work-done by friction along the incline + Work-done by friction on ground level

Wf =

Friction force along the incline = *m*g*cos 10.9 deg

Friction force on level surface = *m*g

Wf = Force*displacement*cos

= Angle between Friction force and displacement = 180 deg

So,

Wf = Ff*d*cos 180 deg = -Ff*d

Wf = -(*m*g*cos 10.9 deg*L + *m*g*x)

L = distance traveled along the incline = 190 m

x = distance traveled on level plane before coming to stop = ?

So,

0 + m*g*hi - (*m*g*cos 10.9 deg*L + *m*g*x)  = 0 + 0

divide by (m*g)

hi - (*cos 10.9 deg*L + *x)  = 0

x = hi/ - L*cos 10.9 deg

x = 35.928/0.0750 - 190*cos 10.9 deg

x = 292.5 = distance traveled by skier before coming to stop

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