**Note: Friction acts on the way down the hill as well as on the level, and even

Question

**Note: Friction acts on the way down the hill as well as on the level, and even though
the coefficient is small, it is the sole reason this skier eventually coasts to a stop.** A skier starts from rest at the top of a hill that is inclined at 11.5 with the horizontal. The hillside is 169 m long, and the coefficient of friction between snow and skis is 0.0702. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier move along the horizontal portion of the snow before coming to rest?
Answer in units of m

Explanation / Answer

work done by gravity + work done by friction = 0

mgh   +   work by friction on slope + work by friction on level part = 0

mgLsin@   - u mgcos@ L    -  umg d = 0

Divide everything by mg

L sin@ - u L cos@ - u d = 0

Solve for d, the horizontal distance

   d = L sin@ / u   - L cos@ = L (sin@ / u   - cos@ )   =

     = 169 (sin11.5 / 0.0702 - cos11.5) =   314 meters

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