Work and Energy Problem

Greetings to all. I hope that someone here will be able to shed some light
on my questions. For those of you who have the book, Serway & Faughn,
College Physics 5th Edition, I am referring to question 5.39.

"A skier starts from rest at the top of a hill that is inclined at 10.5
degrees with the horizontal. The hillside is 200m 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 move along the horizontal portion of the snow before coming to
rest?"

Possible solution:

The Fn of skier is mgcos10.5 while on hill and Fn of skier is mg on
horizontal portion.

Using F * s = Wnc = (KEf + PEf) - (KEi + PEi) and defining the bottom of
hill as zero level for potential energy, I used F * s =
(0.0750)(mgcos10.5)(200) = (1/2)m(v)^2 + 0 - 0 - mgh (h = 200sin10.5).

I determined v = 19.02 m/s.

Using F * s = Wnc = (KEf + PEf) - (KEi + PEi), I used F * s =
(0.0750)(mg)(s) = 0 + 0 - (1/2)(m)(19.02)^2 + 0.
s = 246.1m.

The answer in the back of the book is s = 289m.

Did I do the problem wrong? If so, can I solve it this way?

Thanks,
Will