Posted by smh on .
A skier starts from rest at the top of a hill that is inclined at 12.0° with the horizontal. The hillside is 180.0 m long, and the coefficient of friction between the snow and the 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?
I kniw the answer is 323m but I don't understand how to get there!
The friction force during the motion along the inclined plane is
F(fr)1 = kN = k•m•g•cosα.
The work of this force is
The law of conservation of energy gives
m•g•h = m•v^2/2 + k•m•g•cosα•s.
m•v^2/2 = m•g•h - k•m•g•cosα•s.
For the motion along the horizontal path
m•v^2/2 = k•m•g•s1
s1= m•v^2/2• k•m•g