If the coefficient of kinetic friction in the previous problem was actually 0.11 and the slope 30 degrees, to the nearest meter how far up the hill does he go?
**the problem prior was: A skier traveling at 31.9 m/s encounters a 12 degree slope. If you could ignore friction, to the nearest meter, how far up the hill does he go? (my answer was 250m up the hill)**
Sum of the forces in the y-axis = Fn - Fgy = 0
Sum of the forces in the x-axis = Ff + Fgx = ma
The Attempt at a Solution
I figured acceleration = Fnet / mtotal
therefore, acceleration = ((.14*mg*cos30)+(mgsin30)) / m
then the m's would cancel and it would just be..
a = (.14*g*cos30)+(gsin20)