A skier with a mass of 56 kg starts from rest and skis down an icy (frictionless) slope that has a length of 70 m at an angle of 32° with respect to the horizontal. At the bottom of the slope, the path levels out and becomes horizontal, the snow becomes less icy, and the skier begins to slow down, coming to rest in a distance of 145 m along the horizontal path.

Question

A skier with a mass of 56 kg starts from rest and skis down an icy (frictionless) slope that has a length of 70 m at an angle of 32° with respect to the horizontal. At the bottom of the slope, the path levels out and becomes horizontal, the snow becomes less icy, and the skier begins to slow down, coming to rest in a distance of 145 m along the horizontal path.
(a) What is the speed of the skier at the bottom of the slope?
(b) What is the coefficient of kinetic friction between the skier and the horizontal surface?
μk =

I am really lost have no clue please help

bobpursley · Answer

Force down the slope: mg*SinTheta acceleration= force/mass so figure that. Vf^2=Vi^2+2ad solve for Vf Use that same equation, with Vf=0 now, solve for a. Then a= mu*g, solve for mu.

Michelle · Answer

What does Vi^2 stand for and for you u what value do i use...n for d is it 70 or 145 Thank you

Anonymous · Answer

I kept trying but still i cant come up with the right answer :/ this is what i did for the speed 56(9.8)sin(32) =290.82 a=290.82/56 5.2 m/s^2 Vf^2=0^2+2(5.2)(145)=38.8 can you please help