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December 20, 2014

December 20, 2014

Posted by **Michelle** on Thursday, February 16, 2012 at 5:02pm.

(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

- Physics -
**bobpursley**, Thursday, February 16, 2012 at 5:08pmForce 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.

- Physics -
**Michelle**, Thursday, February 16, 2012 at 5:14pmWhat does Vi^2 stand for and for you u what value do i use...n for d is it 70 or 145

Thank you

- Physics -
**Anonymous**, Thursday, February 16, 2012 at 6:43pmI 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

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