Kinetic friction - toughie! (1 Viewer)

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kinetic friction -- toughie!

A skier weighing 90 kg starts from rest down a hill inclined at 17 degrees. He skis 100 m down the hill and then coasts for 70 m along level snow until he stops. FIne the coefficient of kinetic friction between the skis and the snow. What velocity does the skier have at the bottom of the hill?

I can find the position of the skier down the hill as a function of velocity. I can also find the position of the skier along the level ground as a function of velocity (I think I have this part of the problem right). I get x = (v^2)/(g sin 17 - uk cos 17) for the motion on the hill and x = -v^2/(uk g) for the motion on the level snow. But I'm struggling with how to solve for v at the bottom of the hill and uk (coefficient of kinetic friction)! Please help! Your input is GREATLY appreciated :)

According to the answer given in the book, I should get uk=.18 and v at bottom of hill = 15.6 m/s.


Homework Helper
If we assume that friction acts during his descent as well, then the equation would be:

Energy at bottom of hill = Initial Potential Energy - Wfriction(during descent) = Wfriction(70m level: energy it took to stop him)

Do you agree?

Would you now like to try expressing all three terms in terms of the variables we do know or are of our interest? (i.e. mass, height ~ Lsin17, coefficient of friction) You will find easily that there is only 1 unknown in this equation =)
bump to an old problem,

but i am missing something big here.

energy at the bottom of the hill is 1/2 * m * v^2 right?

Initial Energy is mgh or 25787 or so, right?

w = F * d, or uk*distance, right?

plugging in the answers, energy at bottom of the hill is 10,951.2

10951.2 = 25787 - uk(100meters)

uk = 148.36

i am rusty, really really rusty.....what am i missing?

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