Finding the Coefficient of Friction

  • Thread starter xcsyl
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Homework Statement


An ice skater moving at 10.0 m/s coasts to a halt in 1.0 x 10^2 m on a smooth ice surface. What is the coefficient of friction between the ice and the skates?


Homework Equations


coefficeint of friction = kinetic friction/Normal force
(µk = Fk/Fn)
Σ Fy = may
Σ Fx = max

The Attempt at a Solution


I've tried many different approaches, using motion equations, and somehow end up confused in all of them. Maybe I picked the wrong equations and this problem is simpler than I'm making, but I'm not sure. I'm having problems with incorporating motion equations with force equations. Here's one of my attempts. Thanks! (I know the answer is 0.051, but I can't get to that number)

My attempt:
⌂x = 100 m
Vi= 10 m/s

Σ Fy = Fn - mg = may

⌂x = .5(Vf + Vi)⌂t
100 = .5(0+10)⌂t
⌂t= 20 s

Vf^2=Vf^2 + 2a⌂x
0=100 + 2a(100)
-100 = 200a
a=-.5

[EDIT] I'm pretty sure I found my mistake! I still would like to see this worked out, but the answer is .051 not .51, sorry I'll change that.
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
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You know the acceleration, so what is the magnitude of the resultant force on the skater?
This resultant force is equal to the friction. The friction is proportional to Fn. Fn=mg.

ehild
 
  • #3
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Ahhh I wrote down the answer wrong once again.. sorry. It's .051 not .51
That's what I can't quite figure out past getting acceleration. Again sorry! I'm getting my numbers jumbled


Ooh I also found in my book what some objects on surfaces have as coefficients of friction and "steel-on-ice" is .05. I still want to be able to work this out, and thanks ehild for that. I'm using Fn=mg as well, thanks!
 
Last edited:

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