Iceskating up a hill

  • Thread starter Gurkis
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  • #1
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Homework Statement



An iceskater weighing 69kg is gliding up an incline hill at 20 degrees.
How high up does the skater make it when her initial velocity is 8,9m/s and the friction between her skates and the ice is 0,97N.

Homework Equations



mg(y) = m*g*cos(20)
mg(x) = m*g*sin(20)

The Attempt at a Solution





F(friction) = [itex]\mu[/itex] * mg(y)

F[itex]\rightarrow[/itex] = F(f) + F(mg(x))

a=F[itex]\div[/itex]M[itex]\rightarrow[/itex]F[itex]\div[/itex]m=232,7[itex]\div[/itex]69 = 3,34m/s^2

s = (v^2 - v(o)^2) / 2a = 11,8m

sin(20)*11,8=4,05...[itex]\approx[/itex] 4,1m

the answer should be 4,0m. What am I doing wrong?
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
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Physics looks good, maths is off.....redo the math: you should get something like 4,02 which rounded to two sig figs is 4,0
 
  • #3
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I see, thanks. I thought my problem was that I was thinking about it all wrong. I redid the math and it checks out.

Appreciate the help!
 

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