# Iceskating up a hill

1. Dec 18, 2013

### Gurkis

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. The attempt at a solution

F(friction) = $\mu$ * mg(y)

F$\rightarrow$ = F(f) + F(mg(x))

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

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

sin(20)*11,8=4,05...$\approx$ 4,1m

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

2. Dec 18, 2013

### PhanthomJay

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. Dec 18, 2013

### Gurkis

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!