# Kinematics skier problem

1. Oct 23, 2009

### Fera09

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

Part A. A 64.0-kg skier starts from rest at the top of a ski slope of height 70.0 m.
If frictional forces do −1.00×104 J of work on her as she descends, how fast is she going at the bottom of the slope?
I already got the right answer for this one.. I have problems in this second part.

Part B. Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.19. If the patch is of width 69.0 m and the average force of air resistance on the skier is 140 N, how fast is she going after crossing the patch?

2. Relevant equations

Vinitial= 32.5 (got it from part A)
Vfinal= ???
Distance = 69.0 m
Fnormal= mg = 627.84N
coefficient of friction = 0.19
m= 64.0 kg

Ffriction= Fnormal * coefficient of friction

Wfriction= -Ffriction*Distance
Wairresistance= -Fairresistance*Distance (is this right?)
Change in kinetic energy= (1/2)mvfinal2-(1/2)mvinitial2

3. The attempt at a solution

So I got the equation..
Wfriction+Wairresistance=(1/2)mvfinal2-(1/2)mviinitial2

and i plugged in..

-8222.59 + -9660 = (1/2)(64)vfinal2-(1/2)(64)(32.5)

I know I'm doing something wrong somewhere because I can't take the square root of a negative number, but i don't know where.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 23, 2009

### Fera09

just noticed i put this in the wrong section, sorry!

3. Oct 23, 2009

### Staff: Mentor

No, I moved it here from Advanced Physics. This belongs here in Intro Physics. Check your PMs.

4. Oct 24, 2009

### alphysicist

Hi Fera09,

In your last line, did you forget to square the 32.5 (or is that just a typing error)?