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Fera09
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Homework Statement
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?
Homework 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
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.