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kdizzle711
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Homework Statement
A skier of mass 59.0kg starts from rest at the top of a ski slope of height 70.0m .
Homework Equations
The Attempt at a Solution
If frictional forces do −1.08×10^4J of work on her as she descends, how fast is she going at the bottom of the slope?
(1/2)mV^2=mgh + (1/2)mV^2 + -1.08 x10^4
0=59*9.8*70+(1/2)(59)(v^2)+-1.08x10^4
v=31.4
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is = 0.180. If the patch is of width 65.0m and the average force of air resistance on the skier is 170N , how fast is she going after crossing the patch?
(1/2)*59*31.7=59*9.8*70+(1/2)(59)(v^2)+11.7+170
935-11.7-170=40,474+29.5v^2
V^2=
v=
Where an I going wrong with this?
C. After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 3.00m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?