An extreme skier, starting from rest, coasts down a mountainthat makes an angle 25.0^\circ with the horizontal. The coefficient of kinetic friction be

Line 2020-11-03 Answered
An extreme skier, starting from rest, coasts down a mountainthat makes an angle 25.0 with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 11.9 m before coming to the edge of a cliff. Without slowing down, she skis offthe cliff and lands down hill at a point whose vertical distance is 4.20 m below the edge. How fast is she going just before she lands?
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Expert Answer

falhiblesw
Answered 2020-11-04 Author has 97 answers

First find the skier's final speed when she clears the cliff:
Before we can use the equation vf2=vo2+2ad we need her acceleration.
If you drew a free body diagram, you should come up with thefollowing equation:
mgsinθFiction=ma
imageFiction=mgcosθ
mgsinθmgcosθ=ma
the masses cancel and you have:
a=g(sinθcosθ)=9.81(sin(25)cos(25)0.2)=2.37ms2
Now use acceleration to find her final velocity before leavingthe cliff:
vo=0
a=2.37 ms2
d=11.9m
vf=(2)(9.81)(11.9)=7.51ms
She leaves the cliff at this speed at 25 deg to the horizontal, so
V=7.51cos(25)=6.80ms
Voy=7.51sin(25)=3.17 ms
Now we just need to find Vfx and Vfv
, and then take their magnitude to get the final answer.
Vfx=6.8 ms
Vfy2=Voy2+2ad=3.172+2(9.81)(4.2)=92.45 ms
vf=vfx2+vfy2=6.82+92.4511.8 ms

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