An extreme skier, starting from rest, coasts down a mountain slope that makes an

Question

An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25.0° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts down a distance of 8.9 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 4.20 m below the edge. How fast is she going just before she lands? m/s

A 33-kg girl is bouncing on a trampoline. During a certain interval after she leaves the surface of the trampoline, her kinetic energy decreases to 200 J from 410 J. How high does she rise during this interval? Neglect air resistance.

Explanation / Answer

Applying Newton’s 2nd law,

Fnet = mgsin – uk*mgcos = ma

a= gsin – uk*gcos

a= 9.8*sin25 - 0.20*9.8*cos25 = 2.36 m/s^2 ….directed downward along the slope

thus a= -2.36 m/s^2

vbottom2 = vi^2 + 2ad = 0^2 – 2*2.36*-8.9

gives, vbottom = 6.48 m/s ………………directed downward along the slope

When the skier skies off

vix= vbottom = 6.48 m/s

viy= 0m/s

vfx= vix= vbottom = - 6.48 m/s ……since no acceleration along horizontal

vfy^2= viy^2 –2gh=

vfy^2 = (0^2) -2*9.8*-4.20

gives,

vfy = - 6.34 m/s

Thus vf = ( 6.48 i - 9.07 j) m/s

vf= sqrt(6.48^2 + 9.07^2) = 11.15 m/s

PE = KE

mgh = KEi – KEf

mg(hf – hi) = KEi – KEf

33*9.8*(hf – 0)= 410 – 200 => hf = 0.65 m

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