A skier of mass 60.0kg starts from rest at the top of a ski slope of height 61.0

Question

A skier of mass 60.0kg starts from rest at the top of a ski slope of height 61.0m.

a) If frictional forces do -1.09 x 10^4 J of work on her as she descends, how fast is she going at the bottom of the slope?
Take free fall acceleration to be g=9.80m/s^2.

b)Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is uk= 0.150. If the patch is of width 67.0mand the average force of air resistance on the skier is 160N, how fast is she going after crossing the patch?

c)After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.60minto it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Explanation / Answer

(A) First you find the frictional force by dividing -1.09*10^4 J by 61. W=Fd so F=W/d=(( -1.09*10^4)/61)=(-178.69). Then find the Fp which will be the force the skier exerts going down the slope by multipling mass and gravity. Fp=(60)(9.8)= 588 N. Now the Fnet=Fp-Ffr so Fnet=588-178.69=409.31. Fnet also equals mass times acceleration (Newtons Second Law) Fnet=ma, so you can find the skier's acceleration by dividing 409.31 by 60 (409.31/60)=6.82. Now you can use the kinematic equation V^2=Vo^2+2ad. Vo=0 so you can just say (2)(6.82)(61)=832.04...., then you take the square root of that answer to get your final velocity which will equal 28.845 m/s. (B) Find the normal force (Fn) by multiplying skier's mass and gravity (9.8)(60)=588 and multiply that by the coefficient of friction (uk) to find the frictional force (Ffr) so Ffr=(0.15)(588)=88.20. You know the Fp=588 so now you find Fnet again by saying Fp-(Ffr+ Air Resistance). 588-(88.20+160)=339.80 Now you find the skiers new acceleration by dividing that answer by the mass. (339.80)/(60)=5.663 m/s^2. Now you use the same equation from A to find the final velocity by taking the square root of 2ad, so (2)(5.663)(61)=690.886... and after you take the square root of that, the new velocity is 26.285 ....m/s C) You know the net force acting on the skier from earlier was 339.80 Newtons, so the average force of the snowdrift should be a (-301.772)

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