A skier of mass 57.0 starts from rest at the top of a ski slope of height 62.0 .

Question

A skier of mass 57.0 starts from rest at the top of a ski slope of height 62.0 . Now moving horizontally, at 28.8 m/s the skier crosses a patch of soft snow, where the coefficient of friction is = 0.190. If the patch is of width 66.0 and the average force of air resistance on the skier is 140 , how fast is she going after crossing the patch? After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.70 into 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 -1.07E4 by 64. W=Fd so F=W/d=((-1.07E4)/64)=(-167.18). Then find the Fp which will be the force the skier exerts going down the slope by multipling mass and gravity. Fp=(58)(9.8)= 568.4 N. Now the Fnet=Fp-Ffr so Fnet=568.4-167.18=401.22. Fnet also equals mass times acceleration (Newtons Second Law) Fnet=ma, so you can find the skier's acceleration by dividing 401.22 by 58 (401.22/58)=6.91. Now you can use the kinematic equation V^2=Vo^2+2ad. Vo=0 so you can just say (2)(6.91)(64)=885.45...., then you take the square root of that answer to get your final velocity which will equal 29.75 m/s. (B)Find the normal force (Fn) by multiplying skier's mass and gravity (9.8)(58)=568.4 and multiply that by the coefficient of friction (uk) to find the frictional force (Ffr) so Ffr=(0.17)(568.4)=96.628. You know the Fp=568.4 so now you find Fnet again by saying Fp-(Ffr+ Air Resistance). 568.4-(96.628+170)=301.772. Now you find the skiers new acceleration by dividing that answer by the mass. (301.772)/(58)=5.202 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)(5.202)(64)=665.97... and after you take the square root of that, the new velocity is 25.806....m/s (C)You know the net force acting on the skier from earlier was 301.772 Newtons, so the average force of the snowdrift should be a (-301.772)

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