::A skier of mass 63.0kg starts from rest at the top of a ski slope of height 62.0m .A skier of mass 63.0kg starts from rest at the top of a ski slope of height 62.0m . A) If frictional forces do −1.00×104J 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/s2 . B) Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.210. If the patch is of width 64.0m and the average force of air resistance on the skier is 180N , 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 2.90m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her? m=63.0kg g= 9.8m/s2 H=62.0m What I've done: A) U= mgH= (63.0kg)(9.8m/s2)(62.0m)= 38278.8 N-m NET FORCE= 38278.8-1000= 37278.8 37278.8= (1/2)mv2 37278.8= (31.5)v2 so v= 34.401 Mastering physics says that is WRONG. I've checked my math over and over again. Where did I go wrong? tips?