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A horizontal force of 280 N is exerted on a 2.0 kg
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Answer #1

1.Fc=m*v^2/r solve for v
v=sqr(Fc*r/m)
v=11.83m/s

2.Total mechanical energy of skier = mgh = (56)(9.8)(230) = 126224 J
KE of skier at botttom of trail = 1/2mV² = (0.5)(56)(11)² = 3388 J
Energy dissipated by friction = 126224 - 3388 = 9236 J

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