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An 80.0 kg rollerblader is at rest at the top of a 100 m hill at a 7o incline. The coefficient of friction on the hill is 0.110. What is the roller blader's kinetic energy at the bottom of the hill?

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h/d=sin7 d=100/sin7=820.6m v^2=2gd(sin7-0.11cos7)=2*9.8*820.6*0.0127 v=sqrt(204.264) v=14.3m/s k.e at bottom=1/2*80*14.3^28179.6joules

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