A 62 kg skier starts from rest at the top of a ski slope 70meters high, if the f

Question

A 62 kg skier starts from rest at the top of a ski slope 70meters high, if the frictional forces do - 13.9 kj of work on heras she descends, A) how fast is she moving at the bottom of the slope inm/s? B) Once at the bottom the skier is moving horizontally andcomes across a patch of soft snow, where the coefficient of kineticfriciton is 0.14. If the patch is 86 meters wide and theaverage force of air resistance on the skier is 189 newtons, howfast is she going after crossing the patch? must do part Abefore part B! A 62 kg skier starts from rest at the top of a ski slope 70meters high, if the frictional forces do - 13.9 kj of work on heras she descends, A) how fast is she moving at the bottom of the slope inm/s? B) Once at the bottom the skier is moving horizontally andcomes across a patch of soft snow, where the coefficient of kineticfriciton is 0.14. If the patch is 86 meters wide and theaverage force of air resistance on the skier is 189 newtons, howfast is she going after crossing the patch? must do part Abefore part B!

Explanation / Answer

PE + KE + Friction = PE + KE mgh + 1/2mv^2 + Friction = mgH + 1/2mV^2 62(9.8)(70) + 0 - 13.9 *10^3 = 0 + 1/2(62)V^2 42532 - 13900 = 31V^2 28632 = 31V^2 V^2 = 923.613 V= 30.391 m/s Kinetic friction force = Normal Force = (0.14)(62) = 8.68 =ma F = ma friction + air resistance = ma - 8.68 - 189= 62a 62a = -197.68 a = -3.188 now using kinematics x = 86 = vt+ 1/2at^2 86 = (30.391)t + (0.5)(-3.188)t^2 86 = 30.391t-1.594t^2 t = 3.456 seconds v final = v + at = 30.391 + (-3.188)(3.456) = 30.391 - 11.019 =19.372 m/s Hopefully this helps!

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