Homework Help Forum: physics

Posted by sharon on Saturday, October 31, 2009 at 2:46pm.

an extreme skier starting from rest, coasts down a mountain that makes an angle 25.0¨¬with the horizontal. The coeffiecient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 8.4m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 2.80m below the edge. How fast is she going just before she lands?

• physics - MathMate, Saturday, October 31, 2009 at 11:45pm

  Frictional force, F = μmg cos(θ) Work done to overcome kinetic friction = F*D = 8.4F Difference in elevation, H = 8.4 sin(&theta)+2.8 m Energy left (in the form of KE) = mgH-8.4F Equate Kinetic energy with potential energy (1/2)mv² = mgH-8.4(μmg cos(θ)) All quantities are known except v. Solve for v (9.7 m/s) Check my thinking and calculations.

  
  

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