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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 -

    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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