An extreme skier, starting from rest, coasts down a mountainslpe that makes an a

Question

An extreme skier, starting from rest, coasts down a mountainslpe that makes an angle of 25 degrees with the horizontal. Thecoefficient of kinetic friction between her skis and the snow is0.200. She coasts down a distance of 10.4m before coming to theedge of a cliff. Without slowing down, she skis off the cliff andlands downhill at a point whose vertical distance is 3.50m belowthe edge. How fast is she going just before she lands? I don't know where to start this problem. An extreme skier, starting from rest, coasts down a mountainslpe that makes an angle of 25 degrees with the horizontal. Thecoefficient of kinetic friction between her skis and the snow is0.200. She coasts down a distance of 10.4m before coming to theedge of a cliff. Without slowing down, she skis off the cliff andlands downhill at a point whose vertical distance is 3.50m belowthe edge. How fast is she going just before she lands? I don't know where to start this problem.

Explanation / Answer

   for this initially we use the work energytheorem if W is the work doine by the net external force    that acts on the skier then    W = (1 / 2) m vf2 - (1/ 2) m vo2    the work is also given by    W = F cos    so we get    (m g cos 65o) s + (fkcos180o) s + (FN cos90o) s = (1 /2) m vf2 - (1 / 2) mvo2    the kinetic frictional force is given by    fk = kFN    the normal force will be    FN - m g cos25o =0    FN = m g cos25o    substitute in the work energy equation and solvefor vf    vf = ......... m / s    in the net part that is in the free fall theonly force that is acting on her is her weight mg    again the work energy theorem gives    W = (1 / 2) m vf2 - (1/ 2) m vo2    (m g cos) s = (1 / 2) mvf2 - (1 / 2) m vo2    from the drawing s cos = 3.00m    solve for vf    vf =.........m/s       from the drawing s cos = 3.00m    solve for vf    vf =.........m/s   

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