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Posted by on Friday, October 12, 2012 at 12:08am.

A skier starts from rest at the top of a hill that is inclined at 9.5° with respect to the horizontal. The hillside is 225 m long, and the coefficient of friction between snow and skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?
m

  • physics - , Saturday, October 13, 2012 at 5:16pm

    h = 225*sin9.5 = 37.1 m. = Ht. of hill.

    V^2 = Vo^2 + 2g*d.
    V^2 = 0 + 19.6*37.1 = 727.2
    V = 27 m/s. = Skier,s velocity at bottom
    of hill.

    u = -a/g = 0.075.
    -a/9.8 = 0.075
    a = -0.735 m/s^2. = Acceleration of skier.

    d = (V^2-Vo^2)/2a.
    d = (0-(27)^2)/-1.47 = 496 m.

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