Physics

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A 59-kg person on skis starts from rest down a hill sloped at 36° from the horizontal. The coefficient of friction between the skis and the snow is 0.16. After the skier had been moving for 5.0 s, the friction of the snow suddenly increased and made the net force on the skier zero.

What is the new coefficient of friction?

How fast would the skier now be going after skiing for another 5.0 s?

  • Physics -

    ma=mgsinα –F(fr) = mgsinα – μmgcosα
    a= g(sinα – μcosα ) = 9.8• (0.59-0.16•0.81)=4.5 m/s²
    The speed after the first 5 sec is
    v=at= 4.5•5=22.5 m/s
    If the net force=0, a=0 =>
    0 = mgsinα – μ₁•mgcosα ,
    μ₁ =sinα/cosα=tanα=0.73.
    If a=0 => v=const=22.5 m/s

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