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March 27, 2017

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A skier traveling 12.0 m/s reaches the foot of a steady upward 18° incline and glides 18.2 m up along this slope before coming to rest. What was the average coefficient of friction?

  • Physics - ,

    initial KE= work done on gravity+workdone on friction.

    1/2 m 12^2=mgh/sin18 + mu*mg*sin18*h/sin18

    72=gh/sin18 + mu*g*h

    we know h=18.2/sin18
    so, solve for mu

  • Physics - ,

    3.108

  • Physics - ,

    a skier traveling 12.0 m/s reaches the foot of a steady up ward 18.0 incline and glides 12.2m up along this slop before coming to rest. what the average cofficient of friction?

  • Physics - ,

    the answer is shake your money maker

  • Physics - ,

    KE = PE + W_friction
    1/2 * mv^2 = mgh + mu*mgcos(theta)d
    ----- m is cancel out
    --- sin(theta) = heigh/ distance( inclide) ===> h = d*sin(theta)

    1/2 v^2 = g (d*sin(theta)) + mu* (g*cos(theta))*d

    now you can solve for mu ^^ Good luck.

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