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March 28, 2017

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A skier slides horizontally along the snow for a distance of 11.9 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is 0.0458. Initially, how fast was the skier going?

I do not know how to start this. Thank you.

  • Physics(Please help) - ,

    ΔKE =KE2 –KE1 = 0 -m•v²/2.
    W(fr) =μ•m•g•s•cosα,
    where α is the angle between the friction force and displacement.
    α =180º, cos α = -1.
    -m•v²/2 = - μ•m•g•s,
    v = sqrt(2•μ• g•s).

  • Physics(Please help) - ,

    So for v would I do sqrt(2*11.9*9.8*0.0458) ?

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