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    Newton's 2nd law

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    Kathy and her friend go sledding. Out of curiosity, she measures the constant angle, , that the snow-covered slope makes with the horizontal. Next, she uses the following method to determine the coefficient of friction, k, between the snow and the sled. She gives the sled a quick push up so that it slides up the slope away from her. She waits for it to slide back down, timing the motion. The time it takes for the sled to slide back down from its stopping point is two and one-half times as long as it took to get to the stopping point going up the slope. In terms of the known angle , , what is the coefficient of friction?

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    Solution Preview

    When sled goes up, upward acceleration along the plane will be:
    a = -(g*sin(Q) + mu_k*g*cos(Q))
    Here, read Q as theta.
    mu_k = coeff of friction

    Let the initial velocity given to sled is u. Hence, by using Eqn.,
    v = u + a*t, we get,
    0 = u - (g*sin(Q) + mu_k*g*cos(Q))*t1
    where, t1 is the time taken before ...

    Solution Summary

    The solution is given as an equation with an attached diagram.