A 20 kg girl slides down a playground slide that's 3.2 m high. When she reaches the bottom of the slide her speed is 1.3 m/s. Assume any loss of energy is due to friction. If the slide is inclined at 20 degrees to the horizontal, what is the coefficient of kinetic friction between the fuel and the slide? (65.3N)

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  1. Well how much energy was lost to friction?
    Potential at start = m g h = 20 * 3.2 * g = 64 g = 64*9.81 = 628 Joules
    How much Kinetic energy was left at the bottom?
    (1/2) m v^2 = .5 * 20 * 1.3^2 = 16.9 Joules
    628 - 17 = 611 Joules went into heating up the slide.
    That is the friction force * the slide distance
    611 = Ff * 3.2/sin 20 = Ff * 3.2 /.342
    so Ff = 65.3 Newtons (but that is not the answer. It asked for the coef.)
    Ff = mu m g cos 20
    65.3 = mu * 20 * 9.81 * .94
    mu = 0.354

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