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A movie stunt man (mass 82.3 kg) stands on a window ledge 5.69 m above the floor. Grabbing a rope attached to a chandelier, he swings down to grapple with the movie’s villain (mass 72.3 kg), who is standing directly under the chandelier. (Assume that the stunt man’s center of mass moves downward 5.69 m. He releases the rope just as he reaches the villain) If the coefficient of kinetic friction of their bodies with the floor is uk = 0.66, how far do they slide?

Give your answer in meters to the second decimal place.

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Charlie

  1. First get the kinetic energy of the stunt man at the crash.
    KE = decrease in potential energy = m g h = 82.3*9.81*5.69 (you do the numbers)use that to get the speed of the stunt man
    (1/2) m Vs^2 = KE where I am calling Vs speed of stunt man
    Then CRASH
    and the two stick together (vital fact for conservation of momentum
    so
    momentum stunt man = 82.3*Vs
    momentum of two together = (82.3+72.3)Vt
    where Vt is speed together after collision
    the momentum before = momentum after, so solve for Vt
    Now get the KEnew of both together
    (1/2) * total mass * Vt^2
    NOW we slide to a stop
    friction force FF = (82.3+72.3)(9.8)(.66)
    work done by friction = FF * distance slid = loss of KE = KEnew

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    Damon

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