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7 May, 00:53

The tibia bone in the lower leg of an adult human will break if the compressive force on it exceeds about 4*105N (we assume that the ankle is pushing up). Suppose that someone step off a chair that is 0.40 m above the floor. If landing stiff-legged on the surface below, what minimum stopping distance does he need to avoid breaking his tibias? Assume that the mass of the person is 64 kg

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  1. 7 May, 04:16
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    Distance = 6.27 x 10⁻⁴m

    Explanation:

    The speed just before hitting the ground may be expressed as

    v = √2gh, where g is the acceleration due to gravity and h is the height

    v = √ (2 x 9.81 x 0.4) = 2.8 m/s

    Using the work - energy theorem, we know that any work done on the object produces a change of kinetic energy in the object. Hence

    Net Work = Final Kinetic Energy - Initial Kinetic Energy

    The initial condition is stationary hence the initial Kinetic Energy = 0

    Net Work may be written as

    Net Work = Force x Distance = Final Kinetic Energy

    4 x 10⁵ x d = (1/2) x mass x velocity²

    d = { (1/2) x 64 x 2.8²} / (4 x 10⁵) = 6.27 x 10⁻⁴m

    6.27 x 10⁻⁴m is the minimum stopping distance to avoid breaking his tibias.
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