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8 September, 13:04

Suppose a 78 kg person is bouncing on a bathroom scale which has a force constant of 1.5 * 106 n/m. 50% part (a) what is the maximum velocity of the person, in meters per second, if the amplitude of the bounce is 0.115 cm?

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  1. 8 September, 14:22
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    m = mass of the person bouncing = 78 kg

    k = spring constant of the bathroom scale = 1.5 x 10⁶ N/m

    v = maximum velocity of the person

    A = maximum compression of the spring of scale = 0.115 cm = 0.00115 m

    h = height gained at maximum speed = A = 0.00115 m

    Using conservation of energy

    spring potential energy = kinetic energy + gravitational potential energy

    (0.5) k A² = (0.5) m v² + m g h

    k A² = m v² + 2 m g h

    (1.5 x 10⁶) (0.00115) ² = (78) v² + 2 (78) (9.8) (0.00115)

    v = 0.0538 m/s
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