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12 November, 22:15

A person weighing 0.6 kN rides in an elevator that has a downward acceleration of 1.3 m/s 2. The acceleration of gravity is 9.8 m/s 2. What is the magnitude of the force of the elevator floor on the person? Answer in units of kN.

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  1. 12 November, 23:37
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    520.41 N.

    Explanation:

    From newton's second law of motion,

    The force of the elevator floor on a person is given as,

    R = m (g-a)

    R = W-ma ... Equation 1

    Where R = Force of the elevator floor, m = mass of the person, a = acceleration of the elevator, W = weight of the person.

    But,

    W = mg,

    where g = acceleration due to gravity.

    m = W/g ... Equation 2

    Given: W = 0.6 kN, = 600 N, g = 9.8 m/s²

    m = 600/9.8

    m = 61.22 kg.

    Given: W = 0.6 kN = 600 N, m = 61.22 kg, a = 1.3 m/s²

    Substitute into equation 1

    R = 600-61.22 (1.3)

    R = 600-79.586

    R = 520.41 N.
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