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4 October, 00:47

A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8m by a cable in which the tension is 11,000 N. The speed vi of the elevator at the beginning of the 8 m descent is most nearly?

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  1. 4 October, 01:10
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    Vi = 4.38 m/s

    Explanation:

    Given

    m = 1000 kg;

    y = 8m;

    T = 11000 N;

    ∑Fy = m*a

    W - T = m * a

    (1000kg * 9.8 m/s²) - 11000N = 1000 kg * a

    a = - 1.2 m/s²

    Using the equation from parabolic motion

    Vf ² = Vi² + 2*a*y

    Solve to Vi

    Vi = √ 2 * 1.2m/s² * 8 m

    Vi = √19.2 m²/s²

    Vi = 4.38 m/s
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