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10 December, 15:38

An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.7 s. A passenger in the elevator is holding a 5.1 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates

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  1. 10 December, 18:06
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    Answer: Tension = 53.6N

    Explanation:

    Given that

    Height h = 1 m

    Time t = 1.7 s.

    Mass m = 5.1 kg

    From the equation of the motion we can get the acceleration of the elevator:

    h = X0 + V0t + at2/2;

    Th elevator starts from rest with a constant upward acceleration. Initial velocity Vo = 0, also Xo = 0; thus

    a = 2h/t2 = 2 * 1/1.7^2

    a = 0.69 m/s2.

    Then we can find the tension in the cord by using the formula

    T = mg + ma

    = 5.1 (9.8 + 0.69)

    = 5.1 * 10.5

    = 53.6N
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