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23 January, 22:05

A 2.00-kg box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to v (t) = (2.00m/s2) t + (0.600m/s3) t2. Part A What is the tension in the rope when the velocity of the box is 15.0 m/s?

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  1. 23 January, 22:35
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    T = 27.92 N

    Explanation:

    For this exercise let's use Newton's second law

    T - W = m a

    The weight

    W = mg

    The acceleration can be found by derivatives

    a = dv / dt

    v = 2 t + 0.6 t²

    a = 2 + 0.6 t

    We replace

    T - mg = m (2 + 0.6t)

    T = m (g + 2 + 0.6 t) (1)

    Let's look for the time for the speed of 15 m / s

    15 = 2 t + 0.6 t²

    0.6 t² + 2 t - 15 = 0

    We solve the second degree equation

    t = [-2 ±√ (4 - 4 0.6 (-15)) ] / 2 0.6

    t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2

    We take the positive time

    t = 3.6 s

    Let's calculate from equation 1

    T = 2.00 (9.8 + 2 + 0. 6 3.6)

    T = 27.92 N
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