A2.00-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 ?

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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i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.