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?

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