A 5.00-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. What is the tension in the rope when the velocity of the box is 9.00 m/s?

T = 74.3N

Explanation:

We are given;

v(t) = (2.00m/s²)t+(0.600m/s³)t²

So, when v = 9 m/s;

9 = 2t + 0.6t²

0.6t² + 2t - 9 = 0

Solving this quadratic equation,

t = -5.88 or 2.55

We'll pick t = 2.55 s

Now, kinematic acceleration will be the derivative of the acceleration.

Thus, a = dv/dt = 2 + 1.2t

So, acceleration at that time t = 2.55s is; a = 2 + 1.2(2.55) = 5.06 m/s²

Since the rope is subject to both acceleration and gravity, Tension is;

T = mg + ma

T = m(g + a)

T = 5(9.8 + 5.06)

T = 74.3N

answer; the correct answer is technician a; ///100%good luck