Respuesta :
The given question is incomplete. The complete question is as follows.
A 8.00-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 on the rope when the velocity is 15 m/s.
Explanation:
It is known that relation between acceleration and time is as follows.
      a = [tex]\frac{dV}{dt}[/tex]
      a = [tex]2 + 2 \times 0.6t[/tex] ............ (1)
As it is given that velocity is 15 m/s.
       15 = [tex]2t + 0.6t^{2}[/tex]
        t = 3.604 sec
Substitute the value of "t" in equation (1) as follows.
      a = [tex]2 + 2 \times 0.6t[/tex]
      a = [tex]2 + 2 \times 0.6 \times 3.604 sec[/tex]
       = 6.325 [tex]m/s^{2}[/tex]
Now, we will calculate the tension as follows.
        T = mg + ma
         = 8[9.81 + 6.325]
         = 129.08 N
thus, we can conclude that tension on the rope is 129.08 N when the velocity is 15 m/s.
