Answer
The tension in the rope is 29.7 N.
Work Step by Step
$v(t) = (2.00~m/s^2)~t+(0.600~m/s^3)~t^2$
$a(t) = (2.00~m/s^2)+(1.200~m/s^3)~t$
We can find the time $t$ when $v = 9.00~m/s$
$v = (2.00~m/s^2)~t+(0.600~m/s^3)~t^2 = 9.00~m/s$
$(0.600~m/s^3)~t^2 + (2.00~m/s^2)~t- 9.00~m/s = 0$
We can use the quadratic formula to find $t$.
$t = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{-(2.00)\pm \sqrt{(2.00)^2-(4)(0.600)(-9.00)}}{(2)(0.600)}$
$t = -5.88~s, 2.55~s$
Since the negative value is unphysical, the solution is $t = 2.55~s$.
When t = 2.55 s:
$a = (2.00~m/s^2)+(1.200~m/s^3)(2.55~s)$
$a = 5.06~m/s^2$
We can use a force equation to find the tension $T$ in the rope.
$\sum F = ma$
$T - mg = ma$
$T = ma + mg = (2.00~kg)(5.06~m/s^2)+(2.00~kg)(9.80~m/s^2)$
$T = 29.7~N$
The tension in the rope is 29.7 N.
