Answer
$v=6.62m/s$
Work Step by Step
We can determine the required speed as follows:
We know that
$\Sigma F_y=0$
$\implies N-W=mg$
$\implies N=W=mg=(2000)(9.81)=1962N$
and $\Sigma F_x=0$
$2T-\mu_sN=0$
$\implies \mu_sN=2T=2(400t^{\frac{1}{2}})$
$\implies 800t^{\frac{1}{2}}=(0.5)(1962)$
$\implies t=1.504s$
Now we apply the principle of impulse and momentum in the $x$-direction
$mv{x_1}+\Sigma \int_{t_1}^{t_2} F_x dt=mv_{x_2}$
$\implies mv_{x_1}+\int_4^{1.504} Tdt+\int_{1.504}^4 \mu_kN dt=mv_{x_2}$
We plug in the known values to obtain:
$0+\int_{1.504}^4 400 t^{\frac{1}{2}}dt-(0.4)(1962)(4-1.504)=200v$
This simplifies to:
$v=6.62m/s$
