Answer
$v=7.65m/s$
Work Step by Step
We can determine the required speed as follows:
$\Sigma F_y=0$
$\implies mg-N=0$
$\implies N=500\times 9.81=4905N$
and $\Sigma F_x=0$
$\implies F-\mu_s N=0$
$\implies F=\mu_sN=0.5\times 4905=2452.5N$
We apply the principle of impulse and momentum
$mv_1+\Sigma \int_{t_1}^{t_2}F_x dt=mv_2$
$500(0)+2\int_{2.476}^3 200t^2dt+2\int_3^5 1800dt-\int_{2.476}^5 \mu_k Ndt=500v$
We plug in the known values to obtain:
$(\frac{400}{3}t^3)|^3_{2.476}+(2\times 1800\times (5-3))-[0.4\times 4905\times (5-2.476)]=500v$
This simplifies to:
$v=7.65m/s$