Answer
$-5.68m/s$ and $21.14m/s$
Work Step by Step
The required speed of the crate can be determined as
$F=150+(\frac{450-150}{6-0})t$
$\implies F=150+50t$
According to impulse momentum principle
$mv_1+\Sigma \int _{t_1}^{t_2} F_y dt=mv_2$
$\implies 40(-10)+2\int _{t_1}^{t_2}(150+50t)dt-\int_{t_1}^{t_2}W dt=40v$
$\implies -400+2(150t+25t^2)|^t_0-(40\times 9.81)\Delta t=40v$
$\implies v=-10-2.31t+1.25t^2$
Now when $t=3 s$, then $v=-10-2.31(3)+1.25(3)^2=-5.68m/s$
and when $t=6s$, then $v=-10-2.31(6)+1.25(6)^2=21.14m/s$
