#### Answer

$2.128ft/s$

#### Work Step by Step

It is given that $F=e^{2t}$, but $F$ is equal to the weight, so that the crate can start motion $\implies F=e^{2t}=34$
$\implies t=1.763s$
Now, applying the impulse momentum principle, we have
$mv_1+\Sigma \int ^{t_2}_{t_1} F_y dt=mv_2$
$\implies \frac{34}{32.2}(0)+\int ^{2}_{1.763} e^{2t} dt-W\Delta t=(\frac{34}{32.2})v$
$\implies 0+(0.5 e^{2t})|^2_{1.763}-34(2-1.763)=(\frac{34}{32.2})v$
$\implies v=2.128ft/s$