Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.2 - Principle of Linear Impulse and Momentum for a System of Particles - Problems - Page 251: 24

$2.128ft/s$

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 $mv_1+\Sigma \int ^{t_2}_{t_1} F_y dt=mv_2$ $\implies \frac{34}{32.2}(0)+\int ^{2}_{1.763} e^{2t} dt-W\Delta t=(\frac{34}{32.2})v$ $\implies 0+(0.5 e^{2t})|^2_{1.763}-34(2-1.763)=(\frac{34}{32.2})v$ $\implies v=2.128ft/s$