Engineering Mechanics: Statics & Dynamics (14th Edition)

$v=6.62m/s$
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: $v=6.62m/s$