## Engineering Mechanics: Statics & Dynamics (14th Edition)

$v=29.4ft/s$
We can determine the required velocity as follows: According to the impulse momentum principle in the $y$ direction $m(v_y)1+\Sigma \int_1^2 F_ydt=m(v_y)2$ We plug in the known values to obtain: $\frac{20}{32.2}(0)+N\Delta t-Wcos30\Delta t=\frac{20}{32.2}(0)$ This simplifies to: $N=17.32lb$ Now we apply the impulse momentum principle in the $x$ direction $m(v_x)1+\Sigma \int_{t_1}^{t_2} F_xdt=m(v_x)2$ $\frac{20}{32.2}(2)+Wsin30\Delta t-\mu_k N\Delta t=\frac{20}{32.2}v$ We plug in the known values to obtain: $\frac{20}{32.2}(2)+20sin30(3-0)-(0.25\times 17.32(3-0))=\frac{20}{32.2}v$ This simplifies to: $v=29.4ft/s$