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

$3.48ft/s$, $0.376ft$
We can determine the required speed and distance as follows: According to conservation of linear momentum $\Sigma m_1v_1=\Sigma m-2v_2$ $\implies W_bv_{b_1}+W_{w}v_{w_1}=W_bv_{b_2}(\frac{4}{5}+W_wv_{w_2})$ We plug in the known values to obtain: $0.03(1300)(\frac{12}{13})+0=(0.03)(50)(\frac{4}{5})+10v_w$ This simplifies to: $v_w=3.48ft/s$ Now, according to the conservation of energy of the box $T_1+\Sigma U_{1-2}=T_2$ $\implies \frac{1}{2}mv^2-\mu_k Wd=0$ We plug in the known values to obtain: $\frac{1}{2}(\frac{10}{32.2})(3.48)^2-(0.5)(10)d=0$ This simplifies to: $d=0.376ft$