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

$d=36.2ft$
According to the principle of work and energy $\frac{1}{2}mv_A^2-WS+\frac{1}{2}kS^2=\frac{1}{2}mv_B^2$ We plug in the known values to obtain: $0-10(3)+\frac{1}{2}(100)(2)^2=\frac{1}{2}(\frac{10}{32.2})v_B^2$ This simplifies to: $v_B=33.09ft/s$ From the equations of kinematics $S_y=S_{y_{\circ}}+v_{y_{\circ}}t+\frac{1}{2}a_ct^2$ We plug in the known values to obtain: $-3=0+(33.09)(\frac{3}{5})t-\frac{1}{2}(32.2)t^2$ $\implies t=1.369s$ Now $d=S_x=S_{x_{\circ}}+v_{x_{\circ}}t$ We plug in the known values to obtain: $d=0+33.09(\frac{4}{5})(1.369)$ This simplifies to: $d=36.2ft$