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

$S_B=5.7m$
The required distance can be determined as follows: We know that $3v_A=-v_B$ $\implies v_B=-3(3)=-9m/s$ According to the conservation of energy $T_{A_{\circ}}+T_{B_{\circ}}+V_{A_{\circ}}+V_{B_{\circ}}=T_A+T_B+V_A+V_B$ $\implies 0+0+0+0=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2+W_A S_A+W_BS_B$ We plug in the known values to obtain: $0=\frac{1}{2}(20)(3)^2+\frac{1}{2}(30)(-9)^2+(20)(9.81)(\frac{S_B}{3})-(30)(9.81)(S_B)$ This simplifies to: $S_B=5.7m$