Chapter 8 - Potential Energy and Conservation of Energy - Problems and Conceptual Exercises - Page 247: 46

$1.8m/s$

We know that $v_f^2=\frac{gd(m_2-\mu_k m_1)+\frac{1}{2}v_i^2(m_1+m_2)}{\frac{1}{2}(m_1+m_2)}$ We plug in the known values to obtain: $v_f^2=\frac{(9.8m/s^2)(0.5m)(1.8Kg-0.45\times 2.4Kg)+\frac{1}{2}(1.3m/s^2)(1.8Kg+2.4Kg)}{\frac{1}{2}(1.8Kg+2.8Kg)}$ $\implies v_f=\sqrt{3.37m^2/s^2}$ $v_f=1.8m/s$