Chapter 5 - Newton's Laws of Motion - Problems and Conceptual Exercises - Page 146: 84

Please see the work below.

(a) We know that the mass of the second body is given as $m_2=\frac{F_c}{a}$ $\implies m_2=\frac{4.45N}{1.20m/s^2}$ $m_2=3.71kg$ Now $m_1=M-m_2$ We plug in the known values to obtain: $m_1=16.66Kg-3.71Kg=12.95Kg$ (b) We know that $m_2=\frac{F_c}{a}$ We plug in the known values to obtain: $m_2=\frac{4.45N}{1.20m/s^2}$ $m_2=3.71Kg$