Answer
(a) The total mass of the system is 1.25 kg.
(b) $a_{cm} = (1.50~m/s^3)~t$
(c) $F = 5.63~N$
Work Step by Step
(a) $y_{cm} = \frac{m_1y_1+m_2y_2}{m_1+m_2}$
$y_{cm}~(m_1+m_2) = 0+m_2y_2$
$m_1 = \frac{m_2y_2-m_2~y_{cm}}{y_{cm}}$
$m_1 = \frac{(0.50~kg)(6.0~m)-(0.50~kg)(2.4~m)}{2.4~m}$
$m_1 = 0.75~kg$
The total mass of the system is 0.75 kg + 0.50 kg, which is 1.25 kg.
(b) $v_{cm} = (0.75~m/s^3)~t^2$
$a_{cm} = \frac{dv_{cm}}{dt} = (1.50~m/s^3)~t$
(c) At t = 3.0 s:
$a_{cm} = (1.50~m/s^3)~t$
$a_{cm} = (1.50~m/s^3)(3.0~s)$
$a_{cm} = 4.50~m/s^2$
$F = ma$
$F = (1.25~kg)(4.50~m/s^2)$
$F = 5.63~N$
