Answer
$$m_{O} =30.58 k g $$
$$F_{P}=37.0018 N$$
Work Step by Step
using second Newtons law for the object on earth's surface:
$$
\begin{aligned} F_{g} &=m_{O} \cdot g \\ 300 N &=m_{O} \cdot 9.81 \frac{m}{s^{2}} \\ \Rightarrow m_{O} &=\frac{300 N}{9.81 \frac{m}{s^{2}}} \\ m_{O} &=30.58 k g \end{aligned}
$$
$$F_{P}=m_{O} \cdot a_{P}$$
$$F_{P}=30.58 k g \cdot(4 \cdot 0.3048) \frac{m}{s^{2}}$$
$$F_{P}=30.58 k g \cdot 1.21 \frac{m}{s^{2}}$$
$$F_{P}=37.0018 N$$
