Answer:
Assume that [tex]g = 10\; \rm N \cdot kg^{-1}[/tex]. The net force on this object will be [tex]250\; \rm N[/tex] (downwards.) The acceleration of this object will be approximately [tex]8.3\; \rm m \cdot s^{-2}[/tex] (also downwards.)
Explanation:
Net force
The object is falling towards the ground because of gravity. The size of the gravitational force on this object depends on its mass and the strength of the gravitational field at its location.
Near the surface of the earth, the gravitational field strength is approximately [tex]10\; \rm N \cdot kg^{-1}[/tex]. In other words, approximately [tex]10\; \rm N[/tex] of gravitational force acts on each kilogram of mass near the surface of the earth.
The mass of this object is given as [tex]m = 30\; \rm kg[/tex]. Therefore, the size of the gravitational force on it will be:
[tex]W = m \cdot g \approx 30 \; \rm kg \cdot 10\; N \cdot kg^{-1} = 300\; \rm N[/tex].
Near the surface of the earth, gravitational forces point towards the ground. On the other hand, the direction of air resistance on this object will be opposite to its direction of motion. Since this objects is moving towards the ground, the air resistance on it will be directed in the opposite direction. That's exactly the opposite of the direction of the gravitational force on this object. The net force on this object will be:
[tex]300\; \rm N - 50\; \rm N =250\; \rm N[/tex].
Acceleration
Let [tex]a[/tex] denote the acceleration on this object. Apply Newton's Second Law of motion:
[tex]\begin{aligned} a &= \frac{F(\text{net force})}{m} \approx \frac{250\; \rm N}{30\; \rm kg} \approx 8.3\; \rm m \cdot s^{-2}\end{aligned}[/tex].
Note that the acceleration of this object and the net force on it should be in the same direction.
