Answer:
The object-Earth system is open
[tex]F_n=ma=5\ (7.2)=36\ N[/tex]
Explanation:
Accelerated Motion
When an object is released in free air (with no other forces than the gravity), it describes a free-fall motion and the formulas include the acceleration of gravity as part of the calculations. But when there is another external force, then the acceleration is not the gravity, but the result of the net force exerted on the mass of the object.
By definition, an open system includes the exchange of energy from and to the surroundings, that is why all systems surrounding our planet are considered as open systems. In our case, the object is interacting with the planet's gravity and there is some other external force, which will be computed later. The object-Earth system is open.
If the object starts from rest, its initial speed is zero, and
[tex]v_f=a\ t[/tex]
where a is the acceleration and t is the time. The distance traveled is given by :
[tex]\displaystyle y=\frac{a\ t^2}{2}[/tex]
From the two above equations, we find that:
[tex]v_f^2=2ay[/tex]
Solving for a
[tex]\displaystyle a=\frac{v_f^2}{2y}[/tex]
[tex]\displaystyle a=\frac{12^2}{2\ (10)}[/tex]
[tex]a=7.2\ m/s^2[/tex]
It means the net force is
[tex]F_n=ma=5\ (7.2)=36\ N[/tex]
The object's weight is
[tex]W=5\ (9.8)=49 N[/tex]
This means there is some external force acting upwards delaying the object's fall of a magnitude of
[tex]F_e=49-36=13\ N[/tex]
