I keep getting the sign mixed up. I tried looking online but it is just more confusing.
I know the gravitational potential energy of a body is normally of the form $$V_g = -mgz$$ where $m$ is the mass, $g$ is the gravity and $z$ is the positive (?) height from the zero potential energy.
Now, sometimes it's positive, sometimes it's negative, however it seems very arbitrary. I thought it depended on the choice of the coordinate system, but it doesn't seem so. Consider this 
Where $z$ is the distance of the mass from the floor. Then in my book here we have $mgz$ positive!! Why is so?
Indeed in the case of a projectile thrown in air, the potential energy is negative, but the $z$ axis (or the distance from the floor) is still pointing upwards! How do I, once and for all, decide the sign of it? Having a fixed choice of axis, how do I decide $+$ or $-$?
Best Answer
The value of the potential energy is totally arbitrary until you define a position where the potential energy is zero when in effect you are dealing with differences in potential energy.
Your confusion stems from this statement because it lacks detail.
In the diagram below I have defined a positive $z$ direction.
With the direction of $z$ increasing defined if you move a mass $m$ upwards by a distance $z$ then the change in potential energy is $(-mg)(-z) = mgz$.
This comes from the fact that an upward external force (-mg) has moved the mass upwards a distance (-z).
As expected it is an increase in potential energy but note that going up means that z becomes more negative and that is why it is minus $z$ in the evaluation of the change in potential energy.
If you want an actual value of potential energy then you must define a zero of potential energy.
If you had chosen position $C$ as your zero of potential energy then after lifting the mass up to position $A$ from position $C$ the potential energy of the mass is $(-mg)(-L) = mgL$.
On the other hand if you had chosen the zero of potential energy at position $A$ the potential energy of the mass at position $C$, noting that to go from position $A$ to position $C$ the mass is moving in the positive z-direction, is $(-mg)(+L) = -mgL$.
The change in potential energy in going from position $C$ to position $A$ is the same irrespective of where the zero of potential energy was defined.
final potential energy at $A$ - initial potential energy at $C$ is equal to
$mgL - 0 = 0 -(-mgL) = mgL$
You could have chosen the positive z-direction as upwards and used as the change in potential energy $mgz$ noting that to lift the mass up the external force is in the positive z-direction.
If you had chosen position $C$ as your zero of potential energy then after lifting the mass up to position $A$ from position $C$, the potential energy of the mass is $(+mg)(+L) = mgL$, the same as before.
On the other hand if you had chosen the zero of potential energy at position $A$ the potential energy of the mass at position $C$ is $(+mg)(-L) = -mgL$, again the same as before.
So the advice is, define a positive direction and if you want to find the potential energy define a position where the potential energy is zero.