Why is potential energy negative?

AI Thread Summary
The discussion centers on the definition of total energy for a classical non-rotating object, specifically questioning why potential energy is often represented as negative in the equation. It highlights that total energy is typically expressed as \frac{1}{2}mv^{2}+mgh, with potential energy being positive when measured from a reference point like the Earth's surface. The confusion arises from differing interpretations of potential energy based on the chosen reference height, h. The conversation also touches on the Lagrangian formulation, which can represent potential energy as negative. Ultimately, the sign of potential energy depends on the reference point and the direction of the conservative force field.
Hertz
Messages
180
Reaction score
8
Total energy is \frac{1}{2}mv^{2}-mgh=C for a classical non-rotating object right? C can be determined based on an arbitrary initial condition that suits the problem, right? How come it's not \frac{1}{2}mv^{2}+mgh=C? This seems like a more suitable definition of "total" energy. I'm sure there's a reason it's minus and not plus, but what is the reason? O.o

[edit]
Does it have something to do with the direction of the conservative force field that the object is in?
 
Physics news on Phys.org
It just depends on the definition of h. If h is upwards, it is +mgh, otherwise it is -mgh.
 
Hertz said:
Total energy is \frac{1}{2}mv^{2}-mgh=C for a classical non-rotating object right?

According to who? That's not what I see in the textbooks that I've used, which all have a + sign, and define h as increasing in the upwards direction.
 
Hertz said:
Total energy is \frac{1}{2}mv^{2}-mgh=C for a classical non-rotating object right?

If you have taken the surface of the Earth as the reference point for your potential energy, then the potential energy of a body of mass m at a height h (small compared to Earth's radius) is mgh. It is a positive quantity with respect to the reference point. The total energy in such a case is \frac{1}{2}mv^{2}+mgh. The Lagrangian can be \frac{1}{2}mv^{2}-mgh=L.
 
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top