# Tag Info

## New answers tagged potential-energy

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Yep, borrowed from thermodynamics. For a system that is constrained under fixed pressure, fixed composition, and fixed temperature, the Gibbs free energy is minimized. You're right to question the link between Gibbs free energy and potential energy (or kinetic energy or total energy ...). In fact we know that Gibbs free energy is defined by $$G = E - ST + ... 1 but when they bounce off each other again, the energy does not return to being kinetic The potential energy actually does go back to kinetic energy. But often it only partially goes to the translational motion of whole body, while other part of the energy comes into internal motion of atoms. People normally call the latter thermal energy. 2 Sounds like you're getting at the "coefficient of elasticity," which is a value in [0,1] which represents what percent of the pre-collision kinetic energy is found after the collision. In homogeneous materials, the remainder of the energy is typically lost to deformation or heat (phonons) as you suggest. you could imagine, for the sake of argument, a steel ... 0 Now I am not getting it physically why should there be any Coulomb energy in this case as there is no another charge to provide the electrostatic coulomb potential energy? You're right. For point charge the formula$$ \int \frac{1}{2}\epsilon_0 E^2 d^3\mathbf x $$gives infinite value and is thus unusable - infinite energy would mean one cannot do ... 1 The field itself carries energy. This is, in fact, a vital point because it can be shown that, if the momentum and energy carried by the fields isn't accounted for, electromagnetism would blatantly violate Newton's third law (and this doesn't have anything to do with special relativity per se). 1 The potential energy of a particle with charge q in a conservative electric field \vec E is$$U = q\phi$$where the electric potential \phi is related the electric field by$$\vec E = -\nabla \phi$$Thus, the electric potential is defined, up to a non-physical constant, by the associated electric field - no test charge enters the picture. The ... 0 The value of electric potential difference (aka voltage) is independent of the choice of test charge, including its sign. When determining the electric potential difference between two points, you can imagine either a positive or negative test charge, which ever one tickles your fancy. Note that the same cannot be said about electric potential energy. As a ... 3 I think you're confusing "electric potential" and "electric potential energy". I don't blame you, it's an unfortunate bit of terminology. The electric potential energy is defined for a particular charge (or distribution of charges), and so you need to explicitly put in the charge of the point or distribution of interest. That is, we don't talk about the ... 0 From a kinetic theory of gases perspective, or equilibrium statistical mechanics perspective, potential of ideal gases will strictly be a function of the position of the particles (in fact the relative position of the particles). temperature is just one third of the mean square velocity. For ideal gases, in general U = f(\bar{r}_1,\bar{r}_2 ... \bar{r}_n) ... 1 For an ideal gas there is no potential energy, by definition. For a real gas there is a potential between the molecules, and the average value of the potential energy will have a (very complicated) dependency on temperature. 1 Your second therm is still referring to the conservative force acting on the object. For an ideal spring this would mean: \omega_0^2=\frac{k}{m}. However in the case of a damped oscillator there will be both conservative and not conservative forces acting on the object with mass m. But this does not mean than the potential energy of the conservative ... 1 First of all don't insert actual numbers until the end. It makes it much easier to keep track and check whether your units check out. This problem is easier if you invoke conservation of energy. Simply equate: At t = 0. -Potential gravitational energy. At the end. -Kinetic energy of the toolbox. -Dissipated energy due to friction. You will find that ... 2 Yes, that equation will still give the correct value for the energy of the oscillator system at any point in time, assuming of course that you know dx/dt and x at that time. If there is an external dissipative force on the system (damping) you will find that the value of E decreases with time. But the energy of the oscillator itself is still the sum of ... 2 As far as I understand this problem, "GPE seems to exist only when I introduce the ball into the gravitational field" is a correct statement. As You said "You introduce the ball", so first You do the work that is converted to GPE and then GPE does work in accelerating the ball. I don't think there is a way in which a ball can appear somewhere within the ... 0 It depends where your energy starts. Isolating two cases should give you the idea. The first is if the energy is already at height h, in the other we'll assume it starts ground level. Case 1: h_0 = h [already at height h] E = mc^2. The gravitational potential energy was stored previously in energy. Energy is not immune to gravity. To create the ... 1 Think logically. Assume that you want to create a mass on the earth, where h=0 (assumption). Therefore:$$E=mc^2$$You as well need to consume some work to take the mass from 0 \to h. So the energy needed is the energy you need to create it plus the one you need to "lift" it. So:$$\sum E = E - W_{W(spent)} = E - (-mgh) = mc^2 + mgh = m(c^2 + gh)$$... 5 Your teacher's explanation is incorrect. A simple counterexample can be constructed to illustrate this by considering what happens when the role of your arm is replaced by that of a rubber band. When a weight is suspended from the ceiling by a rubber band, the band stretches and its polymer chains become more ordered, in exact analogy to your teachers ... 5 Think about the work-kinetic energy theorem, which states that the net work done on an object is equal to its change in kinetic energy:$$W_{net}=\Delta\mathrm{KE}. You are right that when lifting an object of mass $m$ by a height $h,$ in a uniform gravitational field, the work you do is $W_{you}=mgh$ (assuming, as you said, that you're applying a force ...

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The electrostatic potential energy of a system of point charges is defined as the work required to be done to bring the charges constituting the system to their respective locations from infinity. Suppose we have a configuration of point charges. If the potential of the energy of the system is negative, this means work is positive. Consider two point ...

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With work you also have to take care in specifying what is doing the work. If it is the work required to move a particle or the work done by the electric field, these have different signs. If the work done by the electric field on a point charge is positive it means it is moving in the direction of the electric force acting on the point charge, therefore, $W ... 2 Rule of thumb for working it out: If you imagine letting a charge go, the direction it tends to move is toward lower potential energy. The opposite direction is toward higher potential energy. This is independent of the choice of where the zero of energy is. 1 Have to be a bit careful with potential energies, as the 0 point of potential can be arbitrarily chosen. Only changes in potential are well defined without a choice of 0 point. That said, it is often convenient to choose the 0 point at$\infty$, and this is the typical choice when talking about assembling point charges. With this choice, the potential ... 4 Good question, and the answer is that$W' > W$if$F > mg$. The reason for this is that if$F = mg$then the net force is zero so the particle travels at a constant velocity. That means it's kinetic energy hasn't changed so the only change is the potential energy. However if$F > mg\$ then the net force is positive and the particle is accelerating ...

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It's valid in the sense that it does tell you the rest energy of a 200-pound person, but it does not tell you how much energy you could get by splitting all those atoms. As a matter of fact, most of the atoms in a human body are carbon, nitrogen, and oxygen; splitting these atoms takes energy, it doesn't produce it. Your character would need to tap into a ...

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