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Potential has a negative sign as work has to be done against the external force to bring the mass from infinity to the point (definition of potential )


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Because of a convention wherein zero gravitational potential is said to be at infinity. See Wikipedia: $V(x) = \frac{W}{m} = \frac{1}{m} \int\limits_{\infty}^{x} F \ dx = \frac{1}{m} \int\limits_{\infty}^{x} \frac{G m M}{x^2} dx = -\frac{G M}{x}$ "By convention, it is always negative where it is defined, and as x tends to infinity, it approaches zero." ...


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Hmm. I think there should be a minus sign and it doesn't matter where you set the zero of potential. Gravitational field strength is the negative gradient of the potential. For a spherically symmetric field $$ g(r) = -\frac{dV}{dr}.$$ If $$V = -\frac{GM}{r} + V_0,$$ where $V_0$ is an arbitrary constant, then $$g = -\frac{GM}{r^2}.$$ i.e. your book is ...


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The difference probably is that the graph for the gravitational potential is the one for a spherical mass distribution (or a sphere with a certain mass if you wish) and the electric one is given for a point charge. You could also draw the gravitational potential for a point mass, then it would look equivalent to your electrical potential, or the other way ...


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So the net work on the object is zero and it doesn't gain any energy. The object obviously did gain energy. The object's potential energy increased by $mgh$, and its kinetic energy didn't change. So what's going on? Is the work-energy principle wrong? The answer is no, the work-energy principle is not wrong. The work energy principle merely says that ...


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Gravitational weight is a conservative force so work done against it is entirely equal to the change in potential between the start and end points (in the absence of dissipative forces like friction). The internal energy of the mass does not change as it is moved upwards and your assessment of the energy considerations is correct. Consider alternatively the ...


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Gravity doesn't do -mgh work on the object while it is lifted, gravity converts the potential energy gained while lifting the object (mgh) into kinetic energy (1/2 mv^2) after it's dropped.


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The total work done on the object is the change in kinetic energy: $W_{total} = \Delta E_K$* While the gravitational potential energy of the object is: $U_G = mgh$ So, although it costs energy to lift the object up, the total work done on it is $0$ because both at the beginning and at the end it has no kinetic energy ($v_{i,f}=0 \rightarrow E_{K_{i,f}}=0 ...


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After the elongation of x units in the spring, the spring develops energy in the form of spring compression or elongation energy(the work done in elongation is converted into spring energy), spring tries to get into its mean position. Now, if the system is isolated the spring will perform SHM otherwise friction and other forces will make it stop its motion. ...


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as we know the energy of an elctron increases as it moves away from the nucleus..that means potential energy is directly depending on distance between electron and nucleus..but when we derive an mathematical expression we get energy inversly proportrional to the radius...to compensate diz we add MINUS sign to show that lesser negative enrgy means more energy ...


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I don't think you need quantum mechanics to understand what's going on in dipole-induced dipole interaction. The basic mechanism is quite simple and just the details of the calculations change by switching to a quantum description. Polarizable molecule in an external field So first things first. Let us consider a simple model of polarizable molecule as ...


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response function = susceptibility = (pure or mixed) second derivative of a (Helmholtz, Gibbs, etc.) free energy. Magnetization is not a response function as the free energy is not observable, so one cannot observe the response to a change of some variable.


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The effective potential is the potential of interaction you measure between two (or more) emergent physical objects when you forget (or "trace over" in the jargon) certain degrees of freedom of a more detailed model. If you take two pinned charges in vacuum for instance, they will interact with a "bare" Coulomb interaction in $\sim 1/r$. If you put these ...


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As wiki says "The effective potential (also known as effective potential energy) is a mathematical expression combining multiple (perhaps opposing) effects into a single potential." Basically the concept of the effective potential simplifies the equations of motion and simplifies their analysis.


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I might be able to answer your question in the context of linear response theory: Response function: the power series expansion of the applied field generated by a weak external perturbation. Mathematically speaking, we can relate the average value of an observable $X$_i to the response function $\chi$ via \begin{align} \langle X_i(t)\rangle=\int_0^t dt'' ...


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I suppose you read this passage in the famous Feynman Lectures. I am fairly certain that what Feynman is referring to (and what you are looking for) is a proof that an electrostatic field is conservative. There are a number of equivalent ways of stating that a vector field is conservative, each of which can be taken as a definition. Let $\vec{F}(x)$ be a ...


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I see the viewpoint you're coming from. From the wording of "potential energy", it sounds like the energy might not exist until it becomes kinetic energy. Potential energy is as real as kinetic energy though. The idea is that the energy gets stored in the fields that produce the force, and that the energy is transferred between kinetic energy and field ...


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Well your question was not perfect, but acceptable. the idea of energy may sound easy, but deeply it is a very strange Idea. but the answer to your question: no it doesn't mean it "potentially" gonna gain. I see that you read Feynman's lectures, that's very good, but these speeches actually are for people who accepted the idea of energy blindly, and didn't ...



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