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It is wrong to think potential energy is stored in the object. The earth pulls the object down, but the object pulls the earth up. They share the potential energy. The object fails to fall down because the tabletop pushes it up. The earth fails to fall up because the bottom of the table legs push the earth down. The table pushes up and down because it is ...


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In relativity we only use the rest mass, also known as the invariant mass, of an object. In days past the concept of a relativistic mass was used, but this is now strongly deprecated as it has caused endless confusion. For example an obvious question is whether the increase of relativistic mass with speed can cause an object to become a black hole (tl;dr it ...


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For your second case, you can change the angular momentum, but remember that you have fixed your total energy. You can't make the planet revolve arbitrarily fast or it will have more energy than allowed. By increasing the angular momentum without adding energy, you are circularizing the orbit. To add, you might take a look at the Specific Orbital Energy ...


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[...] when it already has energy, then why doesn't it fall off from the table top onto the ground by itself? Because it is being held back. It wants to fall straight downwards, but the bookshelf applies a normal force to hold up the book, which is stronger than the downwards force (gravity). Just as the rubber band holds back the spring from elongating, ...


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I think the question is posed from the wrong side... Since you ask for an intuitive reason, it's probably rather the force that is the basic concept, and the potential is derived (hehe, derived by an integral, this sounds contradictory :)) from it (if the force is conservative, i.e. there is a unique value of the integral between two points, independent of ...


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The velocity does not double if the acceleration is doubled. The relevant SUVAT equation is: $$ v^2 = u^2 + 2as $$ where in this case $u=0$ so we get: $$ v = \sqrt{2as} $$ A doubling of acceleration means that the velocity would double if the travel time was kept constant. However in this case it's the travel distance that is held constant. The greater ...


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This answer is only about Where does the stored energy stay in the object, and why does it only convert into vertical motion and not horizontal motion? because I think your other questions have been well-addressed, but this one has only been answered in highly technical terms that may not have clarified anything for you. Think about what happens if ...


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Your first line is fine, everything else is wrong (except the one time you repeated something from the first line). How wrong? You don't even have the right units, so pretty much as wrong as you can be. It's like if I asked for the surface area of a house and you said 5m or 8s or 20N or 80K. For a differentiable function $f$ the directional derivative in ...


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Your understanding is actually perfectly correct. In the general case we have indeed something like $E_m = K(r) + V(r)$ for a central force say. Now, as you rightly point out, the valid positions are those that satisfy $K(r) \geq 0$. This implies that $E_m - V(r) \geq 0$ If $V(r)$ is an ever increasing function of $r$ that has an asymptote at $V_0$, then ...


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The difference is only how you define $\theta$ and the zero of potential energy. The $\cos \theta$ expression takes the zero of potential energy to be when $\theta = \frac \pi 2$ whereas you derivation with $\sin \theta$ in it takes the zero of potential to be when $\theta = 0$.


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The first thing is to note that the gravitational potential energy is associated with both the object and the Earth. You may think that only the object has the potential energy because when you drop the object you see it accelerate downwards and gain kinetic energy. At the same time the Earth is accelerating upwards at a rate of $\frac {\text {mass of ...


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A force field is called conservative if its work between any points $A$ and $B$ does not depend on the path. This implies that the work over any closed path (circulation) is zero. This also implies that the force cannot depend explicitly on time. Consider for instance a time decaying force on a straight line. Choose a long closed path. The magnitude of the ...


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Potential energy depends upon the presence of a force, and the actual configuration of objects. So a rock at the top of a cliff has potential energy $U=mgh$, due to $m$, the mass of the rock, $g$, the acceleration due to gravity, and $h$, the height of the cliff. Inertia, as described by Newton's First Law of Motion, is that property of matter which, in ...


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Potential energy and inertia are related via Einstein's famous equation $E = mc^2$. A compressed spring has more potential energy and therefore more inertial mass compared to an uncompressed one, making it more difficult to accelerate because of its increased inertia.


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This time you're wrong. You are overcounting the terms when multiplying by 4. Potential energy is given for two particles at a time, so you should be multiplying by 2, which is the same as multiplying by 4 and then dividing by 2.


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As Giorgio says, you are overcounting the terms when multiplying by 4. I would like to elaborate a little bit, just to show how this is more than just a math error. It is also a conceptual error: You mention in your reasoning "The total potential energy = the sum of the PE of each of the 4 charges." But this is false. Not only that, but there is no such ...


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Let $X$ be a sphere of radius $R$ with charge $+Q$ on it. The potential of sphere $X$ is $+\frac{kQ}{R}$. Let $Y$ be another sphere of radius $6R$ with charge $+5Q$ on it. The potential of sphere $Y$ is $+\frac{k5Q}{6R}$. The potential of sphere $X$ is larger than that of sphere $Y$ so if the spheres are connected with a wire electrons will flow from ...


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You will get infinity because in addition to $kQ_1Q_2/d$, it also includes the self-energy of the two point charges, which is infinity.


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Relation between Electric field and potential The relationship between electric field $\bf E$ and scalar potential $\varphi$ is given as $$\mathbf E= -\mathbf \nabla\,\varphi$$ where $\mathbf \nabla \equiv \textrm{gradient operator}\;.$ I am unable to understand from this - sign comes. It is worthy to quote from Purcell: The minus sign came in ...


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I would say the system gains the potential energy in that example. The ball does work on the spring equal to the ball's initial kinetic energy. Consider the direction of the force and displacement of each. The force the ball applies on the spring and the balls displacement as it applies it are in the the same direction so the work done is positive - it ...


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The potential energy possessed by the system of two charges is $$U=\dfrac{Q_1Q_2}{4\pi\varepsilon_0r}$$ You can rearrange the equation into$$U=\frac{1}{2}\dfrac{Q_2}{4\pi\varepsilon_0r}Q_1+\frac{1}{2}\dfrac{Q_1}{4\pi\varepsilon_0r}Q_2$$ which then becomes$$U=\frac{1}{2}Q_1V_1+\frac{1}{2}Q_2V_2$$ where $V_1$ and $V_2$ are respectively the potentials created ...


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You are not defining your systems clearly enough. If the system is the mass and the Earth then in the situation that you have described the external force doing work on that system increases the potential energy of the mass-Earth system. Now look at the system which is the mass alone. Then there are two forces acting on the mass. The force as described ...


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By work-KE theorem, $$\Delta \text{KE} = \text{Work done by NET force}$$ $$=\text{Work done by} F_1+\text{Work done by} F_2+\cdots$$ Now if a certain force $F_i$ is conservative, you have the choice of defining its corresponding PE so that $$\text{Work done by}F_i=-\Delta \text{PE}_i$$ and MOVE it to the left hand side so that you have ...


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Must admit this got me really worried, and I wondered if the problem was maybe in the use of $\bf{\tau} = \bf{r \times F}$ and identifying the force with the (negative) gradient of some potential. Then, I really got worried because it seemed to me that the same problem would occur if you were to make similar calculations with determining the torque on an ...


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As far as I know, the potential energy depends on the properties of the particle itself (its mass, its charge) No, not really. Think of a classical mass-spring system. The potential energy is $$ V(x)=\frac{1}{2}kx^2 $$ which is independent of the properties of the mass. In some other cases, such as a charged point particle, the potential energy could be ...


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$V(x)$ is a potential energy function for the system of a particle or particles interacting with a set of constraints. These constraints can be thought of as fields which produce a force on the particle(s) of interest. In the infinite square well (ISW), we examine a particle which has no interaction at all until it gets to some impenetrable constaint, i.e., ...


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I guess you mean $$V=\frac{1}{r}$$ instead of $V=1/x$ (if $\partial_y V=\partial_z V=0$ all your integrals diverge). If this is right, then $$V=\frac{1}{r}\Longleftrightarrow \rho\propto \delta(\boldsymbol r)$$ and not $\rho\propto \frac{1}{r^3}$. Anyway, your example $V=1/r$ is highly pathological: it diverges too fast at the origin (the integrals blow up ...



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