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“The square of the velocity minus some constants divided by radius”. Let’s write down the law of conservation of energy: kinetic plus potential energy is constant. $$K.E. = \frac12 m v^2$$ $$P.E. = -\frac{G m M}{r}$$ Where $m$ is the mass of the planet, $M$ the mass of the sun it is orbiting, $G$ the gravitational constant, and $r$ the distance. Add them, ...


4

There's nothing wrong with this mathematically. The intuition behind setting a boundary condition that the gravitational potential goes to zero asymptotically (rather than at the surface of the earth) is that the gravitational effect of the earth should become arbitrarily small as you go very far away from the earth. If you like, you can imagine that if the ...


2

So first of all realize that you are dealing with a definition within Electrostatics, which means there is no time involved, no dynamics, everything is to be thought as frozen in time. So the question, "if we assume none of the charges to be fixed" makes no sense in this context because we are looking at one snapshot in time and calculating if you ...


2

The negative is just a convention that essentially makes the total mechanical energy $E=K+U$ rather than $E=K-U$.


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Actually, gravitational potential energy is constant with reference to a certain point. For example, in the above query, you found out that energy is suddenly increasing when you move it over to the right. This is not the case. The P.E of the body would be the same even if you hadn't moved it to the right In fact, there is no correct definition for Potential ...


2

There are a few conceptual issues in your question. I went on an computed its energy ( kinetic + potential) and it was constant. Why was it constant? This way of putting it creates or betrays some conceptual confusion. The ability to compute the potential energy already presupposes knowledge that potential energy is well-defined for this situation. But ...


2

This is Hooke’s law for a torsion spring, or the angular equivalent of Hooke’s law, $$\tau = -k \theta$$ so it really makes more sense talking about the torque and not a linear force. Upon inspection you can note that torque is directly proportional to the angular displacement and the potential energy stored in the spring is given by $$U = - \frac{1}{2} k \...


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Yes! If they ask you $U(x)$ you simply have to integrate in the variable $x$. So you can consider $a$ as a constant. If they had asked you $U(x,a)$; you know that: $$\vec{F}(x,a)=-\vec{\nabla}U(x,a)$$ so you would have two variables in the integration. Also, it wouldn't be so easy to evaluate the integral: $$\int sin\bigg(\frac{1}{a}\bigg)da=asin\bigg(\frac{...


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We could. But it's just an extra term to drag around. In the end, what matters are changes in GPE, so that extra $Bm$ constant term would just end up cancelling with itself. The reason that GPE is negative is because we always need a reference point -- we must define the potential energy relative to 'something.' In this case, we chose the reference point to ...


2

The internal energy $U$ of a given mass of an ideal gas depends only its temperature. It is the same whether the gas is compressed or not. What the compressed gas has is more of is free energy. This is defined as $F=U-TS$ where $T$ is the temperature and $S$ is the entropy $$ S= NR(\ln V + constant). $$ Here $N$ is the number of moles of gas present ...


2

You've asked a bunch of questions, albeit all related, so I will just try to break them down. From what I understand, energy is defined as the ability to do work Essentially yes. But the term capacity is generally used instead of ability. What does this actually mean and how is it different from work itself? Work is the transfer of energy from one thing ...


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Absolutely you can do this. It is how a Van-de-Graaff generator works. There is an energy cost to carrying the charges to the dome, so you are not getting enegy fro free, but once inside you can deposit the charge onto the dome and so build up enough charge on the dome, until the voltage is sufficiently high that a long spark occurs and discharges it.


1

What is actually different about the charge carriers on these two lengths of wire? Nothing, they are the same carriers with same properties. What is different is that electric potential at the + terminal is higher than electric potential at the - terminal. Electric potential is function of position. It depends on distribution of electric charge everywhere ...


1

If the bottom section were horizontal and the datum for potential energy were taken as the bottom, then the total potential energy would be $$U=\rho gA \frac{(H+y)^2}{2}+\rho gA \frac{(H-y)^2}{2}=\rho gA(2H^2+y^2)$$where H is the equilibrium height; and the total kinetic energy would be $$K=\frac{M}{2}\left(\frac{dy}{dt}\right)^2$$So the total mechanical ...


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The potential energy $U$ is a function of the displacement $y$ from the liquid's equilibrium position. Although the final answer will be symmetric in $y$, for the sake of definiteness let's take $y$ to be positive when the liquid in the right hand side of the tube is raised by a distance $y$. If the absolute potential energy at the equilibrium position is $...


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You can evaluate the electric energy of a sphere $U_{el}$ in this way: first consider that the charge is $Q=\rho \frac{4}{3}\pi R^3$. Let's start from the general formula: $$U_{el}=\frac{1}{2}\int_{Volume}VdQ$$ Then, using Gauss'Law you can calculate the electric potential of the sphere for radiuses smaller then the radius of the sphere: $$V(r<R)=\frac{\...


1

Your first formula represents the work done (by integrating the force over distance) to move a small mass from radius 1 to a different radius 2 relative to a much larger mass. This gives the change in potential energy. To define the potential energy at a point, you must choose a reference point where it is zero. In this case, chosing radius 1 as the ...


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Problem is a bit ill-defined so if the problem is really solvable I'd guess this is for a system of particles where you can get use of Virial theorem. You have the potential of $$V = x^4+4x^2y^2+4x^3y-2y^4$$ and $F = -\nabla V$ therefore $$<T> = \frac{1}{2}<\nabla V\cdot \textbf{r}>$$ $$\nabla V = ((4x^3+8xy^2+12x^2y)\hat{i}+(8x^2y+4x^3-8y^3)\hat{...


1

Starting from the general equation $E=T+U$ and taking the average on both sides we get $\langle E\rangle=\langle T+V\rangle$; now the total energy is a constant, as you said, and the text says the average kinetic energy is just called $T$, so $E$ is given by $E=T+\langle U \rangle$. To find the average of the potential, if I had a finite domain I would take ...


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The first statement is correct. The second is wrong. A simple definition of the potential energy of a system is the amount of work it can do because of the relative positions of its parts. So if a system has negative potential energy with respect to when the parts of the system are separated to infinity, then work must be put in by an external force in order ...


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Yes, That is correct. It is indeed assumed that the stretch is not sufficient to change the tension. The relevent displacement is always the bit causing the stretch, not the sideways displacment that you call $\Delta A$ as that is perpendicular to the string, and hence perpendicular to the tension force, for small displacements. I'll call the transverse ...


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Potential energy function sign depends on exact potential form definition. Usual convention is to use such sign of potential energy, so that field accomplished work by moving a particle from higher potential energy area to lower potential energy area, should be positive. So for example gravitational potential energy near Earth surface is defined as : $$ E = \...


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