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2

You make a good point. In a battery a chemical reaction (a redox reaction) creates the potential difference, and the potential difference is calculated assuming the electrons start and finish in well defined energy states. If electrons returning to the cathode have some residual kinetic energy then the could affect the reaction and change the EMF. However we ...


0

Are you wondering why your two equations $W=\frac{1}{2}kx^2$ and $W=Fx$ don't match? That's because the latter equation, $W=Fx$, is only true for a constant force. The more general expression is $W=\int \vec F \cdot d \vec x$, similar to what you wrote in differential form $dW=F\,dx$. The expression with $\frac{1}{2}$ in it is correct. The expression with ...


0

Gravity is an acceleration. No negative involved. However, when you use acceleration to find a velocity, since velocity is a vector quantity, you must describe a direction. It is convention that anything that accelerates up, is described as a positive(+) like "The ball accelerates at 20m/s^2", whereas gravity describing a downward acceleration is described ...


-2

It is because gravitational force is attractive and work is done by gravitational force itself. When system does work itself energy is taken as negative and when work is done by external agency on system energy is take as positive.


1

If you look at the 2 graphs, the force varies directly as the slope of the potential energy curve ($F\propto $ slope of $PE$)(or is it the other way around?). There is a reference to attraction in the PE graph because the atoms will stay together if the distance keeps the PE negative. If the distance were to decrease so that the PE is positive, the atoms ...


2

What does negative potential energy mean here? Not much. The particular value of potential energy isn't important at all in classical physics. But changes in potential energy are. You could shift everything up so that $U>0$ everywhere, and you'd still get the same physics. (Why, you ask? Well, would the force change?) So why would people choose to ...


0

In general, the absolute energy of a physical system does not matter for these potential problems. However, for convenience sake, one often choses to make the potential energy term at infinite separation equal to zero. If you were to hold the two atoms infinitely far away from each other, they will first experience an attractive force, because the ...


2

Yes - the coiled spring has a certain amount of (potential) energy. When it gives up the energy to the ball, you could say the ball does negative work on the spring, so it loses (potential) energy.


2

Internal forces are the only contributors to potential energy. Potential energy is the energy associated with the configuration (relative positions) of a collection objects. The potential energy of a single point particle is not defined. As the configuration of the system changes, its potential energy changes according to the definition ...


2

Yes, the two are intimately related. One way, as in QMechanic's answer, is via Wick rotations, but in general there is a lot more freedom once you allow integration contours to go over into the complex plane. In my area, strong field physics, the use of complex time to understand tunnelling problems is everyday bread and butter for many people, and it is the ...


6

Yes, quantum tunnelling in the double well potential can be solved in a Wick-rotated Euclidean formulation $$ S_E[x]~=~\int \! dt_E \left[ \frac{1}{2}\left(\frac{dx}{dt_E}\right)^2 - (-V) \right], $$ see e.g. Ref 1. Here $t_E=it_M$ denotes Euclidean time. The Euclidean action is in turn interpreted as the usual kinetic minus potential term with a potential ...


1

In quantum mechanics velocity is not an easy concept. Here particle motion is replaced by a wave. Momentum, is easier to define in quantum mechanics. A complex wave $\exp(ikx)$ describes a particle with momentum $p=\hbar k$ where $\hbar$ is Plancks constant divided with $2\pi$. It is fundamental in quantum mechanics that momentum cannot strictly be defined ...


1

You need to break the problem into two parts. In part one, you have to calculate the velocity of the man's center of mass at the point his feet leave the ground. You can calculate that by computing the time it takes to rise from h2 to h3, using (h3-h2)=0.5*a*t*t, where a=9.8m/s/s, and then computing the velocity using v=a*t. In part 2, you can calculate ...


1

$\vec{F}=-(\frac{\delta U}{\delta x} \hat{i} + \frac{\delta U}{\delta y} \hat{j}) $ $=Ky \hat{i} + Kx \hat{j}$


0

First, we define the direction of $\vec{\theta}$ between $\vec{\mu}$ and $\vec{B}$ to be positive if $\vec{\theta}$ is coming out of the screen in the sense that $\theta$ is increasing with the right-hand-rule in the counter-clockwise direction. Then, note that the torque $\vec{\tau}=\vec{\mu}\times\vec{B}$ makes the angle between $\vec{\mu}$ and $\vec{B}$ ...


1

Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out Ok, so sum of the forces is 0 and the acceleration is zero. and the book moves with a constant velocity. Spooky. Was the book moving initially? ...after the book has been lifted, and it has come to rest once again. According to ...


7

You say: Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out and the book moves with a constant velocity. so I'm guessing your reasoning is that the net force on the book is zero so the amount of work done on the book is zero. And you are absolutely correct - no work is done on the book and ...


1

Let's say the book starts and stops from rest, as I believe you are assuming. The motion within this interval is unimportant, as you'll see. The increase in gravitational potential energy of the Earth-book system came from your body. You did positive work on the system since your hand force and displacement are in the same direction, resulting in an ...


0

If you lift the book with a force that is exactly equal (but opposite) to the force of gravity acting upon it the book won't go anywhere. After all, that's exactly what a table does when you leave the book on it. To get to book going you need to lift with a force greater in magnitude than that of gravity, resulting in a net force and an associated ...


0

The entropy can be written as (discrete form)$$S=\sum_i p_ilog(p_i)$$ So you must identify what the uncertainty in your problem comes from, you would think that you have (in principle) exact deterministic equations for the evolution of these particles, so there is no uncertainty with respect to that. If you have unknown initial conditions then you could ...


0

I'll make this an answer, even though it is more of a drawn out comment. As I mentioned as a comment, computing the potential energy is trivial. If you want speed, you'll probably want to look at fast methods for long-range interactions. The link takes you to state-of-the-art libraries and methods, but any introductory book on computational statistical ...


0

Correct me if I am wrong, but a potential energy can only be determined for a conservative force field, which means that the force can only depend on position. So, because the charges vary with time you can not determine the potential energy. If the velocity is small compared to the oscillation, such that the displacement during the common period of the ...


3

Let us assume that once the bubble is created, the physics is essentially no different from what is responsible for the work making an Helium balloon rise in the air i.e. you have a macroscopic/mesoscopic object which has a density smaller than that of the fluid surrounding it. Now, we all know that in a static fluid under gravity, pressure is decreasing ...


2

When a bubble is rising up, water is filling up the space behind it. The work done by the bubble rising up is exactly same as the water coming down to fill the space behind it, but with a -ve sign. So the total energy will remain constant.


1

Both masses $m$ are at a distance $R$ of the center of mass. The distance between the two masses is then $2R$. To calculate the potential energy, consider the first mass fixed. The potential energy of the second mass is $\gamma \frac{m^2}{2R}$, with $\gamma$ the gravitational constant. This is also the potential energy of the entire system. The kinetic ...


2

The minus sign is only there by convention. You could replace $\phi$ with $-\phi$ and the minus sign would go away. Note that $\nabla \phi$ points in the direction of steepest ascent for $\phi$, whereas $-\nabla \phi$ points in the direction of steepest descent. Perhaps it seems nice for the force on an object to be pointing in a direction of descent for ...


2

It's total convention. The idea is that one can caluclate, for any path: $$\Delta\left(\frac{1}{2}mv^{2}\right) = \int_{\rm path}{\vec F}\cdot d{\vec s}$$ One can then split the left hand side into a conservative bit and a nonconservative bit. (with the definition of conservative being ${\vec \nabla}\times{\vec F} = 0$). Then, since we know that any ...


0

Because this is a two particle system, you can do this exactly. For a given mass $m$ and distance between objects $2r$, you can compute the required angular velocity $\omega$ (to remain in stable orbit) and thus the kinetic energy ($\frac12m(r\omega)^2$). For that same configuration, you can compute the gravitational energy by seeing how much it takes to ...



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