# Tag Info

1

I asked a somewhat different, yet similar question.Hope this helps! Why is an $LC$ oscillator lossless, but $C V^2 / 2$ energy is lost to a capacitor connected to an ideal voltage source?

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The system is subject to a non-zero net force in the horizontal direction and no friction, so it will experience constant acceleration (of the center of mass). Superimposed on that motion with be the anti-symmetric oscillation of the two masses on the spring. If the masses are both $m$ and the spring is characterized by constant $k$ the angular frequency of ...

0

Yes it will. The spring behaves like friction here. When the first block is pushed, it transfers the energy to the spring which converts the kinetic energy to its potential energy. Once the second block overcomes its inertia, it will also start to move. Think of two blocks on a surface with friction without a spring in between. Pushing the first block ...

1

You are in your reasoning overlooking something. Look at the diagram below: $-q_1,+q_2$ are two point charges at distance $r$. Coulomb's Law dictates that the attractive electrostatic attraction force between them is: $$F=k_e\frac{|q_1q_2|}{r^2}$$ And the electrostatic potential $U(r)$: $$dU(r)=F(r)dr$$ $$U(r)=-k_e\frac{|q_1q_2|}{r}$$ Assume now that ...

1

You can describe the electric force it terms of potential energy, because it is a conservative force. In doing so you actually replace the concept of work done by this force by the concept of potential energy. So you can not longer use both descriptions simultaneously. If you describe the electric force as doing work, then you made positive work and the ...

1

Potential energy is the energy due to configuration of the system. If you keep three charges very, very far from each other, then the potential energy of the system is very effectively zero. But when you bring them close together to a specified coordinate, then the potential energy of the system increases from $0$ to a positive value given by $$U= ... 0 Potential Energy is calculated of a system, ie, a system possesses potential energy and the capacity to do work with it. If you a raise a ball of weight mg to a height h above the surface of the earth, then the total potential energy of the ball will be PE = mgh as you have done work against the gravity of the earth and thus the ball possesses a certain PE. ... 0 Everything you say is correct in the steady state. The problem you run into is that when you remove charge from a charged capacitor to an uncharged capacitor, there is a potential difference. And somehow, you have to remove the energy from the electron that moves from one to the other. It turns out, as you calculated, that you in fact remove half of the ... 2 What is gravitational potential? Usually a potential is defined as the potential energy per mass or per charge or similar. This is most often seen in relation to electricity or chemistry and less often to gravity. GPE is gravitational potential energy. GP is gravitational potential energy per mass:$$GP =\frac{GPE}{m} $$Is it defined for the system ... 0 There's a problem in this equation:$$ d\vec{l}=dy \cdot -\vec{j} $$here dy needs a minus sign. It is easier to see this writing the path. The path of integration is parameterized by:$$ \vec{l}=\vec{P}+(\vec{N}-\vec{P})\frac{y-y_P}{y_N-y_P} = \vec{P}+\hat{j}(y-7a) \\ y_p \le y \le y_N $$Therefore$$ d\vec{l}=\hat{j}dy $$and$$ \int_P^N E_y \cdot ...

2

Potential energy is wrong. Even in Newtonian Mechanics it only works if the force is 1) purely a function of position and also 2) is conservative. Magnetic forces depend on velocity so they fail. And electric forces are not conservative if you aren't in statics. What you really have is kinetic energy and rest energy for the charged particle, some energy ...

1

Don't forget about the rest energy, $E=mc^2$. Since a particle's intrinsic spin cannot be changed, it doesn't make sense to distinguish "intrinsic spin energy" $\frac12 I\omega^2$ (as you would compute for, say, a spinning flywheel) and the rest energy. A particle which is moving in an electric field will see a motional magnetic field. Charged particles ...

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Your copy of Verma has already defined gravitational potential energy previously in (11.3) The gravitational potential energy of a two particle system is $$U(r) = -\frac{Gm_1m_2}{r} \tag{11.3}$$ where $m_1$ and $m_2$ are the masses of the particles, $r$ is the separation between the particles and the potential energy is chosen to be zero when the ...

1

Your teacher is wrong. The gravitational potential $V(x)$ is generally defined as potential energy per unit mass i.e. $V(x) \equiv \dfrac{U(x)}{m}$. So for the points where $U(x)$ is zero, $V(x)$ is zero and vice-versa by definition. EDIT: After you added the comment and a snapshot of the book, I realized your book has defined Gravitational Potential in a ...

2

This is a very basic problem: (a) $F=-\frac{d U(x)}{dx}$. (b) You need to find the minimum of $U(x)$; (c) Sketch a graph. For all the questions you can use wolframalpha. If I understood your problem correctly then (a) and (b,c) .

0

Yes there is! First, electric potential is measured in volts (V) and electric potential energy is measured in joules (J). Now if it sounds familiar is that both tell you about an energy quantity (V=J/C). Indeed, in electromagnetism, the potential is seen as the electric field, multiplied by the distance between the source (for example a point charge) and the ...

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@DilithiumMatrix has talked a bit about the physical meaning behind your answer, and the important distinction behind the gravitational and electromagnetic binding forces. I want to point out that your gravitational calculation is not actually correct (though it might be good enough—see the end of the post.) Why your answer isn't actually correct: The ...

1

Asteroids are held together by a combination of gravity and cohesive (electromagnetic) forces (the same forces which hold rocks together on earth). For small asteroids (smaller than about 1 km), the gravity is negligible, while for larger bodies (larger than 10s to 100s of km) the cohesive forces become negligible*. If you're only interested in the larger ...

-1

Let me try to answer my own question, helped by this post on the binding energy of neutron stars (many thanks DilithiumMatrix for pointing that out). Indeed the example of the neutron star, though not conceptually different from the example of the black hole which I used in my original question, is easier because we do not get distracted by the problems ...

1

The question, as written, has no need for calculus. Find the volume of water needed to fill the frustum. http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html Find the weight of that water Find the work done to pump all that weight up 6 feet over the top of the wall of the tank, letting it splash down into the tank. Done.

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We know that the gravitational potential energy of an object is $mgh$. Initially, $h=0$ for the water before it is pumped in, so the potential energy is 0 as well. From here, this is a related rates problem, the rates you want to be relating is the rate of increase of the gravitational potential energy of a horizontal layer of the water filling the tank to ...

0

Gert's answer states that heat cannot flow from lower temperature to higher temperature because it is obvious. I suggest that your first question could be answered with the concept of entropy. To put it simpler, particles and their energy levels would tend to a more probable state. For example, when you throw 6 dices, the probability of it getting all six ...

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Now I propose the following thought experiment: consider the system of an object of mass $m$ at some far distance from a black hole of mass $M$. The mass of the whole system is $m+M$. It's only because they are so far away from each other that the mass from even farther away looks so close to $m+M.$ So that's not entirely a trivial thing. Now let ...

0

But in electrodynamics, the electrons move from negative terminal of the battery which is at lower potential to the positive terminal of the battery which is at higher potential. This entirely due to a naming convention. Had we decided to call the charge of the electron positive instead of negative the electrons would have moved from the positive ...

2

The total energy at the top is $$T = \frac{1}{2} m v^2 + m g L$$ The total energy at some other point is $$B = \frac{1}{2} m f^2 + m g L \cos\theta$$ Energy is conserved so $$f = \sqrt{v^2 + 2 L g (1-\cos\theta) }$$

0

GPE of a system of two masses is calculated by $$GPE = U_g=\frac{-Gm_1m_2}{r_{12}}.$$ This arises from the definition of potential energy associated with a conservative force: $$U=-\int_{r_0}^{r_f} \vec{F}\cdot d\vec{r},$$ where for gravitational systems $r_o$ is $\infty$, $r_f$ is the separation distance between two masses and $\vec{F}$ is the Newtonian ...

3

Consider an arbitrary potential energy $V(x)$; take $x_0$ to be an equilibrium point, that is, $V'(x_0)=0$. Next, Taylor expand $V(x)$ for $x$ close to $x_0$: $$V(x)\approx V(x_0)+(x-x_0)V'(x_0)+\frac{1}{2}(x-x_0)^2 V''(x_0)$$ The first term is just a constant (ie, irrelevant for energies), and the second one is, by definition, null; therefore we find ...

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