# Tag Info

1

The formula $mgh$ only works close to the Earth's surface where $g$ can be assumed constant. That is not the case here. The appropriate formula is more like $$E = m \int g(r)\ dr$$ where $g(r)$ is gravity as a function of radius from the centre of the Earth. This is the work done in moving the mass $m$ in the gravitational field. If $g$ were constant and ...

0

Q1: I am not sure what you mean, physically it means the maximization of the the distances, but you already knew that. Q2: it is equivalent, so it is still an open problem. Q3: the formula does not apply, you are correct, for n=3 the solution is that the electrons reside at the vertices of an equilateral triangle about a great circle. But reading the ...

0

It is important to be clear what is doing work on what. In the diagram below, consider a charge starting at rest at A. We release it, and it accelerates down the potential gradient. Here, the field is doing work on the charge - it is accelerating it and therefore giving it kinetic energy. As a result, the charge will travel through the potential minimum at ...

1

if we want to move a charge in an electric field then we need a work of an external force to move it from a low potential energy to a high one Not true. The electric field itself exerts a force to move the charge. Moving the charge from high to low does require work and this work is done by the electric field itself. Any force that causes movement is ...

0

Take the example of gravity. Gravitational potential energy is defined as $$U_g=-\frac{GM}{r}$$ This means that $$U_g=-\int_{r_1}^{r_2}F_g \cos \theta dr$$ Because $F_g=mg$, $$U_g=-\int_{r_1}^{r_2}mg \cos \theta dr$$ Integrate and you find $$U=-mg\cos\theta \left (\left.r\right|_{r_1}^{r_2}\right)$$ and $$U=-mg\cos\theta (r_2-r_1)$$ and if the object is ...

0

Sorry, not sure what you mean by neutral state, but I think you need to concentrate on thinking about forces - the normal to the blocks won't 100 % balance out gravity. Suggestion for how to start... I would draw a diagram and put on the force of the spring, which gets compressed by the bar falling and the force of gravity and then try to balance the ...

0

http://en.m.wikipedia.org/wiki/File:HookesLawForSpring-English.png. I think this is also because of the spring constant which is I think is the gap present between the spring when it is coiled where the energy or the potential energy is stored and I don't think the atoms get affected

4

Firstly, you can deform material permanently..spring is no exception. On the atomic level, you are working against Coulomb forces that bind the material id est, that form the lattice. One primitive cell is well defined by the conditions of minimal energy. You can describe this potential as a quadratic, so you get harmonic forces, but it is not truly ...

8

You are asking two questions really 1) How is PE actually stored in a steel spring at the atomic level? The explanation for this lies in quantum mechanics 2) Could you explain in detail how/where potential energy is actually stored in a steel spring and why the material never surrenders to the bending forces taking a new shape? Replying to 1) ...

0

horizontal spring exerts a force $F = (−kx, 0, 0)$ that is proportional to its deflection in the $x$ direction. The work of this spring on a body moving along the space curve $s(t) = (x(t), y(t), z(t))$, is calculated using its velocity, $v = (vx, vy, vz)$, to obtain $$W=\int_0^t\mathbf{F}\cdot\mathbf{v}\mathrm\,{d}t =-\int_0^t kx v_x \mathrm\,{d}t = ... 0 If you look a little bit lower where he sets KE = PE, you will see that$$\text{total_} = (m+M)$$1 Posting another view on the already nice answers. Conservation of (mass-)energy is a principle in physics. Feynman used to say, (Feynman lectures on physics) that when various processes are studied, one finds that energy is not conserved, but then looks under the carpet or in waste bin and finds another form of energy which when taken into account makes the ... -1 You do actually need gravitational potential energy for this problem, and it is central to the concept of this problem, which is energy. The gravitational potential energy at h=100m is mgh = 980000J, as you calculated. This total amount of energy will not change (assuming friction does not apply) because in a closed system, mgh+(1/2)mv^2 = C, a constant; ... 0 Recall that if the mechanical energy is conserved:$$K_0+ U_0 = K_f + U_f.$$In your case you calculated U_0 = 980,000 J. Since K_0 = 0 you have that:$$U_0 = \frac{1}{2}mv^2_f + mgh = m\left(\frac{v^2_f}{2}+gh\right) \Rightarrow \frac{U_0}{mg}-\frac{v^2_f}{2g} = h

1

It would appear evident that its KE has been drained out by g and definitely destroyed. Is this correct? This is from a different perspective than the other answers and is not so much an answer as an extended comment on the above quoted question. It occurs to me that creation and destruction are, in some sense, absolute. In your thought experiment, ...

6

Consider the case in which we shoot an electron up in the stratosphere, it travels up to a certain height and then it stops when its KE = 0. We say, according to that principle, that lost energy is stored as PE. This has been experimentally verified of course, as in falling back it gains the kinetic energy it lost going up. The concept of potential ...

-2

The KE is converted to PE. The photon-planet system still has potential energy since there is gravitational red shift.

-5

(5 downvotes ... oops sorry ... 6 downvotes :) and not a single comment why? None dared to formulate counter-arguments? All one gets on a scientific forum are just NOs? Why I am not surprised? :) Again - as the subject is fundamental - all meaningful comments are heartily welcome) bobie, notice one thing. Gravitation as such is energy from nothing. Where ...

11

The idea of partitioning energy into different forms like "mechanical energy" or "chemical energy" and such is actually arbitrary. More or less by definition, energy is that which is conserved unter time translations by Noether's theorem. If what you call "mechanical energy" has changed, then there is another term in the Noetherian energy that has changed ...

Top 50 recent answers are included