# Tag Info

1

Holding a rubber band stretched is the same as holding a weight above the ground. You aren't adding any energy to the weight, just maintaining its position. However, your muscles' actin and myosin require energy input just to maintain a force. This energy ends up heating the muscles, and is lost.

2

Holding the rubber band at a constant stretched position/length causes you to contract your muscles, but the energy you expend (from food) just heats your muscles instead of doing mechanical work on the rubber band.

8

Visualization The difference in height $h$ is always the same (here 10 m)! Remark This is of course only true if $g$ is constant, e. g. $h$ does not change "much". See also Wikipedia.

5

Potential energy is given only as a difference of energies at different heights. So, if you want to know just how much does the person gain energy (or rather loose by friction in their muscles and joints) by walking down the hill, you might just use their height of their heels on the top of the hill and under the hill. But remember, you always have to use ...

4

Theoretically, their center of mass is what you're looking for. It's somewhere near the stomach. High jumpers bend their body when they are jumping so that their center of mass can travel just above (or sometimes below) the bar which allows them to use the least energy to jump the highest: https://en.wikipedia.org/wiki/High_jump When a body is rotating or ...

1

I am puzzled at your leap: quantum bound state of electron to hydrogen, to the earth potential classical bound state. Bound classically and bound quantum mechanically are two different frameworks. The electron is bound quantum mechanically to the hydrogen atom and does not "see" the gravitational coupling quantum mechanically due to the very small value of ...

0

One usually goes to the continuum because of its nice mathematical properties; lattice QFT is a hint of how hard a quantum theory becomes if we break the symmetries. For scattering theory, you usually want to be able to apply Lorentz invariance, or its classical counterpart, Galilean, and this already implies you are working with the continuum spectra of ...

3

For example, the escape velocity of a particle from the galaxy is about 400 km/s and in most conceivable circumstances (unless you are basically on top of the event horizon of a black hole or on the surface of a neutron star), escape velocities will be far, far below relativistic speeds (here defined as $3\times10^4$ km/s). So basically, if a particle has a ...

1

I propose redefining this problem as follows (because I'm not sure it has a solution the way the OP has defined it). Let $y=f(x)$ be some symmetrical (around $y$) function like $x^2$. Let the point mass experience a friction force acc. to the usual simple model $F_f=\mu F_N$, with $F_N$ the Normal force acting on the point mass in the point $(x,y)$ ($N$ ...

0

All of this is all right, but the problem is that I'm taking a course on electrodynamics and the teacher said that the work $W_{\mathrm{ext}}$ done by one force external to the system is $$W_{\mathrm{ext}} = \Delta K + \Delta U,$$ that is the change in the total energy of the system. I don't know where this comes from It follows from the work-energy ...

1

Technically, the energy of a system is conserved. This is a subtle, but important, distinction from the energy being constant. Conservation is different from constancy in that energy can move in and out of a system and can change forms, but it is neither created nor destroyed. An expanded formula would look like this: ...

0

I believe this is the "missing link", stated in a less abstract fashion than in the above comment: https://en.wikipedia.org/wiki/Quadratic_form . Some programs in physics cover that in undergraduate algebra courses, some leave it for later. Notably, this method doesn't apply just to tensors, it's a general connection between symmetric matrices (of spaces ...

4

The energy is defined as $$E = \frac{p^2}{2m} + V(\vec r)$$ where the first term is the kinetic energy and the second term is the potential energy calibrated so that $V(\vec r)=0$ for $|\vec r|\to\infty$. Consequently, you may say that the energy in a given state (an analogy of an orbit in classical physics) is equal to the kinetic energy $T_\infty$ ...

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