New answers tagged potential-energy
8
Of course it has something to do with the liquid water entering the gas phase just above the cup of tea, but how does that give the bag of tea a directed motion to one side?
Nope.
The teabag is dangled by a string. Remember that the string is made of wound up threads:
Now, the threads stay wound up because they fit well and they have a knack of ...
0
If you take $U_{\infty}=0$ then $U_{sheet}=\infty$ (for infinite sheet)
1
"So what if it is an attraction force? How this should influence our calculations
Because you have written the work--energy relation in an incomplete shorthand. The correct version, $$W = \int \vec{F} \cdot d\vec{x} \quad,$$ depends on the relationship between the direction of the force and the direction of the path.
This relationship is the source of ...
1
NOTE: This comment was too long so I'll make it an answer.
I would assume that the amount that the spring contracts is negligible compared to the distance that the mass has fallen.
But anyway, regarding your amplitude problem: consider that when you have a mass on a level surface connected to the spring, there is an interplay between kinetic and spring ...
3
I think you just forgot that the $\int_A^B F\,dl$ is not a scalar expression. Rather it should be written in a form $\int_A^B \vec{F}\cdot d\vec{l}$. Then it comes to the sign of the scalar product:
$$\vec{F}\cdot d\vec{l}=F\,dl\,\cos\theta$$
where the angle $\theta$ is taken between the vector $\vec{F}$ and the direction of the tangent to the integration ...
-1
The minus sign got put in there when we created mechanical energy.
Take the $\int \vec{F}\cdot\vec{dr}$. This an indefinite integral. If you substract this integral to itself, it gives a constant, because the indefinite integral is defined up to a constant.
Thus,
$c = \int \vec{F}\cdot d\vec{r} - \int \vec{F}\cdot d\vec{r}$
Calculate the first term ...
0
Let's take it from the start . What is Kinetic Energy ?
For one particular particle , it is $\frac{mv^2}{2}$. And for a system of particle it is $\Sigma\large \frac{mv_i^2}{2} $ for all $_i$ particles .
Now how to change Kinetic Energy of a body ? For that you'll have to change $|v|$ and for that you have to apply Force .
then how to calculate after ...
0
It is simply a matter of definition. It is defined in a way such that in infinite distance the potential energy is 0, therefore as you get closer, the potential energy is expressed in a form of kinetic energy and the amount of potential energy "available" decreases. Just definition.
0
First things first: the total mechanical energy is always kinetic energy plus potential energy. So if your answer sheet actually said $KE - PE$, it's wrong. But what I suspect it really said is that the potential energy is negative, so the formula you wind up with is
$$\underbrace{\frac{1}{2}mv^2}_{KE} \underbrace{- \frac{Gm_1 m_2}{r}}_{PE}$$
Now, the ...
3
Imagine your rod is made up from lots of little bricks stacked on each other. Then the total potential energy is the sum of the potential energy of each brick.
The diagram shows the rod and one of the bricks of size $dx$ and at a height $x$. If $\rho$ is the mass per unit length, then the mass of the brick is $\rho \space dx$, and the potential energy is ...
3
You always consider centre of mass. The height $h$ in the equation $U = mgh$ is the height of the centre of mass. In your case, the rod is of height $h$, so the centre of mass is at $\dfrac{h}{2}$
3
You are talking about relativity and gravity together so the question can only be answered in the context of general relativity, but concepts like gravitational potential energy and gravitational force acting over a distance are Newtonian and do not really carry over to general relativity.
However, the gravitational field does contribute to total energy and ...
3
Depends on what you're doing. General relativity handles it for you, in the sense that the Einstein field equation links geometry to the non-gravitational stress-energy tensor. That general relativity is non-linear can be interpreted in part as gravity itself contributing to gravity, but it's generally not even possible to localize gravitational energy in a ...
11
The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles $\vec{p}_1$ (located at $O_1$) and $\vec{p}_2$ located at $O_2$. Their potential energy of interaction is:
\begin{equation}
U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = ...
1
$$U_{g,i}=-G\dfrac{M_{moon}.m}{\underbrace{R_{moon}}_{(\ distance \ from \ moon\ center) }}-G\dfrac{M_{Earth}.m}{\underbrace{(60R_{earth}-R_{moon})}_{(initial \ distance\ from\ Earth)} } $$
You missed $U_i$ due to Earth!
0
The initial kinetic energy needed is whatever it takes to get from the start to the maximum potential between the start and destination.
0
Gravitational Potential is a scalar quantity so can be added algebraically directly for both(or more) bodies.
Also GPE is just Gravitational potential times mass. $$E=\underbrace{\big(\sum P\big)}_{\text{due to all bodies in vicinity}}\times m$$
Now , rest of your aproach is allright ! Continue using this.
0
I think I've understood it now .
$ds=dr$ .
but $dr<0$ and $|dr|=-dr$
Because dr is a small position vector and position vector is directed along field .
Now why I can't use ds directly is because the limits in the integral , (the upper and lower limit in integral notation) are in terms of position vector and not the displacement .
Had they been in ...
2
New version
The problem in your demonstration is when you write down $\vec{A}\cdot\vec{B} = ||\vec{A}||\,||\vec{B}||\,\cos\theta$. More exactly, in your case $||d\vec{r}||\neq dr$ because $dr<0$ when you go from $\infty$ to $r$ and a norm is positive by definition. So the sign error is introduced from 3rd to 4th line.
Old version
The demonstration on ...
1
Just to be clear, the potential energy of a particle of charge $q_2$ at a distance $r$ from a source of potential (supposidely at zero) of charge $q_1$ is the work that an external operator has to provide to bring the particle from infinity to $r$ at constant velocity.
This reads then:
$\int_{\infty}^r \vec{F}_{op}\cdot \vec{ds}$
As people have said, the ...
2
When you calculate work, you do so along a given path. Here, that path has tangent vector $d\mathbf s$. This is a vector with direction; the minus sign will ultimately come from choosing the path's orientation--inward or outward.
Edit: Aha, I think I've found the unintuitive part. The key is in the use of the coordinate $r$ to parameterize the path, in ...
-1
$$\mbox{d}\vec s = \mbox{d}r$$
Therefore,
$$\vec F\cdot \mbox{d}\vec s= F\mbox{ d}r\mbox{ }\cos\theta=F\mbox{ d}r\mbox{ }\cos\pi=-F\mbox{ d}r$$
Edit: sorry for the error where I forgot to put the magnitude sign. I did mean the magnitude sign.
$$\left|\left|\mbox{d}\vec s\right|\right| = \mbox{d}r$$
1
Potential is negative of work done per unit charge by electrostatic force
1
$\mathbf{r}$ is a position vector and $\mathbf{s}$ is a displacement vector between two points, let say A and B. In general case, they are not equal, but they can be if we properly choose the origin of the coordinate system: A={0,0,0} or B={0,0,0} The sign depends on at which point A or B the origin is placed.
1
Use $$\vec F =-\dfrac{dU}{dr}$$ where r is the radial direction.
From graph you can see the slope of $\brace {U-r}$ graph will give: $\underbrace{-\vec F}_{\text{ in radial direction.}}$
2
The electric field is a conservative vector field which implies that there exists a function $V$ for which
$$
\mathbf E = -\nabla V
$$
We call this function $V$ the electric potential. There is no mathematical need to first define potential energy. One can then physically interpret $V$ in terms of a "potential landscape" to get intuition for what it ...
3
Edge effects. After the electron leaves the capacitor, the electric field winds up slowing it back down.
Let's assume the capacitor is infinitely-massive and that the acceleration of the electron is small enough that we can ignore radiation.
Then if you were to idealize the electric field of the capacitor, treating it as a uniform field between the plates ...
2
When we're calculating the energy stored in a capacitor we normally assume it is isolated i.e. there are no other charges nearby to affect it. This makes the calculation nice and simple: the energy is proportional to $Q^2$ and the energy is stored in the electric field around the capacitor.
However in your question you are introducing another charge, your ...
2
The energy is of course coming from the electric field of the capacitor. The energy of any capacitor is always stored in it's electric field. If an electron is initially positioned very far away and then moves close to the capacitor, it's being pulled by the field and that means energy is being transferred. The electric field get's a little weaker - loosing ...
1
Capacitor is losing energy, potential has changed as field is created even by this charge which is moving under the influence of force between capacitor plates .
Take the point charge's potential , and then assume distance between capapcitor plate is d, now as -ve charge approaches +ve plate, it decreases the potential of the +ve capacitor plate more than ...
2
Let's generalize your idea and see if it can be more propellant-efficient, at least in principle.
Call your two orbits $\mho_1$ and $\mho_2$. Both have semi-major axis $a_2$ and $a_2$ and inclinations $i_1$ and $i_2$ respectively. As per the problem,
$$a_1<a_2$$
$$i_1=i_2$$
and any phasing issues may be ignored. Also,
$$r_{p1} = r_{a1}$$
$$r_{p2} = ...
6
Yes the free body moves outward, but there are two critical things you have to know to interpret this statement correctly.
First, this is the effective potential, taking into account gravity and centrifugal force. It has this form because we went into the non-inertial frame co-rotating with the two masses. Mathematically, the potential is
$$ ...
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