# Tag Info

## Hot answers tagged potential-energy

7

Simply, it is just a matter of definitions. Being $$a=\frac{\hbar^24\pi\epsilon_0}{me^2},$$ a constant generally called Bohr's radius, you can invert this formula to obtain $e^2/4\pi\epsilon_0$ instead. Then, put it in your equation and you are done.

6

We see that all the natural systems aspire for minimum potential energy state This is not true. For example, the earth's minimum potential energy state would be a state in which it was inside the sun. What extra knowledge the concept of Entropy serves us in addition to the concept of potential energy? The concrete sidewalk in front of my house was wet ...

4

For just one example: Consider a box full of $N$ particles of an ideal gas in the absence of gravity. There is no useful potential energy here, in a dynamical sense - the particles of the ideal gas are non-interacting except possibly for hard-sphere collisions (no interaction means no interaction potential, and a hard-sphere potential is trivial and ...

4

The notation is misleading - while $\mathrm{d}U_p$ really is the differential of a state function $U_p$, there is no state function $W$ whose differential $\mathrm{d}W$ could be. That is, it is an "inexact differential" or inexact 1-form, which you can evaluate along paths but for which no potential function exists. The "$W$" we usually write on the l.h.s. ...

4

No there isn't any problem with applying a force greater than one required to just lift it upwards but then the object would get accelerated and hence would gain some kinetic energy. Usually when we talk about the force applied by the person (lifting the object) it is necessary to know how the force varies with respect to time or position so as to calculate ...

3

You're missing a negative sign. The gravitational potential energy $U$ of two masses $m$ and $M$ separated by a distance $r$ is given by $$U=-G\frac{mM}{r}$$ As the objects get closer together, the gravitational potential energy becomes more negative, which is another way to say that it decreases. As the potential energy decreases, the kinetic energy ...

3

You are free to exert any magnitude of force you choose and nothing bounds you from doing that.Most of the books use that magnitude of force which is equal to object's weight so that the object is always in equilibrium.In fact most of the books define potential energy in this manner:- The change in potential energy of a system is defined as the total ...

2

The key point to avoid confusion is to have clear in mind the correct attribution of forces, work and potential energy to the corresponding physical systems. When one introduces the potential energy of a body in the gravitational field, the work of interest is the work done by the gravitational force, which, in general is not the same as the work done to ...

2

No, this is not always true. According to conservation of energy, the change in total mechanical energy of a system is determined by the work done by forces external to the system: $$\Delta E=\Delta K+\Delta U=W_\text{ext}$$ What you are proposing is true if $W_\text{ext}=0$, because then as $\Delta K\geq0$ it must be that $\Delta U\leq0$. Since we started ...

2

This is kind of like the analogue of kinetic energy in Newtonian physics. If you add in potential energy as well, then that can have an arbitrary zero.

2

Now the work I've done is distance traveled times force magnitude, hence 20 joules. That is not strictly true. You are neglecting the kinetic energy that the object possesses at 2 meters due to the velocity you initially gave it has when it reaches 2 meters. The total work you did equals the sum of its kinetic energy (which depends on its velocity) and ...

2

However, if you always need two equal and opposite forces on each body in the system to separate them, how does a rocket leave the earth (from the ground and once in space)? It does not "push" against the earth yet in my mind I consider its thrust to be an single external force on the rocket+earth system which, by the previous statement, should not separate ...

2

Assuming that the apple has a diameter $d$ of 5 cm, the bullet takes between $d/v = 83\ \mu s$ and $d/{v_0} = 100\ \mu s$ to cross it. Assuming further that the bullet enters the apple pretty much horizontally, the drop in height due to gravity during the traversal is of the order of $\Delta h \approx g t^2 \sim 100\ \text{nm}$. The variation of ...

2

Energy is a relative quantity. This means that the change in energy $\Delta E$ is of importance. This is also the reason why $A$ can by any constant. By changing $A$, you do not change the energy difference: $$\Delta E = E_2 - E_1 = \frac{kx_2}{2} +A - (\frac{kx_1}{2} +A) = \frac{kx_2}{2} - \frac{kx_1}{2}$$ Therefore, the minimum value of the potential ...

2

As you can see from the above example the potential function can be broken into components just like force functions.In the example demonstrated above the actual axis is the diagonal line and the force function has been broken into two components along two axes inclined at $45^°$ to the diagonal axis,consequently the potential function along the diagonal ...

2

This is a great question. The answer is that the minimum energy principle is fundamentally about entropy. If you don't understand entropy, then you don't understand the minimum energy principle. The explanation goes the opposite of how you are understanding it. To put it in harsh terms, the minimum energy principle suggests that all the air molecules in the ...

2

Note that you are making an error of category. The entropy is always the property of an ensemble of systems, not of a single system, while the potential energy is the property of a single system (and an ensemble of systems only has an average potential energy). We can work with the entropy of a system in macroscopic thermodynamics, because the systems we ...

1

To follow on with @probably_someone's answer replace the "potential energy" in the question by "internal energy", and note that the equilibrium entropy maximum principle is equivalent to the internal energy minimum principle. Does this mean that we can dispense with the entropy concept? And the answer to that is no, because while one can be derived from the ...

1

Look at this example: The position vector to the mass is: $$\vec{R}=\left[ \begin {array}{c} x\\ y \left( x \right) \end {array} \right]$$ with $y(x)=a(\alpha)\,x$ $$\vec{R}=\left[ \begin {array}{c} x\\ a(\alpha)\,x \end {array} \right] \tag 1$$ To obtain the equation of motion I will use the NEWTON method. the generalized coordinate is $x$ and the ...

1

The thing you're missing is that the force of gravity isn't the only force involved. The potential $U(y)=mgy$ is a function of $y$, not directly a function of $x$. In order to get a potential as a function of $x$, we need a function that gives us $y$ as a function of $x$, so we can plug in this function and define $U(x)$ as being equal to $U(y(x))=mgy(x)$. ...

1

You are right to wonder about the possibility of a change in gravitational potential energy. Actually, the problem is poorly worded. The problem appears to apply the work energy theorem which states that the net work done on an object equals its change in kinetic energy. Note my emphasis on "net". Thats because there are two external forces acting on the ...

1

It is the gravitational potential energy of the water and of the dominos. When siphoning water, once started, it flows from a higher point to a lower point. The weight of the water in the tube, below the water level, will pull water up over the edge of the cup. In the case of the dominos, when you stand them up you are adding gravitational potential to them. ...

1

The potential at a point is derived with calculus,you cannot multiply $\frac{GM}{r^2}$ with the difference in positions because the gravitational field varies with distance from centre of mass of the body.Change in potential energy is defined as negative of the work done by conservative force so $$-F_{conservative} \Delta x = \Delta U$$$$F_{conservative}=- ... 1 Non-conservative forces change the total mechanical energy of the system, since$$W_\text{nc}=\Delta E=\Delta K+\Delta U$$assuming all conservative forces are internal to the system. However, nothing from this tells us how the kinetic and potential energies change. More information is needed. For example, with a mass sliding on a horizontal surface with ... 1 if we don't accelerate the body how can we raise the body to height of ℎ in the example? You can either assume that you accelerate it from rest by a very small amount (so small we can ignore it), or you can assume that it magically begins and ends with some constant speed v, so no acceleration is required. but this time we have a problem with W(byus) =... 1 You have a sign error. The potential energy for gravitation is lower the closer two objects are to each other, not higher. The potential energy of two objects separated by r is$$U = U_0 -\frac{GM_1 M_2}{r}$$where U_0 is the potential energy at infinite separation. U_0 is conventionally taken to be zero, but it doesn't really matter what value it ... 1 Since the electric field is a conservative field, we know that the work done by the field is equal to the negative change in potential energy$$W_\text{field}=-\Delta U$$or, per unit charge$$\frac{W_\text{field}}{q}=-\Delta V$$Since the charge starts at X and ends at Y, \Delta V=V_y-V_x. Therefore$$\frac{W_\text{field}}{q}=-(V_y-V_x)=V_x-V_y$$... 1 You are getting mixed up between work done by a single force and the net work done. The net (total) work done on an object is equal to it's change in kinetic energy$$W_\text{net}=\Delta K$$However, it is most likely that in your education of the box lifting example they are talking about the work just done by the lifting force as well as applying the ... 1 Hint #1: conservation of energy . Hint #2: MGH is gravitational potential enrgy 1 I would like to give a general answer for both gravitational and electric systems considered here.Please read the definition below:- The change in potential energy of the system is defined as the negative of work done by the internal conservative forces of the system.$$\Delta U_{system}=-\int \vec{F_{cons}}.d\vec{s} For example:- Take a ball and earth ...

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