- #1

aaronll

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When I accelerate a particle and then I "free it", what happen to its potential energy if the total energy should be conserved?

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- #1

aaronll

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When I accelerate a particle and then I "free it", what happen to its potential energy if the total energy should be conserved?

- #2

anuttarasammyak

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- #3

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Yes, it must. It can only gain KE if it loses PE. That's the nature of a conservative force.If a charged particle moves through a potential difference, it gains kinetic energy but does it also lose potential energy?

- #4

aaronll

- 23

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Yes I know, maybe my question is very wrong, but if I have this apparatus:Yes, it must. It can only gain KE if it loses PE. That's the nature of a conservative force.

Before: particle have a kinetic energy with v1 and so total energy = kinetic energy.

Enter: particle is due to the force, that is conservative, so E is conserved and gain kinetic energy ( loss potential energy).

flow Out: It has only kinetic energy, but is different from when is inside, because there are no more "potential" energy they have relatively to the position in the field.

I think I'm wrong because potential energy have a meaning only when the particle is in a field so there is potential difference ( this is what interest) and energy is conserved because inside the field do a work and convert that work in kinetic energy, is correct?

thanks

- #5

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- #6

aaronll

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Maybe my question was not clear, I try to reformulate with a different situation:

I have a table, 6m tall relatively to the "ground", with a cube on it.

The cube has total energy of 0 ( is at rest and potential energy relatively to the table surface is 0).

So E = 0.

Now, I move the cube beyond the edge of the table, and now it is 6m from the ground ( also before but that is not my doubt) so its energy is the potential energy relatively to the ground, so E > 0.

The value of energy is changed ( value in joules ) but only because I change the "reference", so energy seems to be not conserved, or change value when I change the position of the object but doing no work.

How It is?

- #7

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$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$

Your capacitor at a static voltage is an example. Another one is the gravitational force between some very light object and other very heavy objects at rest.

From Newton's equation of motion,

$$m \ddot{\vec{x}}=\vec{F}(\vec{x}) = -\vec{\nabla} V(\vec{x}),$$

you can then derive the energy-conservation law by multiplying with ##\dot{\vec{x}}## and integrating between two arbitrary times. On the left-hand side this gives

$$\int_{t_1}^{t_2} \mathrm{d} t m \dot{\vec{x}} \cdot \ddot{\vec{x}} = \int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right) = \left [\frac{m}{2} \dot{\vec{x}}(t) \right ]_{t=t_1}^{t=t_2} = T(t_2)-T(t_1),$$

where the kinetic energy at time ##t## is defined as

$$T(t)=\frac{m}{2} \dot{\vec{x}}^2(t).$$

On the right-hand side the above procedure gives

$$\int_{t_1}^{t_2} \mathrm{d} t \mathrm{d} t [-\dot{\vec{x}} \cdot \vec{\nabla} V] = -\int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} V = - \left \{ V[\vec{x}(t_2)] - V[\vec{x}(t_1)] \right \}.$$

Setting this equal to the previous equation and sorting the quantities at each time you get

$$T(t_2)+V[\vec{x}(t_2)]=T(t_1)+V[\vec{x}(t_1)] \; \rightarrow \; E=T+V=\text{const}$$

along the trajectory.

That energy-conservation law is very useful, because you don't need to know the solution of the equations of motion to calculate the energy, provided the force has a potential. For a given force a potential exists if the line integral of the force between two points is independent on the path connecting these two points, and then the potential is given by the line integrals connecting some fixed point with any other point,

$$V(\vec{x})=-\int_{C_{\vec{x}_0 \rightarrow \vec{x}}} \mathrm{d} \vec{x}' \cdot \vec{F}(\vec{x}').$$

Now if you have the situation that ##V(\vec{x}) \rightarrow 0## for ##r=|\vec{x}| \rightarrow \infty## and the particle enters with some velocity ##\vec{v}## from infinity, it gets scattered by the force described by the potential and then goes to infinity again. Then far away from the region, where the potential is non-zero it has some speed ##\vec{v}'##, but since ##V=0## in the beginning and at the end of the scattering process the energy-conservation law tells you that the speed is the same initially and finally ##|\vec{v}'|=|\vec{v}|##.

- #8

anuttarasammyak

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In comparison of before and after, energy ground level should be in common. If you can change energy ground level any time you like, I cannot say anything right for you without knowing your rule.The value of energy is changed ( value in joules ) but only because I change the "reference", so energy seems to be not conserved, or change value when I change the position of the object but doing no work.

How It is?

- #9

- 24,039

- 15,737

The value of energy is changed ( value in joules ) but only because I change the "reference", so energy seems to be not conserved, or change value when I change the position of the object but doing no work.

How It is?

However, the

- #10

jartsa

- 1,565

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You did not draw the important part. I mean the line ends where the particle is not yet free.Yes I know, maybe my question is very wrong, but if I have this apparatus:

View attachment 291013

Before: particle have a kinetic energy with v1 and so total energy = kinetic energy.

Enter: particle is due to the force, that is conservative, so E is conserved and gain kinetic energy ( loss potential energy).

flow Out: It has only kinetic energy, but is different from when is inside, because there are no more "potential" energy they have relatively to the position in the field.

I think I'm wrong because potential energy have a meaning only when the particle is in a field so there is potential difference ( this is what interest) and energy is conserved because inside the field do a work and convert that work in kinetic energy, is correct?

thanks

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