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3

In simple terms the internal energy can be thought of as the sum of the kinetic energy and the potential energy of the molecules. The kinetic energy of the molecules depends on the temperature - a higher temperature means that the molecules have more kinetic energy. The potential energy of the molecules depends on the bonds (interactions) between them - ...


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If you accept that no external work was done, then if there is a change in the state of a system through which the kinetic energy changed, there must be a corresponding change in potential energy. The key to understanding the (rather poorly narrated) video is that the lecturer implies (at T=2:30) that $\Delta E=0$ from which it follows that $\Delta KE= - \...


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A simple pendulum performs simple harmonic motion when it is displaced very slightly. You can say that a simple pendulum performs a periodic motion which can be treated as simple harmonic motion in small oscillation Now lets move forward supposing it to be pure SHM. The time period of a simple pendulum does not depend on how much it is displaced( but it ...


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Your approach is almost correct! To see what's wrong, consider the case $d = 0$. Then for a small change in volume, the change in potential energy of the water is zero (since you're just moving water from the left side to water at the same height on the right side). But the gas has definitely done $p \, dV$ work. The mistake is that you've neglected the ...


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Your intuition that the same amount of fluid goes down and then up by the same amount is incomplete, you are forgetting what happens inside the fluid. It is easier to see using solid blocks as in the figure below: Here you can see that the effect of moving block 1 down is to shift block 2 to the right, and moving block 3 back up the same amount that ...


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My explanation is as follows Force on the spring in stretching it to length x can be written as $F=k(x-L_0)$ where x is the displacement, $L_0$ is the initial length of the spring and k is spring constant. energy stored is $dE=F.dx$ upon integration we will get $E=\frac{1}{2}.k.(L-L_0)^2$ where $L$ is the final length of the spring. This is I think ...


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Precisely, the point is that the slope is not equal to 1. The ratio $\dfrac{F}{\Delta L}=k$. Therefore, $Z=\dfrac{1}{2}(x)*(\dfrac{x}{k})$ $= \dfrac{1}{2k}x^2$. Which looks wrong but is true because in the notation you are using, $x$ is not the elongation but is rather the force acting. So in a more familiar notation $E=\dfrac{1}{2k}F^2$. If you want to ...


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Work is always force times displacement in the direction of the force. The only place where the gas is doing work is at the bottom surface that is moving downward. The force it is exerting there is $PA$, where $P$ is the gas pressure and $A$ is the cross sectional area of the tube. If the lower surface moves downward a differential distance dx, the work ...


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I am answering the question formulated after the "edit" in a newer version of the text because that one seems well-defined. Indeed, a situation with a uniform field $\vec E$ may be said to be "uniform" or translationally invariant in space. Noether's theorem says that this "uniformity" (spatial translational invariance) implies the existence of a conserved ...


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Gravitational field is a vector field and is determined by negative gradient of the gravitational potential. $$\vec g=-\vec\nabla \phi$$ Frome equation above, it is obvious that $|\vec g|=|-\vec\nabla \phi|$ (magnitude of $\vec g$ is equal to magnitude of $-\vec \nabla \phi$) and we know that $|-\vec\nabla \phi|$ is a non-negative quantity. You have made a ...


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First, let's pick a field to work with, because particles act differently in different vector fields. Let's say we're dealing with a charged particle in an electrostatic field. EM fields can be seen as a deformity in spacetime, the field is warping the space in which it is defined. In fact, for advanced EM we use tensors to describe electromagnetic ...


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First of all, we must be talking about a field that would affect (exert a force on) the object (like a charge in an electric field or an object with mass in a gravitational field). Now, what does potential energy mean? It is a measure of "stored energy" in the system. That means, if you released it, this energy would be released. Put a book on a shelf and ...


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From wikipedia: Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity. So, it isn't matter how much work is done by other forces acting on a body. What is matter is work done by the gravitational field.


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It is important to note, that you are totally skipping the kinetic Energy $T$ part. The total energy of a system is given by $E=T+U$. There are several examples of systems which build up potential Energy over time. Some examples are: The mass of a pendulum is constantly cycling the total energy between kinetic and potential energy Objects orbiting a center ...


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Actually, all of your derivation is correct. But, potential energy is the energy due to gravitational field with respect to point at infinity. By your derivation, u at infinity is zero. u at (r,0) is GMm/r . Thus, potential energy at (r,0) is U(r,0) = u(0,inf) - u(0, r) = 0 - GMm/r = -GMm/r


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If you have a proton (mass=$m_p$) and a neutron (mass=$m_n$) and allow them to join via the strong nuclear force the mass of the $^2_1H$ nucleus $m_{np}$ is less than the sum of the masses of the individual particles ($m_{np} \lt m_p + m_n$). During this joining together the proton-neutron system loses potential energy and that energy is called the binding ...


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Mass certainly will change. In fact we have two connected bodies - brick and earth. The Hamiltonian (full energy operator) of a connected particle system is defined as: $H = H(brick) + H(earth) + V$, where $V$ is energy of interaction (potential energy in our case). Since potential energy will increase, mass inevitably increases due to the theory of ...


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No, mass does not increase when you lift an object. Potential energy increases, but that's because of the $\Delta h$ in $U_{\mathrm{grav}} = mg\Delta h$, not the $m$. Mass is, to put it a bit imprecisely, the amount of matter in an object. When you lift something, there doesn't become more of it. (Note that I'm ignoring relativistic effects.)


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This result is consistent with the principle that says "the work is negative whenever the electric potential increases", and the one that states "the electric potential increases whenever a charge moves in the opposite direction of the force applied to it by the electric field of another charge". My question would be whether my answer is wrong and I am ...


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$U$ is the electrostatic potential energy of the system of the three protons. To assemble the three protons coming from infinity external work needs to be done so the electrostatic potential energy is positive. So the textbook answer has computed the external work which needs to be done to bring the protons together. Another approach is to compute the work ...


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I don't think it matters really, it's a matter of convention. In your picture if we start at infinity: $$ W = \int_\infty^r dr' F(r')= C \int_\infty^r \frac{1}{r'^2} = \frac{-C}{r} \\ E_{pot} = -Q\int_\infty^r dr' E(r') = \frac{C}{r} $$ Energy conservation holds:$E_{pot}+W=0.$ Depending upon how you define work i.e. if the the system "carries out the work" ...


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Yes, spring mass will increase but in very very small amount. As per Mass energy equivalence the amount of mass gain will be(E=mc2): m= E/C2 for example if spring have store 1 Joule of potential energy when it is extended then the increased mass will be: m =1/299792458= 3.33564095198152e-9


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Suppose we lift a particle by doing mgh work on the particle. How will this energy be stored in the particle? we normally lift a body (designated by a point on paper-called a particle) of mass m and do some work- It means something is operating to prevent this action /or oppose our action of 'lifting' that's why one has to do work- meaning thereby that ...



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