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When they say "Do not ignore electric force", they mean that there is both a magnetic and an electric force on the electron/positron, and you should not forget the electric force. In other words, you are asked to compute, for the $\vec v_+$, $q_+$ of the positron, the effect on the electron of its $\vec E$ and $\vec B$ field. Fortunately, 5 keV (kinetic ...


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In this case it's because of who's doing the work. The derivation you cite regards the work as being done by the gravitational field. This means in bringing the object from infinity, the gravitational field has lost that energy to the object. Conversely, the work done on the object is the positive value. Everything else works out and when we take the ...


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If we assume that mechanical energy (K+U) is conserved in both the earth frame and the initially co-moving, constant velocity frame, then it's not the differences in velocities which are the same; it's the differences in the squares of the velocities which are the same. $$\frac{1}{2}v_{1e}^2=\frac{1}{2}v_{0e}^2+2gh$$ and ...


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The problem lies in assuming that the plane maintains all its kinetic energy when it turns. How does a plane pull out of a dive? Roughly speaking, through elastic collisions with air molecules! The hidden actor here is the air. When the plane pulls up out of the dive, it's not true that $$\frac 12 v_1^2 = \frac 12 v_0^2 + gh$$ Because some of the ...


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If you take the ground as rest frame then v0 is the velocity of the plane relative to the ground. From this perspective the plane changes its velocity. You can also take the plane as your coordinate origin, then v0 is the velocity of the ground relative to the plane. From this perspective the ground changes its velocity. The change in the relative velocity ...


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The gravitational force acting on the particle as function of radius $r$ is [Gauss's law] $$F(r) = -\frac{GmM(r)}{r^2}$$ where $M(r) = \int_0^r4\pi \rho(r')r'^2dr'$ is the mass contained within a radius $r$. Note that for $r>R$ we have $\rho(r) = 0$ and $M(r) = M(R) \equiv M$ the total mass of the whole sphere. The potential energy is the work needed ...


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I believe we must trash the energy approach. The fundamental assumption in this question is that the forces will balance at the same point that the energy balances. This isn't necessarily true. I would call upon a mental model of a mass displaced on a spring. It starts at rest, but not at the neutral-force position. What happens? It oscillates. The same ...


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The electrostatic potential is a potential just like any other mechanical potential. Since the force on a particle with charge $q$ is $\vec F_\text{stat} = q\vec E$, the potential $\phi'$ that gives the force as $\vec F_\text{stat} = - \vec \nabla \phi'$ is just the electrostatic potential (the "voltage") times $-q$. The electrostatic force is a perfectly ...


1

Momentum is be conserved iff the Hamiltonian has translational symmetry. Usual boundary conditions such as homogeneous Dirichlet or Neumann conditions don't allow for such symmetry. But there still are specific conditions, which do allow the Hamiltonian to have translational symmetry on the bounded domain: Born—von Karman boundary conditions. Thus in the ...


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$U$ in your equation in potential energy, and $W$ is internal work. That is, the work done by forces within the system. The system in question here comprises the object and the earth, and the internal force is gravity. The work that you do to lift an object is external to the system, and does not appear in your formula. In your scenario, lifting an ...


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It would help me, but even more importantly you, if you defined what your symbols are supposed to mean. You did that with $U_i$ and $U_j$. So let's be precise together, as an exercise: Let's explicitly state that we consider a point mass $m$ in a gravity field caused by a much larger mass $M_E$ at the origin of the coordinate system, so that we can assume ...


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This would mean that the eigenbasis of a physical observable is not orthogonal. Is there an error in my derivation, and if not, how can this be understood physically? The set of eigenfunctions of $\hat p$ in the sense $$ \hat{p}\phi = p\phi $$ is sure to be orthogonal if they belong to a subset of $L^2((0,1))$ on which the operator $\hat{p}$ is ...


1

The answer is negative. The gravitational acceleration is the same for all the atoms. Thus, the internal structure of the body is not affected. Using the terminology in your question, the average kinetic energy of atoms inside this object doesn't change. This is the reason for which given a body left free in the gravitational field, we usually consider the ...


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Yes. The kinetic energy from translatoric motion of the object rises as a whole. Any small piece of it contributes with its' own fractional mass, since any small piece also achieves a net translatory motion. Each piece is a fraction of the whole object: $$K_{trans}=½m_{total}v^2=\sum K_{trans,atom}$$


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For a rotating ball you should use for energy $$E=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2+mgh$$ where $I$ is the moment of inertia which for a sphere is $$I=\frac{2}{5}mr^2$$ where $r$ is the radius of the sphere, and $\omega$ it's his angular velocity which is related to the velocity of the center of mass via the equation $$\vec{r}\times \vec{\omega}=\vec{v}$$ ...


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Let me assume for simplicity that we have some big charge(s), and we bring a positive test charge. So, there is a field between the big charge(s) and the test charge. The field encapsulates energy. (For simplicity I will also assume that the big charges are placed on massive bodies.) Now, for out test-charge there are two situations of passing from one ...


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If you move an object, then work $W$ is done when ever you use a force $F$ to move something over a distance $d$. $$W=F \cdot d$$ The electric field that defines the potential difference is a force (per charge) that pulls (or pushes) in the test charge you put in. A force is moving the charge, so work is done. If a force does work on an object, you have ...


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As you quoted, the Law of Conservation of Energy states that Isolated Physical systems have their energy conserved. But, nobody says that a non-isolated system cannot have its energy conserved. The mass above the Earth is acted upon by a field which is conservative, the gravitational field. In a conservative field the total energy of a body is conserved, ...


1

All the known forces conserve energy, but they don't necessarily conserve energy in macroscopic modes. For instance friction takes some of the energy of macroscopic motion and coverts it into an increase in temperature (i.e. energy in microscopic modes). Total energy is conserved but energy that is useful at the human scale is not. Feynman is talking about ...


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There are macroscopic forces that admit no description in terms of a potential, for example, any friction force proportional to the velocity of a moving object as path-dependent integral, and is hence non-conservative. But we know the macroscopic description is not the fundamental description. In terms of the interaction of the constituents of matter, all ...


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Unfortunately, multivariable calculus isn't quite enough! Your goal is to solve a nonlinear system of ordinary differential equations, and that's a high goal. Anyways, there are solutions to this. See an example of an implementation (disclaimer: It's mine) here: http://mathandcode.com/kepler/ The most useful references I found for it were the books ...


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but if you can give me a hint that would be great. Since the force depends on the radial distance only and points towards the origin, the angular momentum (assumed to point along the $z$ axis) is conserved. This suggests that the appropriate coordinates are the spherical polar coordinates $(r, \phi)$ where $$x = r \cos \phi$$ $$y = r \sin \phi$$ In ...


1

I can integrate with respect to time, but that would give $$f(x,y)= \left\langle \frac{GMx}{(x^2+y^2)^{3/2}}t, \frac{GMy}{(x^2+y^2)^{3/2}}t\right\rangle$$ No it doesn't because both $x$ and $y$ are functions of time, so you cannot use this "simple integration" method. What you need to do is go back to the differential equation, $$ m\ddot{\mathbf ...


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You have acceleration, which is a time derivative of velocity. Integrating acceleration with respect to distance doesn't produce a work vector. You can find the potential energy $U(x,y)=GMm/r$ which has the property that its gradient is the force field. But the potential energy is a scalar, not a vector. Your integral with respect to time is not correct, ...


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Your derivation of $V$ is fine, the $E$ not so good. But for the potential you did something better. In fact, you computed that potential for any height $d$ above the rod. So compute $V$ with a $d=0$, then compute it with your actual $d$ and then use the two to get an electric potential difference. Use the charge of the small sphere and the electric ...


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From http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml, the electric field is: $$\begin{align}E_p&=\frac{Q}{4\pi\epsilon_0}\left(\frac{1}{b}+\frac{1}{L+b}\right)\\ &=\frac{Q}{4\pi\epsilon_0}\left(\frac{1}{0.5}+\frac{1}{2}\right)\\ &=\frac{2.5Q}{4\pi\epsilon_0}\\ \end{align}$$ Now, for the ...


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In the same way that the basketball on a shelf could move faster (meaning, could reach a higher speed) if it was placed on a higher shelf and rolled over the edge then at a lower shelf, the atoms have a longer distance to travel when they are further apart and so could move faster (could reach a higher speed if the were set free) than if they where closer. ...


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Your statement "but there needs to be KE for it to fall" is wrong. What it need is a force in that direction to fall. That force on it comes from the negative of the gradient(change with respect to distance in the maximum increase direction) of potential energy which is always present as long as the potential energy is changing with distance.


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An object at rest will start moving whenever there is acceleration. And there is acceleration whenever there is non-zero net force, according to $$\sum \vec F=m \vec a$$ As @ACuriousMind mentions in the comments, kinetic energy is a "result" of velocity, but velocity is not caused by kinetic energy. When you (in the comments to the question) say as ...


0

Writing down the relation between KE and PE and taking the time derivative yields: $$\frac{m\dot{z}^2}{2} - mgz = 0 \Rightarrow$$ $$m\dot{z}\ddot{z} - mg\dot{z} = 0$$ If we assume $\dot{z}$ is non-zero, we conclude that $\dot{z} = g$. However, if $\dot{z}$ is truly zero, this equation says absolutely nothing about what the body should do. Indeed, a totally ...


1

The answer was in fact covered by the Curious Mind, but for you to see the process how the potential energy transforms into kinetic, here is an elementary elaboration. The equation of motion in the gravitation field says that $ \ (1) \ h_0 - h = \int _0^t v(t) \ \text d t $ Multiplying this equation by $mg$ which is constant $ \ (2) \ E_P(0) - E_{P}(t) = ...


0

If the spring is stretching by an amount of $x$, its centre of mass is displaced by an amount of $x/2$.


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$x$ is the extension of the spring. If you extend the spring by a distance $x$, keeping the other end fixed, its centre-of-mass moves by $\frac{x}{2}$.



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