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This is a very basic problem: (a) $F=-\frac{d U(x)}{dx}$. (b) You need to find the minimum of $U(x)$; (c) Sketch a graph. For all the questions you can use wolframalpha. If I understood your problem correctly then (a) and (b,c) .

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Potential energy is wrong. Even in Newtonian Mechanics it only works if the force is 1) purely a function of position and also 2) is conservative. Magnetic forces depend on velocity so they fail. And electric forces are not conservative if you aren't in statics. What you really have is kinetic energy and rest energy for the charged particle, some energy ...

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What is gravitational potential? Usually a potential is defined as the potential energy per mass or per charge or similar. This is most often seen in relation to electricity or chemistry and less often to gravity. $GPE$ is gravitational potential energy. $GP$ is gravitational potential energy per mass: $$GP =\frac{GPE}{m}$$ Is it defined for the system ...

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The total energy at the top is $$T = \frac{1}{2} m v^2 + m g L$$ The total energy at some other point is $$B = \frac{1}{2} m f^2 + m g L \cos\theta$$ Energy is conserved so $$f = \sqrt{v^2 + 2 L g (1-\cos\theta) }$$

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The question, as written, has no need for calculus. Find the volume of water needed to fill the frustum. http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html Find the weight of that water Find the work done to pump all that weight up 6 feet over the top of the wall of the tank, letting it splash down into the tank. Done.

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Asteroids are held together by a combination of gravity and cohesive (electromagnetic) forces (the same forces which hold rocks together on earth). For small asteroids (smaller than about 1 km), the gravity is negligible, while for larger bodies (larger than 10s to 100s of km) the cohesive forces become negligible*. If you're only interested in the larger ...

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Potential energy is the energy due to configuration of the system. If you keep three charges very, very far from each other, then the potential energy of the system is very effectively zero. But when you bring them close together to a specified coordinate, then the potential energy of the system increases from $0$ to a positive value given by $$U= ... 1 You can describe the electric force it terms of potential energy, because it is a conservative force. In doing so you actually replace the concept of work done by this force by the concept of potential energy. So you can not longer use both descriptions simultaneously. If you describe the electric force as doing work, then you made positive work and the ... 1 You are in your reasoning overlooking something. Look at the diagram below: -q_1,+q_2 are two point charges at distance r. Coulomb's Law dictates that the attractive electrostatic attraction force between them is:$$F=k_e\frac{|q_1q_2|}{r^2}$$And the electrostatic potential U(r):$$dU(r)=F(r)drU(r)=-k_e\frac{|q_1q_2|}{r}$$Assume now that ... 1 The system is subject to a non-zero net force in the horizontal direction and no friction, so it will experience constant acceleration (of the center of mass). Superimposed on that motion with be the anti-symmetric oscillation of the two masses on the spring. If the masses are both m and the spring is characterized by constant k the angular frequency of ... 1 I asked a somewhat different, yet similar question.Hope this helps! Why is an LC oscillator lossless, but C V^2 / 2 energy is lost to a capacitor connected to an ideal voltage source? 1 Your teacher is wrong. The gravitational potential V(x) is generally defined as potential energy per unit mass i.e. V(x) \equiv \dfrac{U(x)}{m}. So for the points where U(x) is zero, V(x) is zero and vice-versa by definition. EDIT: After you added the comment and a snapshot of the book, I realized your book has defined Gravitational Potential in a ... 1 Your copy of Verma has already defined gravitational potential energy previously in (11.3) The gravitational potential energy of a two particle system is$$U(r) = -\frac{Gm_1m_2}{r} \tag{11.3} where $m_1$ and $m_2$ are the masses of the particles, $r$ is the separation between the particles and the potential energy is chosen to be zero when the ...

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Don't forget about the rest energy, $E=mc^2$. Since a particle's intrinsic spin cannot be changed, it doesn't make sense to distinguish "intrinsic spin energy" $\frac12 I\omega^2$ (as you would compute for, say, a spinning flywheel) and the rest energy. A particle which is moving in an electric field will see a motional magnetic field. Charged particles ...

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