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In one dimensional motion, sign really does mean direction for vectors. You can say positive is upwards and negative is downwards or vice versa. Now, come to your case. You seem to have mathematically understood. Then start with making your intuition. As stated by other answer, suppose you are very,very far away where gravitational force of the earth ...


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Here is a simple model as explanation: Imagine yourself far away from any gravitational field - your potential energy is zero. As soon as you are entering into a gravitational field, you are accelerating and winning kinetic energy. The origin of this energy is that you "borrowed" some energy which is the potential energy you are losing. When you want to ...


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Remember the definition of work: $\vec F.d\vec x=Fx\cos\theta$. In your case $\theta=180^o$, thus $\cos\theta=-1$. That is the minus sign you were missing.


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In your formula you don't calculate the potential energy, but the difference between the potential energy of the body at the lower level $P_E(h_0)$, and that at the higher level $P_E(h_1)$. Indeed, since $P_E(h_0) < P_E(h_1)$ you get $P_E(h_0) - P_E(h_1) < 0$ as expected. Phenomenologically, starting from $h_0$, you have to invest some work to raise ...


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I do not see any other possibility than doing the integral along the path. Do your telemetry data include the horizontal position? because if does you can calculate an approximation to the integral. Of course, there is no guarantee that the error will be larger than the effect you want to measure. You should have to do some pre-tests (using situations that ...


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It may help: suppose we are close to the earth and at height $h$. So $$\Delta V=Gm_1m_E(\frac{1}{R}-\frac{1}{R+h}) $$ where $R$ is the radius of the earth and $h \ll R$. Now we approximate this relation and it's turn out that $$\Delta V=Gm_1m_E(\frac1R-\frac1R+\frac{h}{R^2})$$ By calling $g=\dfrac{Gm_E}{R^2}$, we find $\Delta V=m_1 gh$. Even if we don't ...


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You have misconcieved the expression for work in your question as you've assumed $W$ as the total work on the object. It is actually $W_g=\Delta K + \Delta {P.E}$ ,where $W_g$ is the net gravitational work on the object. You see, $W=0$ while $W_g=-mg\Delta h$ in your case. And thus, $\Delta P.E=-W_g=mg\Delta h$


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If we estimate the size of the Earth to be a perfect sphere (which it isn't, but as a first approximation it will do), then you may apply the shell theorem. It states that the gravitational field of a spherically symmetric body appears as if it is concentrated in the center of mass of the body. Knowing how much Earth weighs ($5.97219\times10^{24}\,kg$), you ...


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Let us assume, you lifted the book infinitesimally slow and the velocity is zero so is the acceleration making change in kinetic energy zero but the work is never zero. It is mgh(m-mass of book, g- acc. of gravity, h-height to which the book is lifted). The work done by you carrying the book around, that will be zero since no change in height or speed of ...


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There are several possibilities for your confusion - and you bring up several related concepts (Work, potential energy, kinetic energy). When this happens I think it's helpful to isolate one thing which you know must, absolutely be true. The definition of work (for a constant force) is $$W=\vec{F}\cdot \Delta \vec{x}=F\Delta x \cos\theta.$$ First ask "what ...


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Doing work on an object will increase its kinetic energy, but it could also increase its potential energy and therefore not increase its $KE$. To raise a $4kg$ object by $25 meters$ you have to counteract the force of gravity. If your doing this with a force equal to $F_g$ then when you've lifted the object there will be no increase in $KE$. To raise a ...


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This is not quite a full answer to your homework problem here (our homework policy actually forbids detailed answers), but it should be enough to guide you to your answer. Note first that, formally, the solution to $$ cx^2+ca^2-2cx^2=0 $$ is $$ x=\color{red}{\pm} a $$ This indicates that there are two local minima/maxima points. Inserting these into ...


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The are no means to hold kinetic energy in MASSLESS spring. So, the potential energy of extension, will instantly converted to infinite speed of contracting spring. If spring is absolutely rigid itself, it will contract to equivalent minimal length and will start to extending back. Once it extends to initial length, it will start to contract back, and so ...


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Relation between Forces and Potential Energy Can you explain why can't we define potential energy corresponding to a non-conservative internal force? In order to examine the relation between two terms we must consider the definitions of each term: a) forces: 1) internal forces are those that act inside a body (note that in engineering also a ...


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The potential $V(r)=\int \frac{F(r)}{m} dr$ as integral of gravitational force $F(r)$ divided by mass $m$ is only determined up to a arbitrary constant. A satellite coming from infinity will get kinetic energy from falling to the gravity center. So it is a convenient convention to say the potential $V(r) = 0$ and therefore also the potential energy ...


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What they are basically telling is that the gravitational potential at an infinite distance from the fixed mass is zero. The total energy of the satellite would be K.E. + P.E. Since, $$V(r)=-\int{E(r)}$$ $$V(r)=\frac{-GM}{r}$$ With M being the mass of the fixed object. As r tends to infinity, it is clear that the potential of the system rapidly decreases, ...


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Work: $$ W = \int_\gamma\mathbf F\cdot\mathbf{dr} = \int_{t_0}^{t_1}\mathbf F(\mathbf r(t))\cdot\mathbf{r}'(t)dt $$ If it is conservative then $\mathbf F = -\nabla U$. Therefore: $$ W = \int_{t_0}^{t_1}\mathbf F(\mathbf r(t))\cdot\mathbf{r}'(t)dt = \int_{t_0}^{t_1} -\nabla U(\mathbf r(t))\cdot\mathbf{r}'(t)dt = -\int_{t_0}^{t_1}\left[U(\mathbf ...


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The relationship between the potential energy and what you call "conservative" force is: $${\bf F} = \nabla U = {\bf n} \frac{\partial U}{{\partial\bf n}}\tag{1} $$ where ${\bf n}$ is the direction normal to the equipotential surface passing through some point $P$. Then, the mechanical work $dW$ done by the force ${\bf F}$ along WHATEVER path $dL$ that ...


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If you have an example in mind I could point out what is the problem. But you can prove mathematically that if a force is not conservative then it cannot be written as the gradient of a potential. And the reverse is true too. You can show mathematically as a theoren that is a force is conservative, it can be written as the gradient of some potential.


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Potential Energy is negative of the work done by conservative forces. The reason why we define potential energy is so that we can get the result:: Work done on a body = change in potential energy of the body (if only conservative forces act on the body) this is proved by the very definition of potential energy. However let us say we were to define ...


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You can use a flywheel. When spinning in a vacuum, with magnetic bearings, flywheels lose very little energy over time. While they are probably best for large-scale applications, e.g. as backup for other power sources like solar energy, they have even been proposed as a power source for cars - although I assume there could be serious problems caused by the ...


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You can use compressed air energy storage.


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Depending on size, you could have a suspended weight and pulley system connected to a small generator ie use gravitational potential energy. Another would be an ultracapacitor topped up by a solar cell. If it is a very small power device you could consider energy harvesting technology - just about everything is listed in the wiki article


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The formula $mgh$ only works close to the Earth's surface where $g$ can be assumed constant. That is not the case here. The appropriate formula is more like $$ E = m \int g(r)\ dr $$ where $g(r)$ is gravity as a function of radius from the centre of the Earth. This is the work done in moving the mass $m$ in the gravitational field. If $g$ were constant and ...


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Q1: I am not sure what you mean, physically it means the maximization of the the distances, but you already knew that. Q2: it is equivalent, so it is still an open problem. Q3: the formula does not apply, you are correct, for n=3 the solution is that the electrons reside at the vertices of an equilateral triangle about a great circle. But reading the ...


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It is important to be clear what is doing work on what. In the diagram below, consider a charge starting at rest at A. We release it, and it accelerates down the potential gradient. Here, the field is doing work on the charge - it is accelerating it and therefore giving it kinetic energy. As a result, the charge will travel through the potential minimum at ...


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if we want to move a charge in an electric field then we need a work of an external force to move it from a low potential energy to a high one Not true. The electric field itself exerts a force to move the charge. Moving the charge from high to low does require work and this work is done by the electric field itself. Any force that causes movement is ...


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Take the example of gravity. Gravitational potential energy is defined as $$U_g=-\frac{GM}{r}$$ This means that $$U_g=-\int_{r_1}^{r_2}F_g \cos \theta dr$$ Because $F_g=mg$, $$U_g=-\int_{r_1}^{r_2}mg \cos \theta dr$$ Integrate and you find $$U=-mg\cos\theta \left (\left.r\right|_{r_1}^{r_2}\right)$$ and $$U=-mg\cos\theta (r_2-r_1)$$ and if the object is ...



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