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Potential and potential energy are defined for pairs of objects, not individual objects. It's meaningless to say "the potential energy of A". One must say "the potential energy of the system consisting of A and B". There is only one potential and potential energy in your problem. Perhaps the confusion comes from the way potential is introduced in ...

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I think your problem is in how you understand the gradient. Define your z-axis. Lets say it points up. Now $$V(z) = mgz$$. Thus gradient of V(z), that is $$\nabla V(z) = mg \hat z$$. Here, $$\hat z$$ is an unit vector pointing up. So the force will be a vector of length mg, pointing down, because of the minus sign.

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Because of the Work-Energy Theorem. The work done on an object by an external force $\vec{F}$ is $$W=\vec{F}\cdot \Delta \vec{x},$$ where $\Delta \vec{x}$ is the displacement. The Work-Energy theorem says that the work is equal to the change in kinetic energy: $$\Delta K=W.$$ So the reason that the kinetic energy is changing for the particle is that ...

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The question is a bit vague, but the answer is yes. In Thermodynamics and Physics in general, due to the conservation of energy, work is always a transfer of energy between different systems. So we have to talk of what systems we are thinking of. Also, let's start with a simpler situation: consider just flexing your arm through your elbow, just like ...

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Put on your ice-skate go on some ice-sheet, hold your book and ask a friend to give you a push. Your friend will have to provide some work to make you move but once it is done you are sliding at a constant speed. This is an inertial motion: no force involved no acceleration constant speed. A displacement at constant speed do not require any force, only ...

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The reason is mainly in order to be able to write the total energy of the system as $$E=T+V$$ where $T$ is the kinetic energy of the system. It is far more useful to choose the signs of the terms in the total energy to be $+1$ once and for all, rather than setting the force equal to plus the gradient of the potential. The link between the conventions is ...

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The $dx$ is actually a vector $d\vec{x}$ (because $\vec{x}$ is actually a vector). A vector's sign depends on your coordinate axes, e.g. if you pick right to be positive then left pointing vectors are negative. In general, the work done $W$ is given by $$W = \int \vec{F} \cdot d\vec{x} = \int Fdx \cos(\theta)$$ where $\theta$ is the angle between ...

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This is because $P(t)$ as stated is the instantaneous power as a function of time and $W =\mathbf F\Delta \mathbf x$ holds only for constant forces. More generally, recall that a definition of work is the integral: $$W = \int_C\mathbf F(x)\mathrm d\mathbf x$$ Where $C = C(x,t)$ is some curve in space/time. Expressing in terms of time gives: $$W = ... 1 For a force to do work, it must act upon an object as that object moves some distance (with a component along or opposed to the direction of that force). Holding an object stationary, but carrying the strength of a given force, adds no more work done. Because of this, we say that$$W=\int_C\vec{F}\cdot d\vec{r}$$where C is the curve in space along which ... 1 Think about the problem from the point where you neglect the wire and just analyse the motion. You have gravity and at any other point the force has to balance gravity that you end up in a circular motion. In A its towards the origin whereas in B there is no force (except gravity). Different from the classical space-orbit or car-in-a-curve problem is ... 1 You should use 1. Conservation of energy and 2. Centripetal force to the problem. I think you can figure out that when the angle subtended by the bead along with x-axis is between 0 and sin^{-1}(2/3), the normal force acting on the bead due to the wire is inwards. Hence, the normal force on the wire would be outwards. It is an easy problem, so I am not ... 1 I came up with a solution seeing John Rennie's comment. The centripetal force, \vec F= -F \hat r  so infinitesimal work done by centripetal force,$$dW=\vec F.d \vec r= -F \hat r.d\vec r$$but, \hat r⊥d \vec r so$$dW=0$$is this correct ? 1 There is no conflict here. Let the two charged particles (M,Q) be the system with no external forces acting. Momentum is conserved and so for all time M_BV = M_A V_{Af} + M_BV_{Bf} Energy is also conserved and so for all time \frac 12 M_B v_B^2 + \dfrac{kQ^2}{R_i} = \frac 12 M_B v_{Bf}^2 + \frac 12 M_A v_{Af}^2 + \dfrac{kQ^2}{R_f} The ... 1 Using the first 2 statements or conditions, apply the principle of conservation of energy and you will be able to calculate the resistive force offered by the target material. Next use the second condition and again apply the principle of conservation of energy to get the required answers, mainly the velocity of the bullet after penetrating the target. 1 Power = \vec F \cdot \vec v = Fv if \vec F and \vec v are in the same direction. In this case the power at a given time is instantaneous force \times instantaneous velocity. In the first case you are finding the average power which you will note is equal to half the final instantaneous power which is second case value. 1 The point in which the friction acts is not actually moving, but rather as the object rolls past any given point, a new point of friction takes over. See in the animation below how a point on the circle touches briefly: 1 There are some missing words in your description, or the descriptions you've seen. In moving horizontally, gravity does no work because the force and displacement are perpendicular to each other. If you start from rest and apply a force with your hand, then your hand does work as long as your hand is applying a force. Once you reach constant velocity, ... 1 To back up CuriousOne. A good example would be you sumo wrestling with your friend. Lets say your friend is far bigger/stronger than you. So you begin pushing one another. His force on you is greater and so you are obviously moving in what you would call your -x direction. Both of you move in the -x direction, but of course your exerting a force on him too ... 1 I know this answer is pretty late, but I'm hoping it'll help at least a little. The idea here is, of course, using Green's Theorem:$$\oint \mathbf F \cdot \mathrm{d}\mathbf r = \int_A (\mathbf \nabla \times \mathbf F ) \; \mathrm{d}\mathbf a $$Here, you're asking about the line integral around a closed path. When using this equation, you end up making ... 1 How is KE accumulated and thus increasing? This is my primary concern. Also, what's happening with velocity? Is it increasing? Because if KE is increasing then velocity also should increase, right? Between x=0 and x=x_1, a net force, say F, is acting on the body. With Newton's Second Law we get:$$F=ma, with $a=\frac{dv}{dt}$ the acceleration ...

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I think what is missing is the definitions of $F$ and $V$. Consider first the definition of potential energy. The potential energy at a point relative to another point is the work done by a external force (eg the force exerted by you on the mass, $\vec F_{my}$) in taking the mass from the first point to the second point. That force which you exert on the ...

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