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Potential energy can be thought as the amount of work that the force can potentially do on the point because of its position. $$W=-\Delta U=U_{initial}-U_{final}$$

A positive work done by a force translates into a negative variation of potential energy. That sounds ok, given the interpretation of $U$ stated above. If a force does some work, then the "potentiality" of doing more must decrease.

But the equation says also that any time the force does a negative work, the potential energy increases. Why does this happen, in the light of such interpretation of $U$?

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    $\begingroup$ It's like bookkeeping: you debit the one, you must credit the other in order for the accounts to balance. In this case we want total energy to be unchanged. Energy is like money: it's an abstract system for keeping the books balanced. $\endgroup$ – Peter Diehr Apr 13 '16 at 21:54
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If a force does negative work, it is in fact trying to work against another force, doing positive work.

When you lift up a book from the floor, gravity does negative work on the book, while you do positive work. And the books rises higher up, so $U$ increases.

  • Negative work just means "receiving" instead of "giving" away energy. Which basicly is the same as saying that something else is doing work on it.

There is not more to it than that.

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It would be better to say that potential energy is the amount of work that a system can do. Say you have a system consisting of two masses - a brick and the Earth. As the brick moves down U decreases, the force pulling the brick and the Earth together acts downwards on the brick and it can do some positive work. On the other hand, if brick moves up, U increases but now the force of gravity on the brick acts in the opposite direction to its motion so it does negative work. You could imagine a stretched spring pulling the brick up. The brick would do negative work on the spring causing it to lose its U.

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It's a convention. The real reason is so that we can have:

$$\begin{align} \Delta {\rm KE} &= W_{\rm ext}\\ \Delta {\rm KE} &= W_{\rm con} + W_{\rm noncon}\\ \Delta{\rm KE} &= -\Delta {\rm PE} + W_{\rm noncon}\\ \Delta{\rm KE} + \Delta {\rm PE} &= W_{\rm noncon}\\ \Delta {\rm E} &= W_{\rm noncon} \end{align}$$

Which doesn't work without the sign choice. So it's really just an arbitrary convention to make it so you can add potential energy to kinetic energy in the work-energy theorem.

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  • $\begingroup$ You should add that it the making W=-delta P.E. makes sure that the law of conservation of energy is complied with when the work done by non-conservative forces is zero. $\endgroup$ – MrAP Oct 16 '16 at 14:32
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The decrease or increase in potential energy is converted as mechanical work either to increase or decrease it's kinetic energy (we are considering here conservative systems). Suppose you have a system at rest at some height above the ground level, say, a ball placed on the top of a mountain. The work done in order to bring the ball of mass m to a height h on the top of the mountain is given by mgh which is stored as the potential energy of the earth-ball system. However, the displacement of earth towards the ball (by Newton's third law) is so small that you can't possibly imagine. It will be less than the size of a proton!!!. Can you imagine that? Some huge body of radius 6000 km displacing about a distance comparable to the size of a proton. So we say that the work is entirely stored in the ball's potential energy.
Such a body is said to be in unstable equilibrium. The potential energy is maximum which indicate a less stable state as the potential energy is something that refers to a strain. Any disturbance will hence drive off the body to a state of lesser potential energy, thereby attaining stability. But the ball cannot destroy the excess energy. All it can do is to convert it's potential energy to kinetic energy, which makes the body move. Hence the potential energy decreases and the kinetic energy of the ball increases. An increase in kinetic energy means that a positive work is done on the ball by the force of gravity since both shares a common direction (or 0≤θ≤90). You can measure the kinetic energy of the ball by measuring the the work done on the ball (which is the work-energy theorem). So here we have a decrease in potential energy which corresponds to a positive work. This is very well valid with the equation.
Now we consider the second case. A ball is now raised to a height of h meters. Initially, the body was in stable equilibrium with a minimum of potential energy. Now we have increased it's potential energy. For that, we have to do work against the force of gravity. The force of gravity is acting vertically downwards while the displacement is vertically upwards(or 90≤θ≤180). So here the work done by the gravitational force on the ball is negative. But here we have increased the ball's potential energy, which is again valid as compared to the equation.
So an increase in potential energy corresponds to a negative work on the body while a decrease in potential energy corresponds to a positive work on the body. You should note that the sign convention is all based on the direction of displacement when compared to the force of gravity acting on it.

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It is just a convention. If we would use the opposite convention, we would get for conservative forces $$\vec F=+\vec\nabla U.$$ You can easily see (think in the one dimensional case) that the particle would move to points of maximal potential energy. This only sounds strange because we are used to the opposite. The mechanical energy principle would retain its form, $$\Delta E=W_{nonc},$$ as long as we redefine $$E=K-U.$$

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Potential energy of a body is its capacity to do work by virtue of its position in a conservative force field.So if the body is free to move it will do so in such a way as to reduce its potential energy and the reduction in the potential energy will be equal to the positive work done by the body which could result in the raising of a weight or displacement of an object. So -(change in U of the body)=Work done by the body.

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Work done is $\vec F \cdot \Delta \vec x$.
If $\vec F$ and $\Delta \vec x$ are in the same direction then the work done is positive.
If $\vec F$ and $\Delta \vec x$ are in opposite directions then the work done is negative.

Consider a spring fixed at one end as a system and an external force $\vec F$ stretching the spring.

If the external force $\vec F$ stretches the spring a little more then the work done by the external force is positive and the elastic potential energy in the spring has increased.

On the other hand with that external force $\vec F$ acting on the spring if the spring length is reduced then the external force has done negative work and the elastic potential energy of the spring has decreased.

Suppose that the external force was produced the gravitational attraction of a mass at the end of the spring.
Now assume that the mass and the Earth are the system.
The spring is producing an external upward force on the mass and Earth system.

If the spring contacts the spring has done positive work and the gravitational potential energy of the mass and Earth system has increased.

If the spring stretches the spring has done negative work and the gravitational potential energy of the mass and Earth system has decreased.

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protected by Qmechanic Apr 14 '16 at 0:59

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