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Potential energy is defined like this. $ΔP.E=-W_{AB}$. This means that the potential energy at point A minus potential energy at point B should equal the negative of the work done by a conservative force from A to B. Take gravity (The localized one) for example, Let's define potential energy at A (a point on the ground) to be zero. Let's say you lift a ball of mass from the ground to a height of h. The work done by gravity is $-mgh$ due to the opposite direction. and the negative of that is just $mgh$. this should equal $P_a-P_b$ since $P_a$ is zero $-P_b=mgh$ so $P_b=-mgh$. How can this be? isn't it supposed to be $mgh$?

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  • $\begingroup$ what does 𝑊𝐴𝐵 equal? $\endgroup$ Nov 30, 2023 at 5:43
  • $\begingroup$ The force is the force you apply to the ball i.e. the upwards force $mg$. And since the displacement is also upwards the work done by you is $+mgh$ not $-mgh$. $\endgroup$ Nov 30, 2023 at 5:46
  • $\begingroup$ #PhysicsDave WAB means work done from point A to Point B $\endgroup$
    – Hammock
    Nov 30, 2023 at 6:59
  • $\begingroup$ #JohnRennie the work done by me is mgh but work done by gravity is -mgh $\endgroup$
    – Hammock
    Nov 30, 2023 at 6:59
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    $\begingroup$ The change in potential energy is the final potential minus the initial potential energy, or +mgh. That’s the negative of the work done by the conservative force, gravity $\endgroup$
    – Bob D
    Nov 30, 2023 at 11:31

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Without writing the explicit meaning of formulae, comments, and answers may only increase the original confusion.

The work done by a force ${\bf F}$ when a point-like body is displaced along any path from a point $A$ to a point $B$ is defined as $$ W(A \rightarrow B) = \int_A^B {\bf F}({\bf r})\cdot d{\bf r}. $$ With this definition, the class of conservative forces is the particular class of forces, such that the integral depends only on the initial and final point and not on the exact path followed for the displacement. As a consequence of the path-independence, the integral can be written as the difference of a function $\Phi({\bf r})$ between the two points: $$ \int_A^B {\bf F}({\bf r})\cdot d{\bf r} = \Phi(B)-\Phi(A). $$ For historical reasons, the potential energy has been defined as $$ U({\bf r})=-\Phi({\bf r}). $$ The final result of the two definitions (work and potential energy) is that $$ W(A \rightarrow B) = \int_A^B {\bf F}({\bf r})\cdot d{\bf r}=U(A)-U(B). $$ It is easy now to see that if we choose $A$ as the reference point ($U(A)=0$, the work done by the gravitational force when a mass $m$ is displaced at a height $h$ larger than the reference level, work is $-mgh$, and consequently $U(h)=mgh$.

A note about the possibility of defining work in terms of an additional external force ${\bf F}_{ext}$ instantaneously counterbalancing the force ${\bf F}$. From the algebraic point of view, it is perfectly permissible to define work in this way. Since by definition ${\bf F}_{ext}=-{\bf F} $ we can rewrite the previous formulae in terms of ${\bf F}_{ext}$. However, from the conceptual point of view, this is an undue complication: we want to define something that is a property of the force ${\bf F}$, and we can do it without needing an additional force.

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Your definition of potential energy is wrong. Yes, $ΔPE_{AB}=-W_{AB}$, but $W_{AB}$ is not the work done by the conservative force from $A$ to $B$. It is the work done by an external force (against the conservative force) to bring the object from $A$ to $B$, keeping the kinetic energy unchanged. This external force can be the force of your hand or any arbitrary agent. In your case, the work done by your hand to lift the ball a height $h$ is $+mgh$, which is indeed the change in potential energy.

Generally, the work done by the external force will be negative of the work done by the conservative force if there is no change in kinetic energy from $A$ to $B$.

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  • $\begingroup$ The potential energy is defined in terms of the work done by conservative forces and not of some other force that may be (or not) equal and opposite to the Conservative force. $\endgroup$
    – nasu
    Nov 30, 2023 at 11:56
  • $\begingroup$ @nasu They are equivalent definitions. See the accepted answer. It is up to the reader to decide which is more intuitive for their understanding. $\endgroup$ Dec 1, 2023 at 5:59

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