Why is the potential energy equals to the negative integral of a force? I am really confused with this negative sign. For example, why there is a negative sign in the gravitational potential energy and what does it mean?

I read that the negative sign means you do the same force but in the opposite direction. Doesn't that mean the object shouldn't move?

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    $\begingroup$ More on sign conventions and potential energy. $\endgroup$
    – Qmechanic
    Commented Aug 1, 2016 at 14:50
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    $\begingroup$ I#m not exactly sure for what kind of explanation you're looking, since in the end, the signs are mostly a consequence of definition: If we define potential energy such that everything moves towards the lowest potential energy, then you must introduce a minus sign because the derivative points away from minima, so its negative points towards it. Is that what you're looking for or something else? $\endgroup$
    – ACuriousMind
    Commented Aug 1, 2016 at 14:54
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    $\begingroup$ So the potential goes down when a ball rolls down a hill. $\endgroup$
    – knzhou
    Commented Aug 1, 2016 at 15:44
  • $\begingroup$ "I read that the negative sign means you do the same force but in the opposite direction. Doesn't that mean the object shouldn't move?" Not really relevant to the answer (or in a way, it is), but please remember that a zero net force means the object should move with a constant velocity, not that it shouldn't move. $\endgroup$
    – JiK
    Commented Aug 1, 2016 at 21:19
  • $\begingroup$ Another related (duplicate?) question: Why is gravitational potential energy negative, and what does that mean? $\endgroup$ Commented Aug 1, 2016 at 22:14

5 Answers 5


When you do conservative work on an object, the work you do is equal to the negative change in potential energy $W_c = - \Delta U$. As an example, if you lift an object against Earth's gravity, the work will be $-mgh$. Gravity is doing work on the object by pulling it towards the Earth, but since you are pushing it in the other direction, the work you do on the box (and therefore the force) is negative. The field does negative work when you increase a particle's potential energy.

Mathematically, it is just that $F=\frac{dW}{dx}$, which means that if the work is conservative, then $F=\frac{-dU}{dx}$, since $W_c = - \Delta U$. Then $-dU = Fdx$, so $U = - \int F dx$.

We can also say that work is negative when the force and displacement are in opposite directions, since $W = \vec F \cdot d\vec x = Fdxcos\phi$. When $\phi=\pi$, then $\cos\phi = -1$. An example of this conceptually is friction. An object sliding down a plane has kinetic friction acting on it. The friction is in the direction (up the ramp) opposite to the object's motion/displacement. So we say that the friction force is doing negative work.

  • $\begingroup$ So the work done by a field is always positive and any external agent in the opposite direction is negative? If so, y did we thought of it in that way, i mean the directions? Plus, why potential energy is the negative of work done by gravity? Sorry for disturbing you. $\endgroup$
    – Omar Ali
    Commented Aug 1, 2016 at 15:33
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    $\begingroup$ Good questions. Saying the field always does positive work can create some misconceptions, such as when dealing with the signs of electric charges, or just realizing that this applies to elastic forces (where we don't usually consider an elastic field). We say that negative work increases potential energy because as you lift the object, your mechanical energy is being turned into gravitational potential energy, so you are losing energy. It's purely about considering each force, the work that force does, and the direction of $\vec F$ and $d\vec x$. $\endgroup$
    – zh1
    Commented Aug 1, 2016 at 15:55
  • $\begingroup$ We say that negative work increases potential energy because as you lift the object, your mechanical energy is being turned into gravitational potential energy, so you are losing energy. It's purely about considering each force, the work that force does, and the direction of F⃗ and dx⃗ . I am having trouble understanding this line...Can't we just say that the negetive work is the additional energy required for the potential to increase ie Ui+Work=Uf $\endgroup$
    – user184271
    Commented Feb 24, 2018 at 1:09
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    $\begingroup$ @ZackHutchens I came across this answer as it was cited elsewhere. I can't agree with the following statement: "You do negative work when you increase a particle's potential energy". When you lift an object the force you apply is in the same direction as the displacement of the object. The work is positive. At the same time, however, gravity does an equal amount of negative work on the object since its force is in the opposite direction of the displacement. Gravity takes the energy you supplied to raise the object and stores it as gravitational potential energy of the earth/object system. $\endgroup$
    – Bob D
    Commented Jul 30, 2019 at 14:00
  • $\begingroup$ @BobD you are correct. I have edited my response. $\endgroup$
    – zh1
    Commented Jul 30, 2019 at 19:19

It is all about finding or constructing conserved quantities.

When an object is under forces, in general the KE of the object is no longer a constant. But can we add something to it so that we have a conserved quantity again?

People derived that by Work-KE theorem $$\Delta KE = \int_{t_i}^{t_f} \textbf{F}_{net} \cdot \textbf{v} dt$$ where $$\textbf{F}_{net}=\textbf{F}_1+\textbf{F}_2+\cdots$$ is the net force acting on the object.

Then we found that for some force $$\int_{t_i}^{t_f} \textbf{F}_{k} \cdot \textbf{v} dt = \int_{\textbf{r}_i}^{\textbf{r}_f} \textbf{F}_{k} \cdot d\textbf{r}$$ which is path independent and called conservative forces. Forces don't satisfy this property are called non-conservative forces.

So we want to "move" these terms to the LHS and we have $$\Delta KE - \int_{\textbf{r}_i}^{\textbf{r}_f} \sum_{conservative} \textbf{F}_{k} \cdot d\textbf{r} = \int_{t_i}^{t_f} \sum_{nonconservative} \textbf{F}_{k} \cdot \textbf{v} dt$$

So we have the minus side because we "moved" them to the other side of the equation.

Now if we define $$PE_k(\textbf{r}_f)-PE_k(\textbf{r}_i)=-\int_{\textbf{r}_i}^{\textbf{r}_f} \textbf{F}^{conservative}_k\cdot d\textbf{r}$$ then we have $$\Delta KE + PE_1(\textbf{r}_f)-PE_1(\textbf{r}_i) + PE_2(\textbf{r}_f)-PE_2(\textbf{r}_i)+\cdots = \text{Work done by nonconservative forces}$$

If there are no nonconservative forces, or when the nonconservative forces do no work, then we have the conservation of energy, where total energy is defined as the sum of KE and PE.

Note that if you are fine to accept the total energy being KE - PE, then it is completely fine to define PE without the minus sign.

As for your last question, you can imagine that you apply a force which is just "slightly" larger than the conservative force. Then the object will move very slowly. When it is close to the final position, reduce your force so that it is just "slightly" less than the conservative force so that the object will slow down.


The potential energy (PE) is the kinetic energy (KE) that you could get out of an action if it were to go ahead and happen. The energy you put into a ball to move it up 10 feet is the same amount of energy you could get back out if you let it return to its starting point (under similar conditions).

***Note the words put in and get out. One of these actions will be considered negative, and the other positive. It doesn't matter which as long as you stay consistent with your signs.

From this example, the work done on the ball to move it up is the amount of energy it took to get it up there. And by definition, work equals the integral of force over distance. -> If we say that the energy we put in to the ball is positive work, then energy that we can get out of the ball will be negative work.

Or, instead of saying negative work, we could say its the work done by an opposite force to the original. Which leads to the integral of the negative force.

I hope this helps!

edit: slight wording adjustment


This question is very common and I was not satisfied with the other answers. Here is my take:

For a vector field $F$, a scalar potential $U$ can be defined in two ways:

  • $F=\nabla U$
  • $F=-\nabla U$

Both are valid and there are references which uses both forms. I will give you an example to illustrate why it makes sense to write this potential function with a negative sign.

Suppose that $F$ is a one-dimensional force, for example, the weight of an object, acting between $x_i$ and $x_f$:

enter image description here

Notice that $F$ is negative because we orient the height axis ($x$) upwards.

Now let's try to find an expression for the scalar potential $U$ using the first form, that is, $F=\nabla U$:

$$F=\nabla U$$ $$-mg=\frac{dU}{dx}$$

Integrating both sides from $x_i$ to $x_f$ gives us:


Notice that the left side is positive because $x_i>x_f$ (check the figure above). This implies $U(x_f)>U(x_i)$.

Now we got two choices:

  • Define $F=\nabla U$ and find $U(x_f)>U(x_i)$
  • Define $F=-\nabla U$ and find $U(x_i)>U(x_f)$

Potential energy is energy associated with the configuration of a system. Clearly, the configuration at $x_i$ has more potential energy (higher altitude), compared to the configuration at $x_f$. So it makes sense to use the negative sign convention, so that $U(x_i)>U(x_f)$.


Potential energy is defined as the work which we have to do on the system to bring it into a certain configuration at rest from another configuration at rest in which it has zero potential energy - ie in which there are no forces acting. If you think in terms of this definition you will know which sign to use, and what the sign means.

So if we have to push against a repulsive force - eg to bring one +ve electric charge close to another - then the force $F$ which we supply is in the same direction as the displacement $dx$. The work we have done is +ve and the potential energy of the system is +ve. If the force varies with distance then we have to integrate $F dx$. There is no minus sign here. The potential energy is +ve.

However, in the case of an attractive force - eg between 2 masses, or between +ve and -ve charges - we have to push in the opposite direction to the displacement in order to keep the objects from crashing into each other as we allow them to approach. Now the work done, and the potential energy, are -ve. Instead of us having to do work on the system, the system has done work on us.


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