# Why is U(x) = -du/dx for an object in motion?

• negation
In summary, the conversation discussed the relationship between potential energy U(x) and force F, where F is the negative gradient of U(x). This relationship is expressed mathematically as F = -dU/dx, with U being the negative integral of F. The minus sign is necessary for conservation of energy to work. The conversation also touched on the difference between conservative and non-conservative forces and the importance of properly explaining symbols when discussing physics concepts.
negation
I couldn't manage to find substantial contents on the web explaining why U(x) = -du/dx.
But in thinking about why U(x) = -du/dx, I constructed an x-y axis where the x represents the displacement and y represents the potential energy U(x).

Suppose I started with an object moving down slope where m = -ve at an Θ where Θ< 90°. Then as time progress the object continues to move in the +ve x-direction while potential energy decreases. The converse holds true if the object were to move up wards where m = +ve at the same angle. Potential energy U(x) would increase as it moves in the +ve x-direction.

From this, it can then be seen mathematically that U(x) = -du/dx for an object in the first case. In the second case, the negative sign in U(x) would be replaced with a positive.

But what happens when the object is fell at an angle Θ = 90° to the x-axis? It is fairly obvious the the answer would be a real number/ 0 but would this scenario make any mathematical sense?

OK, U(x) is potential energy. What is u?

jtbell said:
OK, U(x) is potential energy. What is u?

u is the force?

Edit: thinking further it does make sense. Force is the derivative of PE.
U(x) =du/dx
If this is true, all that's left is to understand the existence of the -ve sign

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negation said:
Force is the derivative of PE.

Then it should be F = -dU/dx.

U is the (negative) integral of F: ##U(x) = - \int {F(x) dx}##

If this is true, all that's left is to understand the existence of the -ve sign

The minus sign is needed in order to make conservation of energy work. As the force accelerates the object (increasing its kinetic energy), the potential energy must decrease, and vice versa.

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Well, it may be a good a idea to figure out what "u" is.
If it's a force then the relationship is not true so it does not make sense to try to understand it.

You may be mixed up.
There is a relationship between force and potential energy. The force is the negative gradient of the potential energy. Is this what you had in mind?

nasu said:
Well, it may be a good a idea to figure out what "u" is.
If it's a force then the relationship is not true so it does not make sense to try to understand it.

You may be mixed up.
There is a relationship between force and potential energy. The force is the negative gradient of the potential energy. Is this what you had in mind?

Yes, what you suggested was one of the mental bits I had in mind. As the gradient of the PE is negative, force is positive-and vice versa. Why does this relationship holds? Could this be expressed in mathematical language ?

jtbell said:
Then it should be F = -dU/dx.

U is the (negative) integral of F: ##U(x) = - \int {F(x) dx}##

The minus sign is needed in order to make conservation of energy work. As the force accelerates the object (increasing its kinetic energy), the potential energy must decrease, and vice versa.

And then it makes sense. As an object travels down slope, towards +ve x-direction, PE decreases as KE increases.
And should I want to express the change in energy in terms of KE instead of PE.
It would then be F = dK/dx which suggests that as the object tends towards the +ve x-direction as it slopes down, KE increases.

On a tanget, the difference between conservative forces and non-conservative forces is that the former path independent. In other words, the way I think about it, it is dependent on displacement rather than distance as non-conservative force depends on.

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In the future, please never simply write down a symbol without giving proper explanation of what the symbol represents. This is not only a bad practice here in this forum, but a bad practice when you are trying to learn something.

Zz.

ZapperZ said:
In the future, please never simply write down a symbol without giving proper explanation of what the symbol represents. This is not only a bad practice here in this forum, but a bad practice when you are trying to learn something.

Zz.

Understood.

## What is the meaning of U(x) = -du/dx?

U(x) = -du/dx is an equation used in calculus to represent the potential energy of a system. U(x) represents the potential energy at a specific position x, while du/dx represents the rate of change of potential energy with respect to position.

## How is U(x) = -du/dx used in physics?

In physics, U(x) = -du/dx is used to calculate the potential energy of a system, which is the energy stored within a system due to its position or configuration. This equation is commonly used in fields such as mechanics, electromagnetism, and thermodynamics.

## Can U(x) = -du/dx be negative?

Yes, U(x) = -du/dx can be negative. This indicates that the potential energy of the system at a specific position is decreasing as the position increases. It is important to note that the absolute value of potential energy is not physically meaningful, only its differences between different positions are.

## What is the relationship between U(x) = -du/dx and the force on a system?

According to the fundamental theorem of calculus, the force on a system is equal to the negative derivative of the potential energy with respect to position, or F = -dU/dx. This means that if the potential energy is known, the force on a system can be calculated.

## Can U(x) = -du/dx be used for systems with more than one dimension?

Yes, U(x) = -du/dx can be used for systems with more than one dimension. In this case, the equation becomes U(x,y,z) = -grad(U) where grad(U) is the gradient of the potential energy function. This allows for the calculation of potential energy in systems with multiple variables or dimensions.

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