What Does -∂V/∂x Represent in Newton's Law?

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SUMMARY

The discussion centers on the interpretation of the expression -∂V/∂x within the context of Newton's laws of motion, specifically relating to potential energy. The equation F = ma is equated to -∂V/∂x, indicating that the force acting on an object is the negative partial derivative of potential energy V with respect to position x. The term -∂V/∂x represents the rate of change of potential energy as position changes, holding other variables constant. Understanding this concept is crucial for grasping the relationship between force and potential energy in classical mechanics.

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  • Understanding of Newton's laws of motion
  • Familiarity with derivatives and partial derivatives
  • Basic knowledge of potential energy concepts
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  • Explore the relationship between force and potential energy in classical mechanics
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rb120134
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Homework Statement
What does -∂V/∂x mean?
Relevant Equations
F= ma
In one of my textbooks about quantum mechanics, they mention a vehicle moving in a straight line along the x axis. With Newtons first law they take the second derivative from a which is
d^2x/dt^2 and that should be equal to
-∂V/∂x. What exactly does -∂V indicate?
The complete equation becomes
F=ma md^2x/dt^2= -∂V/∂x. It is a partly derivative. What does " -∂V/∂x" indicate, which units belong to -∂V and
∂x? They were talking about potential energy in the textbook.
 
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rb120134 said:
Homework Statement: What does -∂V/∂x mean?
Homework Equations: F= ma

In one of my textbooks about quantum mechanics, they mention a vehicle moving in a straight line along the x axis. With Newtons first law they take the second derivative from a which is
d^2x/dt^2 and that should be equal to
-∂V/∂x. What exactly does -∂V indicate?
It doesn't indicate anything! The "-∂V" alone means nothing without the accompanying "/∂x".

The complete equation becomes
F=ma md^2x/dt^2= -∂V/∂x. It is a partly derivative
"partial derivative", not "partly derivative"
What does " -∂V/∂x" indicate, which units belong to -∂V and
∂x? They were talking about potential energy in the textbook.
As I said above, "-∂V" and "∂x" have no meaning separately. "-∂V/∂x" is the "negative of the partial derivative with respect to x. You titled this "second derivative of Newton's Law" and mentioned "partial derivative". Can we assume that you know what "derivatives" and "partial derivatives" are? The derivative of a function is instantaneous rate of change of the function. If a function, like V, depends on more that one variable (here V is defined in "three space" so depends on the variables x, y, and z and might depend on time, t) then the partial derivative of V with respect to x is the instantaneous rate of change of V as x changes, all other variables held constant.
 
Hi! sorry for sending this so late, I hope it eventually gets to you but for the term "delV/delx" "V" would be the function of for the potential energy of the system with its partial derivative taken with respect to "x" representing it's position.
 

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