Principle of Physics: Derivation of -dU/dx=F

In summary: U(x)=-\frac{m}{2} \int_a^b \left [\frac{m}{2} \dot{x}^2 + U(x_0) \right]$$In summary, the definition of work is different from the usual integral version. To calculate the derivative, you take a limit as the force gets smaller and smaller.
  • #1
mark2142
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Hi, Everyone! This is the page(first image) from Principle of physics by resnik.
I want to ask the definition of work(##W=F(x) \Delta x##) by variable force here is somewhat different from the usual integral version. I don't understand how is this valid definition?
Secondly, how did they reach to the derivative F(x)=-dU/dx by limit?
(There is another approach to proving this by taking potential energies and differentiating them which results in the field force involved. This is given in my another book which is to understand(second image)).
 

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  • #2
This is indeed a bit confusing. At best the conclusion is valid for 1D motions along a fixed "##x## axis", because you can define
$$V(x)=-\int_{x_0}^x \mathrm{d} x' F(x').$$
Supposed ##F(x)## is continuous, then ##F(x)=-V'(x)##, i.e., the force ##F(x)## always has a potential.

This doesn't hold true for 3D motion/forces. There the existence of a potential is constraining the kind of forces more. In this case you must necessarily have ##\vec{\nabla} \times \vec{F}=0## if there's a potential such that ##\vec{F}=-\vec{\nabla} V##. If this constraint is fulfilled within a simply connected region of space, then there's a potential, and it's calculated by taking the line integral from any fixed position ##\vec{x}_0## to the point ##\vec{x}## within this region along an arbitrary path, and the potential is independent of the choice of this path.
 
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  • #3
vanhees71 said:
This is indeed a bit confusing.
Okay!
vanhees71 said:
At best the conclusion is valid for 1D motions along a fixed " axis", because you can define

Supposed is continuous, then , i.e., the force always has a potential.
You mean we can define a potential as ##U(x)=-\int F(x) \, dx + U(x_0)## and then differentiate it we get ##-\frac{dU}{dx}=F(x)##.
But how did you reach to the conclusion that ##U(x)=-\int F(x) \, dx + U(x_0)## where ##U(x_0)## is constant?
 
  • #4
You just use the fundamental theorem of calculus, i.e., that integration is the "inverse" of differentiation.
 
  • #5
mark2142 said:
Okay!

You mean we can define a potential as ##U(x)=-\int F(x) \, dx + U(x_0)## and then differentiate it we get ##-\frac{dU}{dx}=F(x)##.
But how did you reach to the conclusion that ##U(x)=-\int F(x) \, dx + U(x_0)## where ##U(x_0)## is constant?
If ##U(x)## represents potential energy and the work done by a position-dependent force is ##F(x)dx##, then $$\int_a^b F(x)dx = KE(b) - KE(a)$$I.e. the integral represents the change in Kinetic Energy. Then, by conservation of total mechanical energy (KE+PE) that integral must also be the negative change in PE. So that $$\int_a^b F(x)dx = U(a) - U(b)$$ And now we can apply the fundamental theorem of Calculus to get $$F(x) = -\frac {dU}{dx}$$
 
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  • #6
PS note that this only defines the potential energy up to a constant as the derivatives of ##U(x)## and ##U(x)+ U_0## are equal.
 
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  • #7
PeroK said:
by conservation of total mechanical energy (KE+PE) that integral must also be the negative change in PE. So that And
But how did you reach to the conclusion that ##U(x) - U(x_0) =-\int F(x) \, dx ## or ##\Delta U=-W##?( In book its given $$W=\int_{x_0}^x F(x) \, dx$$
We need to use a conservative force such as Gravity-
$$\int_{x_0}^x (-mg) \, dx$$
So, $$-mg \int_{x_0}^x \, dx$$
$$-(mgx-mg{x_0})$$
So we define mgx as potential energy )

Is this the proof of ##-W=\Delta U## ?
 
  • #8
If there is only PE and KE in a system and energy is conserved then ##\Delta KE = -\Delta PE##.
 
  • #9
PeroK said:
energy is conserved then ##\Delta KE = -\Delta PE##.
Energy conservation comes after we have ##W=\Delta Ke## and ##-W=\Delta U## ?
 
  • #10
mark2142 said:
Energy conservation comes after we have ##W=\Delta Ke## and ##-W=\Delta U## ?
You just need:$$E = KE + PE$$ where ##E## is constant. That is conservation of total mechanical energy.
 
  • #11
PeroK said:
That is conservation of total mechanical energy.
I know what it is. And that is derived in the book as ##W=\Delta KE## and ##W=-\Delta U##. So, ##\Delta KE + \Delta U=0##. So ##KE+U=constant##.

But where has ##W=-\Delta U## come from?
Is it so obvious or done experimentally?
 
  • #12
Just by definition you have
$$U(x)=-\int_{x_0}^x \mathrm{d} x' F(x') \; \Rightarrow \; F(x)=-U'(x).$$
Now you use it in the equations of motion
$$m \ddot{x}=-U'(x).$$
Multiply with ##\dot{x}## you get
$$m \dot{x} \ddot{x}=\frac{m}{2} \frac{\mathrm{d}}{\mathrm{d} t}(\dot{x}^2) =-\dot{x} U'(x)=-\frac{\mathrm{d}}{\mathrm{d} t} U(x).$$
Combining both sides of the equation gives
$$\frac{\mathrm{d}}{\mathrm{d} t} \left [\frac{m}{2} \dot{x}^2 + U(x) \right]$$
or, integrated,
$$\frac{m}{2} \dot{x}^2 + U(x)=E=\text{const}.$$
 
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  • #13
PeroK said:
If there is only PE and KE in a system and energy is conserved then ##\Delta KE = -\Delta PE##.
O yes! My mistake. I misunderstood it. The Energy is always conserved. And so there is a potential energy defined in the process. But potential energy is defined only if the force is conservative in nature. Why?
 
  • #14
mark2142 said:
O yes! My mistake. I misunderstood it. The Energy is always conserved. And so there is a potential energy defined in the process. But potential energy is defined only if the force is conservative in nature. Why?
The concept of potential energy is useful when the force is conservative. Otherwise, it cannot be a function of position but also depends on the path taken.

You could try to define a potential for the friction force and see what you get.
 
  • #15
vanhees71 said:
Just by definition you have
$$U(x)=-\int_{x_0}^x \mathrm{d} x' F(x') \; \Rightarrow \; F(x)=-U'(x).$$
Now you use it in the equations of motion
$$m \ddot{x}=-U'(x).$$
Multiply with ##\dot{x}## you get
$$m \dot{x} \ddot{x}=\frac{m}{2} \frac{\mathrm{d}}{\mathrm{d} t}(\dot{x}^2) =-\dot{x} U'(x)=-\frac{\mathrm{d}}{\mathrm{d} t} U(x).$$
Combining both sides of the equation gives
$$\frac{\mathrm{d}}{\mathrm{d} t} \left [\frac{m}{2} \dot{x}^2 + U(x) \right]$$
or, integrated,
$$\frac{m}{2} \dot{x}^2 + U(x)=E=\text{const}.$$
As I understand you defined a potential energy and showed that the KE + PE=constant which it should be.
 
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  • #16
mark2142 said:
O yes! My mistake. I misunderstood it. The Energy is always conserved. And so there is a potential energy defined in the process. But potential energy is defined only if the force is conservative in nature. Why?
If the force cannot be written as the gradient of a potential, then there is no well-defined potential energy. Then all you have is the "work-energy relation",
$$T=\frac{m}{2} [\vec{v}(t_2)^2-\vec{v}^2(t_1)] = \int_{t_1}^{t_2} \mathrm{d}t \vec{v}(t) \cdot \vec{F},$$
where ##\vec{x}(t)## is the trajectory of the particle, i.e., the solution of the equations of motion.
 
  • #17
PeroK said:
The concept of potential energy is useful when the force is conservative. Otherwise, it cannot be a function of position but also depends on the path taken.

You could try to define a potential for the friction force and see what you get.
If I understand it correctly then,
From ##\Delta KE= W##
##\Delta KE= -\mu mg \Delta x## where ##\Delta x= x- x_0##
Since Change in kinetic energy ##\Delta KE## is -ve. So, it decreases.

Now ##W=-\mu mg \Delta x##
##W=-(\mu mg x - \mu mg x_0)##
##-W=\mu mg x - \mu mg x_0##
We define ##\mu mg x=U## potential energy.
##-W=\Delta U##
-ve work causes increase in potential energy which depends on the ##\mu## of the path and x.
 
  • #18
mark2142 said:
If I understand it correctly then,
From ##\Delta KE= W##
##\Delta KE= -\mu mg \Delta x## where ##\Delta x= x- x_0##
Since Change in kinetic energy ##\Delta KE## is -ve. So, it decreases.

Now ##W=-\mu mg \Delta x##
##W=-(\mu mg x - \mu mg x_0)##
##-W=\mu mg x - \mu mg x_0##
We define ##U=\mu mg x## potential energy. -ve work causes increase in potential energy which depends on the ##\mu## of the path and x.
That's not potential energy in any shape or form.
 
  • #19
PeroK said:
That's not potential energy in any shape or form.
Can you tell me why?
 
  • #20
Indeed the potential of the force, ##U##, usually is also called "potential energy", and
$$E=T+U=\text{const}.$$
For a constant force ##F## you indeed get for the potential
$$U(x)=-\int_0^x \mathrm{d} x' F=-F x.$$
 
  • #21
mark2142 said:
Can you tell me why?
An object at rest on a flat frictional surface has no potential energy dependent on its position. There is certainly not a linear energy profile.
 
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  • #22
PeroK said:
An object at rest on a flat frictional surface has no potential energy dependent on its position. There is certainly not a linear energy profile.
Ok! According to the ##W=\mu mg \Delta x## the work by the friction force is also dependent on the surface(##\mu##) i.e. Non conservative in nature so we cannot define a potential energy function just depending on the position of the body. If we can't define the potential energy then the mechanical energy is not conserved for non conservative force. Yes?
 
  • #23
mark2142 said:
If we can't define the potential energy then the mechanical energy is not conserved for non conservative force. Yes?
Yes, that's the whole point.
 
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  • #24
mark2142 said:
But where has ##W=-\Delta U## come from?
Is it so obvious or done experimentally?
In the work-energy theorem, you can separate the work done by conservative forces, which allow for potential energies, from the work done by non-conservative forces, which don't, i.e.,
$$\Delta KE = W_c + W_{nc}$$ Because the work done by conservative forces only depend on the endpoints, we move it over to the lefthand side of the equation and introduce the concept of potential energy.
$$\Delta KE + \underbrace{(-W_c)}_{\Delta U} = W_{nc}$$ So ##W_c = -\Delta U## is pretty much a matter of definition.
 
  • #25
Thank you guys...
 

FAQ: Principle of Physics: Derivation of -dU/dx=F

1. What is the principle of physics behind the derivation of -dU/dx=F?

The principle of physics behind the derivation of -dU/dx=F is the conservation of energy. This principle states that energy cannot be created or destroyed, only transferred from one form to another. In this case, when a force F acts on an object, the potential energy U of the object changes, and this change is equal to the negative of the work done by the force F.

2. How is the principle of conservation of energy related to -dU/dx=F?

The principle of conservation of energy is related to -dU/dx=F because it governs the relationship between potential energy and work. The change in potential energy (-dU/dx) is equal to the negative of the work done by a force (F). This means that the total energy of a system remains constant, as energy is transferred between potential and kinetic forms.

3. What does the negative sign in -dU/dx=F represent?

The negative sign in -dU/dx=F represents the direction of the force and the displacement. In this equation, the force F and the displacement x are in opposite directions. This negative sign indicates that the force is acting in the opposite direction of the displacement, resulting in a decrease in potential energy.

4. How is the principle of conservation of energy applied in the derivation of -dU/dx=F?

The principle of conservation of energy is applied in the derivation of -dU/dx=F by considering the work-energy theorem. This theorem states that the work done by a force is equal to the change in kinetic energy of an object. By applying this theorem, we can equate the work done by the force F to the change in potential energy (-dU/dx), thus deriving the equation -dU/dx=F.

5. Can the principle of conservation of energy be applied in all scenarios involving forces and potential energy?

Yes, the principle of conservation of energy can be applied in all scenarios involving forces and potential energy. This principle is a fundamental law of physics and is applicable in all scenarios, regardless of the specific forces or potential energy involved. It is a universal principle that governs the behavior of energy in all physical systems.

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