Work-KE theorem and net force....

[fog37]

Hello,
Just refining my understanding of the work-KE theorem and seeking some validation:

• The work-kinetic energy theorem states that the mechanical work done by a force (be it conservative or nonconservative) is always equal to the change in kinetic energy of the body.
• The net force in a dynamic problem is a force given by the vector sum including both conservative and non conservative forces. A sum of just conservative forces produces a net conservative force but a sum of conservative and nonconservative forces must certainly be a nonconservative net force, correct?
• ONLY in the case of conservative forces, the change in kinetic energy $\Delta{KE}$ can be equal to the negative change in potential energy $\Delta{PE}$: $\Delta{KE}= \Delta{PE}$, correct?
• A nonconservative force, when it performs nonzero work, can ONLY change the kinetic energy of the body but not the potential energy so it does not make sense attributing a change in $PE$ to a nonconservative force. Is that all there is not understand?
• Example: let's consider a rocket climbing vertically up at constant speed. There are two forces acting on the rocket: gravity pointing down and thrust $F_{thrust}$ pointing up. The thrust force pushes a rocket upward and performs work (force times distance). Gravity also performs work. But neither works causes a $\Delta{KE}$.
• The net force is zero, the net work is zero, so $\Delta{KE}=0$. The thrust force, when separately analyzed, does produce work but not a change in $KE$ nor $PE$. However, gravity produces no change in $KE$ but a change in $PE$.
• Potential energy, which is energy of configuration (relative position between system parts), is therefore a form of energy that can only be affected by conservative forces. Potential energy $PE$ only depends on the mutual distance between the system's components. So while kinetic energy is frame of reference dependent, $PE$ is not since the mutual distance between system parts is not affected by choosing different reference frames. Could we even talk about the potential energy of a system (not the change in $PE$) if there were no conservative forces acting on the system but only nonconservative forces?

Thank you!

 

[Doc Al]

• A nonconservative force, when it performs nonzero work, can ONLY change the kinetic energy of the body but not the potential energy so it does not make sense attributing a change in $PE$ to a nonconservative force. Is that all there is not understand?
• Example: let's consider a rocket climbing vertically up at constant speed. There are two forces acting on the rocket: gravity pointing down and thrust $F_{thrust}$ pointing up. The thrust force pushes a rocket upward and performs work (force times distance). Gravity also performs work. But neither works causes a $\Delta{KE}$.
• The net force is zero, the net work is zero, so $\Delta{KE}=0$. The thrust force, when separately analyzed, does produce work but not a change in $KE$ nor $PE$. However, gravity produces no change in $KE$ but a change in $PE$.

I find these statements confusing. Example: I lift a book. The positive work I do increases the PE of the book. It's also true that since there's zero net work done on the book (in the idealized case), there's no change in KE.

And if I just drop the book, then gravity is certainly increasing the KE of the book. Or you can just say that gravitational PE is being "converted" to KE.

Realize that when you talk about gravitational PE you are already including the effect of gravity, so you don't then also talk about the work done by gravity. (Those are two ways of describing the same thing.)

 

[fog37]

Thank you Doc Al.

In summary:

a) A force does work $W$ when it manages to produce displacement $\Delta s$ for the body.
b) The work-KE theorem states that work $W = \Delta KE$ but this is true only for the net force $F_{net}$.
c) For conservative forces ONLY, the change in $\Delta KE=-\Delta PE$ so $W_{conservative}=-\Delta PE$ as you mention.
d) For nonconservative forces, $W = \Delta KE$ and we cannot talk about $PE$ or its change $\Delta PE$ in the context of nonconservative forces.
d) Regardless of the involved forces performing work $W$, the work-KE truly applies to the net force $F_{net}$, not the individual force. The net force work is equal to the change in $KE$. In the constant velocity example, forces may be performing work but they don't cause a change in kinetic energy. The net force is zero, hence its work is zero, and $\Delta KE=0$.

 

[jbriggs444]

Thank you Doc Al.

In summary:

a) A force does work $W$ when it manages to produce displacement $\Delta s$ for the body.

A force does work when it is accompanied by a displacement for the body. Cause does not enter in.

 

[fog37]

Yes, there can be displacement without having a force or forces involved (for example, constant velocity motion).

I am trying to make sure that I correctly understand that nonconservative and conservative forces, even in the presence of nonzero displacement, don't necessarily produce a change $\Delta KE$ for the body unless these force give rise to a nonzero net force $F_{net}\neq 0$.

 

[jbriggs444]

Yes, there can be displacement without having a force or forces involved (for example, constant velocity motion).

I am trying to make sure that I correctly understand that nonconservative and conservative forces, even in the presence of nonzero displacement, don't necessarily produce a change $\Delta KE$ for the body unless these force give rise to a nonzero net force $F_{net}\neq 0$.

Right. Conservative versus non-conservative is irrelevant. You find the net force. You find the incremental displacement. If the dot product of the two is zero, there is no change in KE (for a point-like object). If the dot product of the two is non-zero, there is a change in KE.

The caveat about "point-like objects" is to dodge the complexities associated with non-rigid or rotating bodies.

 

[fog37]

Great, thank you. Why do you specify "for a point-like object"? What if we were dealing with a rigid extended object? Wouldn't the work-KE theorem not apply?

 

[fog37]

Also, a nonconservative force is one for which work depends on the path taken and the work done by a nonconservartive force changes (adds or removes) the mechanical energy $ME = KE+ PE$ of the body. Work done by a conservative force does not change $ME$.

What has been confusing me is: we cannot associate a potential energy $PE$ to a nonconservative force (but we can to a conservative force). That led me to assume that a nonconservative force should have no effect on the mechanical energy $ME=KE+PE$ of a system since $ME$ involves $PE$ and a nonconservative force has not saying as far $PE$ goes...

 

[jbriggs444]

Also, a nonconservative force is one for which work depends on the path taken and the work done by a nonconservartive force changes (adds or removes) the mechanical energy $ME = KE+ PE$ of the body. Work done by a conservative force does not change $ME$.

What has been confusing me is: we cannot associate a potential energy $PE$ to a nonconservative force (but we can to a conservative force). That led me to assume that a nonconservative force should have no effect on the mechanical energy $ME=KE+PE$ of a system since $ME$ involves $PE$ and a nonconservative force has not saying as far $PE$ goes...

It's just book keeping.

If you have a conservative force, it can go on the left hand side of the equation: $$W_{nc} + W_{c} = \Delta KE$$
Or you can negate it and move it over on the right hand side: $$W_{nc}=\Delta KE + \Delta PE$$

 

[jbriggs444]

Great, thank you. Why do you specify "for a point-like object"? What if we were dealing with a rigid extended object? Wouldn't the work-KE theorem not apply?

You can apply two equal and opposite forces (net force = zero) to a spinning disk whose center of gravity is motionless and increase its rotational kinetic energy.

 

[sophiecentaur]

The work-kinetic energy theorem

I have looked for this theorem, as you call it. All I can find is the Work Energy theorem, which makes total sense.
Have you a good reference to your version? The reason you seem to be confused is that the theorem you quote doesn't actually make sense without significant qualifications.

 

[A.T.]

... this is true only for the net force $F_{net}$...
... the work-KE truly applies to the net force $F_{net}$, not the individual force...
...The net force work is equal to the change in $KE$...

... these force give rise to a nonzero net force $F_{net}\neq 0$...

Why are you messing around with the net force? Just use the net work, which is the sum of the work of each individual force. You compute the work for each force and then sum it, not sum the froces and then compute work. The net work can be non-zero, even if the net force is zero, as @jbriggs444 shows in post #10.

 

[fog37]

Thank you.

In essence, the total work is $W_{total}= W_c+W_{nc}=\Delta KE_{total}$.

Since $W_c = -\Delta PE=\Delta KE_c$ and $W_{nc}= \Delta KE_{nc}$, we get $$W_{nc}= \Delta KE+\Delta PE$$

 

[jbriggs444]

In essence, the total work is $W_{total}= W_c+W_{nc}=\Delta KE_{total}$.

Since $W_c = -\Delta PE=\Delta KE_c$ and $W_{nc}= \Delta KE_{nc}$, we get $$W_{nc}= \Delta KE+\Delta PE$$

You've make up a notation: $\Delta KE_c$. What is it supposed to denote?

 

[fog37]

Just the contribution to the kinetic energy change due to conservative forces while the other term, $\Delta KE_{nc}$, the contribution due to nonconservative forces...maybe not an appropriate and correct notation...

 

[jbriggs444]

Just the contribution to the kinetic energy change due to conservative forces while the other term, $\Delta KE_{nc}$, the contribution due to nonconservative forces...maybe not an appropriate and correct notation...

There is a name for the energy transferred due to forces: "Work".

 

[fog37]

Cool. Since we are on the topic on conservative forces (line integral is constant and does not depend on path length shape) and nonconservative forces, I recall that conservative forces can only be represented by position-dependent functions. But not all position-dependent forces are conservative. For sure, time or velocity dependent forces are never conservative which means we can never have a time dependent $PE$ function...However, I may have seen a time dependent potential energy function in electromagnetics...

 

[vanhees71]

A force is called conservative if it depends only on position (not time nor velocity) and is the gradient of a scalar, which is called the potential:
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
If this holds in a simply connected region there exists a uniquely defined $V$ if for any closed curve $C$ within this region
$$\int_C \mathrm{d} \vec{x} \cdot \vec{F}=0,$$
and the potential can be calculated by fixing a point $\vec{x}_0$ and then using an arbitrary path $C_{\vec{x}}$ connecting $\vec{x}_0$ with $\vec{x}$ within this region,
$$V(\vec{x})=-\int_{C_{\vec{x}}} \mathrm{d} \vec{x}' \cdot \vec{F}(\vec{x}').$$
The result is independent of the shape of this path but of course depends on the choice of the initial point, i.e., the potential is determined only up to an additive constant, which however is physically unimportant, because all we need is the force, and that obviously doesn't depend on the additive constant.

 

[fog37]

Thank you vanhees71.

What would be the simple explanation for the reason that any time-dependent function can never be a potential function $V$ for a conservative force which would be, in turn, also time dependent? The line integral would also change with time and the mechanical energy $ME=KE+PE$ would also become a function of time...What is the problem with that?

Thanks again.

 

[vanhees71]

You can have a time-dependent potential, but then energy wouldn't be conserved, because then you have the equation of motion
$$m \ddot{\vec{x}}=-\vec{\nabla} V(t,\vec{x}).$$
Now take the time derivative of the KE
$$\mathrm{d}_t \left (\frac{m}{2} \dot{\vec{r}}^2 \right)=m \dot{\vec{x}} \cdot \ddot{\vec{x}}=-\dot{\vec{x}} \cdot \vec{\nabla} V(t,\vec{x})=-\left [\mathrm{d}_t V(t,\vec{x}) - \partial_t V(t,\vec{x}) \right]$$
or
$$\mathrm{d}_t E=\mathrm{d}_t \left [\frac{m}{2} \dot{\vec{x}}^2 + V(t,\vec{x}) \right]=\partial_t V(t,\vec{x}),$$
i.e., if the force has a potential energy is conserved if and only if the potential is not explicitly time-dependent.

 

[fog37]

Thank you.

Also, mechanical work $W$, regardless of having the same unit as energy, represents a different concept. I would say work is the mechanism through which forces can add or subtract energy from the system...

Forces do not "cause" work. I guess it would be more correct to say that forces "produce" work if they act while the system is undergoing displacement? Forces can cause displacement though...or should I keep the vector quantities of displacement and force as independent, not cause-effect related quantities?

 

[gmax137]

I think you are getting too wrapped up in words; "mechanism," "cause," "produce," etc. Recognizing that energy and work are mental constructs used for "bookeeping," I think it is best to stick to the mathematical representations (which are well defined).
$dW = \vec F_{net} \cdot d\vec s$

should I keep the vector quantities of displacement and force as independent, not cause-effect related quantities?

$\vec F_{net} = m \ddot {\vec s}$

 

[fog37]

Thank you gmax137.

As far as the two main types of energy go, i.e. kinetic and potential energy, kinetic energy $KE$ is frame of reference dependent quantity since it depends on the squared speed $v$. What about potential energy $PE$? Potential energy is associated to the "system", whichever we define to be the system and its composing parts. $PE$ depends only the configuration of the system, i.e. the mutual position of the various parts so I don't think $PE$ is frame dependent like $KE$ because the mutual distance between the parts does not change with reference frame. $PE$ always requires more than one entity/part/component to be involved, i.e. a multi-component system, while $KE$ can be associated to a single component system, at least I think...

Is it correct to think that whenever a system possesses nonzero $PE$, whatever its type, that implies that conservative forces are acting or have been acting or were acting on the system?

Thanks!

 

[jbriggs444]

There is not much to be gained by rigidly defining the exact circumstances where a notion of potential energy can be defined.

Potential energy does not need two bodies to exist. A single body in a conservative vector field can have a defined potential. One contrived example would be a freely floating object in a rotating frame. The sum of its kinetic and potential energies in this frame is constant. [The centrifugal force has an associated potential in a uniformly rotating frame].

Note that this is also an example where potential energy is frame dependent.

 

[fog37]

thank you jbriggs444. I am not fully following unfortunately.

My basic understanding is that potential energy is a property of a system and its parts and depends only on the system's parts configurations. for example, we can talk about the gravitational potential energy $GPE$ associated to an object of mass $m$ AND planet Earth with mass $M_E$ . In regards to the single object of mass $m$, we then derive the expression $mgy$ where y is the distance above the Earth's surface...

 

[fog37]

From Britannica:

"...Potential energy, stored energy that depends upon the relative position of various parts of a system....
...Potential energy is a property of a system and not of an individual body or particle; the system composed of the Earth and the raised ball, for example, has more potential energy as the two are farther separated.
...Potential energy arises in systems with parts that exert forces on each other of a magnitude dependent on the configuration, or relative position, of the parts. In the case of the Earth-ball system, the force of gravity between the two depends only on the distance separating them..."

 

[A.T.]

From Britannica:

A physics book with practical problems, that actually employ the concept, will give you a much better idea of potential energy.

 

[fog37]

After defining what represents the "system" while everything else represents the "surrounding", the total mechanical energy of the system $ME=KE+{tot}+PE_{tot}$ is conserved if the external net force acting on the system is conservative, correct?
What about the internal forces, i.e. the forces internal to the system and acting between the system's parts? By Newton's third law, the net internal force is always zero so internal forces can be either conservative or nonconservative. Therefore it does not seem to matter if the internal forces are conservative or not. In summary, to keep the $ME$ constant, either the net external force is nonzero AND conservative or the net external force equals zero. Is that correct?

 

[jbriggs444]

Therefore it does not seem to matter if the internal forces are conservative or not.

Two sand blocks in a closed box interacting via friction. Is mechanical energy conserved?
Two pucks on an air hockey table connected by a spring. Is mechanical energy conserved?

 

[fog37]

OK, the pucks on the air table...mechanical energy is conserved while it is not for the blocks with friction...

So, it IS important for the internal forces to be conservative for ME to be conserved!
Therefore, for ME to be conserved, two conditions must be satisfied:

a) The internal force pairs must be conservative
b) The net external force is either zero or conservative

 

[jbriggs444]

OK, the pucks on the air table...mechanical energy is conserved while it is not for the blocks with friction...

So, it IS important for the internal forces to be conservative for ME to be conserved!
Therefore, for ME to be conserved, two conditions must be satisfied:

a) The internal force pairs must be conservative
b) The net external force is either zero or conservative

A zero force does count (trivially) as conservative.

 

[A.T.]

Therefore, for ME to be conserved, two conditions must be satisfied:

a) The internal force pairs must be conservative
b) The net external force is either zero or conservative

Isn't that simply all forces then?

 

[fog37]

I guess so :)

 

[fog37]

In general, when discussing conservation of mechanical energy for a system, the system is considered isolated which implies that no mass or energy can enter or exit the system. Isolated means that the system does not interact with the environment in any fashion. Lack of external interaction means lack of external force. The net external force $F_{net}$ must be automatically zero if the system is isolated since the system cannot receive or lose energy. I see how a net force can add or subtract energy to the system. How could an external force add or subtract mass to the system?

For an isolated system, energy can only transfer between different parts within the system itself...