Is this work-energy theorem for a system correct?

[Arthur Smyth]

Homework Statement

[/B]
A 4 kg mass slides 2 m over a horizontal surface with force of kinetic friction of 2N, initial velocity is 5 m/s.
Find it's final velocity.

My problem... we've been asked to solve this problem using the Work-Energy theorem, and we've been given it in the form of:

W = ∆K + ∆U + ∆Einternal

I find I can make it work if only assume that the change in internal energy is zero, or the work done by friction is zero... and I think that they should equal to each other.

What is my error in understanding... or is the formula wrong

Homework Equations

W = ∆K + ∆U + ∆Einternal

The Attempt at a Solution

Here is one solution if the work done is 0...

0 = ∆K + ∆U + Einternal
0 = (Kf - Ki) + 0 + Fdcosθ
0 = ( 1/2 mvf^2 - 1/2 mvi^2) + 0 +Fkdcosθ
0 = (1/2 (4)(vf)^2 - (1/2 (4)(5)^2) + 0 + (-2)(2)(cos 180˚)
0 = 2vf^2 -50 +0 +4
vf^2 = 23
vf = √ 23 m/s

 

[TSny]

Welcome to PF!

Your worked out solution looks correct. Here, you are considering the block and surface together as your system. The friction force is internal to the system. There are no external forces doing any work on the system. Thus, you are right to set W = 0. The change in internal energy includes the increase in internal energy of both the block and the surface.

Using work-energy concepts with friction can be confusing and there has been quite a bit of discussion in the literature pointing out that many textbooks do not actually treat this correctly.

See https://brucesherwood.net/?p=134 and https://brucesherwood.net/wp-content/uploads/2017/06/Friction1984.pdf

 

[Arthur Smyth]

TSny... thank you!... I'm not going mad, after all. I do have one question though...
I understand that the two friction forces contained in the system of tabletop and sliding block cancel out (at least I hope they do) and thus W = 0... but if E_internal = F_kd cosθ, then isn't Fk doing work. I'm still somewhat confused.
Thank you, thank you for your help.

 

[kuruman]

My problem... we've been asked to solve this problem using the Work-Energy theorem, and we've been given it in the form of:

W = ∆K + ∆U + ∆Einternal

Question: When you were given this equation, what was the definition for W?

The traditional way of writing the work-energy theorem has been ΔK = W_net, where W_net is the sum of the works done by all the forces on the displaced mass. Then some people decided to split out the negative of the work done by all the conservative forces from the non-conservative forces and write ΔK + ΔU = W_n.c.. Then others started questioning whether it is conceptually correct to say that friction does work when all it does is increase the thermal (or internal energy) of the mass so they moved the negative "work done" by the non-conservative force of friction to the right side and wrote ΔK + ΔU + ΔE_int = W_{what's left over}. If nothing is left over, then the right side is zero.

 

[Arthur Smyth]

Hi Karuman...

That makes sense... I think I know what's going on now...
except I still don't understand that in my example W=0, yet E_internal is subsitituted with F_kdcosθ...
doesn't F_kdcosθ imply work?

Thank you for your patience...

 

[TSny]

I still don't understand that in my example W=0, yet E_internal is subsitituted with F_kdcosθ...
doesn't F_kdcosθ imply work?

Again, a very good question. The quantity F_kdcosθ does not equal the work done by friction on the block by the surface. In general, the true work done by friction on the block by the surface is very complicated and depends on what's going on at the multiple points of contact of the surface of the block and the surface of the table. For your problem, there would be no way to find the true work done by friction on the block by the surface. The paper https://brucesherwood.net/wp-content/uploads/2017/06/Friction1984.pdf

, which I linked to earlier, gives a nice discussion. But, I think this paper would be a difficult read for most students taking their first physics course. I don't think I would have had the courage to slog through it when I was taking my first physics course. But, if you browse through it you might at least get the gist of it. The other link https://brucesherwood.net/?p=134 shows how one of the authors of the paper had a very difficult time convincing other physics educators of these ideas. ​

The paper shows that the quantity |F_kdcosθ| is not the true work done by friction on the block. But the paper does show that this quantity represents the total increase in internal energy of the block-table system as the block slides a distance d on the surface of the table. This is derived in Section VIII of the paper for an example similar to your problem. (It's the same as your problem if you take the angle of incline in the paper's example to be zero.) So, your method of solution gives the correct answer if you are considering the block-table system for which no external work is done on this system (W = 0).

 

[PeroK]

There is a risk here of overcomplicating an elementary problem. The issue of internal thermal energy is not normally a consideration when studying the motion of a rigid object.

Also, the apparent complexity of the work-energy theorem as given by the OP is partially obscuring a relatively simple problem. If there at no conservative forces and we are not considering internal energy, then the problem reduces to:

$KE_f = KE_i + W$

Where $W$ is The (negative) work done by friction.

In fact, the issue I would have with the problem is how you know the friction force in this case? It would seem more natural to me to be given the velocities and asked to calculate the friction force and work done by it. But, then, I'm no experimentalist!

 

[Master1022]

Homework Statement

My problem... we've been asked to solve this problem using the Work-Energy theorem, and we've been given it in the form of:
W = ∆K + ∆U + ∆Einternal
I find I can make it work if only assume that the change in internal energy is zero, or the work done by friction is zero... and I think that they should equal to each other.

What is my error in understanding... or is the formula wrong

Homework Equations

W = ∆K + ∆U + ∆Einternal

Hi, I know you have solved the problem by now. As a suggestion for the future, I would not try to memorize/use this Work- Energy formula from memory (I know that you were told to use it this time). I personally think that it makes it too complicated and it is much better to go for an intuitive approach.

The method I used to use for these energy problems was simply writing energy changes with arrows. For example, this one would be:
Initial KE --------------> Final KE + work done against friction

You can then write mathematics for each of the terms... you could just as easily include a potential energy term in the chain if needed...

I think this method is subtly different, but I found it to be much better. Hope that is of some help for the future.

 

[kuruman]

In fact, the issue I would have with the problem is how you know the friction force in this case? It would seem more natural to me to be given the velocities and asked to calculate the friction force and work done by it. But, then, I'm no experimentalist!

You don't have to be an experimentalist to realize that if the kinetic energy of an object increases, the Joules involved must have come from somewhere through some mechanism. The same idea applies if Joules leave the object. The question is "where do the Joules come from (or go to) and how can one account for them?" When conservative forces do work, the task is easy, it's just the work done on the object by these forces. What about non-conservative forces that change the internal energy? There are recipes for that. Some examples: If the internal energy increases because of kinetic friction the recipe is $\Delta E_{int.} = \mu_kF_N.$ If the internal energy of an ideal gas changes then $\Delta E_{int.} =Q+W$ where Q and W have their usual meanings in the context of the first law. If the internal energy of an object changes because another object collides and sticks to it, then the change in internal energy is $\Delta E_{int.}=\frac{m_1m_2}{2(m_1+m_2)}v_{rel.}^2$ which is just the kinetic energy before the collision in the CM frame ($v_{rel.}$ is the relative speed). Note that in this last case of perfectly inelastic collision only some of the mechanical energy lost goes into raising the temperature of the object. The rest goes into sound, material deformation, bond breaking, to name just a few. If there is no recipe for the amount of transformed energy, we can only lump the changes under a common name such as "change in internal energy" or, if this name is deemed inappropriate, find another.

Mechanical energy conservation is (in my heretical opinion) overrated and not very useful. The important concept is total energy conservation and its statement is quite simple: Total energy is always conserved with no exceptions. Having made this statement, I subscribe to Feynman's additional comment (paraphrased) that one has only to find its hiding places and account for all of them. The work energy theorem is the equation describing such an accounting. It contains terms like $\Delta E_{int.}$, $\Delta U$ a.k.a. $-\Delta W_{cons.}$, etc. on the same side of the equation as $\Delta K$. It is an equation where the left hand side is just the sum of the energy changes in all of Feynman's hiding places and the right side is zero. I adopted this view after one student asked me "But in chemistry we were taught that energy is always conserved. How come it is not conserved in the case of inelastic collisions?" I did not have a good answer to give him.

 

[PeroK]

You don't have to be an experimentalist...

How would you practically measure the frictional force on a decelerating block, without first measuring the deceleration?

The problem states that the frictional force is 2N. How is that value experimentally determined?

 

[kuruman]

How would you practically measure the frictional force on a decelerating block, without first measuring the deceleration?
The problem states that the frictional force is 2N. How is that value experimentally determined?

The answers to these questions are obvious, one needs to determine the values that you mentioned experimentally if one were doing some kind of experimental exercise in which they are needed as input to an underlying theory. However, this problem, including many others in textbooks and on PF, contains implied hypotheticals. Rephrased it reads "Find the final velocity of an object if its mass is 4 kg and the object slides slides 2 m over a horizontal surface, if the force of kinetic friction is 2N and if the initial velocity is 5 m/s." In other words, the formulator of the problem meets the answerer half way and says "here are some numbers, never mind how I got them, just show me how you use them to show me that you understand the underlying theory." Better yet, skip the numbers and use symbols, "Mass $m$ slides distance $d$ over a horizontal surface with force of kinetic friction $f$ and initial velocity $v_0$. Find its final velocity." After all, the point of the exercise is to derive an equation with the unknown quantity isolated on the left hand side and an expression involving only given quantities on the right hand side.

 

[PeroK]

The answers to these questions are obvious, one needs to determine the values that you mentioned experimentally if one were doing some kind of experimental exercise in which they are needed as input to an underlying theory. However, this problem, including many others in textbooks and on PF, contains implied hypotheticals. Rephrased it reads "Find the final velocity of an object if its mass is 4 kg and the object slides slides 2 m over a horizontal surface, if the force of kinetic friction is 2N and if the initial velocity is 5 m/s." In other words, the formulator of the problem meets the answerer half way and says "here are some numbers, never mind how I got them, just show me how you use them to show me that you understand the underlying theory." Better yet, skip the numbers and use symbols, "Mass $m$ slides distance $d$ over a horizontal surface with force of kinetic friction $f$ and initial velocity $v_0$. Find its final velocity." After all, the point of the exercise is to derive an equation with the unknown quantity isolated on the left hand side and an expression involving only given quantities on the right hand side.

I know all that. I have the scientific and democratic right personally not to like that sort of question. An opinion I expressed in post #7.

 

[kuruman]

I know all that. I have the scientific and democratic right personally not to like that sort of question. An opinion I expressed in post #7.

Of course you do and it was never my intention to question that right. It seems that I misunderstood the spirit of your last sentence in #7.