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Suppose I take a box with some masses and pull it along the table at a constant speed. In this case, there is a force pulling to the right (the string) and a frictional force pulling to the left.

Since the net force is zero, there is no change in momentum. Everything is fine.

. . . .◆Work-Energy Principle◆. . . . The work-energy principle says that the work (W) done on a point mass is equal to its change in kinetic energy. Work is done by a force moving a certain distance. Actually, it's only the force in the direction of motion that matters. As an equation, it looks like this $W=F d \cos (\theta)$

Here θ is the angle between the force and the displacement. If the force is "pushing backwards" you can have negative work. For the kinetic energy, it depends on the mass and the velocity $KE=\frac{1}{2}m v^2$

For the block being pulled along the table with friction the two forces are the force from the string pulling to the right and friction pulling to the left. The total work on the block would be zero, and it would move at a constant speed.

BUT WAIT! There is a problem. What if you measure the temperature of this block before and after you pull it? It's not a huge increase in temperature, but it did indeed warm up. If I slide the block over a larger distance (or back and forth), That is an area where the table increases in temperature—the block also gets hotter.

But if the block gets warmer, that means it increases in energy. In this case, it would be an increase in thermal energy. So, how can the block increase in energy if there is zero work done on the object? How is it possible for there to be zero work AND an increase in energy?

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    $\begingroup$ Friendly comment: your post is too long. I skipped down to the last two paragraphs. Hint: what exactly is $s$ in the definition of work? $\endgroup$ – garyp Sep 22 at 17:45
  • $\begingroup$ Yup, it is long but I've included each and every minute thing i can include, you can read it fast ! Here s is the distance simply covered by my example $\endgroup$ – Vishal Kumar Sep 22 at 17:47
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    $\begingroup$ Please mark up your math using mathjax. The question is pretty incoherent. You present lots of numerical data, but it's not clear what the point is. $\endgroup$ – Ben Crowell Sep 22 at 17:52
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    $\begingroup$ what do you think friction energy does, the table does not move, yet it absorbs friction energy as ____? $\endgroup$ – Adrian Howard Sep 22 at 18:09
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    $\begingroup$ So what is your question? StackExchange is a question/answer site, not a discussion site. $\endgroup$ – Jon Custer Sep 23 at 13:21
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But if the block gets warmer, that means it increases in energy. In this case, it would be an increase in thermal energy. So, how can the block increase in energy if there is zero work done on the object? That is indeed a mystery. How is it possible for there to be zero work AND an increase in energy. Thanks friends for reading it out , i would like to hear your thoughts about this ,

The work energy principle states: the net work done on an object equals its change in kinetic energy.

The key term is net work. Work can be positive (force in same direction as displacement) or negative (force in opposite direction of displacement). Friction work is always negative because the friction force opposes the motion of an object.

So lets say a constant external force $F$ moves a box a distance $d$ on a surface with friction. This work is positive. But the friction force $f$ acts in the opposite direction so its work is negative. By the work energy principle

$$W_{net}=Fd-fd=\Delta KE$$

If the applied force equals the friction force, the net work done on the object is zero (no change in velocity) and $Fd-fd=0$, or $Fd=fd$.

The friction force is $f=μ_{k}mg$ so the friction work is $μ_{k}mgd$

This tells us that the magnitude of the work done by the external force equals the magnitude of the work done by the friction force. The energy of the work done by friction elevates the temperature of the contacting surfaces and eventually transfers to the surroundings as heat.

Bottom line: There is no increase in energy. Although the net work done is zero, meaning there is no change in kinetic energy (velocity is constant) the work done by friction is not zero. It is simply negative work. Friction work take the energy supplied to the object by the positive work done by the external force and performs an equal amount of negative work dissipating the energy done by the external force as heat. Energy is conserved.

UPDATE: This is in response to your answer to your question.

Jake pulls a box with a mass of 22 kg. The rope makes an angle of 25 degrees with respect to the horizontal. The coefficient of kinetic friction is 0.1. Find the work done by Jake and the work done by friction for the case where the box moves along the ground a distance of 144 meters.

Bad. Bad question. You could indeed calculate the force of friction, but you can't calculate the work done (unless you also know some stuff about the changes in thermal energy).

It is indeed a bad question, but not for the reasons you are citing. You say you "could indeed calculate the force of friction". But the fact is you can't. You need to know the magnitude of the force applied by Jake to calculate the friction force, the friction work, and the work done by Jake. Here's why, starting with the friction force.

The kinetic friction force, $f$, is given by

$$f=μ_{k}N$$

Where $μ_{k}$ is the coefficient of kinetic friction and $N$ is the normal force exerted by the contact surface. That normal force is usually $mg$, the weight of the box, but in this case it is not. That's because Jake is pulling the box at an angle with respect to the horizontal. The vertical component of Jake's force reduces the normal force. So the friction force becomes:

$$f=μ_{k}(mg-Fsin25)$$

where $F$ is the magnitude of the force applied by Jake. Without the friction force, you cannot determine the friction work which is.

$$W_{friction}=μ_{k}(mg-Fsin25)d$$

where $d$ is the horizontal distance moved.

The only work done by Jake is due to the horizontal component of his force (there is no vertical movement of the box so no vertical work done. So the work done by Jake is

$$W_{Jake}=Fdcos25$$

Since Jake's work is positive and friction work is negative, the net work done is

$$W_{net}=W_{Jake}-W_{friction}=d(Fcos25-μ_{k}(mg-Fsin25))=\Delta KE$$

Everything is known except $F$.

If you calculated the work done by friction as the frictional force multiplied by the distance the block moves, how would you account for the increase in thermal energy of the block (and floor)?

If you knew the magnitude of the force applied by Jake, you can calculate the friction work. If the net work done by Jake and friction is zero, all that means is that friction has taken the energy provided by Jake due to his work and converted it into heat. Energy is conserved.

SUMMARY:

I think the main problem you are having is thinking that net work equal to zero means that no work has been done. That is simply not the case. Jake did work in moving the box. Just ask Jake. Jake's work is positive since his force is in the same direction as the movement of the box. Friction did work, as evidenced by the increase in temperature of the contact surfaces. Friction's work is negative since the friction force is in the opposite direction of the movement of the box. If the magnitude of Jake's work equals the friction work, the net work done is zero. That only means there is no change in kinetic energy of the box. It doesn't mean no work was done.

Hope this helps.

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  • $\begingroup$ Yup, that's what I'm saying, if the frictinal force and applied force are just equal and hence opposite the net work done would came out to be zero and then by using the work energy theorem the velocity would simply came out to be constant/zero but because since we've considered the system consisiting of block and table and as i pull it along it has no change in kinetic energy but however the thermal energy increases. $\endgroup$ – Vishal Kumar Sep 23 at 17:50
  • $\begingroup$ And therefore my friend came with an analogy which I've mentioned below please read that out, it means it doesn't necessarily means fiction does the same amout of work using his example. $\endgroup$ – Vishal Kumar Sep 23 at 17:51
  • $\begingroup$ @VishalKumar I'm afraid I can't visualize what is going on in your example of the brushes, so I can't comment on the validity of you and your friend's conclusion. All I know is that total energy must be conserved. If the hand does work moving the brush and the brush begins and ends at rest, there is no change in kinetic energy of the brush. If all the movement is horizontal there is no change in potential energy. $\endgroup$ – Bob D Sep 23 at 19:31
  • $\begingroup$ The bottom line is the work done by the hand has to be accounted form. It has to be dissipated as heat due to friction. How else will you account for the energy expended by the hand? At this point I have no more time to invest in this discussion, so I wish you good luck in resolving your issues! $\endgroup$ – Bob D Sep 23 at 19:31
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Here's what my friend texted me today . . . Suppose that I rub two brushes together instead of a block and a table. Watch what happens. Notice that as the brush is pulled, there are two forces that do work. My hand does work (positive work), and the brushes do work (negative work). But look closely. Notice that as the brush (and my hand) move to the left a certain distance, the brushes bend. This means that the force the bottom brush exerts on the top brush moves over a shorter distance than the hand moves. Even if the force of the brush is the same magnitude as the force of my hand, the brush does less work because it moves over a shorter distance. That means the total work done on the brush is NOT zero joules but some positive amount. Of course the brush is an analogy for friction. We like to think of friction as this nice and simple interaction, but it's not. For the block sliding on a table, the frictional force is an interaction between the surface atoms in the block and the surface atoms on the table. It's not so simple. Physics textbooks like to treat a block as a point object—but it is not a point object. It's a complicated object made of countless atoms. In the case of friction, you can't forget that and just treat a block as a point object. It doesn't work.

but you could do this problem with the momentum principle and it wouldn't be a problem. Remember that the momentum principle deals with forces and time, not distance. So even though the frictional force acts over a different distance, the time is the same for both the frictional force and the force pulling the string.

Well, here is the problem. The main goal in physics is to build models that agree with real-life experiences. These models could be a big idea like the work-energy principle—and that's great. Let's consider an example with another model. What about a globe? It's a model of the Earth. It even shows the location of the continents and everything. But what if I want to use this globe and measure its mass and volume so I can determine the density of the real (full-sized) Earth? That wouldn't work, because the globe isn't actually Earth. The same is true with the work-energy principle. It's great for some things, but you can't just use it wherever you like.

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  • $\begingroup$ but we could do this problem with the momentum principle and it wouldn't be a problem. Remember that the momentum principle deals with forces and time, not distance. So even though the frictional force acts over a different distance, the time is the same for both the frictional force and the force pulling the string. $\endgroup$ – Vishal Kumar Sep 23 at 2:13
  • $\begingroup$ First of fall, you should ask your question in short way secondly, you should mention what condition are you assuming like in high school physics in India, while solving Newtonian mechanics problem they assume there is no loss of energy due to friction $\endgroup$ – yuvraj singh Sep 23 at 4:25
  • $\begingroup$ If I'd have assumed any thing I've already mentioned it in . Secondly I've mentioned all things and makes things crystal clear , that's why my discussion is long but worth to read !and yes that's what I'm trying to focus on that we all are finding it wrong in all these years , no one ever questioned it also! $\endgroup$ – Vishal Kumar Sep 23 at 6:41
  • $\begingroup$ @VishalKumar See the update to my answer in response to your own answer. Hope it helps. $\endgroup$ – Bob D Sep 23 at 14:21
  • $\begingroup$ @VishalKumar I've added a summary to my answer that I think gets to the core of the problem you are having. $\endgroup$ – Bob D Sep 23 at 15:06

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