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According to the/a physics textbook by Douglas C. Giancoli, the net work done on a $15$ $kg$ backpack carried upwards $10$ m by a hiker is $0$ J since the work done by the hiker is $147$ J and the work done by gravity is $-147$ J.

But the potential energy of the backpack is increased by $147$ J , so what's going on? (to make it short)

Edit: The work is $+/-$ $1470$ J, not $+/-$ $147$ J , I'm sorry for this error. (It does not change the issue though).

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    $\begingroup$ Current question on the same topic: physics.stackexchange.com/q/677889/123208 $\endgroup$
    – PM 2Ring
    Nov 18 '21 at 19:35
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    $\begingroup$ Several answers explain that the work does not end up in the backpack but in the gravitational energy field. An example to compare this more intuitively could be the movement of two objects attached by a string. When we pull the objects away from each other then the energy 'ends up' in the spring and not in the objects. $\endgroup$ Nov 19 '21 at 10:43
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    $\begingroup$ Which text book? "Physics: Principles with Applications"? "Physics for Scientists and Engineers"? "Physics for Scientists and Engineers with Modern Physics"? $\endgroup$ Nov 19 '21 at 11:01
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    $\begingroup$ By the way, if you took care to use Giancoli's wording the question would look less absurd: His wording is "Net work done. (c) The net work done on the backpack is W-net = 0". Note the conspicuous absence of the word "hiker" ;-) : As explained in my answer, the hiker of course does work on the backpack. That question is covered in the section before the one you are discussing, and is aptly worded "To calculate the work done by the hiker on the backpack"... $\endgroup$ Nov 20 '21 at 13:38
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But the potential energy of the backpack is increased by 147 J, so what's going on? (to make it short)

The positive 147 J of work done by the hiker transfers energy to the backpack. The negative -147 J of work done by gravity takes away the energy the hiker transferred to the backpack and stores it as gravitational potential energy (GPE) of the Earth-backpack system. The fact that the net work done is zero simply means that the backpack does not have kinetic energy after lifting it 10 m.

The underlying principle is the work energy theorem which states that the net work done on an object equals its change in kinetic energy. Since the backpack presumably starts at rest and ends at rest after moving up 10 m its change in KE is zero. All of the work done by the hiker ends up as GPE.

Hope this helps.

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    $\begingroup$ Thank you for your quick answer. But why would the kinetic energy "belong" to the backpack but not its potential energy? (its motion is also relative to Earth's coordinate system, so not absolute either) - And overall, considering the entire system of hiker, backpack and Earth, chemical energy was converted into GPE (via muscle work) so I still fail to see that the net work done here is zero. $\endgroup$ Nov 20 '21 at 16:57
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    $\begingroup$ @FelixTritschler While the KE of the backpack is relative to different inertial reference frames (frames moving at different constant velocities), it’s the same with respect to all locations in the same inertial reference frame. Considering the hiker’s locations as the same inertial frame, then whether I am standing at the beginning of the hikers climb, 5 m higher, or at the end of the climb (10 m), any single object I observe in motion from those three locations will have the same KE. On the other the GPE of the object will be different for the different locations. $\endgroup$
    – Bob D
    Nov 20 '21 at 18:59
  • $\begingroup$ Regarding your second point, I’m not sure what you are getting at. But the source of the hikers energy (chemical potential energy) is irrelevant with respect to work energy theorem and the net work done on the object. If you are considering the hiker, backpack and Earth as an isolated system, then this work would be internal to the system. No work would be done on the system as a whole if it is isolated. $\endgroup$
    – Bob D
    Nov 20 '21 at 18:59
  • $\begingroup$ Regarding your 1st comment, actually no, the ΔGPE would be the same, regardless of the observer-position. Regarding your 2nd comment: Consider 2 scenarios: 1. with gravity (as before), 2. no gravity --> backpack flies with ~14 m/s after being pushed with ~147 N along a 10 m stretch resulting in ΔKE=ΔGPE (in scenario 1). In both cases this is the result of the conversion of the same amount of chemical energy into the respective energy forms, but this is irrelevant for case 1 and relevant for case 2? That cannot be,can it?The same goes for friction heat vs KE (other comments).Thx in adv. $\endgroup$ Nov 21 '21 at 14:54
  • $\begingroup$ @FelixTritschler First, I didn’t say $\Delta GPE$ is not the same. I said GPE is not the same. GPE has no meaning unless the location of zero GPE is assigned. KE does not depend on a reference point in the same inertial frame. Second, I don’t understand scenario 2. You say there is no gravity then you say $\Delta KE$=$\Delta GPE$(???). Third, since I don’t understand your scenario 2, I don’t understand your point about the conversion of chemical potential energy. It can certainly be the energy source for changing the KE of an object or for changing the GPE. $\endgroup$
    – Bob D
    Nov 21 '21 at 16:18
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The keyword here is net work.

Earlier in the exercise the author computes the work (no "net") the hiker has done on the backpack, which is the expected amount depending on altitude difference and mass. That's what the hiker did.

Now for the net. The hiker can attest that gravity is a bitch: It counteracted the hiker's force on the backpack the entire time! Witness to that is that the backpack didn't accelerate (if for simplicity we consider a stretch in the middle of the hike). No net forces, no change in kinetic energy.

The work the hiker did didn't end up in the backpack, it ended up in the gravitational field, in what Newtonian physics calls "potential energy". A similar situation would arise if a worker pushed a box across a cement floor, with constant velocity. They certainly perform work, also in the physics sense; but none goes into the box: The forces on the box cancel each other out! All the mechanical work is transformed into heat.

An opposite example would be the same scenario, but without gravity: The hiker displaces a weight with a constant force F over some distance s. Because there is no gravity, we have a net force: It is exactly F, and it all goes into kinetic energy, which is simply F*s if the force pointed in the direction of travel. (Because we have a net force, the backpack also accelerates all the time, as opposed to the gravity scenario.)

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  • $\begingroup$ Thank you for you quick answer as well. In addition to my comment on Bob's answer, I would like to say that for the case of a box pushed around against frictional force, I don't understand either why the net work done would be zero because the temperature of the box and of the floor increased. $\endgroup$ Nov 20 '21 at 17:06
  • $\begingroup$ @FelixTritschler: Peter isn't saying the overall net work done by the person pushing the box would be zero. He's saying the net work done "to the box" is zero because it moves at constant velocity. This is the distinction Giancoli is making, too. $\endgroup$ Nov 20 '21 at 21:52
  • $\begingroup$ @FelixTritschler Regarding your comment here, the net work done on the box in pushing it at constant velocity against the kinetic friction force is zero because its change in KE is zero. In this case, unlike gravity where the equal negative work it does is stored as GPE, the negative work done by the kinetic friction force is stored as an increase in the internal energy of the materials rubbing together, increasing their temperature and subsequently resulting in heat transfer to other cooler areas. $\endgroup$
    – Bob D
    Nov 20 '21 at 22:22
  • $\begingroup$ In effect, friction converts the macroscopic KE the box would have gained into an increase in microscopic KE, i.e. and increase in the KE of molecules of the materials. $\endgroup$
    – Bob D
    Nov 20 '21 at 22:23
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As the backpack is carried up, the net force, work, and change in kinetic energy are all zero (as it would be if it were carried back down). The potential energy is the work that can be done by the field (alone) in moving the object from its current position to a chosen reference position. (If the pack were tossed back down, there would be a change in the kinetic energy equivalent to the work done by the hiker in taking it up.)

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  • $\begingroup$ Thank you as well for your quick answer. I commented on Bob's and Peter's answer. In addition to that you actually implicitely said what I have a problem with here: "If the pack were tossed back down there would be a change in the kinetic energy equivalent ..." - This shows that the box' energy increased, so the net work done beforehand should not be zero. (!) $\endgroup$ Nov 20 '21 at 17:11
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The work-energy theorem says that the total work done on a body is equal to the change in its kinetic energy. Since the pack starts at rest and ends at rest no net work has been done on it. The total force on the pack (The lifting force minus the weight) was slightly upwards at the beginning to get it started moving, and slightly downwards at the top to stop it moving.

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  • $\begingroup$ Thank you as well for your quick answer. I wrote comments to some of the other answers - please note what I wrote regarding the chemical energy involved here, if you are interested. $\endgroup$ Nov 20 '21 at 17:22
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Gravitational potential energy is basically another name for the work done by gravity.

So you can describe the scenario as

  • you providing energy via work that is stored as potential energy, or as
  • you doing work that is balanced out by work done by gravity.

The net work, meaning the work done by all involved parties is indeed zero in your case when potential energy is considered as work. If you aren't losing any energy to heat and aren't speeding something up (the backpack is assumed moving at constant speed, I reckon) then this will have to be the case due to the energy conservation law. All energy supplied has been taken from somewhere else.

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  • $\begingroup$ Thank you as well for your quick answer. I wrote comments to some of the other answers - please note what I wrote regarding the chemical energy involved here, if you are interested. $\endgroup$ Nov 20 '21 at 17:22
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Would like to add that you can understand this sitaution mathematically by modelling the system of the bag and the Earth as a non-isolated one (for energy).

The reduced equation for the conservation of energy then gives: $$\Delta K + \Delta U + \Delta E_{\text{int}} = \sum T \rightarrow \Delta K + \Delta U_g = W_{\text{ext}}$$

Now assuming that the bagpack comes to rest after the hike, we must have $\Delta K = 0$, and hence, the increased potential energy (gravitational) of the bag-Earth system ($\Delta U_g = mg(\Delta y )$ > $0$) comes from the work done on the bag by the hiker which is external to the system.

If you model only the bag as the system, then there is no $U_g$ term (as there are no interacting objects) and the conservation of energy gets reduced to the familiar work - kinetic energy theorem.

Again, we find that $$W_{\text{ext, net}} = \Delta K = 0$$ which proves the assumption in the first assertion, and also gives the meaning of the bag having no net work done on it ; the bag has gained no additional kinetic energy.

Hope this helps.

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  • $\begingroup$ Thank you for your quick answer as well. In addition to my comments to above answers: Your equation of the entire system shows precisely what I see as problematic here: Considering the entire system (Earth, hiker, backpack), chemical energy was transformed into potential energy (via muscle work). The former is not zero, the latter is neither, hence the net work should not be zero. (!) $\endgroup$ Nov 20 '21 at 17:19
  • $\begingroup$ If you model the system as containing all of the three entities, then the system becomes isolated. There are no external agents doing work on the bag, hiker or the Earth! In this case, the equation reduces to $\Delta U_g + \Delta U_{\text{chem}} = 0$. As $\Delta U_g > 0$, we must have $\Delta U_{\text{chem}} < 0$, which makes sense ; The increase in gravitational potential energy of the earth-bag system came with a reduction in the chemical potential energy of the hiker. $\endgroup$
    – Cross
    Nov 21 '21 at 17:44
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I think you probably got your answer from the previous ones. I just want to add something.

Potential energy of a system changes only when when the internal conservative forces in a system do work. In this case the force applied by the man on the bag is not conservative but gravitational force between bag and the earth is conservative, thus this causes change in the potential energy of the bag.
$$\Delta U = -W_c = -\int\vec F_c.d\vec s$$

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