B Formal definition of work done

[physicals]

Energy is quite an ambiguous topic in physics. Considering this, I have been trying to get a formal definition for what work done really means, a definition that truly defines it not just in terms of math but also theoretically. The closest |(partial) definition for work done ive come to is "Energy is transferred when work is done against a force or resistance". This definition holds true for most cases but my biggest confusion lies in the differentiation of force and resistance. Aren't they the same thing? For example and example of resistance might be resistance, if work is done against inertia energy is transferred to an objects kinetic energy store. In this case could resistance and force be considered as the same thing.

 

[PeroK]

The simplest way to describe things is that work done is the amount of energy that gets transferred in some way. Total energy is like a set of individual bank balances. And work done is like transfers from one account to another.

Resistance might mean electrical resistance, as opposed to mechanical friction. Athough, it all boils down eventually to the fundamental forces ot nature.

 

[jbriggs444]

I do not consider the notion of "resistance" to be important at all. It is enough that a force exists.

There are at least three general notions of "work". These are the definitions that I use:

Mechanical work:

The dot product of an applied force and the displacement of the material of the contacted object at the point of application.

If the motion or the force are not constant, one may have to evaluate an integral.

If the area of application is extended, one may have to evaluate an integral.


Center of mass work:

The dot product of an applied force and the displacement of the center of mass of the contacted object.

If the motion is not in a straight line under a constant force, one may have to evaluate an integral.


Generalized work:

The amount of non-thermal energy transferred into a system across an interface.

 

[Dale]

There are at least three general notions of "work".

Those are the three I focus on also. It is important to know that the first is included in the third, but the second is not. The second is also called “net work” by many authors.

Words have more than one meaning.

 

[kuruman]

Energy is quite an ambiguous topic in physics. Considering this, I have been trying to get a formal definition for what work done really means, a definition that truly defines it not just in terms of math but also theoretically. The closest |(partial) definition for work done ive come to is "Energy is transferred when work is done against a force or resistance". This definition holds true for most cases but my biggest confusion lies in the differentiation of force and resistance. Aren't they the same thing? For example and example of resistance might be resistance, if work is done against inertia energy is transferred to an objects kinetic energy store. In this case could resistance and force be considered as the same thing.

"Work done" is itself ambiguous if one does not specify who is doing the work and on whom this work done. The sign of this work is also important.

Let's consider a simple illustration of a block of mass $m$ initially moving with initial speed $v_0$ on a horizontal frictionless surface. At some point in time a hand exerts constant force $F$ in the same direction as the block's velocity. The block is displaced by $\Delta x$ under constant acceleration $a$ and reaches final speed $v_{\!f}$ at which point the hand is removed.

With this information we write the kinematic equation relating the acceleration, displacement and speeds at the beginning and end of the displacement. $$\begin{align} & 2a\Delta x=v_{\!f}^2-v_0^2. \end{align}$$Note that we can do this without describing how or why the block changes its speed. To do that we bring in Newton's laws.

1. The first law tells us that since the block accelerates, there must be an unbalanced external force acting on the block. We identify the net force as the force $F$ exerted by the hand.
2. The second law tells us that the external force exerted by the hand is equal to the mass of the block times the acceleration $F=ma.$ This is the link between what is outside the system (pushing hand) and what happens to the motion of the system (change of velocity.)
3. The third law tells us that the block exerts force of magnitude $F$ on the hand but in the opposite direction. This force is always there and can be viewed as the "resistance" that you mentioned in post #1.

Now for some math. We multiply both sides of equation (1) by $\frac{1}{2}m$ to get $$\begin{align}
& ma\Delta x=\frac{1}{2}m\left(v_{\!f}^2-v_0^2\right)\nonumber\\
& F\Delta x=\frac{1}{2}m\left(v_{\!f}^2-v_0^2\right).
\end{align}$$The left side in equation (2) is, by definition, the work done on the block by the hand. It is positive work because the force and the displacement are in the same direction as stated in the problem. It follows that the right side of equation (2), the change in kinetic energy, is also positive. This makes eminent sense, if you push something in the same direction that it is already moving, you increase its speed.

Note that if the hand pushes in a direction opposite to the displacement of the block, the left side of equation (2) will be negative which means that the kinetic energy change will also be negative.

Force $F$ enables the energy transfer across the boundaries separating the hand from the block. The amount of energy transferred is the work done by the hand on the block. To extend @PeroK's analogy, the work that the hand does on the block in Joules is like money transferred from the hand's bank account to the block's bank account and could be positive or negative. Force $F$ is analogous to the bank clerk who mediates the transfer from one account to the other. If the bank charges a fee every time there is a transfer, that is analogous to a dissipative force, e.g. friction or air resistance, that reduces the amount transferred no matter which way it moves.

 

[A.T.]

The closest |(partial) definition for work done ive come to is "Energy is transferred when work is done against a force or resistance".

Resistance is completely irrelevant here.

Work is transfer of energy.
Just like:
Impulse is transfer of momentum.

Or, specifically for mechanical work:

Work is force applied over a distance.
Just like:
Impulse is force applied for a time.

 

[kuruman]

Resistance is completely irrelevant here.

If you mean electrical resistance, yes it is completely irrelevant.

Here, I think that it is to be interpreted here as "resistance to acceleration" i.e. the reaction force exerted by the system on the external agent that applies the force. It's always there as long as the system has inertial mass. That's what I tried to illustrate in post #5.

 

[A.T.]

Here, I think that it is to be interpreted here as "resistance to acceleration" i.e.

I interpreted it as resistance to motion, like drag. Which just goes to show ambiguous the term and any definition based on it is.

... the reaction force exerted by the system on the external agent that applies the force. It's always there as long as the system has inertial mass.

I disagree. We use massless elements in idealized scenarios all the time. And they obey Newton's 3rd Law just fine. The only condition is that the net force and net torque on them must be zero all the time. They can also have work done on them.

 

[kuruman]

I interpreted it as resistance to motion, like drag. Which just goes to show ambiguous the term and any definition based on it is.

I disagree. We use massless elements in idealized scenarios all the time. And they obey Newton's 3rd Law just fine. The only condition is that the net force and net torque on them must be zero all the time. They can also have work done on them.

Can you provide an example of such an idealized scenario of a massless particle on which work is done? Does it accelerate as a result? If "yes", how can it be since the net force and torque "must be zero all the time"? If "no", what happens to the work that crosses the system boundary?

It is easy to imagine a massless mathematical point on a massive object and talk about the acceleration of the massive object at that point. It is not easy, at least for me, to imagine a massless element that can be considered the "system" and obeys Newton's laws.

 

[Dale]

An example would be a hand pulling a cart via a massless rope. Positive work is done on the rope by the hand and negative work is done on the rope by the cart.

Another example would be a massless spring in the same scenario. In that case the positive work done by the hand may not equal the negative work done by the cart.

 

[kuruman]

An example would be a hand pulling a cart via a massless rope. Positive work is done on the rope by the hand and negative work is done on the rope by the cart.

I understand that, as an approximation of a real situation, we can ignore the mass of the rope and consider essentially an action-at-a-distance force from the hand to the rope. However, the approximation that the rope is massless does not refute my argument. I wrote

Here, I think that it is to be interpreted here as "resistance to acceleration" i.e. the reaction force exerted by the system on the external agent that applies the force. It's always there as long as the system has inertial mass. That's what I tried to illustrate in post #5.

In terms of this example, the reaction to the action of the hand is exerted by the cart on the hand via the massless rope. This reaction can be construed as a "resistance" force to the pulling force of the hand. The rope in-between is immaterial, literally and figuratively.

 

[jbriggs444]

Here, I think that it is to be interpreted here as "resistance to acceleration" i.e. the reaction force exerted by the system on the external agent that applies the force. It's always there as long as the system has inertial mass.

In the case of interaction forces, the "reaction force" is automatically there by Newton's third law. We need not mention inertial mass. Yes, this reaction force is always there. For interaction forces.

In the case of pseudo-forces such as the centrifugal force, there is no reaction force. Nor is there any "resistance to acceleration". Yet pseudo-forces do work.

 

[A.T.]

In terms of this example, the reaction to the action of the hand is exerted by the cart on the hand via the massless rope. This reaction can be construed as a "resistance" force to the pulling force of the hand.

That's a short cut you can take, but don't have to. You can just as well consider the work done on/by the massless rope, and it works just fine

The massless spring makes it even clearer. It can store energy so obviously you can do work on it, since even the net work is non zero.

 

[Dale]

However, the approximation that the rope is massless does not refute my argument.

I wasn’t intending to either support or refute your argument. Just give an example.

In terms of this example, the reaction to the action of the hand is exerted by the cart on the hand via the massless rope

Why would that be? The reaction to the force of the hand on the rope is the equal and opposite force of the rope on the hand. The reaction of the force of the rope on the cart is the equal and opposite force of the cart on the rope.

As you yourself indicated, this is an approximation to a real situation. In that real situation the forces are as I described. That doesn’t change as the mass gets arbitrarily small and then goes to zero.

 

[Nugatory]

Energy is quite an ambiguous topic in physics.

Less so when we use math instead of natural language with its built-in ambiguities…. but if you haven’t seen https://www.feynmanlectures.caltech.edu/I_04.html you may want to give it a try.

 

[physicals]

Thanks @Dale @kuruman @A.T. @jbriggs444 , sorry for the confusion caused here. My definition for resistance was something like inertia, according to some sources including (Chatgpt) inertia isn't a force, so It can be called as resistance. In my original post I have noticed that my wording was quite ambiguous. To clarify, this is what I meant: Energy is transferred to the kinetic energy store of an object if work is done against its inertia (considering no friction or external forces), for example to move an object that is currently at rest "work must be done" against inertia (resistance) in order to transfer energy. In this case can inertia be considered to be a force? Or is inertia an exception to the work done definition?

 

[PeroK]

Or is inertia an exception to the work done definition?

Inertia is another name for mass. It's not a force. Newton's second law describes the relationship between force, mass and acceleration:
$$F = ma$$

 

[A.T.]

Energy is transferred to the kinetic energy store of an object if work is done

That is just one special case. The most trivial one. You can't base the definition of a general concept like work on that.

In general the object can also pass on the energy instead of gaining kinetic energy, or store it internally as deformation. The inertia or inertial mass of the object is irrelevant then.

against its inertia

Just call it 'mass'.

 

[physicals]

So its agreed: Inertia or "mass" is an exception to the work done definition right?

 

[weirdoguy]

So its agreed: Inertia or "mass" is an exception to the work done definition right?

No:

The inertia or inertial mass of the object is irrelevant then.

Besides, what would that even mean? You need force to accelerate a thing, and the bigger mass the bigger the force for the same acceleration. That is all that "inertia" is needed for.

 

[A.T.]

So its agreed: Inertia or "mass" is an exception to the work done definition right?

No, the general definition, which doesn't mention 'inertia, or 'resistance', applies regardless if the work done goes into acceleration or not.

The definition based on inertia that you provided is simply not the general one.

 

[PeroK]

So its agreed: Inertia or "mass" is an exception to the work done definition right?

This is what happens when you try to learn physics from ChatGPT!

 

[Dale]

So its agreed: Inertia or "mass" is an exception to the work done definition right?

As others have said, no.

My preferred definition is that work is the amount of non-thermal energy transferred into a system across an interface.

Mass is not an exception to that definition. And I don’t really see how it would be an exception to any other definition either.

 

[kuruman]

Why would that be? The reaction to the force of the hand on the rope is the equal and opposite force of the rope on the hand. The reaction of the force of the rope on the cart is the equal and opposite force of the cart on the rope.

I think we agree but are talking past each other so let me do some math to explain what I am saying with equations instead of just words. Shown on the right are two FBDs: (A) for the massive rope plus cart and (B) for just the massive rope. Eventually, we will let the mass $m$ of the rope to go to zero. The double subscript label as in $F_{\text{HR}}$ indicates the force exerted by the Hand on the Rope.

From FBD (A) we get the common acceleration $$a=\frac{F_{\text{HR}}}{M+m}.$$From FBD (B), $$F_{\text{Net}}^{(\text{rope})}=F_{\text{HR}}+(-F_{\text{CR}})=\frac{mF_{\text{HR}}}{M+m}\implies F_{\text{CR}}=\frac{MF_{\text{HR}}}{M+m}.$$In the limit $m=0$ we get $F_{\text{CR}}=F_{\text{HR}}$. I think that we both agree so far. At this point you have argued that when the rope is massless, $F_{\text{HR}}$ does positive work whilst $F_{\text{CR}}$ does an equal amount of negative work, therefore the net work don on the rope is zero, therefore there no change in the kinetic energy of the rope.

When I said (without equations),

n terms of this example, the reaction to the action of the hand is exerted by the cart on the hand via the massless rope.

I meant that, in the case of a massless rope, it doesn't matter that $F_{\text{CR}}=F_{\text{HR}}$ and that the two forces do equal and opposite work on the rope. The change in a massless rope's kinetic energy is zero because, well, its mass is zero therefore its kinetic energy is always zero no matter what work is done on it.

Arguably, the length of the rope is irrelevant; it can be reduced to zero and the rope be ignored altogether without affecting the dynamics of the cart. A string with zero length and zero mass is no string at all.

It follows that the hand pulling a cart of mass $M$ through a massless rope can just as well be FBD (C) shown on the right. There is of course the reaction counterpart of the force exerted by the cart on the hand, $F_{\text{CH}}$, which does not belong inside the FBD.

To summarize, when a massless, inextensible, string is attached to a system on one end and to a force $F$ on the other, one can always consider force $F$ applied directly on the system in the direction of the tension. That's what is implied when we say the "tension is the same at any point along the string."

 

[jbriggs444]

To summarize, when a massless, inextensible, string is attached to a system on one end and to a force $F$ on the other, one can always consider force $F$ applied directly on the system in the direction of the tension. That's what is implied when we say the "tension is the same at any point along the string."

All well argued and completely true.

None of this is in conflict with the fact that the mechanical work done by the hand on the end of the rope is given by the dot product of the force applied by the hand times the displacement of the rope end.

The inertial mass of the rope does not enter in. The extensibility of the rope does not enter in. The mass of whatever the rope is attached to does not enter in. If we know force and displacement, we simply compute their dot product.

We can all agree that in the case of a massless inextensible rope subject to no other forces that the work performed by the effort on the one rope end will be equal to the work performed by the other rope end on the load.

 

[physicals]

When we lift an object such that it is further away from the ground it gains gpe. Here what exactly does work? Is it the thing that lifted the object or is it gravity?

 

[kuruman]

When we lift an object such that it is further away from the ground it gains gpe. Here what exactly does work? Is it the thing that lifted the object or is it gravity?

Both. The thing that lifts the object does positive work on it while the gravity does negative work on it. Here, the negative of the work done by gravity is also known as the change in potential energy of the object + Earth system.

 

[physicals]

Is this a general rule for all energy transfers that there must be both positive and negative work done?

 

[PeroK]

Is this a general rule for all energy transfers that there must be both positive and negative work done?

No.

 

[kuruman]

Is this a general rule for all energy transfers that there must be both positive and negative work done?

No. If the thing that does the work pushes down on the object as it falls, both works are positive.

 

[physicals]

Positive and negative work done is pretty confusing. Does positive work done counteract negative work done just like all other physical values? If not, what do we really mean when we say work is done in opposite directions?

 

[weirdoguy]

Postivie work means energy is added to the system. Negative means that it is taken out. Think of energy as of bank account balance, and work is the transfer of money. Negative transfer means that money went out of your account.

If not, what do we really mean when we say work is done in opposite directions?

"Direction" does not fit to work very well, since work is a scalar and has no direction (apart from the direction of energy transfer in-out of the system).

 

[Dale]

some sources including (Chatgpt)

Don’t use any current LLM for physics. They are not adequate. You will spend more effort unlearning the mistakes it makes than actually getting anything useful from it.

 

[kuruman]

None of this is in conflict with the fact that the mechanical work done by the hand on the end of the rope is given by the dot product of the force applied by the hand times the displacement of the rope end.

When mechanical energy is conserved, the work line integral can be linked to either a change in kinetic energy, which requires a mass, or a change in potential energy which requires a change in the spatial configuration of parts of a multicomponent system. To explain this to a novice the usual statement is that "By doing work, forces transform energy from one form to another."

One can certainly calculate the line integral $~W=\int \mathbf F\cdot d\mathbf s$ for one end of the string and then for the other and conclude that the net work done on the string is always zero. I see no conflict anywhere either. I am only expressing my doubts whether the "work done on the rope" is a useful construct when trying to explain energy transformations through work to a novice.

 

[physicals]

Postivie work means energy is added to the system. Negative means that it is taken out. Think of energy as of bank account balance, and work is the transfer of money. Negative transfer means that money went out of your account.



"Direction" does not fit to work very well, since work is a scalar and has no direction (apart from the direction of energy transfer in-out of the system).

According to this idea doesn't it mean that if something causes an object to gain gpe then the same energy it gains through this way is lost since it experiences opposite work done by gravity. Or if this idea is correct then the work done by the thing that caused the gain in gpe would have had to do twice the work as gravity which of course doesn't make sense. So how does it make sense if the work done (both negative and positive) are equal in magnitude? isn't there supposed to be a net work done of 0?

 

[physicals]

When mechanical energy is conserved, the work line integral can be linked to either a change in kinetic energy, which requires a mass, or a change in potential energy which requires a change in the spatial configuration of parts of a multicomponent system. To explain this to a novice the usual statement is that "By doing work, forces transform energy from one form to another."

One can certainly calculate the line integral $~W=\int \mathbf F\cdot d\mathbf s$ for one end of the string and then for the other and conclude that the net work done on the string is always zero. I see no conflict anywhere either. I am only expressing my doubts whether the "work done on the rope" is a useful construct when trying to explain energy transformations through work to a novice.

I understand all of this to some extent, can we say the same for a massless objects that supposedly gains gpe. My idea is that an object would gain gpe regardless of its mass, mass is only needed for energy transfers that involve kinetic energy.

 

[jbriggs444]

my doubts whether the "work done on the rope" is a useful construct when trying to explain energy transformations through work to a novice.

Can you clarify? By "work done on the rope", which of these do you mean:

1. Total external force on the system multiplied by the displacement of some point on the structure?

2. Force on a particular interface multiplied by the displacement of the material point where the force is applied?

3. External force across a particular interface multiplied by the displacement of some other point on the structure?

4. Something else?

For me, #2 above is the essence of mechanical work. It identifies an energy flow. For me, whether that flow involves a transformation from one sort of energy to another is less relevant than that a flow has been identified.

 

[kuruman]

I understand all of this to some extent, can we say the same for a massless objects that supposedly gains gpe. My idea is that an object would gain gpe regardless of its mass, mass is only needed for energy transfers that involve kinetic energy.

When you release an object from rest at some height above round, it trades potential energy for kinetic energy and picks up speed. A massless object will not be attracted by the Earth, has no potential energy to trade and will just hover above ground.

 

[jbriggs444]

When you release an object from rest at some height above round, it trades potential energy for kinetic energy and picks up speed. A massless object will not be attracted by the Earth, has no potential energy to trade and will just hover above ground.

The laws of Newtonian mechanics make no prediction for the motion of a massless object under a zero net force. If $F=ma$ and $m = 0$ then we are assured that $F=0$ but any value for $a$ is consistent with Newton's second law.

If we insist that a prediction be made about the trajectory of a massless object subject to gravity, the usual approach is to take the limit as mass and, consequently, gravitational force go to zero. The resulting prediction is that a massless object will have the same acceleration as an object with mass.

 

[kuruman]

Can you clarify? By "work done on the rope", which of these do you mean:
. . .
4. Something else?

Answer 4. You and I agree. It all started with me trying to understand this in post #10

An example would be a hand pulling a cart via a massless rope. Positive work is done on the rope by the hand and negative work is done on the rope by the cart.

 

[A.T.]

So how does it make sense if the work done (both negative and positive) are equal in magnitude?

That's what transfer of energy means: One object loses it, one gains it.

But sometimes you have dissipation / generation of mechanical energy (conversion into / from other energy forms). So the mechanical energy gained / lost are not always equal in magnitude.

 

[kuruman]

The laws of Newtonian mechanics make no prediction for the motion of a massless object under a zero net force.

Doesn't Newton's first law make the prediction that an object will retain its state of motion unless acted upon by an unbalanced force?

Anyway, I think that this thread-in-a-thread detracts from helping the OP. If you wish to continue it, it can be done PM.

 

[weirdoguy]

According to this idea doesn't it mean that if something causes an object to gain gpe then the same energy it gains through this way is lost since it experiences opposite work done by gravity.

If you use GPE you don't use work done by gravity.
1. If you use $W_{net}=\Delta E_k$ and there is external force acting on a body, namely $\vec{F}$ with the same magnitude as gravity, you have: $(F-mg)\Delta x=0$ so $W_F=mg\Delta x$. I assumed that body moves upwards.
2. If you use $W_{external}=\Delta E_{mechanical}$ you have $W_F=\Delta E_P=mg\Delta x$.

Either way, I think that you should study some textbook on the very basics of this things.

 

[Dale]

I am only expressing my doubts whether the "work done on the rope" is a useful construct when trying to explain energy transformations through work to a novice.

Let me explain my pedagogical approach before responding to some of your previous responses. Hopefully if you understand my pedagogical process you will understand why I take a different approach than yours.

A new student does not have much physics intuition. Intuition comes from experience. So my goal is to give a new student a small number of fixed rules that can simply be applied, without intuition, to consistently get the right answer. Intuition can then build as they gain experience, but they will have standard tools that they can reliably use whenever intuition fails them.

the reaction to the action of the hand is exerted by the cart on the hand via the massless rope

This is, to me, problematic pedagogy because it requires intuition. Ropes are objects that have forces acting on them, not forces themselves. The hand does not touch the cart, it touches the rope. The reaction to the action of the hand is a force acting on the hand, this force comes from the rope, not the cart.

The double subscript label as in $F_{HR}$ indicates the force exerted by the Hand on the Rope.

This type of subscripting is a very useful tool for teaching Newton's 3rd law. The 3rd law pair of $F_{HR}$ is $F_{RH}$, not $F_{CR}$. And $F_{CH}$ doesn't even exist. Whether the rope is massive or massless doesn't change any of that. It is important that it not change any of that so that the students learn just one simple set of rules and don't have to rely on intuition that they have not yet developed.

Contact forces are exerted between the things that are in contact. The hand is in contact with the rope, so there are contact forces between the hand and the rope. The rope is in contact with the cart, so there are contact forces between the rope and the cart. The hand is not in contact with the cart, so there is no contact force between the hand and the cart.

To summarize, when a massless, inextensible, string is attached to a system on one end and to a force F on the other, one can always consider force F applied directly on the system in the direction of the tension. That's what is implied when we say the "tension is the same at any point along the string."

I would never teach this. What I would teach is that you can redraw the system boundaries. You can make the rope part of the cart system or part of the hand system. Since this adds no mass to the larger system, it will not change the acceleration of the larger system. But as long as you are speaking of the rope as its own system, then you need to identify the forces correctly. It is an analytical mistake to define the rope as its own system and have the forces act between the hand and the cart.

The change in a massless rope's kinetic energy is zero because, well, its mass is zero therefore its kinetic energy is always zero no matter what work is done on it.

This is why I included the massless spring as another example. There, the KE is still always zero, but the PE is not. If a student has been properly instructed in analyzing scenarios systematically, then they will have no problem analyzing the massless spring case and will not get tripped up by the shortcuts used in the massless rope case. Then, they will discover that the fact that the positive and negative works are equal is not due to the masslessness of the string, but rather it is due to the string's inextensibility. The masslessness only ensures that the KE is always 0, not that the works always sum to 0.

I am only expressing my doubts whether the "work done on the rope" is a useful construct when trying to explain energy transformations through work to a novice.

I assert that it is a useful construct because the novice needs simple conceptual tools that can arrive at the right answer. Not all ropes are massless or inextensible. Not all connections between hands and carts are ropes.

My preferred definition of work is that work is a transfer of energy by means other than heat. Mechanical work is a specific kind of work that can be calculated with $dW/dt=\vec F \cdot \vec v$ where $W$ is the work done on the system and $\vec v$ is the velocity of the material of the system at the point of application of the force $\vec F$. It becomes instructive to examine the way that energy flows through a machine, which can be done with this type of simple rule. Positive work being done on the rope at the hand end and negative work being done on the rope at the cart end is just a specific case of energy transfer that can be extended to camshafts, hydraulics, and other machine parts. I choose the mental tools that I give students specifically so that they can be clearly applied to as many scenarios as possible. They can develop intuitive shortcuts as they gain experience.

 

[A.T.]

To explain this to a novice the usual statement is that "By doing work, forces transform energy from one form to another."

Why try so hard to avoid what the definition of work actually says? I really don't understand this obfuscatory creativity.

One can certainly calculate the line integral $~W=\int \mathbf F\cdot d\mathbf s$

Yes, that is the formal definition of mechanical work, that the title of this thread asks about. It can be stated informally as: By applying forces over a distance objects transfer energy to other objects.

There is nothing in there about resistance, inertia or converting between different forms of energy.

I am only expressing my doubts whether the "work done on the rope" a useful construct when trying to explain energy transformations through work to a novice.

What is not useful to a novice who asks about the formal definition of work, is to give him just tricks and shortcuts (like ignoring the work done on the rope) or just non-generalizable special cases and pretending they are representative (like a single force accelating a mass).

 

[Dale]

I understand all of this to some extent, can we say the same for a massless objects that supposedly gains gpe. My idea is that an object would gain gpe regardless of its mass, mass is only needed for energy transfers that involve kinetic energy.

The gravitational PE of an object is $mgh$. For a massless object $mgh=0$ regardless of $h$.

 

[PeroK]

The gravitational PE of an object is $mgh$. For a massless object $mgh=0$ regardless of $h$.

And it's kinetic energy is zero, regardless of its speed. It might as well accelerate the same as everything else under gravity. I would expect the mythical massless rope to obey the law of gravity!

 

[Dale]

And it's kinetic energy is zero, regardless of its speed. It might as well accelerate the same as everything else under gravity. I would expect the mythical massless rope to obey the law of gravity!

Indeed. In Newtonian gravity everything accelerates at the same rate, regardless of the mass.

 

[kuruman]

Let me explain my pedagogical approach before responding to some of your previous responses. Hopefully if you understand my pedagogical process you will understand why I take a different approach than yours.

I understand and I thank you for taking the time to explain your approach in such detail. I don't disagree with you. We can't teach everything all at once so we have to make our own decisions about what to put forth and what to sweep under the rug when we make the first pass.