B Formal definition of work done

[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.