Is the Definition of Work Done Applicable to Free Fall?

[Darmstadtium]

For example, if a ball is from a certain height, the work done is 0 as there is no change in total energy the Ef =Ei. However, there is a constant force applied over a certain distance, suggesting work is being done. Which aspect am I forgetting/missing? Or is it that the definition of work done applies only to the change in kinetic energy?

Thanks!

 

[Baluncore]

... , if a ball is ??? from a certain height, ...

What is done to the ball ???

 

[Darmstadtium]

What is done to the ball ???

free fall

 

[Baluncore]

The ball is "dropped" near the Earth, where the acceleration due to gravity is g, and where there is no atmosphere to slow the ball down.

When a ball with a mass, m, is dropped, from height h;
the potential energy of the ball, PE = m·g·h ;
is progressively converted into kinetic energy; KE = ½·m·v².
The total energy will remain constant, it will be conserved.
As the ball falls, the force, F = m·g, accelerates the ball.
In effect, the PE is doing work on the KE.

 

[vanhees71]

It doesn't make sense to say that "the potential energy is doing work on the kinetic energy". I'd rather say that "potential energy is transformed into kinetic energy".

 

[Baluncore]

I agree with you.
But what is "doing work"?
I believe doing work is a flow of energy, which is happening in this case.
The Force is constant, so the rate of energy flow, or energy conversion, is proportional to velocity.

 

[vanhees71]

In the most general case you have the socalled "work-energy theorem", which is not of much practical use though. It's just formally integrating Newton's equation of motion for a particle, on which an arbitrary external force acts:
$$m \ddot{\vec{x}}=\vec{F}(\vec{x},\dot{\vec{x}},t).$$
Taking the line integral of both sides between two times $t_1$ and $t_2$ along the trajectory of the particle, i.e., a solution $\vec{x}(t)$ of this equation of motion,
$$\frac{m}{2} [\dot{\vec{x}}^2(t_2)-\dot{|vec{x}}^2(t_1)]=\int_{t_1}^{t_2} \mathrm{d} t \vec{F}[\vec{x}(t),\dot{\vec{x}}(t),t] \cdot \dot{\vec{x}}(t),$$
which says that the change in kinetic energy is given by "the work done on the particle".

This becomes something really useful, i.e., something helping to solve the equations of motion, if the force is "conservative", i.e., if it is of the form
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}),$$
where $V(\vec{x})$ is the potential of the force, and it's not explicitly time dependent. Then also the right-hand side of the work-energy theorem is given by
$$-[V(\vec{x}(t_2)-V(\vec{x}(t_1)],$$
and then you can rearrange the work-energy theorem to the energy-conservation law
$$E=T+V=\text{const}.$$

 

[malawi_glenn]

Potential energy just means how much work that force can perform. You can think of it as "latent work" or "work that has not yet been done".

A ball with mass m at a height h above the surface of the Earth (h much smaller than Earth radius) has the potential energy mgh. This means that the force of gravity (mg) can perform work equal to mgh on that ball.

According to the work-energy theorem (as written above by vanhees1), work performed on an object is equal to the change of the kinetic energy of that object.

Here is a simpler proof of that theorem, considering the special case of a constant force (which is applicable to gravitational force close to the surface of the Earth).
Work: W = Fh = mgh.
But, because F = ma we can write the work as W = mah. Note that because the force mg is constant, we also have constant acceleration a = g. So, we can use the basic formulas v = v₀ + at and s = (v + v₀)t/2.
Now, use that a = g and that the displacement s is equal to h.
$W = mah = m \cdot \dfrac{v-v_0}{t} \cdot \dfrac{(v+v_0)t}{2} = \dfrac{mv^2}{2} - \dfrac{mv_0{}^2}{2}$ which is the change in the kinetic energy $E_\text{k,final} - E_\text{k,initial}$.@vanhees71 you probably missed the "B" tag :)

 

[Herman Trivilino]

I agree with you.
But what is "doing work"?

In this case it's the force of gravity $mg$ that's doing work.

 

[Baluncore]

In this case it's the force of gravity mg that's doing work.

The work done is the force multiplied by the height of the fall.
Why can you not equally well say it is the height of the fall that is doing the work?

 

[A.T.]

But what is "doing work"?

In Newtonian Gravity it's the Earth that exerts a force on the falling object. But the energy comes from the gravitational field, so it might be better to say that the field is doing the work.

 

[Herman Trivilino]

Why can you not equally well say it is the height of the fall that is doing the work?

Because work, by definition, is done by a force.

 

[sophiecentaur]

Because work, by definition, is done by a force.

I prefer to say that Work is Done (but not by anything, in particular). There is work done due to gravity and also due to the friction (drag) that part of the work moves and heats up the air. The result is that the parachutist gains some KE but less than if there were no air. Work done 'by' or 'to' makes it too personal for me and implies some agency.

I can't make up my mind whether this is just my idle fancy or a relevant distinction.

 

[Herman Trivilino]

I can't make up my mind whether this is just my idle fancy or a relevant distinction.

The work done must be given a sign (either positive or negative) when trying to use things such as the 1st Law of Thermo. Whether one uses "by" or "on" or some other method to keep track of the sign is of course a matter of preference.

So, perhaps it's idle fancy.

 

[A.T.]

I prefer to say that Work is Done (but not by anything, in particular).

Since work represents the transfer of mechanical energy, it is quite relevant what transfers energy to what.

 

[Baluncore]

Since work represents the transfer of mechanical energy, it is quite relevant what transfers energy to what.

An object or particle has state variables of mass, height and velocity.
It also has two reservoirs of energy, potential and kinetic, functions of the above.

When a mass falls, the kinetic energy of the mass is increased, while the potential energy of the mass is reduced. Energy flows between the reservoirs.

The potential field does not do the work. The force does not appear to be raising or lowering the energy of the mass. The force somehow mediates the rate of transfer of energy between the reservoirs of energy associated with the mass.

 

[sophiecentaur]

Since work represents the transfer of mechanical energy, it is quite relevant what transfers energy to what.

That sounds ok as far as it goes and in many circumstances it's easy to say which is 'by' and wjhich is 'on'. But in the case of a car being driven with the handbrake partly on? By dissipating energy, you could say the handbrake is doing negative work - but what 'on'? (That's not really my question but it will be the question of many students.) Why not just say work is done and give the sign and the question disappears.

This topic is only one of many, including the Ohm's 'law' and resistance inconsistency. So many students have problems simply because the wording used by 'experts' demands extra appreciation of a subject. It worries me that you guys all sorted these things out and can cope without a second thought but I have to ask if you really did slip into this understanding with no trouble or was it a problem for you at the start?

 

[vanhees71]

For me there's much hype about this in this thread, which is completely unnecessary. The simple point is that the force has a potential,
$$V(\vec{x})=-m \vec{g} \cdot \vec{x}.$$
Thus you have the equation of motion
$$m \ddot{\vec{x}}=\vec{F}=-\vec{\nabla V}=m \vec{g}.$$
multiplying the equation of motion with $\dot{\vec{x}}$ ("dot product") you get
$$m \dot{\vec{x}} \cdot \ddot{\vec{x}}=-\dot{\vec{x}} \cdot \vec{\nabla} V.$$
Now both sides are total time derivatives, i.e.,
$$\frac{\mathrm{d}}{\mathrm{d} t} \left [\frac{m}{2} \dot{\vec{x}}^2+V(\vec{x}) \right]=0,$$
or
$$\frac{m}{2} \dot{\vec{x}}^2 + V(\vec{x})=E=\text{const}.$$
If you let the body fall from rest, indeed it gets accelerated through the gravitational field of the Earth and it's kinetic energy increases and the potential energy decreases. The total energy stays constant, i.e. (within this approximation with the Earth as "infinitely heavy" as assumed here) no energy is transferred.

All the other lingo about "who does work on whom" seems rather confusing to me.

 

[A.T.]

(That's not really my question but it will be the question of many students.) Why not just say work is done and give the sign and the question disappears.

Why not just answer the question of the students?

 

[A.T.]

An object or particle has state variables of mass, height and velocity.
It also has two reservoirs of energy, potential and kinetic, functions of the above.

Potential energy is not a property the individual object but of the whole system of gravitationally bound particles. This becomes more clear when you have two similar masses.

The potential field does not do the work.

Consider two masses connected with a rubber band under tension. The energy is stored in the rubber band and the rubber band does work on both masses accelerating them. Now replace the rubber band with the gravitational field.

 

[sophiecentaur]

Why not just answer the question of the students?

But they're not all my students and the question / confusion can be expressed in many ways. But my answer would be that you don't actually need to have a subject and object (grammatical terms) involved. As you say, it's the whole system involved.

Consider two masses connected with a rubber band under tension. The energy is stored in the rubber band and the rubber band does work on both masses accelerating them. Now replace the rubber band with the gravitational field.

So which end of the rubber band is doing work on which? Your example totally makes my point. If there is a tank of fuel on one part of the thought experiment then you 'could' treat that part as supplying energy and doing mechanical work 'on' something. The net energy flow could give a clue as to the 'by or on' question but there can be multiple sources of motive power so what then? Do the by and on terms suddenly change places as the sign changes. This is the old acceleration / deceleration problem that tries to do without the sign and we do our best to discourage that use of words.

I am surprised that so many people seem to think there's no problem here. But the problem arises only because of the insistence of using those two short words.

 

[Baluncore]

In the top-down analysis, there are networks or circuits, through which energy flows.

In the bottom-up analysis, the “does work on” or “work is done by” phrases are related to the analyst's viewpoint within the system. That is where the student finds themselves.

Can there be a middle ground ?

 

[A.T.]

But my answer would be that you don't actually need to have a subject and object (grammatical terms) involved.

But you do need to have a subject and object for work to be done, otherwise it's not clear what the direction of energy transfer is.

For kinetic friction between two objects (like in your slipping brake example), you can compute the work done by A on B, and the work done by B on A. The sum of these two will be negative, representing the energy dissipated as heat.

So which end of the rubber band is doing work on which?

The rubber band is doing positive work on both attached masses, while each attached mass is doing negative work on the rubber band.

 

[A.T.]

In the top-down analysis, there are networks or circuits, through which energy flows.

In the bottom-up analysis, the “does work on” or “work is done by” phrases are related to the analyst's viewpoint within the system. That is where the student finds themselves.

Can there be a middle ground ?

Defining work done and its subject & object depends on:
- How you cut the system into parts / bodies
- What reference frame you choose

So you can do the work analysis on different levels of abstraction / coarseness.

Note that at some of those abstraction levels the work analysis might stop to be useful to tell you anything meaningful. But then you should simply not do it, instead of saying: Work is done somehow in there, but don't ask me how or by what on what.

 

[malawi_glenn]

The simple point is that the force has a potential,

Not all forces has a potential though, like friction.

 

[sophiecentaur]

Defining work done and its subject & object depends on:
- How you cut the system into parts / bodies
- What reference frame you choose

So you can do the work analysis on different levels of abstraction / coarseness.

Note that at some of those abstraction levels the work analysis might stop to be useful to tell you anything meaningful. But then you should simply not do it, instead of saying: Work is done somehow in there, but don't ask me how or by what on what.

At least this and much of the above should suffice to prove that it ain't as straightforward as people were claiming further up the thread. This chosen quote is, of course, totally accurate but would it really help a student who feels it absolutely necessary to decide where to put the subject and object in their workings?

Energy flow is a good way to look at most situations but even in a simple mass on spring situation, the 'on and by' description involves continual change of preposition over the cycle. The system can be described easily enough without prepositions so why add confusion for the poor student. This is all just another instance of where Classification adds potential confusion and can reduce confidence for the beginner (and all of us at times).

There are many examples of PF-type responses which can produce unease when they not backed up with helpful comments. Responding to someone who needs help is different from responding to an obviously well informed member with whom you just happen to disagree. I could make a list of typical responses, starting with "Just draw a free-body diagram . . . . . . . . .", followed by a vacuum.

 

[A.T.]

This is all just another instance of where Classification adds potential confusion and can reduce confidence for the beginner (and all of us at times).

The point of the work theorem is not "classification", but relating forces to energy, so you can invoke energy conservation to solve problems. But this requires to be specific about which way the energy is flowing.

 

[sophiecentaur]

The point of the work theorem is not "classification",

Of course it's not but people try to classify blindly when a simple sign would be easier and always give the same answer because there is no feeling of cognitive dissonance. I know you know all this and so do I. I am speaking on behalf of those who are confused by it all.

Let's face it we have enough questions questions about whether or not the brakes are doing any work on anything when there is no motor inside them. That implies that 'they' don't get it. They ask how friction (a mechanism for loss) can be driving a car forward. It's because the rôle of a brake is not to provide energy yet it 'causes' the car to accelerate. If. your post hoc appreciation of it all would work for them, they (as you) would have no problem.