What is the Definition of the Joule and How Does it Relate to Energy?

[ghazzguhk]

Mentor's note: Moved from HW for a better fit​

Homework Statement:: A joule is defined as "the energy transferred to an object when a force of one Newton acts on that object in the direction of its motion through a distance of one meter."
Explain this definition.
Relevant Equations:: 1J = 1N.m

Hello.
This is not a homework but just a question I'm asking myself. I did not found a more appropriate forum, sorry if this one is inappropriate.

I can't grasp the idea of the Joule definition. Here's my current state of understanding and some questions that are bugging me:
The joule represents an amount of 'energy' (I used 'energy' in the common sense).

In the Joule definition, there is mention of "in the direction of its motion", which means the Joule in only defined for moving objects?
Also, if I push a 1kg object for 1m (on a table, for left to right, for example) with a force of 1N, I used a certain amount of energy.

If I push a 10kg object for 1m with a force of 1N in the same setting, I used more energy, as the object took more time to go through this meter and thus I had to apply this 1N force for more time.
I have the impression that this is not taken into account (well, of course it is, but I don't see how) in the Joule definition.

Can somebody help me clarify this?
Thanks.

 

[Ibix]

In the Joule definition, there is mention of "in the direction of its motion", which means the Joule in only defined for moving objects?

No. The "direction of motion" thing comes in because if a body is constrained to move in one direction I can apply a force at an angle and have to work harder to achieve the same effect because I'm wasting energy. For example, this just says that if I have a cart on a track I could push it parallel to the track or I could push diagonally. The cart will move in the same direction in both cases, but will not accelerate as hard in the latter case because I'm wasting some of the force trying to grind the wheels against the tracks.

Also, if I push a 1kg object for 1m (on a table, for left to right, for example) with a force of 1N, I used a certain amount of energy.

This is not a good example because you are working against friction here, so there are two forces at work - your pushing and the table's friction. You need to consider a frictionless case - a mass on ice, for example, which will continue to slide at constant speed when you stop pushing.

Now you can think about your masses. If you apply a force of 1N, what will the accelerations be? After 1m, what will the velocities be and how long did it take? What will the kinetic energies at that velocity be?

I used more energy, as the object took more time to go through this meter and thus I had to apply this 1N force for more time.
I have the impression that this is not taken into account (well, of course it is, but I don't see how) in the Joule definition.

No, the time for which you are applying the force doesn't come into the definition. If you are applying a force to a mass then you have specified its acceleration. You are free to specify its initial velocity, and then you can have any time you like to cover the specified meter - but you will find that the difference in kinetic energies is independent of everything except the distance, the force, and the mass. Edit: corrected stupid typo.

 

[Delta2]

If I push a 10kg object for 1m with a force of 1N in the same setting, I used more energy, as the object took more time to go through this meter and thus I had to apply this 1N force for more time.

Yes the situation here seems counter intuitive but it turns out if we do the math the work will be the same in both cases. The reason is that the rate we offer energy to the object is $F\cdot v$ and the work is the integral of that over time, that is $W=\int_{t_0}^{t_1} \vec{F} \cdot \vec{v} dt$. In the case of the second object, the time interval $\Delta t=t_1-t_0$ is larger but the velocity $v$ is smaller.

 

[ghazzguhk]

I failed to use the quoting tool, so I'll quote by putting in italic.

No. The "direction of motion" thing comes in because if a body is constrained to move in one direction I can apply a force at an angle and have to work harder to achieve the same effect because I'm wasting energy. For example, this just says that if I have a cart on a track I could push it parallel to the track or I could push diagonally. The cart will move in the same direction in both cases, but will not accelerate as hard in the latter case because I'm wasting some of the force trying to grind the wheels against the tracks.
Ok! This part makes sense for me.

This is not a good example because you are working against friction here, so there are two forces at work - your pushing and the table's friction. You need to consider a frictionless case - a mass on ice, for example, which will continue to slide at constant speed when you stop pushing.
Yes, I did not consider friction but forgot to write it, sorry. I agree, we only care about the force we exert here.

Now you can think about your masses. If you apply a force of 1N, what will the accelerations be? After 1m, what will the velocities be and how long did it take? What will the kinetic energies at that velocity be?

Here are my calculation (and they seem correct, as I find the same result, 1J, in both cases):

Formulas reminder:
Newton = kg * meter * second^-2
final speed = initial speed + acceleration * time
distance covered = initial speed * time + 1/2 * acceleration * time^2
kinetic energy = 1/2 * mass * speed^2

Problems information:
initial speed = 0 m.s^-2
distance covered = 1 m

Case of the 1kg object:
° acceleration = 1 m.s^-2
° distance covered = initial speed * time + 1/2 * acceleration * time^2
° 1 = 0*time + 1/2 * 1 * time^2 <=> time^2 = 2 <=> time ≃ 1.41 s
° final speed ≃ 1.41 m.s^-2
° kinetic energy at the end of this meter ≃ 1/2 * 1 * 1.41^2 = 1 J

Case of the 10kg object:
° acceleration = 0.1 m.s^-2
° distance covered = initial speed * time + 1/2 * acceleration * time^2
° 1 = 1/2 * 0.1 * time^2 <=> time^2 = 20 <=> time ≃ 4.47 s
° final speed ≃ 0.447 m.s^-2
° kinetic energy ≃ 1/2 * 10 * 0.447^2 = 1 J

Also you said:
but you will find that the difference in kinetic energies is independent of everything except the distance, the force, and the mass.
Did you mean "and *not* the mass"? If yes I agree, else I don't understand what you meant.

So, for answering both @Ibix and @Delta2, here is my reformulated question:
If I put a weight on a pulley system that gradually use the potential energy of the weight by lowering it in order to push either my 1kg or 10kg objects, they will end up with the same kinetic energy, which is the same as the weight's, when the weight touches the ground (modulo loss and other forces), right?
But in this case, why do I have the feeling that applying a force on the 10kg object for 4.5s would be harder than applying the same force on the 1kg object for 1.4s? Like, if I had to do it with my arm I would rather use the smaller object. I fill like I will have to put the same intensity in both cases (as I apply the same force), but one during 1.4s and one during 4.5s. What do I miss?

 

[jbriggs444]

Like, if I had to do it with my arm I would rather use the smaller object. I fill like I will have to put the same intensity in both cases (as I apply the same force), but one during 1.4s and one during 4.5s. What do I miss?

The relevant measure here is how much energy (aka "work") was transferred to the mass. Not how much ATP was consumed in your muscle fibers to perform it. Nor how much "burn" you feel after exerting the effort.

The human body is not an efficient machine. Holding a one gallon jug of milk at arms length for one minute feels like a lot of effort. And it is. But it does not involve any work at all being done on the jug of milk. It is at the same place it started. You might as well have left it on the table. [Tables are pretty efficient at doing no work with no effort].

Note that there is a quantity that reflects the amount of force that you apply to an object multiplied by the duration for which you apply it. That quantity is "momentum". The momentum ($p$) of a moving object is given by $p=mv$. The amount of momentum transferred by a constant force $F$ acting over a duration $\Delta t$ is given by $\Delta p=F\Delta t$.

 

[Delta2]

What do I miss

I told you already, I ll tell you again, it is the speed of the masses that you don't take into account. The smaller mass will have a larger velocity curve, hence the $F\cdot v$ will be larger, but it will be for smaller amount of time $\Delta t$. The larger mass will have a smaller velocity profile hence $F\cdot v'$ will be smaller but it will be for a larger amount of time $\Delta t'>\Delta t$. The work , in a simplified form, is the product of $F\cdot v$ and the required time interval $\Delta t$(to be more accurate it will be the integral of $F\cdot v$ over that time interval).

So its like having $A\Delta t$ for the work in one case and $A'\Delta t'$ for the work in the other case $A=F\cdot v,A'=F\cdot v'$. It is $A>A'$ but $\Delta t<\Delta t'$ so the products can be equal.

 

[ghazzguhk]

@jbriggs444 Maybe I have a problem with the notion of 'work' then.
I understand that the jug of milk do not gain any energy at being held at arm length.
I still am a bit blurry on these notions. For example, if the human always applies a constant force on the jug to oppose the gravity, it does not involve any work on the jug. But, if the jug of milk is being nudged (upwards) ever so slightly, then the human, by still only opposing gravity, can carry the jug as high as he wants (he has extensible arms), and therefore the work on the jug is as high as the human wants (depending on when he stops)?

 

[ghazzguhk]

@Delta2 Yes, I see, or mainly believe, that the math checks out, but i don't intuitively understand how I should take the speed of the objects into account.

 

[Delta2]

@Delta2 Yes, I see, or mainly believe, that the math checks out, but i don't intuitively understand how I should take the speed of the objects into account.

I see, haven't you met the formula $P=F\cdot v$ before? It is the instantaneous power of the force F applied to a body that is moving with velocity v. Even if the force is the same, the larger the velocity $v$ the larger the instantaneous power P.

 

[Ibix]

Did you mean "and *not* the mass"?

I did - sorry.

But, if the jug of milk is being nudged (upwards) ever so slightly, then the human, by still only opposing gravity, can carry the jug as high as he wants (he has extensible arms), and therefore the work on the jug is as high as the human wants (depending on when he stops)?

So if the human applies a force $mg$ and lifts the object $h$ the total work done is $mgh$, yes.

i don't intuitively understand how I should take the speed of the objects into account.

In terms of energy, you don't have to, it only depends on distance and force. Time and/or speed only comes in when you think in terms of the rate at which the energy is transferred - the power.

 

[jbriggs444]

@jbriggs444 Maybe I have a problem with the notion of 'work' then.
I understand that the jug of milk do not gain any energy at being held at arm length.
I still am a bit blurry on these notions. For example, if the human always applies a constant force on the jug to oppose the gravity, it does not involve any work on the jug. But, if the jug of milk is being nudged (upwards) ever so slightly, then the human, by still only opposing gravity, can carry the jug as high as he wants (he has extensible arms), and therefore the work on the jug is as high as the human wants (depending on when he stops)?

Yes. If the human has telescoping arms that can extend indefinitely then he can slowly lift the jug so that it ends up many kilometers over his head. This will involve work being done on the jug by the man's hand.

I think I see where you are going with this...

The man ends up exerting the same force on the object as before. And he exerts it for the same time interval as before. So he imparts the same momentum as before. We can see that this is the case because in both cases the jug's change in velocity from start of scenario to end is the same (ignoring that from the tiny nudge). Its momentum is unchanged in both scenarios.

But although the force is the same, the distance through which that force was applied is different. So the amount of energy imparted (or work done) is different. We can see this as well. The jug ends up far higher than it started. In the scenario with the nudge, the jug has gained potential energy.

Let us see if we can trace this difference back to the man...

In the first case, the man is rigid. We could replace the man with a steel frame (or table) that supports the jug. Clearly such a steel frame can provide no energy. It is just standing there. No batteries required.

In the second case, the man has extensible arms. They are extending. The act of extending those arms against a resistance unavoidably requires energy. The amount of energy required is given by the force times the displacement. You have to provide batteries. A gas tank. A spring. Some source of energy.If you want to make things interesting, we could place the man on an escalator and start talking about frames of reference and invariance.

 

[ghazzguhk]

@Delta2 Oh, I did not. So the formula $P = F \times \nu$ means, in non-scientific terms, that the difficulty of applying a force (in the same direction as the object is moving) on a moving object is proportional to its speed?
That would make sense, as I find it more difficult to accelerate a moving swing than a stopped swing, although this may also only be because of friction..
And so the biggest error I was making was to believe that the effort I had to do to push an object was only dependent on the force I wanted to apply, but not taking into account the speed on objects, as you said. So the human effort is more describable in Watts than in Newtons.
I find this formula fascinating, as
- it expresses something really not intuitive (at least for me), like why would the power needed to apply a force vary proportionally to the speed of the object and not quadratically?
- It is simply a unit conversion between Newtons and Watts.
This 'truth' is therefore embedded into our (the international) unit system?

 

[jbriggs444]

@Delta2 Oh, I did not. So the formula $P = F \times \nu$ means, in non-scientific terms, that the difficulty of applying a force (in the same direction as the object is moving) on a moving object is proportional to its speed?

Yes. One common way to experience this is on a bicycle with ten (or more) gears. As you go faster and faster, you find yourself forced to higher and higher gears where it takes more work to accelerate.

[There is also increased wind resistance, but the effect we are discussing is in addition to that]

 

[Delta2]

@Delta2 Oh, I did not. So the formula P=F×ν means, in non-scientific terms, that the difficulty of applying a force (in the same direction as the object is moving) on a moving object is proportional to its speed?

Yes exactly.

I find this formula fascinating, as
- it expresses something really not intuitive (at least for me), like why would the power needed to apply a force vary proportionally to the speed of the object and not quadratically?

Well the definition of Work, Power and Velocity are such that the math tell us that $P=F\cdot v$ not $P=F\cdot v^2$ or something else.

- It is simply a unit conversion between Newtons and Watts.

I am not sure what you mean here. Usually when we talk about unit conversion we talk about different units of the same quantity, for example conversion between Joules and KWh for the quantity of Energy.
Here Watts are units of the quantity Power , but Newtons are units of the quantity Force.

 

[ghazzguhk]

@Delta2
Well the definition of Work, Power and Velocity are such that the math tell us that not or something else.
Oops, i wanted to propose P = F * 2v, for example, for that units are still valid.

I am not sure what you mean here. Usually when we talk about unit conversion we talk about different units of the same quantity, for example conversion between Joules and KWh for the quantity of Energy.
Here Watts are units of the quantity Power , but Newtons are units of the quantity Force.
I don't know if 'unit conversion' was the right term, but I meant that, because
Newtons = kg * meter * second^-2 and Watt = kg * meter^2 * second^-3, it is sufficient to take any value of unit meter * second^-1 (which happens to be a speed) and multiply it with a value expressed in Newtons to obtain a value expressed in Watt.
There is no thinking in it, just as if I told you that to obtain a value of unit 'meter^2', you had to multiply two values of unit 'meter'.
This can be differentiated (I believe this could be a pun) from the formula 'kinetic energy = 1/2 * mass * speed^2', where a '1/2' appears, making it more than a 'unit conversion'.
I don't know if what I wanted to express is clear..

 

[ghazzguhk]

Any other questions that I have on the subject are not really fitting in this thread and I believe my initial question has been answered.
Thank you @Ibix , @Delta2 and @jbriggs444 for you answers!