# Confusion between two different definitions of work?

I'm doing physics at high school for the first time this year. My teacher asked us this question: if a box is slowly raised from the ground to 1m, how much work was done? (the system is only the box)

Using the standard definition, $W = Fd\cos(\theta)$, the work should be 0, because the sum of the forces, the force due to gravity and the force of the person, is 0.

However, using the other definition he gave us, $W = \Delta E$, work is nonzero. $\Delta E = E_f - E_i$ , so that would be the box's gravitational potential energy minus zero.

My teacher might have figured it out but class ended. Does anyone have any insight?

• Fis here is Fnet .Fnet is the sum of forces but how can you say that it is zero , if it was zero then how the box had raised to a height. – Prabhat Nov 28 '16 at 7:59
• I believe many misunderstandings of this type are due to the fact that in physics we almost always omit the function parameters. The rigorous definition of work would explicitly state that $W$ is not a constant, but a function which depends on the force, the motion of the object etc. – Bakuriu Nov 28 '16 at 10:21
• The solution is in a proper definition of "the box's gravitational potential energy". I for one would contend that the box proper doesn't "have" any energy (besides its mass's energy equivalent). The energy is in the gravitational field. It is perfectly unclear to me how the teacher meaningfully wants to limit the system to the box if there's obviously essential interaction with the mass of earth. In isolation (no earth etc.), any net force would accelerate the box and create a text book kinetics example. – Peter - Reinstate Monica Nov 28 '16 at 15:55
• The sum of the Forces cannot be zero. If it were, then the box would not have moved. – RBarryYoung Nov 28 '16 at 16:36
• @RBarryYoung That's assuming that the box is initially at rest. – Devsman Nov 28 '16 at 19:04

You have a teacher who knows his/her Physics.

the system is only the box

That statement made by your teacher immediately means that there can be no mention of gravitational potential energy as it is a system comprising of the box and the Earth which has gravitational potential energy.
A system comprising of the box alone cannot have gravitational potential energy.

Acting on the box there are two equal in magnitude but opposite in direction (external) forces: the gravitational attractive force of the Earth on the box and the force exerted on the box due to the person.

The second equation $W=\Delta E$ is the work-energy theorem which states that the work done on a system is equal to the change in kinetic energy of the system.
In the example given the work done on the box is zero and the change in kinetic energy of the box is zero just the result found using the first equation.

• Nice answer, but watch out: the $\Delta E$ in the work-energy relation is equal to the change in kinetic energy only if the system has no internal degrees of freedom. This is implied by your answer, but that one phrase taken out of context ... – garyp Nov 28 '16 at 3:31
• @garyp Quite right but the OP did not seem to necessitate such a qualification as the question was all about how to deal with a specified very simple system. – Farcher Nov 28 '16 at 10:53
• I would be grateful for feedback on the down vote. – Farcher Nov 28 '16 at 10:53
• I didn't downvote, but I can give you some feedback. I disagree with the first part of your answer; it is totally reasonable to model the box as being in a fixed external potential, and in fact this type of model is used a lot. That is not the problem here. You are right that the first equation with the line integral of the force gives the correct answer, but the concept of potential energy can be used to give the answer as well. You can say the work is the line integral of the force from the lifter plus the change in gravitational potential energy... – Brian Moths Nov 28 '16 at 19:44
• However what hughjohnson222 did was to just take the work as being the change in potential energy. This is wrong because the change in potential energy is only the work done by the convservative force associated with that potential. To get the total work, you must also add in the work done by forces not associated with the potential. – Brian Moths Nov 28 '16 at 19:46

F is not the sum of the forces on the block, it is the force which is doing the work. It is either the force provided by the person (if you want to find the work done on the block by the person) or the force of gravity (if you want to find the work done on the block by gravity). You choose.

• For @hughjohnson222: both of your formulae are correct, you just have to be clear about what the involved parameters mean exactly. $W$ is in both cases the work done by a particular force, as Sammy explains here. Not the total force. – Steeven Nov 28 '16 at 0:25
• @Steeven I think that you have missed the point of the teacher's argument. – Farcher Nov 28 '16 at 0:51
• @Farcher : Yes I think we have. – sammy gerbil Nov 28 '16 at 1:43

Work is done by something, on something.

If you put the weight inside a box (so you can't see it), with the rope sticking out of the top, and you pull on the rope, you can say "I am doing work on something in the box". You don't know what the something is - gravity, a gang of minions, a very long spring, a paddle wheel in a bath of treacle, ... and it doesn't matter.

When you look inside the box, you will see that something else is also pulling on the box - but it is pulling in the opposite direction to the motion of the box. So gravity is doing negative work on the box, and we can say that the box + earth gains potential energy.

If you look at you, the box, the earth all together - then no net work was done on the total system (what you would have if you put you, the weight and the earth all in a really big box). No external forces acting on the contents of the box (for the purpose of this explanation) -> no net work. What actually happened is that your work was converted to potential energy of the weight, and the total energy of the system you+weight+earth is unchanged.

• It is not the box which is gaining potential energy, it is the box and the Earth. I think that the teacher was trying to make that distinction and that is why the two forces acting on the box due to the earth and the person were considered to be external forces. So if you have the box as the system which is implied in your second paragraph, the box cannot gain potential energy. If you consider the box and the Earth as the system then work is done on the system by the person exerting the force and that work done results in an increase in the gravitational potential energy of the system. – Farcher Nov 28 '16 at 1:27
• @Farcher thanks for the comment - I hope I have now clarified that phrase. – Floris Nov 28 '16 at 3:25

For a system with no internal degree of freedom (such as a point mass), work is equal to the change in kinetic energy:

$$W=\int_{\mathcal L} \vec F \cdot d \vec x = \Delta E_k$$

The $\vec F$ in the above equation is the net force. This is a very important point.

Let's model our box as a point mass. At the initial time, the box is still ($E_k^i=0$), and at the final time, it is again still ($E_k^f=0$). So,

$$W=\Delta E_k=E_k^f-E_k^i=0$$

Total work is $0$.

But, you can also ask yourself what is the work done by a single one of the two forces. The work done by gravity is

$$W_g=m\int_{\mathcal L}\vec g \cdot d \vec x =- m g \Delta z$$

where $\Delta z$ is the vertical displacement.

Or you could compute the work $W_a$ done by your arm, which is lifting the box. Since total work $W$ is $0$, we have

$$W_a=-W_g=mg\Delta z$$

# Using the wrong model / using the model wrong

You say...

Using the standard definition, $W = F\cdot d\cdot \cos(\theta)$...

This warrants clarification. This is a definition / model of the work done by someone that is dragging a mass where...

• $F$ is the magnitude of the force applied on the mass
• $d$ is the distance that the mass travels
• $\theta$ is the angle between the direction of travel and the direction that the force $F$ is applied.

The usual case when you use this model is when dragging the mass over a surface. In this case $\theta$ is (also) the angle between the ground plane and the force that is being applied, because the direction of travel is in the plane. In that particular case the model holds true even if you assume that $\theta$ is the angle between the plane and the force $F$.

But in your case, this condition is not true. The mass is not travelling in / parallel to the plane. Therefore it is wrong to assume that $\theta$ is $\pi/2$ or $90^\circ$.

In the case of lifting the weight, the direction of travel is straight up. And since the direction of the force $F$ is also straight up, this means that $\theta$ is $0$.

So you have either used the wrong model by defining $\theta$ to be the angle between the ground plane and the force $F$, or you have used the model wrong by assuming that the direction of travel is $\pi/2$, or $90^\circ$ in relation to the force $F$.

So you are using the model wrong in that you are including the counter-force in the formula. This is not how the model was meant to be used, because then the answer will always be $0$. Technically it is correct when you consider both gravity and the one lifting the mass. But the formula then becomes a useless tautology because the result is always nil.

I say again: the model is used to calculate the work done by the one that is dragging the mass. It is not meant to include the work done by that which is providing a counter-force. You can, if you wish, but that is a pointless enterprise.

• The reason that using the formula $W = Fd\cos(\theta)$ gave a value of zero is nothing to do with the angle it is because the net force on the box is zero. – Farcher Nov 28 '16 at 15:11
• @Farcher If we counted it that way, then $W=F\cdot d\cdot cos(\theta)$ would always yield $0$ because when dragging it over the ground the friction force would balance out the dragging force. So — again — the model is used wrong. That model is used to calculate the work done by the one that is dragging the mass. If you suddenly change the model by including the counter-force, then the equation... well it is not wrong but it is useless because the result is always $0$. – MichaelK Nov 28 '16 at 15:16