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?
 A: 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.
A: 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.
A: 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.
A: 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
$$W_a=-W_g=mg\Delta z$$ 
A: 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.
