Work energy theorem: variation of kinetic energy as sum of terms given by the work of single forces Work-energy theorem states that $$W_{\sum F_i}=\Delta K,$$ where I expressed with $W_{\sum F_i}$ the work done by the resultant of forces ${\sum F_i}$ and with $K$ the kinetic energy.
In general it is valid that $$W_{\sum F}=\sum W_{F_i},$$ where I used $W_{F_i}$ for the work done by the single force $F_i$.
Nevertheless I read that it is not possible to divide the variation of $K$ in parts each caused by $W_{F_i}$. I mean that I cannot write $$\Delta K=\Delta K_{W_{F_1}}+\Delta K_{W_{F_2}}+...$$
I don't see the reason of that.
 A: The reason is that work always depends on the actual velocity the body has during its movement, even if you compute the work for only one force out of many.
In other words, $W_{F_i} = \int F_i\cdot v dt$, rather than, as you might perhaps expect, $W_{F_i} = \int F_i \cdot v_i dt$, where $v_i$ is a hypothetical "what $v$ would be if only $F_i$ acted".
Now you could define $W_{F_i}$ in such an incorrect way. After all, Newton's 2nd law tells us $F=ma$, and nothing prevents us from defining hypothetical $a_i =F_i/m$ as "the part of acceleration due only to $F_i$". Then we can split $v$ into $v_i$'s such that $\frac{dv_i}{dt}=a_i$, and define $W_{F_i}$ as the work done by $F_i$, but considering only the velocity "due" to $F_i$. 
Unfortunately, if you do that, the work so defined stops being additive: it is no longer the case that $W_{\sum F_i} = \sum W_{F_i}$. And that makes it pretty useless. I guess that demonstrates why $a_i$ and $v_i$ are pretty useless as well.
If I convinced you that the standard definition of work where actual velocity (to which many forces contributed) is used even for work of one force is the right one, now consider the reason the total work $W$ ends up being the change in kinetic energy.
Fundamentally this is because the power $F\cdot v$ (which work is the integral of), can be presented, due to Newton's 2nd law again, as the rate of change of kinetic energy$$F\cdot v = ma \cdot v = m(a \cdot v) = m \frac{1}{2}\frac{d|v|^2}{dt}$$
The kinetic energy arises here from $a\cdot v$ because they are related in the obvious way: $a$ is the derivative of $v$. If you now look only at the power of $F_i$, that is $F_i\cdot v = m(a_i \cdot v)$, there's no relationship between $a_i$ and $v$ that allows you to present this as derivative of some "kinetic $K_i$". And it has to be $v$ rather than $v_i$ for the reason explained above.
A shorter less precise way of saying the same thing: force splits into $F_i$ because it's linear in units of length ($kg*m/s^2$). Work splits into $W_i$ because we split only the part of it - the part that depends on force - that's linear in units of length, holding $v$ unchanged. Kinetic energy is necessarily quadratic in units of length ($kg*m^2/s^2$), because you have to multiply $v$ by itself. And that's why it cannot split into $K_i$ in a way that preserves additivity.
A: $\Delta K$ means $K_\text{end}-K_\text{start}$. How would you split this in parts?
A: It's not that it's not possible, it's just not necessary. What we seek is not the sum of the changes in kinetic energy. We seek the total change in kinetic energy. Why would the former be of use? (in Newtonian mechanics that is, since my knowledge is limited to it as of now)
If a box is accelerated through the force of $F$ through a distance $d$, from rest, the energy gained by the box is simply the work done on the box. If 10 forces composed that force $F$, the energy that the box gains does not change. So, why calculate the amount of energy each force gave to the box? 
