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[...] the infinitesimal work $\overline{\mathrm dw}$ comes out as a linear differential form of the variables $q_i$: $$\overline{\mathrm dw}= F_1~\mathrm dq_1 +F_2~\mathrm dq_2+ \ldots + F_n~\mathrm dq_n\,.\tag{17.3}$$ This differential form is of prime-importance for the analytical treatment of force. .... the coefficients $$F_1, ~ F_2, \ldots, ~F_n \tag{17.4}$$ are the components of a vector, but it is a vector of $n$-dimensional configuration space and not one of ordinary space. We call this vector the "generalised force" and the quantities $f_i$ the "components of generalised force".

[...] Let us assume that the infinitesimal work $\overline{\mathrm dw}$ is the true differential of a certain function so that the dash above $\mathrm dw$ may be omitted. The function is called the "work function" and is frequently denoted by the letter $U\,.$ Hence, we put $$\overline{\mathrm dw}= \mathrm dU,\tag{17.5}$$ where $$U:= U(q_1,q_2, \ldots, q_n)\,.\tag{17.6}$$ We thus have $$\sum F_i~\mathrm dq_i = \sum \frac{\partial U}{\partial q_i}~\mathrm dq_i\tag{17.7}$$ which gives $$F_i = \frac{\partial U}{\partial q_i}\,.\tag{17.8}$$ The present practice is to use the negative work function and call this quantity $V:$ $$V= -~U\,.\tag{17.9}$$ ... The equation $(17.8)$ can be now re-written in the form $$F_i = -~ \frac{\partial V }{\partial q_i}\,, \tag{17.10}$$ where $V$ is a function of the position coordinates: $$V:= V(q_1,q_2,\ldots,q_n)\tag{17.11}$$

[...] The definition of work function on the basis of the equation $$U:= U(q_1,q_2,\ldots,q_n)\tag{17.6}$$ is too restricted. We have forces in nature which are derivable from a time-dependent work function $U(q_1,q_2,\ldots,q_n; \,t)\,.$ The equation $$\overline{\mathrm dw} = \mathrm dU\tag{17.5}$$ still holds, with the understanding that in forming the differential $\mathrm dU$ the time $t$ is considered as a constant. The equations $(17.7)$ and $(17.8)$ remain valid, but the conservation of energy is lost.

This is excerpted from The Variational principles of Mechanics' Work function and generalised force by Cornelius Lanczos; the author asserts that for a $U$ which explicitly depends on time $t,$ the law of conservation of energy doesn't apply it is "lost".

Why is it so? How did he conclude that?

Also, he said, even though $U$ depends explicitly on $t,$ the equations $(17.7)$ and $(17.8)$ are still true.

Can I say, $(17.9)$ i.e. $V= -~U$ is also true when $U$ depends explicitly on $t$?

Could anyone thus explain me why the conservation of energy doesn't hold true for $U:= U(q_1,q_2,\ldots, q_n; ~t)$ and also whether in this case, $V= -~U$ is still true?

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    $\begingroup$ If U is a function of time, the associated force also depends on time and the force is thus not conservative. The work therefore depends on the path taken and returning to your initial point might result in a net loss or gain in energy. $\endgroup$
    – Paul
    Commented Sep 23, 2016 at 8:45
  • $\begingroup$ @Pau OP knows that, but is trying to understand the detailed statements given in the excerpt. $\endgroup$
    – DanielSank
    Commented Sep 23, 2016 at 8:48
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    $\begingroup$ I think it would make this question much better if OP could focus on a specific conceptual issue. Asking "Why is it so?" is a bit vague. $\endgroup$
    – DanielSank
    Commented Sep 23, 2016 at 8:49
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    $\begingroup$ I am not sure I understand the question regarding the use of $V=-U$, thats just a nomenclature (basically mathematicians like gradient flows and will define energy such that forces are $F=\nabla U$, whereas physicists like their energies to be minimized. $\endgroup$
    – Bort
    Commented Sep 23, 2016 at 10:07
  • $\begingroup$ Is it really that complicated? If U was a function of t, then clearly dU isn't conserved in time. V= -U is just notation. All the equations shown are still valid for U = U(t) - the physics is that whatever causes U(t) to change with time is the extra "work" that accounts for the nonconservation of energy. $\endgroup$ Commented Sep 23, 2016 at 10:37

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