Can a conservative force do not conserve mechanical energy? Let us define a conservative force as being a force whose work is path independent. Then, in particular, a vanishing force is conservative.
If a force acting on a particle can be written from a scalar function, $\vec F=-\vec\nabla U(\vec r,t)$, then the change in mechanical energy is
$$\frac{dE}{dt}=\frac{\partial U}{\partial t}.$$
If $U$ depends explicitly on time, then mechanical energy is not conserved.
Now consider a case where the potential energy depends only on time $U=U(t)$. For example, a charged particle inside a charged conducting sphere whose charge changes with time. The force on the particle is zero, therefore it is conservative. On the other hand,the potential change uniformly with time and
$$\frac{dE}{dt}=\frac{\partial U}{\partial t}\neq 0.$$
Is it true that a conservative force does not imply in energy conservation as this example suggests?
 A: Yes, that is accurate.
Let the position of our particle be given by $\mathbf x(t)$, and let the potential energy function be $U(\mathbf r, t)$.  The total energy of the particle at time $t$ is then
$$E(t) = \frac{1}{2}m|\dot{\mathbf{x}}(t)|^2 + U\big(\mathbf x(t) , t\big)$$
and so
$$\dot E(t) = \dot{\mathbf{x}}(t) \cdot \left[m\ddot{\mathbf{x}}(t)+\nabla U \big(\mathbf x(t),t\big)\right] + \frac{\partial U}{\partial t}\big(\mathbf{x}(t),t\big)$$
Because the force on the particle at time $t$ is given by
$$\mathbf F(t) = -\nabla U\big(\mathbf x(t),t\big) \underbrace{=}_{\text{Newton's 2nd}} m\ddot{\mathbf{x}}(t)$$
the first term always vanishes, whether $U$ depends on position or not.  This leaves us with
$$\dot E(t) = \frac{\partial U}{\partial t}\big(\mathbf x(t),t\big)$$
A: The idea of potential energy you used above is just a shortcut. What is always going on in situations like this is that we have some larger system, say with a small object and a large object. The potential energy is a function of both their configurations. But often we can treat the large object as fixed in place, in which case the potential energy only depends on the configuration of the small object, and we can think of it as belonging to the small object alone. The small object's energy is then conserved. This of course breaks down if you allow the large, fixed object to start moving. 
For example, consider a ball with position $\mathbf{r}$ and the Earth with position $\mathbf{R}$. Let their kinetic energies be $K_b$ and $K_E$. Their potential energy is
$$U(\mathbf{r}, \mathbf{R}) = - \frac{G M m}{|\mathbf{r} - \mathbf{R}|}.$$
The total energy is 
$$E = K_b + K_E + U(\mathbf{r}, \mathbf{R})$$
which is conserved. But for many situations it's good enough to treat the Earth as fixed. In this case, $K_E$ is just zero, and we can plug the fixed position of the Earth into $U(\mathbf{r}, \mathbf{R})$ to get "the potential energy of the ball",
$$U(\mathbf{r}) = m g z$$
which only depends on the position of the ball, and is a conservative force. We can think of the remaining energy as the energy of the ball alone,
$$E_b = K_b + U(\mathbf{r})$$
and this quantity is conserved. This is the starting point of your analysis. 
The problem is that the very next thing you do is start letting the fixed system move. Of course once you do this, the preceding two steps don't make sense anymore, and you have to go back to the full expression for energy, which is conserved. 
A: Can the potential of a conservative force have a dependence on time in the first place? It won't be. And hence, the given example shouldn't work.
