"a conservative force can definitely change the mechanical energy of an object". How?

@annav I don't want strict boundaries. I was just wondering why they do not exist. As you point out with the visible example, these boundaries are not defined by the frequencies themselves but by the physical characteristics of the waves of each portion of the spectrum.

From your answer, I understand that your opinion is that any given EM signal is classified to a different portion of the spectrum depending on its "physical effects of concern and test equipment employed", but not necessarily according to the exact frequencies composing the signal. Do you agree?

I recognize there is a continuous change of the characteristics in the waves as wavelength changes. I believe you are implying that the definition of each portion of the spectrum relies more on these characteristics than the specific wavelengths composing the signal, am I right?

Does this answer your question? How can you push the ground backwards?

@DKNguyen although even if we are talking about tensions in both halves, they do not need to be equal. Consider the case where the pulley has angular acceleration

@DKNguyen the OP is asking about 2 tensions acting on a small piece of the rope, not the tensions on the two halves (at least that is what I was talking about)

@lalit right, that follows from Newton 2nd Law for the normal direction: $N - m g \cos {\alpha} = m \, a_n$ with $m=0$

It is compensated by the normal component of gravitational force (say $m g \cos {\alpha} $, if you know what I mean) but again, if you assume that the rope has no mass, there's no need to take it into consideration

"Very likely you were thinking of ball's c.o.m. But in this case too your statement isn't correct, as the normal reaction of the slope has a (negative) moment wrt to ball's centre." - but isn't the normal reaction parallel to the vector that goes from the c.o.m. to the point where the normal force is applied? Shouldn't this moment be $0$ w.r.t. to the c.o.m.?

I understand. So when you choose a reference point for computing the torques, how do you know if the eq $\vec{\tau} = I \, \vec {\alpha}$ holds for that point? E.g. in this case it works fine for the cm, even though the cm is not fixed to an inertial frame.