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13

As you probably know, Newton thought that energy is linearly proportional to velocity. The second law's original formulation reads: "Mutationem motus proportionalem esse vi motrici impressae" = "any change of motion (velocity) is proportional to the motive force impressed". This law, which nowadays is wrongly interpreted as: $F = ma$ (there is no ...


6

The answer is yes. The de Broglie wavelengths of freely propagating particles (i.e. forget the influence of interactions and gravity perturbations, just consider the Universe as a whole) are redshifted by the expansion of the universe. Another way of saying this is that their peculiar momenta with respect to a co-moving local volume decrease as the inverse ...


5

Gravitational potential energy is usually measured as a negative value. We do this because an object that is so far away from a gravity well that it practically is unaware of it shouldn't be considered as having any potential energy. So as $r\to\infty$, $PE\to0$. As an object falls into a gravity well, it loses potential energy, so gravitational PE is a ...


4

I believe the difference comes from the fact that forces can do different amounts of work in different reference frames. In particular, the normal force by the ramp does no work in the "lab" frame, but does do work in the moving frame (since there is a component of velocity that is now parallel to the normal force). I don't think you accounted for this work ...


2

You've already got some answers, but nobody mentioned Noether's Theorem yet. Noether's theorem maps a conserved quantity to each continuous symmetry. The relevant continuous symmetry needed to prove the conservation of energy is the one that leaves the laws of nature invariant, meaning the laws of physics don't change with time. Each continuous symmetry ...


2

Potential energy for a point mass (and also for a sphere) is not $PE = mgh $ (this is special case for a uniform field) but rather: $$ PE = - \frac{GMm}{r}$$ where G is the gravitational constant, M and m are both masses and r is the distance between the masses. (in the case of a sphere, the distance is to the centre of the sphere) Can you see the answer ...


2

The "missing" kinetic energy is still there ... it's now in the rotation of the yo-yo.


1

Suppose you have a constant angle $\theta$ slope in the original frame (we suppose that the transitions from horizontal movements to the slope is quasi-instantaneous). Call $x$ and $x'$ the horizontal displacements in the original and moving frame. Call $T$ the total time for going to $z=h$ to $z=0$. Then you have $x'= x- v_0 T$ With $ x= h \cot \theta$, ...



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