Does the change in kinetic energy of an object depend on the reference frame? If we consider two inertial reference frames $S$ and $S’$ where one moves with respect to the other at a velocity $v$. If we find that in the reference frame S we have that the change in the kinetic energy of a projectile of mass m is $\Delta K_S=\cfrac{1}{2}m\left(V_2^2-V_1^2\right)$. And using the other reference system it is obtained as $\Delta K_{S'}=\cfrac{1}{2}m\left(V_{2*}^2-V_{1*}^2\right)$. Is it possible for equality $\Delta K_S = \Delta K_{S'}$  to be fulfilled?
That is, I wonder if it is possible that $\cfrac{1}{2}m\left(V_2^2-V_1^2\right)=\cfrac{1}{2}m\left(V_{2*}^2-V_{1*}^2\right)$?
 A: Your analysis is correct. In the case of one-dimensional motion with $0 \leq |v|$ the changes in kinetic energy of a particle traveling at speeds $V_1$ and $V_2$ w.r.t. the reference frame $S$ w.r.t. which the reference frame $S*$ travels at velocity $v$. The changes are $\Delta K_S = \frac{1}{2}m(V_2^2 - V_1^2)$ and $\Delta K_{S'} = \frac{1}{2} m ((V_2 - v)^2 - (V_1 - v)^2)$ with $\Delta K_S = \Delta K_{S'}$ if and only if $V_1 = V_2$ or $v = 0$, that is the trivial case. In the multi-dimensional case, we can similarly obtain the analogous necessary and sufficient condition for equality in the change of kinetic energy calculated using the two different frames as $(\vec{V}_1 - \vec{V}_2)\cdot\vec{v} = 0$. In the non-trivial case, this condition essentially means that the differential $(\vec{V}_1 - \vec{V}_2) \perp \vec{v}$, that is, the differential occurs in a direction perpendicular to $\vec{v}$, thus leading to the interpretation related to the one-dimensional analog.
Notice that the application of the work-energy theorem (work done by external conservative forces on a material particle $W$ equals the change in it's kinetic energy, in the absence of non-conservative forces) would also need to incorporate the change in reference frame of observation. Let us take a brief look at the case in which $S$ is an inertial reference frame, so that $S'$ is one as well because we assume that the latter moves with respect to the former with a constant velocity $\vec{v}$. Abusing the notation, the work done (in the context of the analysis above) calculated using the two reference frames is then obtained as $W_S = \int_1^2 \vec{F}^\text{ext} \cdot \vec{V} \; dt$ and $W_{S'} = \int_1^2 \vec{F}^\text{ext} \cdot (\vec{V} - \vec{v}) \;  dt$. Verifying the statement of the work-energy theorem by evaluating these integrals is a good way to understand that $W_S = \Delta K_S$ is in general not equal to $W_{S'} = \Delta K_{S'}$, with the condition for equality being $(\vec{V}_1 - \vec{V}_2) \perp \vec{v}$.
In the case that either one or both reference frames are non-inertial, application of the work-energy theorem is more complicated than the analysis presented here.
A: It is actually the total "energy-momentum" which is frame independent. Kinetic energy change is not frame independent. It is indeed frame dependant. The kinetic energy depends on momentum, where in units with $c=1$, the total energy is given by $$E^2=p^2+m^2$$ where the kinetic energy is represented by $p^2$.
Note that momentum is a vector and transformations that include rotations will mean the momentum components in one frame will differ in another.
But if the kinetic energy is conserved in one frame where momentum is also conserved, then the kinetic energy will be conserved in another frame.
