If a body is at rest at Earth's surface, can we say that its kinetic and potential energy are $0$ simultaneously? If a body is at rest at Earth's surface, can we say that its kinetic energy $E_{\textrm{kin}}=0$, and its potential energy $E_{\textrm{pot}}=0$ also? Because its velocity $v$ and height above ground $h$ are zero.
 A: Yes, absolutely.
Potential energy is not a measurable physical quantity. What can be measured are differences in potential energy. So, if you compare the potential energies of a given mass at the Earth's surface and $100 \, \mathrm{m}$ above the surface, you cannot choose their difference, because that is governed by the laws of physics. However, you can freely choose to assign any point in space a specific potential energy (for the mass $m$). You could define that $m$'s potential energy at the surface is $0$, or $1234 \, \mathrm{J}$, or any other value.
Kinetic energy of a point particle of mass $m$ is defined by $\frac{mv^2}{2}$, where $v$ is the velocity. "The velocity with respect to what," you may ask. The choice is, again, yours to make; for different frames of references, you get different speeds and kinetic energies, and different notions of "rest". However, as long as $m$ is not accelerating (and $m\ne 0$), there is always a reference frame in which $v=0$ and thus $E_{\textrm{kin}}=0$.
So yes, for a body at rest (with regard to an inertial frame) at the surface (or any other point), you could very well claim that it has zero potential and kinetic energy.
A: It all depends on where you have set your coordinate system. If it is on the earths surface then yes but if you set it say on the sun then no.
A: The kinetic energy of a particle is dependent on the reference frame, so if a particle is at rest in a particular reference frame, the KE is zero. If someone moving with respect to the particle calculates the KE based on the reference frame in whichthey are at rest they will say $K \neq 0$.
If you are using $$U_g = mgh$$
for gravitational potential energy, then the reference position is arbitrary and $h$ is a small deviation from that reference position.  If someone else chooses a different reference position, say 2 meters above your reference position, they will say $U_g \neq 0.$
What's important for the gravitational potential energy in most simple problems is the change in $U_g$, not the actual value.
Always specify the reference frame being used and the reference height.
Also note that for large changes in vertical position one should use $$U_g = \frac{Gm_1m_2}{r_{12}}$$
which rightly recognizes that the gravitational potential energy rightly involves 2 masses. $r_{12}$ is the distance between the centers of the masses. For the simplified $mgh$ form,  $m=m_2$ and $g=\frac{GM_{earth}}{r}$
A: Kinetic energy is due to motion and potential energy is due to height. There is no motion and height. So,there is only inertia in the body.its kinetic and potential is zero
A: Yes of course. A stable body has potential energy... as the definition itself says that "the energy of a body at rest is called potential energy". So as the stable body which is of course in rest will contain potential energy.
