Power when work is 0 Imagine a rocket of mass m that is at a constant altitude.  Gas velocity v.  It is necessary to find the power P of the engine.  I have a problem: to find the power, you need to divide the work by the time, but the work is zero, since the displacement of the rocket is 0, so the power of the engine is 0, but if it were, the rocket would fall to the ground, so the power is not zero.  How to solve this problem? Sorry for my English level.
 A: It is probably a "trick question", with obvious but non-intuitive answer. When the rocket stays at the same position, mechanical power is indeed zero.
There may be another answer if the question is about fuel energy (as opposed to mechanical work) released per unit time. But to get that, we need to know more about the working engine: how much fuel is being consumed per unit time, and how much energy is released per unit mass of the fuel.
A: This question needs more information. Which way is the gas being expelled? Is it exhaust gas from a rocket engine so with the purpose of generating forward motion?
Assuming that we are dealing with rocket engines that are pointed vertically downwards, then they indeed do no work on the rocket because they perfectly balance out gravity and thus keep the rocket stationary, non-moving. Energy is being spent by the rockets by chemical reactions sending gas particles downwards at high velocities - but this energy does not transfer as work into the rocket.
So your work formula indeed should give zero - because you are trying to calculate work done on the rocket and not just any energy spent by the rocket engines.
If you want to calculate the energy that is being spent by the rocket engines, then calculate the force of gravity that must be balanced out. For this you may use Newton's law of gravitation and you will have to take into account any sideways, orbital motion of the rocket that might produce an interfering centrifugal effect. This is then the force that the rocket engines must exert upwards in order to keep the rocket stationary - and it requires energy to produce such a counteracting force. Not energy that is transfered as work to the rocket, but still energy expenditure nontheless.
A: The rocket is subjected to two forces : the gravity $\vec{F}_g = -mg\,\hat{e}_z$, where $\hat{e}_z$ is the vertical unit vector pointing upwards, and the motor traction $\vec{F} = F\,\hat{e}_z$. As the rocket doesn't move, we know that $\vec{F}_g + \vec{F} = \vec{0}$ by Newton's law, hence $F = mg$.
On the other hand, the gas is subjected to the force $-\vec{F}$ by the action-reaction principle $-$ gravity on the gas is neglected $-$, such that it reaches a downwards velocity $v$, hence a power $P_{gas} = -\vec{F}\cdot\vec{v} = (-mg\,\hat{e}_z) \cdot (-v\,\hat{e}_z) = mgv$.
In consequence, the engine power is $P_{motor} = -P_{gas} = -mgv$ by energy conservation. Note that the result is negative, because the engine loses energy, which is transmitted to the rocket, such that $P_{rocket} = -P_{motor} = mgv > 0$.

N.B. The two points to remember are :

*

*A static body has no total power, because the forces compensate, hence no displacement, but the power of individual forces can be non-zero.

*In general, the power of a force $\vec{F}$ on a body $m$ is not its work over time, but the scalar product of the force with the body velocity.

