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enter image description here The diagram shows a block of mass M with two rocket engines of negligible mass attached to it (pointed in opposite directions) and producing 5N of thrust each.

Consider two cases where the mass is moving with constant velocity (since both rockets are equal and in opposite directions) a) 2m/s and b)7m/s

Each rocket produces thrust by burning fuel (of negligible mass compared to M) that has some specific energy. Now I reason that the rate at which work is done on the mass M by the thrust of each rocket, should be equal to the rate at which energy of the rocket fuel is being consumed for conservation of energy.

(although the net work is zero, I'm talking about the work done by the thrust from each rocket, which should be non zero)

Since the thrust produced by the rockets on both cases a) and b) are the same, the rockets must be consuming fuel (and therefore energy) at the same rate on both cases.

power is given by Force$\times$velocity.

power of one rocket in case a) $5N\times2m/s=10W$

power of one rocket in case b) $5N\times7m/s=35W$

Now we have a situation where rate of work being done by the rockets are different for each case, but rate of consumption of fuel are equal for both cases. This seems to be a violation of conservation of energy.

Please help me clarify this.

Qmechanic
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jumpmonkey
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  • What work? The net force on the block is zero, and therefore the work is also zero. – PM 2Ring May 18 '19 at 05:36
  • Not the net work done, just the work done by each rocket. – jumpmonkey May 18 '19 at 05:37
  • If youre travelling at a faster speed, it makes sense that power has increased in the second case. Fuel would be consumed faster if you travel faster. So long as both rockets are using fuel a the same rate, there is no violation. – Vishal Jain May 18 '19 at 05:39
  • But the rockets are not consuming more fuel while travelling faster, as the thrust in all cases in 5N. The rate at which fuel is burnt depends on the thrust produced. – jumpmonkey May 18 '19 at 05:42
  • Power is force times velocity. Maintaining the same force while increasing the velocity travelled at increases the power output of the system. In this case, you are making the system travel faster while maintaining the same thrust, you have increased the power of your system. So the rate of fuel consumption increases. – Vishal Jain May 18 '19 at 08:30
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    It's a good question, but we've had this exact same question many times. I've even answered it personally about five times at this point. Here is one of the ~100 duplicates of this question. – knzhou May 18 '19 at 10:46

2 Answers2

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That is not a violation of conservation of energy. Conservation of energy applies to a system in any one frame. Your question can be reframed in this way:

Consider two frames $A$ and $B$. $A$ is moving away from your rocket (at rest) at a speed of $2 m/s$ and $B$ is moving at a speed of $7 m/s$.

In $A$'s frame the power due to any constant force on the rocket will be $2F$ while in $B$'s frame will be $7F$. How can the power and energy be different? The answer is that energy(and energy per unit time) is frame dependent. But Conservation of energy can be applied to every frame. Energy is conserved in $A$'s frame and in $B$'s frame. But you cannot say that energy in $A$'s frame should be the same as in $B$'s frame.

Manvendra Somvanshi
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Power is given by the dot product of force and velocity. The work done by the two rockets are equal in magnitude but opposite in sign (just calculate ${F}\cdot{v}$ of individual forces to check it), so they cancel each other. So rate of increase of kinetic energy is also zero, as expected.

If the net force is zero, one can burn any amount of fuel in a rocket without increasing kinetic energy of it. That does not violate conservation of energy.

Archisman Panigrahi
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