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Let a particle move in space with constant velocity $v$. Its mass is directly proportional to time: $m=\mu t$, where $\mu$ is a constant with dimension kg $\text{s}^{-1}$. A single force acts on the particle so that it can maintain its constant velocity $v$. No other forces act on the particle.

This is not strange at all: just imagine you are pushing a trolley, and people around you continuously throw stationary things into the trolley. That's how it gains mass.

Now here is the problem:

The particle has no potential energy. Let $E$ denote the total energy inside the system, then $E$ is also the total kinetic energy. On one hand, we have

$$ dE=d(\frac{1}{2}mv^2)=d(\frac{1}{2}\mu v^2t)=\frac{1}{2}\mu v^2dt. $$

On the other hand, we have $$ dE=Fdx=\frac{dp}{dt}dx=vdp=vd(\mu t v)=\mu v^2dt, $$ which differs for the first expression by a factor of $2$. What's wrong? Why energy disappears?

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  • $\begingroup$ Isn’t it a odd question? ”Let a particle **move in space with constant velocity 𝑣... A single force acts on the particle so that it can maintain its constant velocity 𝑣”** In free space the particle will accelerate in the described case? ”Its mass is directly proportional to time: 𝑚=𝜇𝑡.” Where does the part get its mass from? $\endgroup$ – HolgerFiedler May 1 at 5:35
  • $\begingroup$ @HolgerFiedler Not strange at all: just imagine you are pushing a trolley, and people around you continuously throw stationary things into the trolley. That's how it gains mass. $\endgroup$ – Ma Joad May 1 at 7:49
  • $\begingroup$ Thrown things are not stationary. $\endgroup$ – PM 2Ring May 1 at 9:26
  • $\begingroup$ @PM2Ring Thrown things are stationary in the direction of motion of trolley, because they move perpendicular to each other. $\endgroup$ – Ma Joad May 1 at 9:29
  • $\begingroup$ In that case, you aren't just adding mass, you're also adding kinetic energy. $\endgroup$ – PM 2Ring May 1 at 10:21
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The first expression is correct. The second equation does not apply as no work is done. You are adding mass that already moves at the same speed $v$. The work done to get that mass at $v$ is not included in your kinetic energy expression.

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  • $\begingroup$ So when does my second equation apply? $\endgroup$ – Ma Joad May 1 at 10:55

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