# What is wrong with the energy of a particle, of which the mass increase with time?

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?

• 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? – HolgerFiedler May 1 at 5:35
• @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. – Ma Joad May 1 at 7:49
• Thrown things are not stationary. – PM 2Ring May 1 at 9:26
• @PM2Ring Thrown things are stationary in the direction of motion of trolley, because they move perpendicular to each other. – Ma Joad May 1 at 9:29
• In that case, you aren't just adding mass, you're also adding kinetic energy. – PM 2Ring May 1 at 10:21

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.