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Suppose, I'm pushing a block across a smooth table.

The length of the table is $d$, and the force that I applied is $F$. According to an observer at rest, standing next to the table, the work done is $W=F.d$. Since there is no other force contributing to this, I could say this is also the net work done. In that case, if the block was initially at rest, and at the edge of the table, it had a velocity $v$, I could say the total work done is the change in Kinetic energy i.e. $\frac{1}{2}mv^2$.

So we have established $F.d=\frac{1}{2}mv^2$

Now consider a second observer moving past the table at some velocity $u$ in the direction opposite to which I'm pushing the block. This person observes the force that I applied is still $F$ since he is also in an inertial frame. He also sees the length of the table as $d$. He too concludes that the total work done by me is $W=F.d$.

However, the velocity of the block is different with respect to him. According to him, the block starts at a velocity $u$, and at the end of the table, it has a velocity $v+u$. So the change in kinetic energy is not the same as in case of the stationary observer.

So, one one side of the equation, we have a kinetic energy difference which is different from the first observer. This would suggest that the net work done would be the same. However, according to both the observers the force and the distance remain the same, so the work done must be $F.d$.

I can't seem to resolve this apparent paradox. How can the work according to both be the same, and yet the 'net work' or change in kinetic energy be different at the same time. There is no concept of potential energy here, and no other forces seem to work.

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  • $\begingroup$ Relative to the moving observer, the block starts with the velocity $u$. Also, the distance to the observer till it reaches $d$ is $d+ut$. $\endgroup$
    – trula
    Commented Dec 7, 2021 at 12:44

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Trula is right and here are the details.

Relative to the observer moving at $u$, the work done by the force is

$$W=F(d+ut)=F(d+u\times\frac{2d}{v}) = Fd(1+\frac{2u}{v})\tag1$$

since the time for the mass to cover the distance $d$ is $\frac{d}{v/2} = \frac{2d}{v}$ from the average speed of the mass.

The apparent gain in kinetic energy is $$\frac{1}{2}m(u+v)^2 -\frac{1}{2}m(u)^2 = \frac{1}{2}m(v^2+2uv) = \frac{1}{2}mv^2(1+\frac{2u}{v})\tag2$$

and (1) and (2) are consistent.

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  • $\begingroup$ Can you please elaborate, why the distance travelled by the block changes in this new inertial frame. Since the length of the table is fixed ( ignoring length contraction ), I'd expect the distance travelled to be the same. $\endgroup$
    – RayPalmer
    Commented Dec 7, 2021 at 13:29
  • $\begingroup$ Isn't the distance travelled the same as the length of the table ? $\endgroup$
    – RayPalmer
    Commented Dec 7, 2021 at 13:29
  • $\begingroup$ @RayPalmer it's an interesting question and the resolution is that the distance travelled by the force is further from the moving observer's point of view. Let's say we start a clock when the observer is moving past the block, after 1 second the force appears to have moved a distance $u\times 1$ plus the distance travelled by the block along the table, away from them, from their point of view. You are right that the block still seems to go $d$ along the table, but both the table and block have moved further away from the observer. $\endgroup$
    – John Hunter
    Commented Dec 7, 2021 at 13:44
  • $\begingroup$ Ah so what you mean in, relative to the table the block has just travelled a distance $d$, but relative to our new observer, the table is moving, and by extension the block. So, in time $t$ if the block moves $d$ on the table, according to the observer, the table would also have moved by a distance $ut$. Now, in the table's rest frame, the block has moved $d$, and in the observer's rest frame, the table itself has moved $ut$. The only way to reconcile this is if the block has moved $d+ut$ in the observer's rest frame. $\endgroup$
    – RayPalmer
    Commented Dec 7, 2021 at 13:52
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    $\begingroup$ @RayPalmer yes that seems to be a similar situation $\endgroup$
    – John Hunter
    Commented Dec 7, 2021 at 13:56

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