Paradox in the definition of work

Question

I'm trying to understand the Oberth effect and came across this paragraph that seems crazy to me:

when the rocket moves, its thrust acts through the distance it moves. Force multiplied by distance is the definition of mechanical energy or work. So the farther the rocket and payload move during the burn (i.e. the faster they move), the greater the kinetic energy imparted.

It makes sense to me that a force acting against an immovable object doesn't impart any change in kinetic energy- the fixed force of gravity can push the stones of Stonehenge into the ground for thousands of years, but if they don't move there's no energy changing hands.

What makes no sense at all to me is that the change in kinetic energy is directly proportional to the distance the object moves. What if the object just coincidentally happens to be moving anyway? Say you have two objects; one is at rest, and the other is coasting through space at 1000km per second. You impart a tiny little 1 newton force on both objects for a second or so. The first object moves a meter so you've increased its kinetic energy by one joule. But just because the second object happens to be traveling at hypervelocity already, you've imparted the kinetic energy of a stick of dynamite? How do those definitions of work and kinetic energy make sense?

Come to think of it, the Earth is moving around the Sun at 30 km/s so if I stand on one of the stones of Stonehenge around sunrise, is the weight of my body dumping 17 megawatts of energy into it? It seems that the definition of work contains some kind of caveat that the distance the object moves has to be related to the force somehow but that's completely unclear to me.

Also, how does the object moving through space at a constant velocity even "know" that it's moving? There is no universal inertial frame of reference that everything moves relative to; who's to say that the fast-moving object isn't at rest and we are the ones moving at 1000 kilometers per second?

Answer 1

So, you have discovered something that is actually very important, and if I could have my way, it would be present in every starting mechanics textbook.

It is that energy is only a pseudo-stuff.

If you have conventional stuff in a box, then it has two properties:

1. if the amount of stuff in the box changes, it has to be that stuff has either flowed through the sides, or teleported in there from some other “outside” (so, it is “conserved”), and
2. if you set that box into motion, every observer agrees how much stuff is in the box (it is “frame-independent”).

Energy obeys the first, it is conserved, so you can think about it as a sort of "stuff." But it is not frame-independent, so it is not quite a stuff in the way that, say, sugar molecules are a stuff. If you transition into a reference frame where you are parachuting down on to Stonehenge, it appears to be moving upwards at constant velocity, and thus has a kinetic energy. That kinetic energy is not changing, because it exists in a state of force balance, so it is not accelerating at all, but it's also not zero.

When forces are not balanced, everyone agrees on changes in velocity. But kinetic energy is quadratic in velocity, so one in the same change in velocity maps to a much larger kinetic energy change, in those reference frames where the thing is already moving in that direction really fast, $$ \frac12 ~m~ (u + v)^2 = \frac12 ~m~ u^2 + m~u~v + \frac12~m~v^2.$$ So a train starts moving faster by $v=1\text{ km}/\text{hr}.$ In the co-moving frame where $u=0$ the change in kinetic energy is some amount $E=\frac12mv^2$. But in the frame where that train is moving forward at $u=100\text{ km}/\text{hr}$, that quadratic law means that it has gained $201 E$ of kinetic energy. The baseline energies are not the same (it started with $10000~E$ of energy), and also the changes are not the same.

The solution is to multiply each force by the velocity that it's acting on, to get the rate that it is changing the kinetic energy. This is called the power that that force exerts at any instant, and it is a dot product, so there is also a cosine of the angle between the force and the velocity, $P = \vec F \cdot \vec v = F~v~\cos\theta.$

If you integrate the net power exerted by all forces, over time, you get the total change in kinetic energy. But if the force is constant, then you are just multiplying a velocity times a time, and that gives you the displacement of the object.

So the work-energy theorem still applies in the new reference frame, it's just that the changes in kinetic energy are all different, so the works must also be different for the math to still equate the work with the change in kinetic energy.

Answer 2

You have noticed something very important about work, something that is almost never explicitly stated in introductory physics classes. Specifically: work is frame variant. Meaning that the amount of work done depends on the reference frame.

So, the immediate reaction to this observation is usually something along the lines of “but if work depends on the reference frame then energy can only be conserved in certain reference frames”. The key to understanding is to remember that momentum is also conserved, and to work through the consequences of momentum conservation.

Suppose you have a person firing a rifle from the bed of a pickup truck. For simplicity we will say that the truck weighs 1000 kg and has frictionless wheels, the bullet weighs 5 g, and after firing the bullet has a velocity of 200 m/s in the frame where the truck and bullet were initially at rest. In that frame the bullet has gained 100 J of energy, and has momentum of 1 kg m/s. Due to conservation of momentum the truck has a momentum of -1 kg m/s and therefore a negligible velocity change of -1 mm/s.

Now, consider the same situation in the frame where the truck is initially moving at 20 m/s. The bullet starts with 1 J and finishes with 121 J. So almost 20 J seem to have appeared out of nowhere since the chemical energy is still just 100 J. However, looking at the truck, we see that the truck started with 200 kJ, but even though the velocity change is only -1 mm/s, that small difference in velocity reduces the KE of the truck by 20 J. So the source of the additional KE of the bullet is actually the KE of the truck, transferred to the bullet by the conservation of momentum.

Notice a couple of important points. To work this out you need to consider both energy and momentum. Because energy is proportional to velocity squared, a small change in velocity starting from zero gives negligible change in energy, but the same change in velocity starting from a higher velocity is a larger change in energy. Finally, note that the KE is different in different frames, but energy is still conserved in each frame.

Answer 3

Here is a quantitative example. We need two inertial reference frames, $K$ and $K'$, oriented in the standard way (the $x$-$x'$ axes parallel, etc.). $K'$ moves with a speed $u = 1,000$ m/s along the $x$-axis of $K$.

An object of mass $m = 1$ kg is initially at rest in the $K'$ frame. An observer in $K'$ applies a force $F = 1$ N in the $x'$ direction to the object for a time $\Delta t =$ 1 s. We want to calculate the work done by $F$ as measured in the $K'$ frame and in the $K$ frame.

In $K'$:

The acceleration:

$$ a = \frac{F}{m} = 1 \textrm{ m/s}^2.$$

The change in velocity:

$$\Delta v' = a\Delta t = 1 \textrm{ m/s}.$$

The displacement:

$$\Delta x' = \frac{1}{2}a\Delta t^2 = \frac{1}{2} \textrm{ m}.$$

The work done by F:

$$W' = F\Delta x' = \frac{1}{2} \textrm{ J}.$$

The change in $KE'$:

$$\Delta KE' = \frac{1}{2}mv_{f}'^{2} = \frac{1}{2} \textrm{ J}.$$

In K, $a$, $F$, and $\Delta t$ are all the same as in $K'$, per Galileo.

The change in velocity:

$$\Delta v = \Delta v' + u = 1,001 \textrm{ m/s}.$$

The displacement:

$$\Delta x = \Delta x' + u\Delta t = 1,000.5 \text{ m}.$$

The work done by $F$:

$$W = F\Delta x = F\Delta x' + Fu\Delta t = \frac{1}{2} \textrm{ J} + 1,000 \textrm{ J} = 1,000.5 \textrm{ J}.$$

The change in KE:

$$\Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mu^2 = \frac{1}{2}m\left(v_f^2 - u^2\right) = 1,000.5 \textrm{ J}.$$

So it seems we have 1,000 J of energy 'created' merely by changing reference frames, but that is exactly how we should interpret it. $KE$ is a frame-dependent quantity. But the work-energy theorem is Galilean invariant.

For your Stonehenge example, let $m$ be the mass of the Earth, 5.972 $\times$ 10$^{24}$ kg. In $K'$, the rest frame of the Earth, $a$, $\Delta v'$, $\Delta x'$, $W$, and $\Delta KE$ all become approximately zero. This would seem to lead to a contradiction, since in frame $K$, the rest frame of the Sun, $\Delta KE$ is zero, but it appears $W \neq 0$, since Stonehenge moves an appreciable distance in 1 second. However, $W$ in the work-energy theorem is the net work done, and we must include all the forces acting on Stonehenge-- which when added will yield zero net force parallel to its direction of motion.

Answer 4

Also, how does the object moving through space at a constant velocity even "know" that it's moving?

No, it does not.

And the object does not know it's own kinetic energy.

If you apply a force of 1 Newton during the time 1 second to the object, the energy of object would change. But how much it changes depends on the frame of reference. It may increase, increase very much or decrease. Or even remain constant (if the velocity of the object changed direction).

The energy is not something like an amount of money on bank account :) The energy of object depends on the frame of reference. You just need to stick to some frame of reference and calculate energy using it. Than the energy would remain constant.

Come to think of it, the Earth is moving around the Sun at 30 km/s so if I stand on one of the stones of Stonehenge around sunrise, is the weight of my body dumping 17 megawatts of energy into it?

I have not verified the calculations, but looks like yes. You produce work on the stone. The Earth produce work on you. Your energy, the energy of the stone and the energy of Earth remain constant. It's not a very convenient frame of reference, but it's possible to use it.

It seems that the definition of work contains some kind of caveat that the distance the object moves has to be related to the force somehow but that's completely unclear to me.

For me it was counter-intuitive as well. But not any more :)

To apply some force to a staying car is much easier than applying the same force to a car which is moving fast.

Answer 5

What makes no sense at all to me is that the change in kinetic energy is directly proportional to the distance the object moves.

I think this is not right.kinetic energy change is propotional to the distance travelled only when the object is continuously accelerated(i.e. its velocity changes continnuously).Kinetic energy change depens on change in velocity.

What if the object just coincidentally happens to be moving anyway?

In this case,by using equations of linear motion if the force you applied is the only force acting on the body, the product of the force applied and the displacement of the object under the action of the force gives the change in kinetic energy of the object.