# change of energy in changing frame of reference

Let's imagine a car that can jump onto, or off a moving train. The train moves at 10m/s. The car, on a road next to the train, accelerates to the same 10m/s, jumps off a ramp and lands on the train, in a relative standstill in relation to the train.

To achieve that, a car of 1 ton needs to burn 50,000 Joules worth of gasoline. ($E=0.5mv^2 = 0.5 \cdot (1000kg) \cdot (10m/s)^2$). With 46MJ/kg of gasoline energy density that's about 1 gram of gasoline burnt and turned into the car's kinetic energy.

Now, we forget about the ground. We're on a massive train moving at a constant speed. The car accelerates in relation to the train. Again, to gain 10m/s relative to the train, from its relative 0, it needs 50,000J of energy, and burns another gram of gasoline.

It then reaches a ramp on one of the train cars, jumps off it, and now it moves at 20m/s over the road, all this at cost of 2 grams of gas.

Now let's try to reach 20m/s without the train. $E=0.5mv^2 = 0.5 \cdot (1000kg) \cdot (20m/s)^2 = 200,000J$. We need to burn 4 grams of gasoline to achieve this speed.

What happened? Where is the fallacy? What do I miss?

## 1 Answer

The train is doing work on the car, that is why its able to accelerate it to a kinetic energy of 200k Joules from the frame of reference of the ground without the need of the car engine burning more gasoline. Notice that you said that the train is already moving at 10 m/s, so this implies the train already has kinetic energy, and some of this kinetic energy is being transferred to the car. That is why you need to put a seat belt on, because the car is doing work in your body, and the moment the car stands still, you already have the kinetic energy given to you by the car.

• I'm still not sure how that works. The train's engine is off; it's very massive so it loses a minimal amount of speed due to our shenanigans - but how does that help us save fuel? Is the force the train exerts on the car (friction against the tires) somehow "better" than corresponding force Earth exerts when you drive on the road?
– SF.
Commented May 9, 2016 at 1:15
• If you actually work out the energy loss, you'll see that the train still does a finite amount of work (or, if its engine is off, loses a finite amount of energy) no matter how high its mass is. This happens even though you can make its change in velocity arbitrarily small. Commented May 9, 2016 at 1:50
• For example, you can write its kinetic energy as $E = p^2 /2m$. Then if the mass is very high, $\Delta E \approx p \Delta p / m = v \Delta p$, which is independent of mass. Commented May 9, 2016 at 1:51
• @knzhou: So we're tapping into the Train:Earth set energy pool... Spending 50kJ by accelerating on the train we extract another 100kJ of energy from the train... where would that "100" appear in maths of the equations? (besides the obvious, 200-50-50)
– SF.
Commented May 9, 2016 at 2:10