What is wrong with kinetic energy?

Question

I came up with the following problem: A car with mass m=1000kg is cruising at a constant velocity 10m/s (assume zero friction with the road or with air). The car then tries to accelerate to 15m/s. In the frame of the road, this would mean an increase in kinetic energy $\Delta T = 1/2 \cdot m (15^2-10^2) = 62.5kJ $. In the inertial frame where the car is initially at rest, however, the increase in kinetic energy is $\Delta T = 1/2 \cdot m (5^2-0^2) = 12.5kJ $. Now of course kinetic energy depends on the frame of reference, so that´s fine.
But now let´s say the energy powering this acceleration is coming from the engine, which is tremendously well engineered (as it happens in physics textbooks) and has a 100% efficiency, and the potential chemical energy available in the fuel tank is $30kJ$.
Then an observer on the roadside would point out that $30kJ < 62.5kJ$, and conclude that the car does not have enough energy in the tank to make the acceleration, but the driver would object that $30kJ > 12.5kJ$, so clearly it does!
The only way I can see to make it work would be to say that the chemical potential energy also depends on the frame, but I don´t see how that could be physical.
So either Galileo was wrong or I´m an idiot. Can you argue either way?
P.S.: I know Galileo was wrong, but everything in this problem is kind of a lot slower than light, on purpose.

Answer 1

The catch is that the fuel tank and the engine move together with the car.

Indeed, the argument made in the question is based on work-energy theorem (without naming it), that is the change of kinetic energy equals to the work done by the engine (by the force due to the engine): $$ \frac{mv_f^2}{2}-\frac{mv_i^2}{2}=Fd, $$ where $d$ is the displacement in the reference frame of the engine/fuel tank. In other words, the work done by the engine is also dependent on the reference frame.

Answer 2

There is a similar problem with using rocket fuel. NASA had the idea of using a spaceship's motion through the earths magnetic field to generate electricity, but it was pointed out that the resulting drag would slow the ship down. NASA says, "Yes-- but the rate of work done on the very rapidly moving ship by even a small thrust would be very large, so it is more efficient to the use the fuel as propellant than in a fuel cell". At first sight this seems to violate the conservation of energy.

The paradox is that a spaceship is moving at speed $v_x$ while a backward pointing rocket engine exerts a forward force $F_x$ on the ship. The rate at which the rocket is doing work on the ship is $F_xv_x$, where the thrust $F_x$ does not care about $v_x$. Why is the rate at which the rocket fuel is able to give kinetic energy to the ship so much greater when $v_x$ is large?

Suppose we burn fuel at a mass rate of $\dot m$ (kg/sec), then:

a) Standard rocket theory gives us that the thrust on ship is $F_x= \dot m c$ where $v_{x,\rm exhaust}=-c$ is the exhaust velocity with respect to the ship.

b) The KE of the rocket exhaust gas, as measured in the frame of the ship, comes from its chemical energy $\epsilon$ (J/kg), so equating the exhaust KE to that supplied by the chemical energy of the fuel gives an estimate of the exhaust velocity via $$ \frac 12 \dot m c^2 = \dot m \epsilon. $$

Put these facts together with the algebraic identity $$ cv = - \frac 12 (v-c)^2 + \frac 12 v^2 + \frac 12 c^2 $$ to see that the rate that the spaceship gains energy is
$$ F_xv_x = \dot m c v_x\nonumber\\ = \dot m \left( \frac 12 v_x^2 -\frac 12 (v_x-c)^2 + \frac 12 c^2\right)\nonumber\\ = \epsilon \dot m + \dot m \left( \frac 12 v_x^2 -\frac12 (v_x-c)^2 \right)\nonumber $$ The first term is the rate at which the fuel is supplying chemical energy, and the second is coming from the fuel giving up its kinetic energy as a result of its velocity decrease. If $c=v_x$ we recover all the KE given to the fuel during the launch into orbit. If $v_x<c$ then the second term can be negative, so that some of the chemical energy goes into giving KE to the exhaust gas, rather than recovering it.

Answer 3

If there is no air drag or any other mechanical frictional losses (i.e., there are no dissipative losses) associated with the motion of either car, then no fuel need be consumed by the first car moving at a constant speed of 10 m/s. It could just as well be moving with the engine off. Therefore, it would appear that the 100 kJ of kinetic energy associated with the initial speed of the first car is totally irrelevant with respect to the fuel consumed during its acceleration.

Fuel is only required by the first car to achieve the kinetic energy associated with its increase in speed of 5 m/s, namely, 12.5 kJ. This makes the first car completely equivalent to the second car that accelerates to 5 m/s from rest, as long as neither car experiences dissipative losses. Under these conditions, the decrease in fuel in the tank in each car (the amount of fuel consumed) will be the same during the period of acceleration.

In real life, however, the first car will consume more fuel than the second simply because it would be consuming fuel just to maintain its initial kinetic energy of 100 kJ against dissipative losses, whereas the second car would not. The greater the speed, the greater the dissipative losses and fuel consumed to compensate for those loses.

Hope this helps.

Answer 4

I will go other way around. You have maximum of $30kJ$ of chemical energy in the fuel tank. This means that you can speed-up car by maximum of $$\Delta v=\sqrt{\frac{2~E_{chem}}{m}}\tag 1$$ meters per second, given $100\%$ energy conversion effectivity, i.e. $\Delta T/E_{chem} = 1$. Subbing your values, we see that car can speed-up maximally by $7.7~m/s$.

Both observers agrees that car has speed-up by $5m/s$ and it agrees within maximum bounds of allowable range. Case closed.