# Conservation of energy, and Newtonian physics that does not fit

Here is a theoretical example where the energy into the system does not mach the energy output.

There is no friction or air resistance. A sledge with a rocket engine attached to it is lying still on the ground, some distance in front of a ramp. The ramp has a shape such that, if the sledge drives onto the ramp, it will be thrown vertically upwards. The sledge and the engine have a mass of 1000kg, the engine has a power output of 10kN. The fuel usage is 1L/s. The fuel's mass is too low to be included in the total mass.

The sledge accelerates towards the ramp, until it runs out of fuel. It then hits the ramp, and is thrown upwards. The potential energy the sledge will have at the highest point after the jump, will be proportional to the height the sledge will reach (E = mgh). The potential energy at the highest point will be proportional to the kinetic energy the sledge had before it ran out of fuel, which will be proportional to the energy in the fuel in the engine at the start. Therefore the height the sledge will reach will be proportional to the fuel in the engine at the start.

I do two theoretical tests. First I start with 1L of fuel, the with 2L of fuel. I expect that since I double the fuel, I will also double the maximum height.

Test 1. I let the fuel be 1L.
fuel_1 = 1L
fuelPerSec = 1L/s
g = 9.81 m/s^2 ≈ 10m/s^2
m = 1000kg
f = 10kN
v_0 = 0m/s
a = f/m = 10kN / 1000kg = 10m/s^2
t_1 = fuel_1 / fuelPerSec = 1L / ( 1L/s ) = 1s
v_1 = a*t_1 = 10m/s2 * 1s = 10m/s
Ek = 1/2*m*v_1^2 = 1/2 * 1000kg * (10m/s)2 = 50kJ
Ep = Ek = 50kJ
Ep = mgh    i.e    h = Ep / (mg) = 50kJ / (10m/s^2 * 1000kg) = 5m

Test 2. I let the fuel be 2L.
fuel_2 = 2L
fuelPerSec = 1L/s
g = 9.81 m/s^2 ≈ 10m/s^2
m = 1000kg
f = 10kN
v_0 = 0m/s
a = f/m = 10kN / 1000kg = 10m/s^2
t_1 = fuel_1 / fuelPerSec = 2L / ( 1L/s ) = 2s
v_1 = a*t_1 = 10m/s2 * 2s = 20m/s
Ek = 1/2*m*v_1^2 = 1/2 * 1000kg * (20m/s)2 = 200kJ
Ep = Ek = 200kJ
Ep = mgh    i.e    h = Ep / (mg) = 200kJ / (10m/s^2 * 1000kg) = 20m


When I double the energy input, the energy output is quadrupled. The relation should be proportional but it is quadratic.

• Why do you expect a linear relationship? You have $E\propto v^2$ so if $v$ doubles then $E$ quadruples. – Kyle Kanos Feb 14 '15 at 13:43
• You put a specific ammount of energy into the system, this should be proportional with the energy in the system after the fuel runs out. When you double the fuel, you double the energy input, but you quadruple the kinetic energy after the fuel runs out. The total amount of energy should be constant, always, since energy can not appear or disappear. – Markus Fjellheim Feb 14 '15 at 20:18
• You've not doubled energy input, you've doubled the fuel cell container, from 1L to 2L. This allows for a longer burn (compare $t_1$) which allows for larger maximum speed (compare $v_1$). – Kyle Kanos Feb 14 '15 at 20:24
• You double the kinetic energy, but you dont double the velocity. – Christian Feb 14 '15 at 20:26

You assume that if you put the double amount of fuel in the engine, the velocity after the acceleration doubles as well. This is wrong, since when you double the energy input, the velocity only goes up with the square root: $E_{kin} = \frac12 m v^2 <=> v=\sqrt{\frac{2}{m}E_{kin}}$

Edit:

2) is wrong since the effective force is not constant. It's a bit hard to get, but imagine the following:

You may know reduced gravity aircrafts; planes which are basically falling towards the earth resulting in nearly no gravity for the crew. In those planes the gravity acts as well, although the crew doesn't notice that.

Or a different approach: While you accelerate, 2 forces act on you:

1) The force due to the expelled fuel

2) The force you feel due to the simple fact that you accelerate (F=m*a). Those forces are opposed to each other and thus, the effective force is less.

• I know. But why is the kinetic energy quadrupled when the speed is doubled? I double the fuel, double the energy input, then the kinetic energy should double as well? I know the formula says it will be quadrupled, but the energy you en up with, should be equal the energy you start with? – Markus Fjellheim Feb 14 '15 at 20:14
• Since the force output if constant, the acceleration is constant. If you have twice the fuel, the engine will run for twice as long. Then the delta velocity will be twice as large. If I use energy conservation, I get that the speed should be the sqare root of 2 times larger when I double the energy input. If I use newtonian physics I get twice the speed. – Markus Fjellheim Feb 14 '15 at 20:23
• "I double the fuel, double the energy input, then the kinetic energy should double as well?" Yes, the kinetic energy doubles. But not the velocity "If you have twice the fuel, the engine will run for twice as long. Then the delta velocity will be twice as large" No, as you can easily see in my answer. – Christian Feb 14 '15 at 20:24
• I can see that if you use the conservation of kinetic energy, you won't get twice the speed with twice the fuel. But My question is, why don't I get the same answer using a newtonian approach? Which of these steps are wrong: 1 The fuel is twice as large, therefore the burn time is twice as large. 2 Force times time = momentum, since the burn time is twice as large, the momentum is twice as large. 3 Delta momentum / mass = delta velocity, therefore delta velocity is twice as large. – Markus Fjellheim Feb 15 '15 at 2:29
• I have edited my answer – Christian Feb 15 '15 at 8:23

you forgot the acceleration part.. as the volume of the fuel tank increases so the maximum velocity acquired at the moment when fuel runs out is also increased.. as the maximum velocity is achieved all other forces except ( kinetic + potential + acceleration - gravity ) is removed.. then the time it requires to all other forces other than gravity to act on it is also proportionally increased

When I double the energy input, the energy output is quadrupled. The relation should be proportional but it is quadratic.

Let's consider firing a double barrel shotgun. If the gun is massive, it gains a small amount of kinetic energy, if the gun is less massive, it gains more kinetic energy.

If both barrels are fired at the same time, both barrels may think that the gun is less massive than it really is.

When number of simultaneously firing barrels goes from one to two, the gun gains double kinetic energy from each barrel, that's four times more energy compared to one barrel firing.

Note that the motion of the gun is an unwanted vibration of a machine as it's working. And note how inefficient the rocket is: One liter of fuel is needed to accelerate a sledge from 0 to 10m/s.

Summary: When the mass of the rocket fuel is insignificant compared to the mass of the sledge, the efficiency of the rocket powered sledge is insignificant. Efficiency increases as the mass of the fuel increases.

You assumed a great deal about your engine. You assumed that it "uses" fuel at 1L/s and that it produces a force of 10kN. That's not realistic, but it is possible for slow moving sleds, basically it is possible if the way it "uses" fuel is to burn some to get 10kN of thrust and then dumps the rest to make sure it will "use" 1L/s. It technically meets the specifications, but the energy it attains is not the energy of the fuel, because you never asked the engine to be efficient, you asked it to produced 10kN of trust and to empty the fuel tank at 1L/s, so it does it by burning some fuel and dumping the rest. And it's doing that because you told it to do that.

If you'd insisted on making an efficient engine you could have asked that the engine convert 1L/s of fuel energy efficiently into additional kinetic energy. And then you'd find out that like most real engines, the acceleration gets smaller when the object is already going faster.

Ask for an inefficient energy use with a simple acceleration profile, and you get that not all the fuel energy is used efficiently. If instead you ask for an efficient engine that converts fuel energy into kinetic energy then you'll find out that you get a nonsimple acceleration profile (or a nonsimple fuel consumption rate, or both).