# My misconception about the preservation of energy.

1. Nov 19, 2009

### Negatron

I'm looking into studying physics after my affair with comp-sci, but for the time being I have one small issue in particular which has some simple resolution I'm overlooking.

In Newtonian physics, the kinetic energy of a particle, if I understand correctly is the square of it's velocity, but my understanding about acceleration is that applying a fixed force to a particle causes fixed acceleration regardless of velocity.

So if a force is applied to give it an acceleration of 10 units per second, after 1 second of applying that force, the particle will have a speed of 10 units per second. If the force is applied for twice as long (2 seconds), it would seem to indicate that twice as much energy was input into that particle, and if that particle was accelerating for 2 seconds at 10 units per second, after 2 seconds it's velocity would be 20 units per second. However the square law now suggests that this particle traveling at 20 units per second has FOUR times the energy of the particle traveling at 10 units per second, even though as far as I can tell only twice the energy was put into it.

What's the silly thing I am missing here? Thanks.

2. Nov 19, 2009

### mgb_phys

Energy put in is force * distance , not force * time
as the particle accelrates you have to apply the force for a longer distance (although thesame time) which means more energy

3. Nov 19, 2009

### FoxCommander

I know it may seem confusing but to be exact, energy is a force applied over a distance. So when you applied the first force it went from 0 to 10 m/s in 1 second. the distance traveled is (using the equation d= vi(t)+1/2a(t2) ) is 5m. Now lets try that with the second time. so now you went from 0 to 20m/s, plug that into the equation and what do you get? 20m.... which is 4 times as much. So if you think about it you would have to apply the force over a greater distance the faster it goes. So technically you werent adding twice the energy you were adding 4 times. Its confusing I know

4. Nov 19, 2009

### Negatron

Alright, all this makes sense mathematically, but the intuition is lost on me.

What about a practical example then.

I have a toy car powered by a battery. On one run I hold the forward button for 1 second and the car accelerates to 10. On the second run I hold the forward button for 2 seconds and the car accelerates to 20. The car on the second run has four times the energy of the first.

The battery on the second run must have lost four times the energy even though it ran for twice as long. Electronics is also not my strong point, but presumably impedance must have decreased with increasing speed, true?

5. Nov 19, 2009

### mgb_phys

if the car's motor gives out a constant power, then in the second 'second' it won't get twice as fast - it will only get sqrt(2) times as fast.

In a normal car, your do 0-60 in say 10secs, but if you keep the same engine power on for another 10secs you don't go 120mph

6. Nov 19, 2009

### Negatron

I always presumed this to be a result of drag, but this is a whole other ball game. Interesting, there is much to ponder here, thanks.

7. Nov 19, 2009

### FoxCommander

Well another thing that you are forgeting is that the object has mass, the equastion for KE is 1/2MV2 so assuming that your object had a mass of 2 kg then it would have a total of 400 Joules(at 20 m/s). assuming that your velocity units are in m/s. Now lets say that the mass was 4 which means your acceleration was 5 m/s2 then the energies would be still at the same ratio of 1:4. This second object would be going at 10m/s meaning it would have a total KE of 200Joules.
Basically what im trying to get to is that you shouldnt put time into the factors of how much energy is in the system, because it has to do with the distance. Lets say you put brakes on the front of the toy car and you reved the engine for 2 sec but the car does not move. The energy used by the engine will be something, not sure what, but the car will have 0Joules.

Im trying to explain the best I can, so sorry if im confusing you. Heres one more tip. To Me, Energy is defined as the ability to accelerate objects over a distance.

8. Nov 19, 2009

### mgb_phys

(In the real world it is.)

In physics it's best to believe the equation ( E=1/2 m v^2) and see what effect that has on the world - car's accelerates less and less as they get faster.

9. Nov 19, 2009

### Negatron

Okay, I have another conflict here.

An object dropped from twice the height of another would have four times the impact energy.

To raise this object we have a crank lift. Let's say the lift is used to raise a brick to 1 meter and another to 2 meters.

If I understand crank lifts correctly (and apparently I don't), lifting the brick from 1 meter to 2 meters should not be any harder than lifting it from 0 meters to 1 meter. But the brick at 2 meters now has four times the energy on impact as the brick at 1 meter.

Where did the energy come from?

10. Nov 19, 2009

### Staff: Mentor

Why do you think this? Twice the height = twice the energy.

11. Nov 19, 2009

### Negatron

Twice the height would mean twice the impact speed, and twice the speed would mean four times the energy, would it not?

 I see my problem, the impact speed would not double (silly me) (time to impact and speed = Sqrt[distance] I believe)

Alright, I think I'm unifying this whole thing. That should be it for now, much thanks!

Last edited: Nov 19, 2009
12. Nov 19, 2009

### Staff: Mentor

Nope. Twice the height means twice the energy, not twice the speed. The speed will increase by a factor of √2.

13. Feb 21, 2010

### Negatron

Hi, me again!

There are two cars, red and blue. The red car is parked at the blue car is moving towards it at 20km/h.

If energy is velocity squared then the red car is carrying a kinetic energy of 20^2(400), and that is the energy that would be released on impact.

Now if the blue car is moving towards the red car at 10km/h and the red car is also now moving towards the blue car at 10km/h then their collective energies are now 10^2 + 10^2 (200) so it would lead to reason that the energy released on impact would be 200 as this is all the energy in the system.

However from the perspective of the blue car, it is moving towards the red car at 20km/h, the exact same speed as in the first scenario. If the impact speed is the same, why does the amount of energy differ? If the impact energy is the same, where does the energy come from?

Much thanks.

14. Feb 21, 2010

### DrGreg

You also have to consider what the speeds of the two cars will be after impact. Some of the initial kinetic energy may result in kinetic energy after impact.

15. Feb 21, 2010

### Staff: Mentor

I know this is old, but since we're still on cars:
No it isn't, at least not until you get within a decent fraction of the top speed (probably within 10-20%). It is primarily due to the energy required to accelerate the car increasing with speed. Think about it another way: If the torque of the engine is constant and the gear ratio is constant, the power will rise linearly with speed while accelerating. But in the moste efficient case of a real car (one with a continuously variable transmission), the torque of the engine is constant and the gear ratio decreases linearly, so the torque applied to the wheels decreases linearly with speed. So you can expect, even neglecting friction, that if you go from 0-30 in 4 seconds, you'll go from 30-60 in 8 because you only have half the torque available.

16. Feb 21, 2010

### Negatron

For the sake of simplicity lets assume no kinetic energy remains, or a negligible amount.

17. Feb 21, 2010

### Staff: Mentor

You can't do that. You must keep the situations the same except for the change in reference frame.

Careful: When comparing initial and final energy you must stick to a single frame of reference.

For simplicity, assume the two cars are sliding along a frictionless surface.

Viewed from the frame of reference of the surface, we have:
Red car has velocity +v; Blue car has velocity -v. Final velocity is zero. The initial total KE equals mv² and the final KE is zero: ΔKE = mv².

Now view from a frame in which the initial velocity of the blue car is zero:
Red car has velocity +2v; Blue car has velocity 0. Final velocity is +v, not zero! (This is the point DrGreg was making.) This is a result of conservation of momentum. The initial total KE equals ½m(2v)² = 2mv² and the final KE is ½(2m)v² = mv². Once again, ΔKE = mv².

18. Feb 21, 2010

### Negatron

It took some pondering but I finally understand. It's a new energy-conserved world for me. Much thanks again!

Thought is one hell of a tool.

19. Feb 22, 2010

### Negatron

... I think I might have one more conflict.

I'm picturing a solid wall which absorbs all kinetic energy, let's say as heat.
The air is also VERY thin and there is negligible rolling resistance, so a car could move without drag.

There is a car with a full tank of gas which has a total energy of 4.
The car accelerates to a velocity of 1 using up 1/4th of the tank.

The car accelerates to a velocity of 2 using up the entire tank.
The fuel's entire energy of 4 is now converted to a kinetic energy of 4.

The car smashes into the wall losing all it's kinetic energy. The wall gains an energy of 4 as heat.

Am I making sense so far?

If the above is true, the problem I have is that if there is another car moving alongside the first as they both move at a velocity of 1, as the initial car goes up to a velocity of 2 and burns 3/4ths of it's remaining fuel, the other car will see it accelerate from 0 to 1 not 1 to 2 and thus by that car's estimate the amount of fuel that should have been burned is 1/4th from the remaining 3/4ths, not the entire 3/4ths.

I tried to disqualify the observing car's calculation as a frame-of-reference bias, but because I don't think cars would burn dramatically more fuel for a given distance relative to earth if the earth moved more quickly around it's orbit, I'm not sure what to think. What is causing the observing car's misconception?

This leads me to think that I might have the initial assumptions all wrong. Also feel free to replace cars with rockets or whatever else, this isn't object specific... I don't think.

My second thought is that the car's fuel consumption grows linearly despite it's kinetic energy growing quadratically. But if this is the case, only 2 units of energy are released on impact despite the car having a kinetic energy of 4.

Last edited: Feb 22, 2010
20. Feb 22, 2010

### sophiecentaur

It's time to introduce Momentum, I think.
Momentum (P) = mv
In any collision, the total momentum remains the same (unlike the KE)
One moving car hits a 'stationary' one and they're on ice. They will move off, locked together, at half the original car's speed.
If you work out where the energy has gone, the total energy of the car that we said was moving is shared between that lost in the collision and the KE that the two cars still have because they are moving off at half the original speed.
If, alternatively, you work it out for two cars approaching each other at half the speed of the one, the energy in the collision will be due to two lots of a quarter of the first case - which is the same value as is lost in the first collision.

What about the 'extra' energy in the first case? Well, that would be the energy available if the cars hit a 'stationary' brick wall. It would be the same if a brick wall came along at half speed and hit the two stationary 'head-on' cars.
It all depends on the frame of reference you happen to use.

21. Feb 22, 2010

### willem2

If you want to compute the fuel consumption of another car, you would multiply the force delivered by the wheels, with the speed of the tires. This is equal to the speed of the car relative to the road that it's riding on. All observers would agree on what this speed is.

Another problem for you:

I'm on an airplane moving at 250 m/s. I start acceleratimg towards the front at 1 m/s^2
my kinetic energy at the start is (1/2) m v^2 = (1/2) 80 * 250^2 = 2.5 * 10^6 J
(my mass is 80 kg)
Kinetic energy after 1 second: = (1/2) * 80 * 251^2 = 2.52 * 10^6 J
The gain of my kinetic energy is 2*10^4 J = 20 KJ. This means that the power I used to accelerate was 20KJ/1 sec = 20 kW. How is this possible?

22. Feb 22, 2010

### sophiecentaur

Did you miss out something in that problem or is it that you consider 20kW to be a very small amount? There's already quite a lot of power needed to fly at that speed and I guess that 1m/s^2 is quite modest acceleration - you need quite a bit more than that at takeoff - like ten times as much. A 200kW engine would be reasonable on a light aircraft, I should have thought. Did the numbers seem outlandish to you?

23. Feb 22, 2010

### Negatron

I thought this might be a problem. I really don't want any object-specific associations. I probably gave a bad example if the speed of the tires is required to provide a solution. How about we replace the car with rockets? I don't believe energy consumption should have a different order polynomial with different types of propulsion, so maybe we should work with the simplest object possible. Perhaps some 'god' finger that pushes spheres.

By the way I believe I understand the concepts behind the car collisions now so that needs no further explanation, my only problem at the moment is how to resolve the problem of fuel consumption I just described, and how this would fit into reference frames. Essentially each reference frame as far as I understand them would expect the car to consume a different amount of fuel depending on it's relative velocity to the frame because force is applied over a variating number of distances. However the car (or rocket, or anything else, please don't speak of the wheels) cannot consume a variating amount of fuel at the same time.

I can probably resolve the problem on my own if I had one piece of information. If an object accelerates from 1 to 2, will it consume the same amount of energy as when it accelerated from 0 to 1? (no resistances of any sort)

Last edited: Feb 22, 2010
24. Feb 23, 2010

### willem2

Well the idea was that I was that by walking/running in the plane I could increase my
own kinetic energy relative to the ground with 20 KJ in just 1 second. The solution is
that I'm robbing the energy from the plane.

25. Feb 23, 2010

### willem2

Every method of accelerating comes down to pushing against something, wether this is the road for a car, the air for a plane, water for a rowboat, or the rocket exhaust for a rocket.
The amount of energy needed is the force you're pushing with, multiplied with the speed of what you are pushing relative to yourself. Your speed relative to anything else isn't important.