How does kinetic energy work? It doesn't make sense to me.

Factual Questions
1

The same difference in velocity means more KE difference for a faster body. How so?

There’s you, observer A, standing still beside 2 parallel railway tracks. There’s your friend, observer B, on a train doing 100 metres per second. There’s another train (called C)on the 2nd track doing 110 mps in the same direction. C’s mass is 100kg. The formula for KE is e = mass/2 x velocity^2.

B measures C’s velocity as 10mps (relative to B), so if C slows to 100mps (so it’s at rest for B), B will measure the change in C’s KE as 100kg/2 = 50kg x 10^2mps = 50kg x 100mps = 5000 joules. That’s the energy it took C to slow down.
Now, this 5000 joules is a real, fixed quantity. If B could witness C’s fuel consumption he should see 5000j of fuel burned to slow the train. But what do you see, playing the role of observer A?

You see train C with an initial KE of 100kg/2 = 50kg x 110^2mps = 50ks x 12100j = 605,000j. After C has slowed you measure its final KE to be 100kg/2 = 50kg x 100^2mps = 50ks x 10000j = 500,000j. It follows that C needed 605,000 - 500,000 = 105,000j to slow down.
This 105,000 joules is also a real, fixed quantity. It’s the energy C needed to use to slow down. If you could witness C’s fuel consumption you should see 105,000j of fuel burned to slow the train. You should measure the brakes ending up with a higher temperature than B measures. You would see C’s brakes getting white hot and B would see them only getting red hot.

What is it about the world in motion that resolves this contradiction?

2

No matter how the brakes work, the actual force that’s causing the train to slow down is friction between the wheels and the track. The train’s speed relative to the track is not dependent on which reference frame we use. Maybe that has something to do with it. I feel like I’m missing something obvious but I have no idea what that might be.

3

Where you’re going wrong is in putting your “trains” on a track, and looking at fuel consumption and brake colour. B doesn’t only observe C going 10 m/s, he also observes two parallell train tracks comming at him at 100 m/s and has to include that in his calculations of braking.

If you replace your trains with space ships, and all breaking involves pushing reaction mass against the direction of travel, all the formulae add up nicely. (Or so I assume, I’m too lazy to actually do the maths…)

4

I suspect (and could be wrong) the answer to your problem is that you are looking at the trains in isolation. To slow down they must push against something (conservation of momentum). This happens to be the world. So the world changes its rotational momentum by an equivalent amount, and hence its change in kinetic energy must be accounted for. The calculation of this change also depends on the relative speed of the observers.

In a space craft the change in momentum and speed must be induced by throwing something the other way (feul, ions etc). The kinetic energy of these must be accounted for.

5

All velocities are relative to some reference. If A wants to know what kinetic energy that B will compute for slowing C to B’s velocity he has to take into accout that he, A, is moving toward B and C at 100 m/sec.

6

You can’t legitimately switch references in mid problem. The velocities you assumed are relative to the ground and so you have to stick with that. Let the trains get up to their velocities one at a time. B starts first and gets up to 100 m/sec. C is now receding from B at that 100 m/sec. In order for C to match B’s velocity he must gain 100 m/sec of velocity and be given 500000 J of KE. In order to exceed B’s velocity by 10 m/sec he must get to 110 m/sec he must gain 605000 J.

7

Replying to scm1001, the energy C uses to brake is transferred to the earth in the opposite direction (Newton’s 3rd law), but that just means the quantity of energy is conserved. It is not relevant to the calculation of how much energy the train needs to slow down (or speed up).

The Wikipedia article on Kinetic energy begins:
Kinetic energy (SI unit: the joule) is energy that a body possesses as a result of its motion. It is formally defined as the work needed to accelerate a body from rest to its current velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. Negative work of the same magnitude would be required to return the body to a state of rest from that velocity.

The last sentence points out that braking and acceleration are treated identically - one is the negative of the other, so my example could have been equivalently framed using speeding up instead of slowing down. When a body changes velocity (without changing its mass) it loses or gains kenetic energy, which must go somewhere (eg into the earth’s rotation or into rocket exhaust) or come from somewhere eg by burning fuel).

Your point about the spacecraft is interesting. Why is the KE formula half mass time velocity squared? Maybe because of having to make something move in the opposite direction. The spacecraft’s change of motion requires rocket exhaust with the same momentum - so does it also have the same kinetic energy? Then in total there would be an energy production of whole mass times velocity squared?

Replying to naita, “he also observes two parallell train tracks comming at him at 100 m/s and has to include that in his calculations of braking”, if he had to include that in his calculations the textbooks would say so, and they don’t.

8

How about this?
**Before: **One track, two trains, Highlander and Lowlander, both stopped. Both weigh 100kg.

1st scenario: H accelerates to 110 metres per second. Energy acquired = 605,000 joules.

After: Two trains receding from one another at 110mps. Energy taken from fuel = 605,000j.

2nd scenario: H accelerates to 100 metres per second. Energy acquired = 500,000j. L accelerates in the opposite direction to 10mps. Energy acquired = 5000j.

After: Two trains receding from one another at 110mps. Energy taken from fuel = 500,000 + 5000 = 505,000j.

9

It’s actually incredibly relevant. In fact, it’s the main point. Let’s even assume that there is no air resistance and that the wheels on the train are spinning completely frictionlessly. So, while they are moving at a constant speed, there are no forces (other than gravity) acting upon the trains. Now, if you want to define a frame of reference (FoR) as an observer on a train (B), you have to say that (C) is moving 10m/s to the east, and the tracks and ground are moving 100m/s to the west of train (B). (Remember, velocity is speed and a direction - you can’t have a peanut butter cup without both chocolate and PB!)

So, when (C) chooses to slow down, the only thing around for (C) to apply a force to is the tracks below - and they’re moving to the west! The work done by friction will equal the change in KE. Work (here, in the simplest form) is Force times Distance.

I’ll work this out two ways, and you can see that it will be the same. Let’s say that the force of the brakes is 1000N. That force, when applied to the train, gives an acceleration of -10m/s^2 (negative, because it is acting in a direction directly opposite of the velocity). Using v1 = v2 + at, we find that the train slows down either from 110 to 100 or 10 to 0 (m/s) in one second. In the one second, if you are using the FoR from (B), the tracks move under the train 100m, and the train moves 5m (x = vt + 1/2at^2). If you are using (A)'s FoR, the train moves 105m.

In either case, the brakes (friction) act for 1 second (not important), and over 105m (the important part). W = F[sup].[/sup]d, so W = 1000N * 105m = 105,000J.

Nope. 1/2 m *v^2 is the integral of momentum (mv). (Have you done integrals in calculus yet?)

::snort:: muuahahahahahaha!

Textbooks lie, my boy! Well, maybe not lie but they do contain mistakes and can be written in a way that the authors think gets the idea across, but doesn’t.
[size=1]BTW, I am acting on good faith that I am not solving a homework problem for you. You do know that posting HW problems is verboten, ja?[.size]

10

Always, the important part is what are they applying a force to? Einstein showed us that the all time important thing in picking a frame of reference is in looking at the accelerations.

11

Still trying to expose the contradiction, another example.

You are on a spaceship, not accelerating, and an asteroid hurtles past at a tremendous velocity. Then another drifts past at a small velocity, in the same direction - on the same vector. To increase your velocity by 10mps with respect to the first you have to acquire kenetic energy proportional to a tremendous velocity squared. To increase your velocity by 10mps with respect to the second you have to acquire kenetic energy proportional to a low velocity squared. But by doing that you automatically increase your velocity by 10mps with respect to the first!. So you have produced 2 very different energy outputs by the same manouver.

To support this, imagine both asteroids are the same mass as the spaceship. Then adding 10mps with respect to the first can be done equivalently by imparting the same thrust for the same length of time either to the spaceship or the first asteroid. The energy required is the same in each case. The same thing goes for the second asteriod. In both cases you just apply mass/2 x velocity^2.

12

Right, but that was not the original problem. In the original problem train C reached 110 m/sec traveling in the same direction as B, not 10 m/sec in the opposite direction. The question was how much energy difference is there between 100 m/sec and 110 m/sec, not what is the energy difference between 10 m/sec and 0 m/sec or between +100 m/sec and -10 m/sec.

13

Thanks, JustAnotherGeek. It’s not a homework problem - its a philosphical problem that has been bothering me; and you’re helping. I’m not all the way to enlightment yet, though.

The same force of 1000N will decelerate (or accelerate) the train by 10mps in 1 second whatever its initial speed, right? And it will take more energy to change its motion by 10mps if it’s initially going faster. This is what bothers me. It’s the same change of motion so why should it take more energy?

I guess understanding frames of reference in this context is crucial, but where are the frames of reference, plural, in the previous paragraph?

14

It requires differing amounts of energy in different reference frames because energy is force * distance. It will take more energy to increase its speed if it’s already going fast because the force will be spread out over a longer distance.

15

OK lets do the maths in space

rocket 100 kg travelling at 10 m/s burns 10 kg of fuel throwing the fuel back (relative to the rocket) at 90 m/s. By conservation of momentum rocket (now 90 kg) is thrust forward at an extra 10 m/s making a total of 20 m/s.

A stationary observer see a rocket go from 10m/s to 20 m/s with fuel being thrown back at 80 m/s (relative to the observer not the rocket)

Initial energy according to stationary obs = 1/2mv2 = 0.5 x 100 x 100 = 5000 J

after combined KE of fuel and rocket = 0.5 x 90 x 400 + 0.5 x 10 x 6400 = 50000 J. Change in KE =45000 J
Lets do the maths from an observer travelling initially at the same speed as the space ship. After fuel burn he sees ship go up to 10 m/s and fuel shot back at 90 m/s

Initial KE = 0 (as relative speed is 0)
Final KE = 0.5 x 90 x 100 + 0.5 x 10 x 8100 = 45000 J eg. the same !!!

This holds true on the ground where it is the entire earth, not the spent fuel being thrust back.

You must look at the entire system as a whole when calculating KE.

afterwards = 0.5 x 90

16

ignore last line

17

'K. Cool.

It’s not an easy problem to understand. Glad I can be of service.

Again, you have to look at how the acceleration is being applied. And, it’s not the same change of motion. It’s the same difference in momentum, to be sure, but since there is more momentum to begin with, you need more energy to change that momentum. (ala the 105m of stopping distance, vs 5m)
Also, I think you may be confusing your units, somewhat.

F[sup].[/sup]d = W (which is where the change in KE in this problem comes from)
F[sup].[/sup]t = (change in) mv (You may be used to seeing “change in” represented by a triangle pointing up.)

As for what I said about KE being the integral of mv, try this:

mv/t = F, (rearranging the above eqn) so
(mv/t)[sup].[/sup]d = W, right? Wrong. It’s:

([(change in) mv]/t)[sup].[/sup]d = W.

As in, the velocity needs to change with time. Therefore, it is not a simple “algebreic” problem, but now a “calculus” problem; you need to integrate all of the little changes in mv over time. You might be thinking that you can re-arrange the above to get

change in mv[sup]./sup = change in mv[sup]./sup

but, here, there is not a constant v - it’s changing, right? So d comes from

d = v1t + 1/2 a[sup].[/sup]t[sup]2[/sup]

just like (change in) mv = m(v2-v1)

Notice that in these two equations, there’s still a v1 - the initial velocity. That’s what sets the KE at different numbers.
As I said earlier, this is not an easy distinction to visualize. This is the first time I’ve sat down to work out why this would be the case - and I’m a physics teacher! I might just turn this into an extra credit problem.

18

I always liked to think of kinetic energy (and all other forms, too) as a series of buckets that could pass “energy” back and forth between each other. Kinetic, potential, and thermal are all forms of energy that an object inherently possesses. A solid hydrogen sphere in a pure and formless void at absolute zero would have no energy… but other than that perfect case, there is some basic energy present in every object. Usually, at the start of any problem, we redefine zero to exclude energy that does not concern us. Defining the zero point is the most crucial part of doing any physics problem with energy calculations.

When you throw a tennis ball up in the air, it starts with some potential energy based on the height of your arm above the ground and a bunch of kinetic energy derived from the calories in the food you ate for lunch. The highest point the ball will reach is when all of that energy (minus a very small “tax” converted to thermal, from air resistance) is potential energy. Note that I arbitrarily chose the earth’s surface as my “zero” point, so the gravitational potential of the tennis ball doesn’t turn into a five- or six-digit number. I don’t calculate the conversion of calories in my cereal to motion - I assume lossless arm muscles and define my system as a ball moving at a fixed velocity with a known mass, with only gravity acting on it. I could define zero as the center of the earth or the center of the sun if I wanted to go blind on significant figures.

In the case of the trains, it’s crucial to define your zero point reasonably. If you really want to choose Observer B as your frame of reference, then you need to state that he sees the plane – er, train – running on a steel treadmill that’s moving at 100 m/s. Notice also that you say

…that’s not accurate. Some fuel would be burned, and also the brakes would heat up by a finite temperature and pass some of that temperature to the rails. The sum of the energy in the fuel burned plus the brake and rail temperature increases times the specific heat capacity of those brakes and rails should be approximately 5000 Joules. If you want to get pedantic there are probably another fifteen places the system is losing heat (various friction and exhaust heat losses) but they’ll be small compared to the overall change, and we can define the boundaries and zero-points in our problem to exclude them.

19

This is exactly right.

The main problem here seems to be that, given the generic Newtonian formula for kinetic energy (KE=.5mv^2), it takes more energy to increase an already-moving object’s KE by (say) 10 mph than if you start with a stationary object.

This at first doesn’t seem right. The train in the example burns the same amount of fuel to acceletate from 0->10mph as it does from 100->110mph, so why is the energy so much more greatly increased in the second case? In short, speed and velocity are relative, so why not kinetic energy?

The key to solving this apparent dilemma is that we’re not clearly defining what energy is. First off, work is defined in Physics as force acting thru a distance. The force required to accelerate the train by 10mph may be the same whether is starts from a stationary position or is already traveling 100mph, but the distance thru which that force acts is greater in the second case, so there is more work done (hence a greater increase in kinetic energy).

Second, **energy[/B} is always defined in terms of a difference between two levels. For KE to make sense, you must have a defined zero energy point. Ditto that for the chemical energy of coal burned as fuel for the train. The coal that is moving with the train is burned while it is in motion relative to the observer, which is different than coal burned stationary to the observer, hence the unburnt, zero-level of the fuel in the two cases is different.

I’m not saying chemical reactions are different whether or not they’re in motion, merely pointing out that the reference zero levels for the presumed chemical energy are different in the two cases (standing still and in motion). Thus, the energy perceived in their burning (as demonstrated in the train’s change in KE) will be different, much as they were different in the force-over-distance case.

20

Ludovick has got it. Suppose we accelerate C’s 100 kg so as to gain 10 m/sec over a period of 10 sec. This makes the accelerration -

a = 10/10 = 1 m/sec[sup]2[/sup]

if we start from 0 velocity we will have travelled a distance of -

s = at[sup]2[/sup]/2 = a10[sup]2[/sup]/2 = 50 meters.

the force on the mass is -

f = ma = 1001 = 100 newtons

and the energy added is -

fs = 50100 = 5000 J

Now suppose we start at a velocity of 100 m/sec instead of 0.

The distance traveled during the 10 seconds is now -

s = vot +at[sup]2[/sup]/2 = 10010 + 110[sup]2[/sup]/2 = 1050 m

the force is still 100 newtons

and so the energy added is 100*1050 = 105000 J.

You can use any time you want and the result will always be the same.

Ludovics insight about the different distances over which the force acts is the key to understanding the solution to the problem.