Physics question about kinetic energy

Factual Questions
1

Everyone knows kinetic energy equals mass times velocity squared, right? Well maybe not everyone, but I do, and it bothers me. Why do you square the velocity? That seems to suggest that the faster you’re going, the less change in velocity you’ll get from a certain amount of energy, which may be true for cars because of gearing, but isn’t true for rockets in space. You burn a certain amount of fuel, you accelerate a certain amount, no matter how fast you’re going from any point of reference. So unless I’m mistaken (which I must be), you could burn fuel at a linear rate resulting in linear acceleration but the kinetic energy would rise at an exponential rate, even exceeding input energy at some point and violating the law of conservation of energy. I doubt my statements are going to rock the scientific community, so tell me where I’m going wrong.

PS: Try to keep relativistic math out of the explanation. I’m pretty sure KE=m*v^2 has been around longer than the theories of relativity, so surely it can be explained with Newtonian math.

2

Wrong. KE = ½mv[sup]2[/sup]. We can go at this from a units standpoint. Energy is forcedistance or newtonmeters. From F = ma, mass has the units of newton*sec[sup]2[/sup]/meter.

so the units of mv[sup]2[/sup] are (newtonsec[sup]2[/sup]/meter)meter[sup]2[/sup]/sec[sup]2[/sup] which works out to newtonmeters.

3

You can also come up with the equation for kinetic energy by maipulating these equations:

  1. e = F*d

  2. F = ma

  3. d = at[sup]2[/sup]/2

  4. v = at

From 1) and 2) - 5) e = mad

from 3) and 5) - 6) e = ma*at[sup]2[/sup]/2

from 4) and 6) - 7) 3 = mv[sup]2[/sup]t[sup]2[/sup]/2t[sup]2[/sup]

or e = mv[sup]2[/sup]/2

4

Why would kenetic energy rise at an exponential rate? That’s where you made your mistake. I’m not trying to be snarky, but you just pulled the term “exponential” out of the air. **David **showed you how the math works out.

5

Furthermore…

A constant force will produce a constant acceleration, and a constantly increasing velocity. But there is no way a finite amount of fuel will produce an infinite velocity. At some point you will run out of fuel, and then the velocity remains constant until another force is encountered.

6

To expand a little on what John Mace wrote. The kinetic energy equation is a quadratic form. And exponential form has the variable in the exponent.

x[sup]2[/sup] is quadratic

2[sup]x[/sup] is exponential.

The difference is shown here. The quadratic is in red and the exponential in blue.

7

You’re not the only one who’s been bothered by this. In fact, I asked a question in GQ ages back, about why a car accelerating 0-30 along the deck of an aircraft carrier already going at 30 ends up with four times the KE of one accelerating 0-30 on land. Assuming the same amount of energy is required each time, it looks like some energy has materialised from nowhere!

In fact, that is not the case. The extra energy comes from the aircraft carrier, which has to do extra work to prevent the accelerating car from slowing it down. After all, your tyres were pushing back along the deck to accelerate the car, so they were exerting a force to decelerate the carrier as the car accelerated. Or another way to look at it is that the carrier does slow down infinitessimally, and the KE it loses is transferred to the accelerating car.

Part of the problem arises from changing frames of reference. Kinetic energy isn’t absolute - it’s only relative to other objects. You personally have an impressive KE relative to the Sun, but a trivial KE compared to the chair in which you sit. You can regard yourself as stationary or moving as fast as you like - all that matters is how fast things are moving relative to you. This is a tricky concept because we’re used to energy being absolute - a stretched spring, a charged battery, a tank full of gas. But as far as KE is concerned, your “zero” is arbitrary, same as with potential energy. Only changes are important.

If you want to work out how much energy it takes to bring your car from 0-60, normally you assume the car starts off stationary. But what about the rotation of the Earth? What about the motion of the Earth around the Sun? Shouldn’t you actually be working out how much energy it takes to bring your car from 60,000 mph to 60,060 mph? And if you do work that out, it turns out to be some colossal quantity that makes no sense at all.

You don’t actually have to worry about this, and can treat the Earth as a stationary frame of reference, for exactly the same reason you can treat the moving aircraft carrier as a stationary frame of reference. If you treat it as rotating, then the car affects the rotating Earth as it moves, and you find you get the same answers anyway - the “extra” KE comes from the infinitessimal changes in the Earth’s rotation caused by the car pushing on it as it accelerates. Relative to the Earth, you find the KE change works out the same.

You have to apply the same thinking to rockets. Rockets don’t just accelerate themselves - they accelerate the reaction mass they spit out the back. Starting from zero velocity, the rocket is stationary and the reaction mass comes out at its exhaust velocity, say 4 km/s relative to the rocket. After accelerating for some time, the rocket is doing 40km/s and the ejected reaction mass is actually merely slowed down to 36 km/s in the same direction. The additional KE needed to accelerate the rocket at 40 km/s, compared with accelerating it at 0 km/s, is balanced by the “slowing down” of the reaction mass.

Proper physicists will be cursing my name at this point, partly because I used the term “decelerate” (and even worse, “slowed down”) when they use “accelerate” for any velocity change, and partly because they are quite happy to switch frames of reference at their convenience and make all the fiddly stuff disappear. As an example, in the rocket equation (scroll down to “Energy”) they treat the accelerating rocket as the stationary frame of reference even as it accelerates so after a period of acceleration, the rocket is “stationary” with “zero K.E.” and all the work done by the rocket engine has gone into accelerating the reaction mass.

8

When a rocket is moving through the vacuum of space propelled by its fuel, is that constant acceleration? Or, an ever-increasing acceleration since there is no friction (air resistance, etc.)? How do you prove it to yourself and, if a <> k, then the question becomes why isn’t kinetic energy maybe v^3 or more for such a scenario??? :confused:

9

It can be either, depending on how much fuel you’re buring. That’s why I left that part out of that post and just started with “a constant force…”

The equation for KE is independent of whether a is constant or a fucntion of time. If it’s a function of time, the math is a little tricker, and you need to do some Calculus. Instead of E = Fd, you have E = Integral(Fdx).

=Integral(madx)
= mIntegral(adx)
= mIntegral((dv/dt)dx) [a = dv/dt]
= mIntegral(dvdx/dt) [allow me that little math trick w/o explanation]
= mIntegral(vdv) [v = dx/dt]
= (mv^2)/2 [you can add a constant if you want, but that doesn’t change anything]

10

BTW, Jinx, as **David **pointed out in his first post, if you know that Kinetic Energy is a function of mass and velocity, then the equation has to take the form:

E = kmv^2, where k = some dimensionless constant

If you have any other power of “v” or “m”, you don’t get a qunatity with the units of Energy (massdistancedistance/(time*time)). All you really need to figure out is that k = 1/2. :slight_smile:

11

This is a common misconception among new physics students, and I fear the amount of math being thrown around on this thread is obscuring the basic ideas.

The problem here is that you are mixing physical frames of reference. An equation like KE=.5mv^2 hides a lot of physical assumptions which are crucial to understanding the physics; it’s like memorizing the words to Hamlet’s soliloquy, but not realizing that it’s a discussion of suicide. Two things to keep in mind:

[ul]
[li]Energy, like velocity, is always measured with respect to some (often hidden) reference energy level of 0; this base level is defined by the frame of reference used. It’s meaningless to say an object has a kinetic energy of x unless I know what has kinetic energy of zero. A 1000kg car moving at 50 m/s has a kinetic energy of 1,250,000 joules with respect to the road, but 0 joules of kinetic energy with respect to a vehicle driving alongside at the same speed. Increasing the speed by 1m/s will raise the kinetic energy 500,500 joules with respect to the road, but only 500 joules with respect to the parallel vehicle.[/li][li]Energy in physics is equivalent to work, which is force (mass times acceleration) acting thru a distance (W = F*s). This discrepancy between the energy added with respect to the road vs. energy added with respect to the parallel car is explained then as follows: The force of acceleration is acting thru a greater distance with respect to the road (where it likely travels a hundred meters before the acceleration is complete) **than with respect to the parallel car ** (may not even be a car length ahead when the acceleration is complete).[/li][/ul]

Bottom line: It always takes the same amount of energy to accelerate an object by a velocity v with respect to an inertial frame of reference that matches the object’s current velocity (i.e. a frame where object’s current velocity and kinetic energy is zero). This energy seems higher when measured with respect to a different frame of reference because 91) a different value for “zero” on the kinetic energy scale is used, and (2) the force to accelerate appears to act over a longer distance (since the object is already moving).

12

Actually, your explanation was pretty good. At least, it was correct, and I thought that it was reasonably understandable (I’ll let others judge its accessibility to a layman). Deceleration is just a kind of acceleration, but that doesn’t mean that deceleration doesn’t exist. And while physicists are (or should be) comfortable enough with reference frames that we can casually shift between them, it’s still very important to recognize what frame or frames you’re using.

13

I think this comes the closest to explaining it for me, but I’m still not there. I simply don’t understand how you could continue using a constant rate of energy but get an increasing rate of output energy. I mean it doesn’t matter what your reference point is, a kilogram of rocket fuel is a kilogram of rocket fuel. Say a 50 kilogram rocket has to burn a kilogram of fuel to accelerate at 50 m/s^2. Maybe that kilogram of fuel has 100 kilojoules of chemical energy. If you’re accelerating from 0 to 50 m/s, you would get an output of 62.5 kilojoules of energy, which is 62.5% efficiency. If another rocket is already going 100 m/s with the same mass and fuel, accelerating to 150 m/s would give an output of 312.5 kilojoules, giving an efficiency of 313%! I know it doesn’t seem like that from the rocket itself, but from the frame of reference, it doesn’t add up. Well it has to add up somehow.

I know eventually it’ll just dawn on me, like that time I wondered why centrifugal force equals 1/2 radius times velocity squared. (I might have gotten that constant wrong too.) I managed to figure out the logic behind that one after hours of thinking, but I’m not getting anything for the kinetic energy issue. I’m hoping one of you can say something that will light the lightbulb in my head.

14

I don’t get this thread. There’s no treadmill.

15

I’m not sure how you’re doing the math, but you have to look at the change in velocity, not the velocity relative to 0. Are you using 150 m/s for “v” in the second example? If so, why? You only changed the velocy by 50 m/s. If you insist on calculating the energy at 150 m/s, then you have to use the total amount of energy that got you from 0 to 150.

16

… or subtract out the energy you had at 100 m/s, just like you subtracted out the velocity you were traveling at 100 m/s.

All calculations have to be done in the same reference frame. You can’t measure energy in one reference frame and velocity in another reference frame and expect everything to balance out.

17

To further belabor John Mace’s point, the rocket already going 100 m/s had 250 kJ of energy when it started accelerating, so the difference in energy was only 62.5 kJ, just as in the first case.
IOW, KE = 1/2mv[sup]2[/sup] is a simplification assuming a zero starting velocity. A more correct formula would be KE = 1/2m(v[sub]2[/sub]-v[sub]1[/sub])[sup]2[/sup].

All of the above also assumes a constant mass, which is not the case with your rocket example. For better accuracy, you would have to calculate the final energy as KE = 1/2(m[sub]2[/sub]-m[sub]1[/sub])(v[sub]2[/sub]-v[sub]1[/sub])[sup]2[/sup].

18

Formally, what you’re missing in your calculation is the energy in the rocket fuel (first mentioned in matt’s post, reply #6 above). Burning the fuel accelerates the combustion products (“reaction mass”) in the direction opposite the rocket, so some of the energy is “lost” to the kinetic energy of the reaction mass. How much kinetic energy is associated with the reaction mass depends on the initial velocity of the rocket. If you consider the total change in energy of rocket+fuel (using conservation of momentum) you will find that this is the same regardless of the rocket’s initial velocity.

But where is this free energy coming from? It looks like the faster our rocket is going, the higher its efficiency is! The problem here is that if our 50kg rocket is going at 100 m/s, we’ve already done quite a lot of work at accelerating the fuel, not just the payload we care about, to 100 m/s. The later burns by the rocket are essentially “pushing against earlier work done”–i.e., pushing against, and thus decelerating, the fuel we earlier worked on in accelerating–in accelerating the rocket-plus-fuel system. Later parts of the burn have higher “efficiency” as you calculated, but these are balanced by losses in the earlier part of the burn, where a much heavier rocket (still full of fuel) must be accelerated. (To handle this situation formally, you can set up a differential equation to model the rocket’s mass and velocity; the simplest solution, assuming constant exhaust speed, is the rocket equation.)

19

What the others are saying and what wasn’t made clear before is that the formula for kinetic energy computes the energy that is added to the existing energy. The mass is the total mass that is accelerated and the velocity is the velocity that is added to the initial velocity.

That is the velocity that is used in the comutation is the final velocity minus the initial velocity as Rhubarb said. And in the case of a rocket the mass is decreasing, usually at a pretty constant rate, so that the mass that is accelerated is the average mass during the period of acceleration, viz (initial mass + final mass)/2

20

Well then I did my previous calculations wrong. What I did was calculate the energy at 100 m/s, then calculate the energy at 150 m/s, and find the difference. Okay, let me change it up a bit. Why would a rocket accelerating from 0 to 150 m/s have 9 times the energy of one that went from 100 m/s to 150 m/s? It would have only spent 3 times the fuel.

For the sake of simplicity, let’s forget about the change in mass. That’s actually why I said there were two rockets instead of one so that they can start with the same mass but different speeds.