The Mysterious 8.88 ft-lbs: A Man in Space Story

[Slinkie]

A man in space fires a CO2 pellet rifle, the pellet accelerates to 500 fps. The KE is calculated to be 4.44 ft-lbs. His buddy happens to fly by his position at 500 fps with perfect timing and trajectory so that he is flying parallel to the pellet. He reaches out, grabs the pellet, places it in his CO2 pellet rifle and fires the pellet at 500 fps along exactly the same path. From his perspective he has also added 4.44 ft-lbs of energy to the pellet. From his buddy's perspective however the pellet is now traveling 1000 fps with an energy of 17.76 ft-lbs. From where did this mysterious 8.88 ft-lbs of energy appear and in what form to we find it? :grumpy:

 

[Phymath]

I'm not sure but it sounds like the "mysterious 8.88 ft-Ibs" comes from the frame transformation...(2^2 * 4.44 ft-Ibs = 17.76) idk think about that..

 

[Tide]

There's nothing mysterious about it. If you double the speed you quadruple the kinetic energy. The pellet ends up with eactly the same amount of kinetic energy as it would had the first man fired it at 1,000 fps.

 

[Slinkie]

I'm aware of the KE formula,

that's how I got the numbers. What's not clear is why the pellet has 4 times the energy content from a different point of reference. Intuitively since each pellet rifle added 4.44 ft-lbs of energy, you would think you could sum up the amounts. The KE formula yields a different result because it squares the velocity. So did the second pellet gun add an additional 8.88 ft-lbs of energy over and above the perception of the second individual?

I guess what I'm pointing out is that squaring the velocity doesn't make much sense. I do understand how the formulas are derived, just not comfortable with the results.

 

[bruce2g]

Energy = force x distance. From the buddy's perspective, the second force was applied over a larger distance, since the second guy was moving relative to the first.

If you do the math, you'll find that the first observer thinks the bullet goes 3 times farther than the guy on the ship does. So the bullet gets 3 times the energy. You can see this simply because the average velocity of the bullet the first time, assuming uniform acceleration, is 250 fps, whereas the second time its average velocity is 750 fps (as viewed by the first guy), so it goes 3 times as far in the same time period.

Bruce

 

[Eyesaw]

Energy = force x distance. From the buddy's perspective, the second force was applied over a larger distance, since the second guy was moving relative to the first.

If you do the math, you'll find that the first observer thinks the bullet goes 3 times farther than the guy on the ship does. So the bullet gets 3 times the energy. You can see this simply because the average velocity of the bullet the first time, assuming uniform acceleration, is 250 fps, whereas the second time its average velocity is 750 fps (as viewed by the first guy), so it goes 3 times as far in the same time period.

Bruce

Slinkie is asking if a horse pulls a carriage at 10 miles per hour, why should it take 4 horses to pull the carriage to 20 miles per hour and not just two so it kind of misses the point to answer : because you can attach more horses to the carriage given a longer time. Why isn't the carriage moving at 40 miles per hour after you've added 4 times as many horses?

One possibility would be to say that the energy added to move a mass has to also move the energy already attached to the mass. It's as if the horse pulling the original carriage becomes a carriage also so the addition of horses end up pulling more carriages (mv^2 # of carriages) than the previous.

 

[Tide]

Slinkie is asking if a horse pulls a carriage at 10 miles per hour, why should it take 4 horses to pull the carriage to 20 miles per hour and not just two so it kind of misses the point to answer : because you can attach more horses to the carriage given a longer time. Why isn't the carriage moving at 40 miles per hour after you've added 4 times as many horses?

One possibility would be to say that the energy added to move a mass has to also move the energy already attached to the mass. It's as if the horse pulling the original carriage becomes a carriage also so the addition of horses end up pulling more carriages (mv^2 # of carriages) than the previous.

Think of it this way. Suppose those horses apply a constant force to change the speed of an object. They would have to apply that force over 4 times the distance and do 4 times as much work in order to double the speed of the 'payload.'

 

[Eyesaw]

Think of it this way. Suppose those horses apply a constant force to change the speed of an object. They would have to apply that force over 4 times the distance and do 4 times as much work in order to double the speed of the 'payload.'

Again, this analysis provides little insight for me as to why the velocity of the "payload" only doubles. Assuming 1 unit of energy is required to move the mass from "rest" to 2 meters per second, why should it take 3 more units of energy to increase the velocity of the mass only 2 more meters per second? And why isn't the velocity increased to 8 meters per second since 2 X 4 = 8?
Why isn't kinetic energy just mv?

One explanation would be that energy is massive so that to increase the velocity of the mass-energy hybrid would require an exponential addition (mv^2) of energy instead of just v if energy had no mass. A simpler explanation would be that the expression is due to experimental error. To measure the kinetic energy of an mass, you can either add energy to it and relate it with the final velocity or collide it to something else. In adding energy to it, it may not be efficient enough to get to the hypothesized true value (like Carnot's engine), and in colliding it, it's hard to attribute which energies were from which mass, leading to miscalculation.

 

[Slinkie]

There may be a simpler explanation; perhaps we have mis-defined Energy. As long as KE = Force x Distance we do indeed have to square Velocity. If instead we use KE = Force x Time we would arrive at more intuitive results for adding Energy to an object in motion. We would then have Energy = Mass x Velocity (and not Velocity^2).

Another non-intuitive example of KE = MV^2 is a rocket pushing an object in space consumes 1 gallon of fuel per minute to exert a force F on an object. Let's ignore the mass of the rocket (assume whatever fuel needs to be consumed to accelerate the rockets mass is consumed) and assume that all of the energy of the 1 gallon is used to generate the force F for one minute. After 1 minute a certain amount of energy has be transferred to the object. After 2 minutes, 4 times the amount of energy has been transferred to the object! Wow, something for nothing in that second gallon of fuel. This is what suggests we might have a problem in our definition of KE.

Just shrugging our shoulders and saying it's that way because the formula says so doesn't cut it for me. This isn't religion and blind faith should have no place here. Mathematical abstractions are not reality and are only useful to the degree that they model and/or predict reality.

So I put it to you all: Why is energy defined as Force x Distance?

 

[arildno]

Read up on impulse, Slinkie.
In addition, learn the difference between vectors and scalars.
Energy is not misdefined.

 

[Slinkie]

Thank you for that condescending remark. That certainly clears things up a bit. Anyone else care to say "Nuh Uh".

 

[bruce2g]

...
Another non-intuitive example of KE = MV^2 is a rocket pushing an object in space consumes 1 gallon of fuel per minute to exert a force F on an object. Let's ignore the mass of the rocket (assume whatever fuel needs to be consumed to accelerate the rockets mass is consumed) and assume that all of the energy of the 1 gallon is used to generate the force F for one minute. After 1 minute a certain amount of energy has be transferred to the object. After 2 minutes, 4 times the amount of energy has been transferred to the object! Wow, something for nothing in that second gallon of fuel. This is what suggests we might have a problem in our definition of KE.

Well, gravity is a force. Let's say a 40 lb rock falls for a second. Then its velocity rises to 32 ft/sec, and it falls 16 feet.

If you stop it after one second, and you try and put the rock back up where it started, you will need to lift it 16 feet.

But, suppose you let it fall for one additional second. Its velocity rises to 64 ft/sec, and it travels 64 feet. So if you want to put it back, you will need to lift it 64 feet.

So, if something falls for 2 seconds instead of 1 second, you will need to do 4 times the work to haul it back up to its initial position, and trust me, you will expend 4 times the energy and be 4 times as tired when you are done!

Bruce

 

[bruce2g]

here's the real reason

For every action, there is an opposite but equal reaction (Newton's Third Law).

So the energy that propels the bullet also propels something in the opposite direction.

To simplify your example, assume there are two bricks with an explosive charge between them. You ignite the charge, and one brick goes to the left with velocity v, and the other goes to the right with velocity v. Each has energy 1/2 mv**2, so the total energy from the charge is mv**2.

Now, assume there's a train going to the left with velocity v, and there's a guy on the train with another brick. So, he grabs the left brick from the first explosion, slaps his brick next to it, puts a charge in between, and ignites it.

Now, that left brick has a velocity of 2v as observed from the ground, so its energy is 2mv**2. However, the right hand brick from the train now has velocity 0, so its energy is 0. And the charge had energy mv**2.

So, that left brick had energy of 1/2mv**2, the explosion released energy of mv**2, and the right brick gave up energy of 1/2mv**2, and so the left brick's energy is 1/2mv**2 + mv**2 + 1/2mv**2, or 2mv**2.

So the answer to your riddle, which was quite interesting, is that the charge has twice the energy that you thought, and that some object on the train must slow down and also give up energy, and that's why two charges appear to give you four times the energy.

Another way to say it is that when you fire a gun on the ground, the recoil uses up kinetic energy, but if you do it on a moving train, the recoil actually slows the gun (and train) down and releases kinetic energy, and that's why it looks like a 2 for 1.

Bruce

 

[bruce2g]

I think there are a few mistakes in your math. For example, you write "Now, that left brick has a velocity of 2v as observed from the ground, so its energy is 2mv**2 ": From whose point of view is the left brick's energy (2mv)^2? From the ground view, the kinetic energy is of the left brick is : 2 (1/2mv^2) = mv^2; from the train view, the kinetic energy of the brick is 1/2mv^2. .

After the second explosion, the left brick has velocity 2v as viewed from the ground. So, its energy as viewed from the ground is:
1/2m * ((2v)^2 )= 1/2m * 4(v^2 )or 2mv^2, . The symbol 'v' here is the velocity of the left brick after the first explosion as viewed from the ground, and 2v is its velocity after the second explosion, as viewed from the ground.

Then you continue "So, that left brick had energy of 1/2mv**2, the explosion released energy of mv**2, and the right brick gave up energy of 1/2mv**2, and so the left brick's energy is 1/2mv**2 + mv**2 + 1/2mv**2, or 2mv**2. " Not sure how you got this. Perhaps it would help if you made clear which frames each values belong to. From the ground view, the explosion released mv^2 of energy to the left brick if the left brick was at rest with respect to the ground- since the left brick was moving already at v, the actual energy imparted to the left brick by the charge in the train is 1/2mv^2, with respect to the ground. Thus, the final kinetic energy of the left brick with respect to the ground is, as calculated above, mv^2.

And why did you add the kinetic energy imparted to the right brick to the left brick?

I don't follow the rest of your reasoning either, try again.

Everything I said is from the viewpoint of an observer on the ground. Before the second explosion, the observer sees two bricks on the train, both moving to the left with velocity v. After the second explosion, the observer sees one brick going to the left with velocity 2v, and and the other brick is now sitting on the ground with velocity 0. I'm assuming that the bricks are free to move without friction to keep it simple.

So the observer on the ground sees a brick lying there with velocity 0 and energy 0. Before the second explosion, it had velocity v and energy 1/2mv^2. So, the question is, where did its energy go?

The answer is, it went into the other brick! Now, you're probably asking "how did it get transferred to the other brick?" That's a good question, and you need to model the explosion in order to get the answer. If you model the explosion as a coiled spring between the two bricks which is suddenly released, then it should be easy to see. Basically, as the right brick slows down, an observer on the ground sees it as giving its energy to the spring, which then gives its energy to the left brick.

You might object to the spring analogy, but the fact is that in order for the explosion to increase the left brick's velocity to 2v (as viewed from the ground), the explosion must also push against something on the right side. An observer on the train thinks that the explosion gives energy to the object on the right; an observer on the ground thinks that the object on the right adds its energy to the explosion's as it slows down.

To put it another way, an observer on the train thinks that the spring pushes against both bricks and accelerates both of them simultaneously. However, an observer on the ground thinks that the right brick pushes the expanding spring against the left brick, and thereby transfers its energy to the left brick.

Bruce

 

[Eyesaw]

After the second explosion, the left brick has velocity 2v as viewed from the ground. So, its energy as viewed from the ground is:
1/2m * ((2v)^2 )= 1/2m * 4(v^2 )or 2mv^2, . The symbol 'v' here is the velocity of the left brick after the first explosion as viewed from the ground, and 2v is its velocity after the second explosion, as viewed from the ground.


Everything I said is from the viewpoint of an observer on the ground. Before the second explosion, the observer sees two bricks on the train, both moving to the left with velocity v. After the second explosion, the observer sees one brick going to the left with velocity 2v, and and the other brick is now sitting on the ground with velocity 0. I'm assuming that the bricks are free to move without friction to keep it simple.

So the observer on the ground sees a brick lying there with velocity 0 and energy 0. Before the second explosion, it had velocity v and energy 1/2mv^2. So, the question is, where did its energy go?

The answer is, it went into the other brick! Now, you're probably asking "how did it get transferred to the other brick?" That's a good question, and you need to model the explosion in order to get the answer. If you model the explosion as a coiled spring between the two bricks which is suddenly released, then it should be easy to see. Basically, as the right brick slows down, an observer on the ground sees it as giving its energy to the spring, which then gives its energy to the left brick.

You might object to the spring analogy, but the fact is that in order for the explosion to increase the left brick's velocity to 2v (as viewed from the ground), the explosion must also push against something on the right side. An observer on the train thinks that the explosion gives energy to the object on the right; an observer on the ground thinks that the object on the right adds its energy to the explosion's as it slows down.

To put it another way, an observer on the train thinks that the spring pushes against both bricks and accelerates both of them simultaneously. However, an observer on the ground thinks that the right brick pushes the expanding spring against the left brick, and thereby transfers its energy to the left brick.

Bruce

The charge doesn't move inside the ship, the bricks do so it's the charge that imparted kinetic energy to the bricks, either from the ground point of view or from the ship's point of view. I think your example shows more the confusion surrounding Newton's third law than it goes to explain why kinetic energy is not one to one with velocity increase.

 

[bruce2g]

The charge doesn't move inside the ship, the bricks do so it's the charge that imparted kinetic energy to the bricks, either from the ground point of view or from the ship's point of view..

No, from the ground's view, the charge makes the right hand brick slow down, so the energy of the right hand brick decreases.

I think your example shows more the confusion surrounding Newton's third law than it goes to explain why kinetic energy is not one to one with velocity increase.

Basically, I'm trying to illustrate that energy is conserved in the ground reference frame. Before the second explosion, from the ground frame's view, each brick has an energy of 1/2mv^2, for a total energy of mv^2. The charge generates energy of mv^2, and after the charge the left brick has energy 2mv^2 and the right brick has energy 0.

Lastly, I'd like to point out again that if a brick falls for a second, it has velocity 32 ft/sec and it falls 16 ft. If it falls for two seconds, then it has velocity 64 fps and it falls 64 ft. So if you want your brick to have a velocity of 32 ft/sec, you have to carry it up 16 ft. And, if you want it to hit the ground with a velocity of 64 ft/sec (double the velocity), you have to carry it up 64 ft (4 x the height, and 4x the work). These are physical facts.

I think that historically the definition of energy developed from observations like these. Basically, the height that you need to lift something up to in order to attain a given velocity goes up with the square of the velocity (the exact formula in feet and seconds is: height = ((desired velocity)^2)/(2g)): if you can give a better definition for energy I'd sure like to hear it.

Bruce

 

[krab]

A man in space fires a CO2 pellet rifle, the pellet accelerates to 500 fps. The KE is calculated to be 4.44 ft-lbs. His buddy happens to fly by his position at 500 fps with perfect timing and trajectory so that he is flying parallel to the pellet. He reaches out, grabs the pellet, places it in his CO2 pellet rifle and fires the pellet at 500 fps along exactly the same path. From his perspective he has also added 4.44 ft-lbs of energy to the pellet. From his buddy's perspective however the pellet is now traveling 1000 fps with an energy of 17.76 ft-lbs. From where did this mysterious 8.88 ft-lbs of energy appear and in what form to we find it? :grumpy:

I'm going back to the original question because no one has given a satisfactory answer yet. And Slinkie has a really good question here. But it has come up many times before. The thing that is always forgotten is that the platform from which the pellet is fired the second time suffers a change in momentum that exactly accounts for the energy change.

The real question is that even though each CO2 firing expends the same energy, the total pellet energy after 2 firings is not 2 times this energy, but 4 times.

Let m be the mass of the pellet, and v be its speed after being fired from a gun at rest. Each firing releases an amount of energy equal to $mv^2/2$. Let M be the mass of the buddy plus his pellet gun plus whatever else he is riding in. By conservation of momentum, this assembly changes velocity by $\Delta v$, where $M\Delta v=-mv$. So the second firing changes the pellet energy from $mv^2/2$ to $m(2v)^2/2=2mv^2$ for a change of $(3/2)mv^2$. But we know the energy release on firing is only $(1/2)mv^2$, so there is an extra energy of $mv^2$. Now calculate the energy change of the "buddy platform". It is $(1/2)M\Delta(v^2)$. Let $\Delta v=v_2-v$, then $(1/2)M\Delta(v^2)=(1/2)M(v_2-v)(v_2+v)=(1/2)M\Delta v(2v)=-mv^2$. There it is; your missing energy.

 

[Eyesaw]

I'm going back to the original question because no one has given a satisfactory answer yet. And Slinkie has a really good question here. But it has come up many times before. The thing that is always forgotten is that the platform from which the pellet is fired the second time suffers a change in momentum that exactly accounts for the energy change.

The real question is that even though each CO2 firing expends the same energy, the total pellet energy after 2 firings is not 2 times this energy, but 4 times.

Let m be the mass of the pellet, and v be its speed after being fired from a gun at rest. Each firing releases an amount of energy equal to $mv^2/2$. Let M be the mass of the buddy plus his pellet gun plus whatever else he is riding in. By conservation of momentum, this assembly changes velocity by $\Delta v$, where $M\Delta v=-mv$. So the second firing changes the pellet energy from $mv^2/2$ to $m(2v)^2/2=2mv^2$ for a change of $(3/2)mv^2$. But we know the energy release on firing is only $(1/2)mv^2$, so there is an extra energy of $mv^2$. Now calculate the energy change of the "buddy platform". It is $(1/2)M\Delta(v^2)$. Let $\Delta v=v_2-v$, then $(1/2)M\Delta(v^2)=(1/2)M(v_2-v)(v_2+v)=(1/2)M\Delta v(2v)=-mv^2$. There it is; your missing energy.

Actually, if slinkie was interested in the conservation of momentum, a collison would've served as a better example. I think what you have said above is that if K.E. is defined to be 1/2mv^2 then the pellet will have 2mv^2 k.e. after the second firing, that it got this 1.5mv^2 more energy from the second firing, and that because of Newton's third law, momentum gained by the pellet is equal to momentum lost by the platform and if pure conversion to and transfer of kinetic energies are assumed, the 1.5mv^2 kinetic energy gained by the pellet is accounted for by the lost of 1.5mv^2 of kinetic energy by the platform. That's a lot of assumptions to make for one. Another is that even if they are all correct, all you and bruceg2 (whose calculations
were wrong even if his solution was similar to yours) appear to be saying is that if the platform loses 1.5mv^2 of kinetic energy to the pellet, the pellet gains 1.5mv^2 of kinetic energy. I don't consider that a great revelation. [all considerations with respect to the "stationary" observer]

What slinkie wanted to know was why, without defining it first with the equation 1/2mv^2, you need 4 times as much horsepower to move a vehicle
only twice as much distance in the same amount of time? An obvious answer is that an 8 cylinder engine would be more massive than a 2 cylinder so that some of that horsepower is used to move the extra mass. Or more relevant to the case, energy probably has mass.

 

[krab]

... That's a lot of assumptions to make for one. ...

Uh Eyesaw, I presented no assumptions. What I gave was a proof based on the conservation of momentum plus Slinkie's original gedanken. You may need to read it again. Or maybe you do not realize that conservation of momentum has been considered a principle rather than an assumption for over 300 years now.

 

[Eyesaw]

Uh Eyesaw, I presented no assumptions. What I gave was a proof based on the conservation of momentum plus Slinkie's original gedanken. You may need to read it again. Or maybe you do not realize that conservation of momentum has been considered a principle rather than an assumption for over 300 years now.

I probably read more into Slinkie's gedunken than
you guys did because if slinkie was just asking where the "extra" energy
came from, the answer would have been trivial- the switching of inertial frames to calculate the kinetic energy of the pellet. Or more precisely, the
"extra" energy is a consequence of K.E. = 1/2mv^2. Or more facetiously, it came from the firing of the gun.

No, I read into the gedanken as questioning why the expression for kinetic energy is such and not something like K.E. = mv. My guess was confirmed by a later post by slinkie that asked exactly that.

 

[Doc Al]

The conservation of change in momentum may be a principle but conservation of kinetic energy assumes a perfectly elastic collision.
So I think you made many assumptions by modeling the firing of the pellet
as an elastic collision.

Where does krab do anything like this?

 

[Eyesaw]

Where does krab do anything like this?

Sorry, misread again.

How did krab get the 2v in the latter part of the derivation:

$(1/2)M\Delta(v^2)=(1/2)M(v_2-v)(v_2+v)=(1/2)M\Delta v(2v)=-mv^2$

looks like he substituted (v_2+v) for 2v ?

 

[cepheid]

*Sigh*

Ok guys, I worked this problem out during one of my breaks today while on campus. I changed things to SI units, since Imperial units tend to give me the heebie jeebies.

Let's call the original observer A and consider the entire problem in his frame i.e. we will call him the "stationary observer". I decided that the bullet would have a mass of 1.00 gram and the gun a 30.0 cm barrel. So, if the bullet's final velocity is 500 m/s (kind of fast, but ok), then its final kinetic energy is:

$$K_A = \frac{1}{2}mv^2 = \frac{1}{2}(0.00100\ \ \text{kg})(500\ \ \text{m}/\text{s}^2)^2 = 125\ \ \text{N}\cdot\text{m}$$

By the work-energy theorem, this change in the bullet's kinetic energy from zero is equal to the work done on the bullet while in the barrel. I'm not sure whether the explosive gases exert a constant force on the bullet or not, but I can still just calculate the average force exerted on the bullet by the gun over that distance:

$$F_{\text{av}} = \frac{W}{d} = \frac{125\ \ \text{N}\cdot\text{m}}{0.300\ \ \text{m}} = 417\ \ \text{N}$$

(416.6666666666666... rounded to the correct number of sig figs.)

Because we'll need it later on, I'm going to calculate the time needed for the bullet to traverse the chamber (again this assumes more or less constant force on the bullet, and therefore constant acceleration. But that's fine as long as I make this assumption about the gun consistently throughout the problem, right?)

$$d = \frac{1}{2}at^2$$

$$t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2d}{F_{av}/m}} = 0.00120\ \ \text{s}$$

The buddy, observer B, is flying along parallel to the bullet with velocity $v_B = 500\ \ \text{m}/\text{s}$. He grabs the bullet, puts it in his gun, and fires it in a straight line, still along its original path. The buddy's gun is of identical design to the first guy's. The key question is, what is the work done by the second gun on the bullet? Remember that we are doing this entire problem from the point of view of observer A, so the work done will be the average force acting on the bullet times its displacement as seen by observer A . Can I justify this assertion? Well, I can at least show you a statement that it is true, later.

Let's call this displacement $d_B$. Clearly, it is not merely equal to the length of the barrel, but the length of the barrel plus the distance traversed by everything in frame B (ship, buddy, gun, bullet) during the time t:

$$d_B = 0.300\text{m} + v_{B}t$$

$$= 0.300\text{m} + (500\ \ \text{m}/\text{s})(0.00120\ \ \text{s})$$

$$= 0.300\text{m} + 0.600\text{m} = 0.900\text{m}$$

So the work done on the bullet when the buddy fires it is:

$$W = F_{\text{av}}d_B = (417\ \ \text{N})(0.900\ \ \text{m}) = 375\ \ \text{N}\cdot\text{m}$$

The total work done on the bullet is:

$$W = 125\ \ \text{J}\ \ +\ \ 375\ \ \text{J}\ \ =\ \ 500\ \ \text{J}$$

This result is completely consistent with the observation that the final kinetic energy of the object as seen by A is:

$$K = \frac{1}{2}mv^2 = \frac{1}{2}(0.00100\ \ \text{kg})(1000\ \ \text{m}/\text{s}^2)^2 = 500\ \ \text{N}\cdot\text{m}$$

There is no missing energy.

I realize I haven't done anything new here, or stated any priniciple that wasn't already pointed out in the thread. But I just wanted to show the result quantitatively, to drive the point home. I hope that for this reason, you will consider it a useful contribution to the topic. Admittedly, the idea that the work done would be different as seen from observers in two different inertial frames didn't quite sit so well at first, until I dug out my first year textbook and saw this (in bold):

From University Physics by Young and Freedman, 10th ed pg. 170:

Because we used Newton's laws in deriving the work-energy theorem, we can use it only in an inertial frame of reference. The speeds that we use to compute the kinetic energies and the distances that we use to compute work must be measured in an inertial frame. Note also that the work-energy theorem is valid in any inertial frame, but the values of W_total and $\Delta K$ may differ from one inertial frame to another (because the displacement and speed of a body may be different in different frames).

krab has shown us exactly why this is so, and what he did was much more elegant than my number-crunching. So as far as Slinkie's original question goes, CASE CLOSED. Notice, Eyesaw, that krab was responding directly to Slinkie's original post (he even quoted it!), so any whining that he did not address later questions is misplaced. Slinkie was later driven to call into question the definition of work (and therefore KE) because these results were not intuitive to him. This was, misguided. I'm not trying to be mean or condescending in saying that. I'm just suggesting that when you come across a result in Newtonian Mechanics (an area of physics that, as pointed out, has firm foundations going back 300+ yrs) that is stated clearly in any introductory physics text, don't be inclined to immediately revamp Newtonian physics just because the result doesn't agree with your intuition! Be inclined always instead to revamp your intuition.

*phew*

 

[Eyesaw]

*Sigh*

Ok guys, I worked this problem out during one of my breaks today while on campus. I changed things to SI units, since Imperial units tend to give me the heebie jeebies. ...The total work done on the bullet is:

$$W = 125\ \ \text{J}\ \ +\ \ 375\ \ \text{J}\ \ =\ \ 500\ \ \text{J}$$

This result is completely consistent with the observation that the final kinetic energy of the object as seen by A is:

$$K = \frac{1}{2}mv^2 = \frac{1}{2}(0.00100\ \ \text{kg})(1000\ \ \text{m}/\text{s}^2)^2 = 500\ \ \text{N}\cdot\text{m}$$

There is no missing energy.

I realize I haven't done anything new here, or stated any priniciple that wasn't already pointed out in the thread. But I just wanted to show the result quantitatively, to drive the point home. I hope that for this reason, you will consider it a useful contribution to the topic. Admittedly, the idea that the work done would be different as seen from observers in two different inertial frames didn't quite sit so well at first, until I dug out my first year textbook and saw this (in bold):



krab has shown us exactly why this is so, and what he did was much more elegant than my number-crunching. So as far as Slinkie's original question goes, CASE CLOSED. Notice, Eyesaw, that krab was responding directly to Slinkie's original post (he even quoted it!), so any whining that he did not address later questions is misplaced. Slinkie was later driven to call into question the definition of work (and therefore KE) because these results were not intuitive to him. This was, misguided. I'm not trying to be mean or condescending in saying that. I'm just suggesting that when you come across a result in Newtonian Mechanics (an area of physics that, as pointed out, has firm foundations going back 300+ yrs) that is stated clearly in any introductory physics text, don't be inclined to immediately revamp Newtonian physics just because the result doesn't agree with your intuition! Be inclined always instead to revamp your intuition.

*phew*

I never doubted for a minute the definition of kinetic energy was consistent within Newtonian mechanics. I don't think slinkie did either. He made it clear by the second post that he knew how the math worked and was curious as to why velocity should be squared in kinetic energy. So I think I'm the only one who has addressed his question appropriately. Wouldn't you think it's rude if you asked someone why multiplication works and they keep trying to teach you how to multiply two numbers?

 

[cepheid]

that's how I got the numbers. What's not clear is why the pellet has 4 times the energy content from a different point of reference. Intuitively since each pellet rifle added 4.44 ft-lbs of energy, you would think you could sum up the amounts. The KE formula yields a different result because it squares the velocity. So did the second pellet gun add an additional 8.88 ft-lbs of energy over and above the perception of the second individual?

Gee Eyesaw...why are you giving me the third degree? Here is his second post, and with my post, I have addressed exactly his statement: "intuitively since each pellet rifle adds [the same] energy, you would think you could sum up the amounts." I have shown that it is a little more involved than that. Slinkie never suggested in this post that he knew that the work done would be different as perceived in different frames. You knew that, but I wasn't trying to insinuate that you didn't! He said it was not clear why the pellet has four times the energy content from A's frame. I have actually shown that from the point of view of observer A, the second pellet rifle does more work, exactly more enough to account for the additional KE. Talk about being totally unappreciated for one's efforts.

 

[Doc Al]

Sorry, misread again.

How did krab get the 2v in the latter part of the derivation:

$(1/2)M\Delta(v^2)=(1/2)M(v_2-v)(v_2+v)=(1/2)M\Delta v(2v)=-mv^2$

looks like he substituted (v_2+v) for 2v ?

I analyzed things slightly differently, so maybe my perspective may help a bit. First off, since we are talking about firing a gun (setting off an explosion, in effect) mechanical energy (KE) is not conserved. So that is certainly not an assumption.

So how much energy is released by the firing of the bullet? The bullet gets $1/2mv_1^2$, as stated, but momentum conservation tells us that the platform gets pushed back at speed $v_2 = (m/M)v_1$. So the total KE released is that of the bullet plus platform = $1/2mv_1^2 + 1/2Mv_2^2 = (1 + m/M)(1/2mv_1^2)$.

Viewed from the "rest" frame, the initial KE of everything before the second firing is $1/2(m + M)v_1^2$. After firing, the bullet has speed $2v_1$ and the platform has speed $v_1 - v_2 = (1 - m/M)v_1$, so the final KE of everything is: $1/2m(2v_1)^2 + 1/2M(1 - m/M)^2v_1^2$.

Do the arithmetic and you will find that the increase in KE of everything exactly equals the KE released by the firing. There is no missing energy.

 

[Eyesaw]

Gee Eyesaw...why are you giving me the third degree? Here is his second post, and with my post, I have addressed exactly his statement: "intuitively since each pellet rifle adds [the same] energy, you would think you could sum up the amounts." I have shown that it is a little more involved than that. Slinkie never suggested in this post that he knew that the work done would be different as perceived in different frames. You knew that, but I wasn't trying to insinuate that you didn't! He said it was not clear why the pellet has four times the energy content from A's frame. I have actually shown that from the point of view of observer A, the second pellet rifle does more work, exactly more enough to account for the additional KE. Talk about being totally unappreciated for one's efforts.

You left out the part in his post where he states that he knew how the formulas work out. "I guess what I'm pointing out is that squaring the velocity doesn't make much sense. I do understand how the formulas are derived, just not comfortable with the results" . By his third post, he left no doubt as to what he wanted to know in his first question- you guys kept ignoring or misunderstanding his real question so he had to answer it himself.

Now, if Slinkie only wrote his first post, I could understand how most would misinterpret his question. But after he writes a second post telling everyone he didn't want to know how to derive the "extra energy" using the formulas since he knew how that worked, and then a third post giving an answer to his own question that left no doubt he was interested in the metaphysical aspect of kinetic energy and not how to do a homework problem, it just seems a little patronizing imo to keep trying to show him different ways to calculate the kinetic energy. That's not what he wanted to know.

 

[bruce2g]

...the 1.5mv^2 kinetic energy gained by the pellet is accounted for by the lost of 1.5mv^2 of kinetic energy by the platform. That's a lot of assumptions to make for one. Another is that even if they are all correct, all you and bruceg2 (whose calculations
were wrong even if his solution was similar to yours) appear to be saying is that if the platform loses 1.5mv^2 of kinetic energy to the pellet, the pellet gains 1.5mv^2 of kinetic energy. ...

Eyesaw,
If there is any error in my caluculation, please identify it. Otherwise, please retract your statement that my "calculations were wrong."

Bruce2g

 

[Slinkie]

Eyesaw was correct; the point of this thread was not to simply balance equations - that has been done for 300+ years. If I were to present a hypothesis that another form of energy transfer had a similar multiplicative effect, eye brows would certainly be raised. If I had supporting equations that balanced under all analysis (i.e., they are internally consistent within their defined domain), this would be accepted by many as a plausible model. Suppose the energy transfer I modeled was heat transfer: apply a flame for 1 sec. and a certain amount of energy is absorbed, apply a flame for 2 sec. and 8 times the energy is absorbed. This is clearly absurd, yet we accept this type of "definition" of energy transfer to an object in motion from our earliest physics courses and treat it as fact. Certainly there was a time when this was a lively debate and it is clear which side won. The pivotal question was whether to define work and energy by Force x Distance or Force x Time.

Let's alter the gedanken. A new pellet rifle has been created to compensate for the change in momentum; it fires a pellet of equal mass in the opposite direction (the Momentum Compensation Pellet - MCP). Let's call the pellet being accelerated twice the Doubly Accelerated Pellet - DAP. Being aware of the principle of energy conservation and the currently accepted energy equations it is with some trepidation that Buddy1 agrees to perform the experiment again. If in fact 3 times the energy is transferred to the DAP, then it stands to reason that 3 times the energy will be transferred to the MCP. The experiment is repeated and much to Buddy1's relief the MCP becomes stationary relative to his position.

Perplexed, Buddy1 consults his two physicist friends Mr. FxD (force x distance) and Mr. FxT (force x time).

Mr. FxT states that it is clear that the MCP was decelerated for 1 time unit yielding -1 energy unit relative to Buddy1's position hence the stationary position of the MCP. In addition, the DAP was accelerated for 1 time unit by Buddy1 and 1 time unit by Buddy2 yielding a total energy transfer of 2 energy units.

Laughing, Mr. FxD states how cute and misguided that analysis is and proceeds to apply his equations. "By the way, do you realize that there are over 300 years of equation balancing and analysis to support my model?" notes Mr. FxD gleefully, "So you KNOW I have to be correct. Why, just open any introductory Physics textbook and it'll be there." Mr. FxD states that since the MCP was decelerating relative to Buddy1's position, the force was applied over the same distance that Buddy1 used to accelerate the DAP to 1 energy unit's velocity (V1), therefore -1 energy unit was applied and the MCP became stationary relative to Buddy1's position. However, since Buddy2 was traveling at velocity V1, he applied force to the DAP over a much longer distance so much more energy was applied. In fact, Buddy1 added 1 energy unit to the DAP and Buddy2 added 3 energy units to the DAP yielding a total energy transfer of 4 energy units (from Buddy1's perspective).

Being aware of the principle of energy conservation, Buddy1 and Buddy2 both are scratching their heads at this point. Buddy1 and Buddy2 both saw -1 energy unit removed from the MCP. Buddy1 wonders: since Buddy2's velocity remained constant, all of the energy applied to the DAP must have come from the MCP so how can the DAP have 4 times the amount of energy? It doesn't seem like 1 energy unit transferred from the MCP will account for the 3 energy units acquired by the DAP. Where did the 2 additional energy units come from? Buddy2 and Mr. FxT are wondering the same thing.

Mr. FxT and Mr. FxD begin arguing, balancing their respective equations and deriving different results for energy. Buddy1 and Buddy2 begin to wonder what "energy" is and if Mr. FxT and Mr. FxD are talking about the same thing. Clearly they're not; the units they each use to represent "energy" are different as a direct result of how each has chosen to define "energy".

 

[cepheid]

Slinkie,

This is very frustrating. It is clear from your last post that you did not read any of my work. (Post # 23, which Eyesore did not quote in its entirety. Please read the original.) If you had read it, you would not think that the fact that the pellet ended up with four times the inital kinetic energy was "absurd". My analysis did not mention momentum conservation at all. It showed quantitatively that the second gun did indeed do three times more work on the pellet than the first, which negates the subsequent reasons that you had for questioning the definition of work. I will invest no more time in a thread in which other participants are unwilling to consider evidence presented to them. Debate is good, but a discussion in which my valid and quantitative points are consistently ignored or obfuscated in order to lend credence to yours, does not constitute an intelligent debate or physical analysis. Please go over my calculations again. Didn't you see how I arrived at the answer that the second gun does 375 J of work, three times more than the first?

As for your question: what is energy? It is an excellent and very deep question. I am not sure when the concept of work as defined in its precise, physical sense first arose, but consider this: there is some quantity in physics having fundamental units kgm²/s² that as we expand physics beyond the realm of mechanics we find is conserved in every conceivable situation, although it has the capability of being transformed into various forms, some more useful for "doing stuff" than others. Examples are mechanical energy (kinetic/potential), thermal energy, EMR, etc. That is all we can say for sure. SO the definition of work as is provides for the simplest setup with which to analyse this amazing conservation law, for if you changed the definition of work to force X time, the resulting change mv caused by this new work would not be dimensionally consistent with any of the other forms of "energy" identified in other realms of physics...and hence would not qualify as energy. I see no benefit to applying the same name to totally different quantities...the accounting books would not balance. In fact, the quantities as defined in your system would not differ from their current roles in physics. They would differ in name only. What you would be calling 'work' would still have the fundamental properties of what we properly call impulse, and nothing would have changed, except that we would be left with a far more contradictory and cumbersome system of terminology in which so called "energy" would refer to at least two different types of quantities, and would therefore be neither transferable nor universally conserved.

 

[bruce2g]

Being aware of the principle of energy conservation, Buddy1 and Buddy2 both are scratching their heads at this point. Buddy1 and Buddy2 both saw -1 energy unit removed from the MCP. Buddy1 wonders: since Buddy2's velocity remained constant, all of the energy applied to the DAP must have come from the MCP so how can the DAP have 4 times the amount of energy? It doesn't seem like 1 energy unit transferred from the MCP will account for the 3 energy units acquired by the DAP. Where did the 2 additional energy units come from? Buddy2 and Mr. FxT are wondering the same thing.

Do the Buddies and Mr. FxT realize that two shots were fired, one to accelerate the DAP, and another to declerate the MCP. Each shot added energy of $1/2 mv^2$. The MCP added energy of $1/2 mv^2$. It all went to the DAP, to give it energy of $2 mv^2$. Where else could the energy from the shots have gone, assuming you believe in conservation of energy?

Bruce

PS: Thanks for giving me the opportunity to repeat the "error" in my previous calculations here; perhaps you or Eyesaw would be kind enough to point it out this time.

 

[Eyesaw]

I analyzed things slightly differently, so maybe my perspective may help a bit. First off, since we are talking about firing a gun (setting off an explosion, in effect) mechanical energy (KE) is not conserved. So that is certainly not an assumption.

So how much energy is released by the firing of the bullet? The bullet gets $1/2mv_1^2$, as stated, but momentum conservation tells us that the platform gets pushed back at speed $v_2 = (m/M)v_1$. So the total KE released is that of the bullet plus platform = $1/2mv_1^2 + 1/2Mv_2^2 = (1 + m/M)(1/2mv_1^2)$.

Viewed from the "rest" frame, the initial KE of everything before the second firing is $1/2(m + M)v_1^2$. After firing, the bullet has speed $2v_1$ and the platform has speed $v_1 - v_2 = (1 - m/M)v_1$, so the final KE of everything is: $1/2m(2v_1)^2 + 1/2M(1 - m/M)^2v_1^2$.

Do the arithmetic and you will find that the increase in KE of everything exactly equals the KE released by the firing. There is no missing energy.

The argument made here basically is that measured inside the inertial frame of the second firing, there's a quantity of kinetic energy c, which when viewed by an inertial frame moving at -v, obtains the value c-v. But after taking into account of the -v, the second inertial frame also obtains the value c for the kinetic energy. Now if you only admitted this same logic applies to the speed of light, you'd agree there's no need to warp space-time to conserve the quantity c.

 

[Eyesaw]

Do the Buddies and Mr. FxT realize that two shots were fired, one to accelerate the DAP, and another to declerate the MCP. Each shot added energy of $1/2 mv^2$. The MCP added energy of $1/2 mv^2$. It all went to the DAP, to give it energy of $2 mv^2$. Where else could the energy from the shots have gone, assuming you believe in conservation of energy?

Bruce

PS: Thanks for giving me the opportunity to repeat the "error" in my previous calculations here; perhaps you or Eyesaw would be kind enough to point it out this time.

Well, the way you added up the total kinetic energy in your earlier post was
1/2mv^2 + mv^2 + 1/2mv^2 to get 2mv^2. Since you were viewing from the ground frame, in your example the second explosion actually released 3.0mv^2 of kinetic energy, judging by the final velocity of the left brick of 2v, 1.5mv^2 added to the left brick and -1.5mv^2 added to the right brick. The 1.5mv^2 gave the left brick a final kinetic energy of 2.0mv^2 while the 1.5mv^2 subtracted from the kinetic energy of the right brick (since the right brick was moving at v when the second explosion transferred k.e. against the direction of its motion , i.e. -1.5mv^2) leaving the right brick with a final kinetic energy of -1.0mv^2, viewed from the ground.

You thought the right brick came to rest but it didn't since it has a final kinetic energy of -1.0mv^2.

So a few things you went wrong in your calculation. First, you calculated the wrong amount of total kinetic energy of the system as viewed from the ground. Second, I think it's in error in your example to consider the right brick as having lost kinetic energy to the left brick when it's obvious that the explosion is what transferred kinetic energy to both bricks, just as it was obvious when the explosion happened "at rest" on the ground. A negative value in kinetic energy does not automatically mean a loss in kinetic energy- if that were so, I could probably lose weight just by walking backwards than I normally do.

No, in your case, the 1.50 mv^2 gained by the left brick was generated by the explosive charge, it's the same 0.5mv^2 transferred by the charge in the first explosion when at "rest". The second explosion has more kinetic energy than the first because you are viewing the second explosion from a frame that has v less velocity, which when plugged into the kinetic energy equation, accounts totally for the "extra" energy.

 

[Eyesaw]

Slinkie,

This is very frustrating. It is clear from your last post that you did not read any of my work. (Post # 23, which Eyesore did not quote in its entirety. Please read the original.) If you had read it, you would not think that the fact that the pellet ended up with four times the inital kinetic energy was "absurd". My analysis did not mention momentum conservation at all. It showed quantitatively that the second gun did indeed do three times more work on the pellet than the first, which negates the subsequent reasons that you had for questioning the definition of work. I will invest no more time in a thread in which other participants are unwilling to consider evidence presented to them. Debate is good, but a discussion in which my valid and quantitative points are consistently ignored or obfuscated in order to lend credence to yours, does not constitute an intelligent debate or physical analysis. Please go over my calculations again. Didn't you see how I arrived at the answer that the second gun does 375 J of work, three times more than the first?

As for your question: what is energy? It is an excellent and very deep question. I am not sure when the concept of work as defined in its precise, physical sense first arose, but consider this: there is some quantity in physics having fundamental units kgm²/s² that as we expand physics beyond the realm of mechanics we find is conserved in every conceivable situation, although it has the capability of being transformed into various forms, some more useful for "doing stuff" than others. Examples are mechanical energy (kinetic/potential), thermal energy, EMR, etc. That is all we can say for sure. SO the definition of work as is provides for the simplest setup with which to analyse this amazing conservation law, for if you changed the definition of work to force X time, the resulting change mv caused by this new work would not be dimensionally consistent with any of the other forms of "energy" identified in other realms of physics...and hence would not qualify as energy. I see no benefit to applying the same name to totally different quantities...the accounting books would not balance. In fact, the quantities as defined in your system would not differ from their current roles in physics. They would differ in name only. What you would be calling 'work' would still have the fundamental properties of what we properly call impulse, and nothing would have changed, except that we would be left with a far more contradictory and cumbersome system of terminology in which so called "energy" would refer to at least two different types of quantities, and would therefore be neither transferable nor universally conserved.

Yes, you showed that the second firing did 3 times more work, but did you show why 3 times more work didn't add 3 times more velocity to the pellet? No, the only answer you could give was that because K.E. is defined to be 1/2mv^2. If K.E. was undefined and you started from rest and applied a constant force over 1 distance unit to achieve a velocity v for a particular mass, wouldn't it be reasonable to assume that if you applied the same force over 2 distances, you should achieve a final velocity of 2v? Why isn't this the case? Isn't that a more interesting question than whether you applied the formulas correctly or not?

 

[Doc Al]

The argument made here basically is that measured inside the inertial frame of the second firing, there's a quantity of kinetic energy c, which when viewed by an inertial frame moving at -v, obtains the value c-v. But after taking into account of the -v, the second inertial frame also obtains the value c for the kinetic energy.

You seem to mix up speed with kinetic energy.

Now if you only admitted this same logic applies to the speed of light, you'd agree there's no need to warp space-time to conserve the quantity c.

It is not "logic" which allows this calculation of c-v, but an assumption of Galilean relativity, which we know to be inadequate at higher speeds.