My problem with the fromula for KE

[MasterColg]

E = .5*m*v^2

 

I have a problem with this formula.

 

To illustrate lets play some mental catch with a friend.

 

Assume I have a ball weighing 1 Kg and that I can throw it 10mps.

When my friend catches it he will absorb the 50joules of transitional kinetic energy that my arm had imparted to the ball.

 

My arm gets tired and I board a train and my friend remains by the track. (Didn’t I mention that we were playing catch next to a railroad track that goes in a big circle? Well we are…) The train is traveling at 10mps and as I pass my friend I drop the ball so he can catch it. Again when my friend catches it he will absorb the 50joules imparted to the ball by the train.

 

My arm is feeling better now so I throw the ball to my friend as the train passes him. The train is going 10 mps and I throw at 10mps giving the ball a combined speed of 20mps. This time when my friend catches the ball he must according to the formula absorb 200joules of energy imparted to the ball by the train and my arm.

 

The question is, if I imparted 50joules and the train imparted 50joules, where did the other 100 joules come from?

 

Can anyone help me out here?

[swansont]

You are measuring the energy in two different reference frames, and it's not an invariant quanity under that transformation.

 

If you play catch on the train in your scenario, you measure the ball's energy as 50 J, but an observer on the ground measures it as 200 J. But it's the same ball, just measured in two different frames of reference. So saying you imparted X Joules of energy is a frame-dependent statement.

 

Other things to consider: when you throw the ball, the train should slow down from conservation of momentum, but you have constrained the speed to be constant. The train engine does work on the system. Within a given frame of reference, energy will be conserved. But kinetic energy is not itself a conserved quantity; it is only conserved under the special condition of an elastic collision, and this situation is the reverse of a completely inelastic collision.

[MasterColg]

Ill begin by addressing your reply in reverse order:

 

Chemical energy in my arm is translated into Kinetic energy in the ball, which is then translated into thermal energy in the glove. Kinetic energy is not conserved but energy in general is.

 

It is true that I did not take the trains reaction of my throw into account. It would have been minute and there are a lot of other things that I didn’t take into account, Wind resistance and gravity just to name two. None of these things are relevant to the question and I ignored them for simplicity’s sake.

 

The calculation of an objects kinetic energy is completely dependant on frame of reference. However the statement that I imparted 50J of energy to the ball is not. It is based on the referential frame of the thrower and no one in an external frame will observe differently.

 

The scenario works fine if we adhere to a single frame of reference, that of the catcher:

As we stand together by the tracks he will observe the ball approach him at 10mps and he can calculate the Kt to be 50j. As the train passes and the ball is dropped to him he will observe it to be traveling at 10mps and he can again calculate the Kt to be 50j. Again the train passes and the ball is thrown, he observes the ball approach him at 20mps (disregarding external influence) He can then calculate the Kt to be 200j. Here the question arises. He knows the limit of my arm is 10mps or 50j. He knows that the speed of the train is constant and that any reaction of it to my throw would have had a negative affect on the speed and energy of the ball therefore it’s contribution could only have been 10mps or 50j. He knows that energy can be converted from one form to another, but it cannot be created or destroyed and he wonders where the extra 100j came from…

 

I constructed the scenario with frames of reference as an aid to understanding. The formula implies that it requires 4 times as much energy to accelerate an object to 100mps as it does to accelerate it to 50mps. Here is another illustration:

 

Take a rocket in deep space. You place a buoy (A) to provide a frame of reference and burn half of your fuel in acceleration. (Assume that this fuel is energy dense so that the loss of its mass will not unduly affect the next phase, say the ships mass is 1,000,000Kg, it contains 2grams of fuel and it burns 1g in the first phase) Your buoy’s beacon now indicates that it is receding at a rate of 10000mps. You place a second buoy (B) and burn the remaining fuel along the same vector. The signal from B indicates a velocity of 10000mps and the signal from A indicates a velocity of 20000mps, all is well except that according to the formula you should have burned 4 times as much fuel to reach 20000mps as you did to reach 10000mps.

[swansont]

He knows the limit of my arm is 10mps or 50j.

 

 

No, he doesn't. He knows that in his frame (i.e. at rest), it's 50 J, but that value doesn't apply in any other frame of reference.

 

 

Take a rocket in deep space. You place a buoy (A) to provide a frame of reference and burn half of your fuel in acceleration. (Assume that this fuel is energy dense so that the loss of its mass will not unduly affect the next phase, say the ships mass is 1,000,000Kg, it contains 2grams of fuel and it burns 1g in the first phase) Your buoy’s beacon now indicates that it is receding at a rate of 10000mps. You place a second buoy (B) and burn the remaining fuel along the same vector. The signal from B indicates a velocity of 10000mps and the signal from A indicates a velocity of 20000mps, all is well except that according to the formula you should have burned 4 times as much fuel to reach 20000mps as you did to reach 10000mps.

 

The conclusion that the 1g of the fuel changes your speed by 10^4 m/s is only valid in one frame of reference: the one where you are at rest with respect to "A"

[MasterColg]

You have one Throw and two observers:

 

Observer A is standing next to the pitcher on the train.

He records a speed of 10mps or 50J

 

Observer B is standing next to the rail bed beside the train.

He records a speed of 20mps or 200J

 

Trace the origin of the balls kinetic energy for both frames of reference.

 

---------------------------------------------------------------------------

 

 

The rocket is moving at 10,000mps with respect to buoy A after burning 1g of fuel.

It releases buoy B and is now at rest in reference to it.

 

It burns its remaining gram of fuel along what ever vector you like.

 

What is its speed now in reference to buoy B?

[J.branco]

the problem of the train:

for someone outside the train the ball has kinetyc energy corresponde to 50J. equivalent to the 10mps, the it's travelling. When you throw it, you give it more 50 J.

ΔE = 1/2m (vf^2-vi^2) ↔ 50 = 1/2 (vf^2-10^2) ↔

↔ 100 =1/2 (vf^2) ↔

↔ 200^(1/2) =vf

so the final velocity of the ball wont be 20mps but 14,14mps aproximatly. in order to achieve the variation of speed of 10mps, you'd have to give it aditional energy

[J.branco]

now referring to the rocket, the same apllies. Suppposing the second burn of fuel will give the rocket the same amount of energy, the variation of velocity will be smaller : (10000^2 - 0) ≠ (20000^2 - 10000^2) so the final velocity, in the frame of the buoy A won't be 20000mps.

Now in the frame of buoy B, im not sure but since the inicial velocity is 0 i suppose the final will be 10000mps.

ANy mistakes in what i've said?

[MasterColg]

ANy mistakes in what i've said?

 

It doesn’t hold true in the light of empirical evidence.

 

The Pitcher is at rest in reference to the train. If the observer on the train records a velocity of 10mps forward and the train is moving at 10mps forward, then the observer at the rail bed must record a velocity of 20mps vis-a-vis the Galilean transformations.

 

The supposition that a pitcher must exert two separate levels of force in a single pitch to satisfy the requirements of separate frames of reference is patently absurd.

 

The supposition that a pitcher on a train must exert more force, i.e. impart more energy to the ball to achieve 10mps in a forward throw than a rearward throw is likewise patently absurd.

 

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The rocket is moving at 10,000mps with respect to buoy A after burning 1g of fuel.

It releases buoy B and is now at rest in reference to it. Both the rocket and buoy B are now moving at 10,000mps in reference to buoy A. The Rocket now burns 1g of fuel along the original vector and is now moving (as you have already stated) at a velocity of 10,000mps in reference to buoy B. Again this is simple addition of velocity. If Buoy A is stationary, Buoy B is moving at 10,000mps relative to it, and the rocket is moving 10,000mps faster than Buoy B along the same vector, then the rocket must be moving at a velocity of 20,000mps relative to buoy A.

[swansont]

The supposition that a pitcher must exert two separate levels of force in a single pitch to satisfy the requirements of separate frames of reference is patently absurd.

 

Then you need not bother with learning any more physics, because you will find it patently absurd. If a quantity is not preserved under a transformation between frames, you cannot expect that quantity to have the same value in both frames. (Any more than you can expect a given item to cost the same amount in US dollars, Canadian dollars, Euros and Yen.)

[MasterColg]

Then you need not bother with learning any more physics, because you will find it patently absurd. If a quantity is not preserved under a transformation between frames, you cannot expect that quantity to have the same value in both frames. (Any more than you can expect a given item to cost the same amount in US dollars, Canadian dollars, Euros and Yen.)

 

 

The energy has to come from somewhere. In the example of the pitch it was chemical energy converted by muscle cells into kinetic.

 

Do you honestly expect me to believe that our pitcher burns more calories in one frame of reference than another in a single pitch?

[J.branco]

The supposition that a pitcher on a train must exert more force' date=' i.e. impart more energy to the ball to achieve 10mps in a forward throw than a rearward throw is likewise patently absurd.[/quote']

 

No it doesn't need to exert more force. In fact the force will be the same as told by the galilean transformations and as u can see since the variation of momentum is the same. It's the energy required to exert that force that will be different, according to the frame, so as the Kinetyc energy of the ball.

However, for the pitcher, it wiil be the same since, in the frame at rest with him, wherever he throws it, the energy he spends will be the same.

[swansont]

The energy has to come from somewhere. In the example of the pitch it was chemical energy converted by muscle cells into kinetic.

 

Do you honestly expect me to believe that our pitcher burns more calories in one frame of reference than another in a single pitch?

 

If all the energy terms are different in the other frame such that energy is conserved, where, exactly, is the problem?

[MasterColg]

If all the energy terms are different in the other frame such that energy is conserved, where, exactly, is the problem?

 

Elaborate.

[swansont]

Elaborate.

 

Your objection is that energy isn't the same in the different reference frames, but haven't demonstrated that energy is actually not being conserved within any frame.