Kinetic energy

[devrimci_kürt]

Part 1: Kinetic energy is not ½mv².

 

 

A 4kg object dropped 1m (meter) has the same amount of ½mv² as a 1kg object dropped 4m, because force times distance equals ½mv² for an accelerating mass. But a rocket accelerating the masses to those velocities requires twice as much energy as fuel for the large mass as for the small one.

 

Therefore, both masses do not have the same energy; the rocket does not transform energy in proportion to ½mv²; ½mv² is not kinetic energy; and a gallon of fuel does not produce a consistent amount of ½mv².

 

Part 2: Kinetic energy is mv.

 

 

A 4kg object dropped for 1s (second) has the same amount of mv (momentum) as a 1kg object dropped for 4s, because force times time equals mv for an accelerating mass. A rocket accelerating the masses to those velocities uses the same amount of energy as fuel for both masses.

 

Therefore, both masses have the same amount of energy; the rocket transforms energy in proportion to mv; mv is kinetic energy; and a gallon of fuel produces a consistent amount of mv.

 

All logic and evidence of energy points to the same conclusion. The logic created the need to derive the mathematical proof.

 

About ninety percent of physics is corrupted by the error.

 

 

 

excerpt:

http://nov55.com/ener.html

 

I didn't understand

İs not Kinetic energy ½mv².

Edited November 16, 2009 by devrimci_kürt

[ajb]

By looking at the units you see that it is possible that [math]E \sim mv^{2}[/math] (mass times speed times speed.) and not [math] E \sim mv[/math] (mass times speed).

 

The combination [math]mv[/math] is (the magnitude of) the linear momentum.

[D H]

Part 1: Kinetic energy is not ½mv².

 

 

A 4kg object dropped 1m (meter) has the same amount of ½mv² as a 1kg object dropped 4m, because force times distance equals ½mv² for an accelerating mass.

Correct. That is the work energy theorem.

 

But a rocket accelerating the masses to those velocities requires twice as much energy as fuel for the large mass as for the small one.

(1) Incorrect, and (2) non sequitur.

 

Regarding (1) incorrect:

The relevant equation here is the Tsiolkovsky rocket equation,

 

[math]\frac{\Delta v}{v_e} = \ln\left(\frac {m_r+m_f}{m_r}\right)[/math]

 

where [math]\Delta v[/math] is the change in velocity experience by the rocket, [math]v_e[/math] is the exhaust velocity, [math]m_r[/math] is the mass of the rocket after achieving the specified delta-V, and [math]m_f[/math] is the mass of the fuel expended in achieving that delta-V.

 

The problem at hand: Given that propelling a rocket of final mass [math]m_1[/math] to some velocity [math]\Delta v_1[/math] requires a quantity of fuel [math]m_{f1}[/math], how much fuel [math]m_{f2}[/math] is needed to make a rocket with a mass [math]m_1/4[/math] to twice the final velocity? The answer is

 

[math]m_{f2} = \frac 1 2\,m_{f1} + \frac 1 4\,\frac{m_{f1}^2}{m_1}[/math]

 

Regarding (2) non sequitur:

Novak is ignoring the energy in the exhaust.

 

Therefore, both masses do not have the same energy; the rocket does not transform energy in proportion to ½mv²; ½mv² is not kinetic energy; and a gallon of fuel does not produce a consistent amount of ½mv².

Wrong math => wrong conclusion.

 

 

Part 2: Kinetic energy is mv.

Pure nonsense. mv has units of momentum, not energy.

 

[Nonsense elided] About ninety percent of physics is corrupted by the error.

You quoted a crackpot, Kurt.

 

I wrote a tutorial on rocket dynamics for another physics site. See http://www.physicsforums.com/showthread.php?t=199087.

 

 

Bottom line: Someone saying along the lines of "About ninety percent of physics is corrupted by the error" is almost certainly a crackpot. Don't get sucked in by their nonsense.

Edited November 16, 2009 by D H

[CaptainPanic]

The two objects have the same potential energy:

 

[math] E=m\cdot{g\cdot{h}}[/math]

[math] E=m\cdot{g\cdot{h}}=1\cdot{9.81\cdot{4}}=39.24J[/math]

[math] E=m\cdot{g\cdot{h}}=4\cdot{9.81\cdot{1}}=39.24J[/math]

 

Therefore, they will have the same kinetic energy after they fall down. But: the 1 kg object falls much longer (4 meters instead of 1)... so it will have more velocity. In fact, it's velocity will be exactly twice that if the 4 kg object.

 

After falling:

 

[math]E=0.5\cdot{m\cdot{v^2}}=0.5\cdot{1\cdot{8.86^2}}=39.24J[/math]

[math]E=0.5\cdot{m\cdot{v^2}}=0.5\cdot{4\cdot{4.43^2}}=39.24J[/math]

 

You can check the velocity of the object yourself with the formulas found here: http://en.wikipedia.org/wiki/Acceleration