Say I have a charged molecule running along a linear evacuated tube (so no wind resistance). In the laboratory frame, we can measure that the particle is moving at some speed $v_1$. Provided that kinetic energy scales as the square of the velocity, how much energy do we need to impart on the particle to accelerate it to some velocity $v_2$? Why should this depend on $v_1$ and not strictly the difference $v_2 - v_1$?
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1$\begingroup$ kinetic energy is also proportional to mass. Try moving a ton. $\endgroup$– anna vCommented Jul 14, 2012 at 5:20
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5 Answers
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Why should this depend on v1 and not strictly the difference v2−v1?
If I understand your question correctly, you're not so much asking for a mathematical explanation as you are for a physical one?
Remember that KE is essentially the energy stored from work done by a force on the particle and that work is force through a distance.
When there is an initial velocity, the distance through which the force is applied is different than when there is no initial velocity.
For example, let's say that to uniformly accelerate from 0 to $\Delta v$ requires 1 second and a distance of 1 meter.
Now, imagine that there is an initial velocity $v_1$, in the direction of the applied force, and we have the same uniform acceleration for 1 second. The change in velocity is the same but the distance through which the same force is applied is now larger by $v_1 \cdot 1 s$.
More work is done by the force since it is applied through a greater distance. So, the change in KE must depend on the initial velocity, i.e., change in KE must be frame dependent.
answered Jul 14, 2012 at 12:20
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The kinetic energy depends on $v_1$ because it's not a linear function of velocity. You can show the problem with a bit of algebra - nothing too scary!
Take your example of accelerating from $v_1$ to $v_2$, then the energy required is:
$$ \Delta E = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 $$
To make the problem clearer rewrite this as:
$$ \Delta E = \frac{1}{2}m \left( v_2^2 - v_1^2 \right) $$
Can we write this equation as a function of just $\Delta v = v_2 - v_1 $? Well we can factor $(v_2^2 - v_1^2)$ to $(v_2 - v_1)(v_2 + v_1)$ or $\Delta v(v_2 + v_1)$, but we can't get rid of both $v_1$ and $v_2$ in the $(v_2 + v_1)$. The best we can do is write $\Delta E$ as a function of $v_2$ and $\Delta v$ or $v_1$ and $\Delta v$ e.g.
$$ \Delta E = \frac{1}{2}m \Delta v \left( \Delta v + 2v_1 \right) $$
or
$$ \Delta E = \frac{1}{2}m \Delta v \left( 2v_2 - \Delta v \right) $$
answered Jul 14, 2012 at 9:55
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(This answer uses Special relativity, if that is not what your question is concerned then I apologize.)
If $v_1$ is close to $c$ ($v_1 \geq 0.3c$) then it does matter.
as the energy (Using Lorentz transformation) that is required to accelerate an object from $v_1$ to $v_2$ is: $E = (\frac{1}{\sqrt{1-(\frac{v_2}{c})^2}}-\frac{1}{\sqrt{1-(\frac{v_1}{c})^2}})mc^2$
so, it takes more energy to accelerate an object from $0.3c$ to $0.3c+15 [\frac{m}{s}]$ (almost 1.8 billion joules assuming $m=1 [kg]$)
then from 0 to 15 $[\frac{m}{s}]$ (about 112 joules, again with $m=1 [kg]$)
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$\begingroup$ While this isn't wrong you appear to have misunderstood the question. $\endgroup$ Commented Jul 14, 2012 at 15:34
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Kinetic energy: $E=\frac{m v^2}{2}$
Derivative of E towards v: $\frac{dE}{dv}=mv$
Thus, the energy $dE$ required to accelerate a mass $m$ by a given additional velocity $dv$ is proportional to the mass $m$ and the velocity $v$. Or in simpler terms: the higher the velocity the more energy it takes for the same amount of additional velocity.
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Based on my separation energy formula, neglecting relativistic effects by assuming velocities significantly below the speed of light, to accelerate a molecule from rest relative to the tube to velocities $v_1$ and $v_2$ (where $v_2$ > $v_1$), you would require energies given by:
$$ E_1 = \frac{1}{2} \frac{M m}{M + m} v_1^2 $$
and
$$ E_2 = \frac{1}{2} \frac{M m}{M + m} v_2^2 $$
Here, $m$ represents the mass of the molecule, and $M$ is the mass of the Earth. This assumes that the tube is securely attached to the Earth and the molecule rebounds from the Earth.
Given that $m \ll M$, the energies can be simplified to:
$$ E_1 \approx \frac{1}{2} m v_1^2 $$
and
$$ E_2 \approx \frac{1}{2} m v_2^2 $$
These approximations are essentially error-free. The energy required to accelerate the molecule from $v_1$ to $v_2$ is then:
$$ E = E_2 - E_1 \approx \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 = \frac{1}{2} m \left( v_2^2 - v_1^2 \right) $$
