How can kinetic energy be proportional to the square of velocity, when velocity is relative?

The kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed $v$ is $\frac{1}{2}mv^2$.

Let's say you & your bike have mass of 100kgs, then your kinetic energy at 10m/s would be

$$E_a = 1/2 \times 100 \times 10^2 = 5000J = 5kJ$$

If you apply another 5kJ of energy, you don't get to 20m/s though, you only get to:

$$E_b = 10000J =1/2 \times 100 \times V_b^2$$ $$\implies V_b = \sqrt{10000 / (1/2 \times 100)} = √200 = 14.14m/s$$

Let's say you and a buddy are both coasting along at 10m/s though, from their perspective you've just burned 5kJ but only accelerated 4.1m/s, even though you seemed stationary.

Imagine you and your mate are in space drifting along together, at an unknown speed. Your mate fires his burners and accelerates away from you. There's a big screen on his ship showing how many joules of energy he just burned, and you can measure his resulting relative velocity just fine.

The question is, Will 5kJ of energy always produce 10m/s of relative velocity,,assuming 100kg spaceships?

If 5kJ always produces 10m/s, Why does the second 5kJ only produce 4.1m/s? What is going on here?

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Possible duplicate: physics.stackexchange.com/q/535/2451 – Qmechanic Nov 23 '12 at 0:33
the kinetic energy is $E_{k} = \frac{1}{2}m \vec v \vec v$ so is an SCALAR , an scalar does not depend on the reference ssytem – Jose Javier Garcia Nov 23 '12 at 11:12
@JoseJavierGarcia It is a scalar under rotations; but just try a quick boost $v\rightarrow v'=v+a$ and you will see it is entirely frame dependent. Alternatively, if me and you are standing in a room at rest w.r.t each other, you see my KE as zero. Now start walking. What's my kinetic energy in your frame? – kηives Nov 23 '12 at 18:13

How can kinetic energy be proportional to the square of velocity, when velocity is relative?

Without reading the rest of your question, I must first reply that one has nothing to do with the other.

Kinetic energy is frame dependent, just as velocity is.

Momentum is proportional to velocity and is frame dependent too, just as velocity is.

Now, looking at the body of your question:

Imagine you and your mate are in space drifting along together, at an unknown speed.

Unknown speed relative to what? Unknown speed relative to Earth? Unknown speed relative to the solar system? Unknown speed relative to the CMB?

Assuming 100kg spaceships, will 5kJ of energy always produce 10m/s of relative velocity?

Relative to what? Relative to the initial inertial frame of reference before the acceleration? Or relative to some frame of reference in some arbitrary relative motion?

(The point of all these questions is to prompt you to think more clearly about your question in the hope that you'll come to the answer yourself...)

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 Relative to you, when initially you were travelling together. If they burn 5kJ of fuel, all of which goes into propulsion, will they be travelling 10m/s away from you? – geelen Nov 23 '12 at 0:59 According to frame A, initially at rest with respect to you and your buddy, if your KE is increased by 5kJ, in that frame, and your mass is unchanged, your speed, relative to A, is objective. – Alfred Centauri Nov 23 '12 at 1:04

Hmm, what's actually happening here is that nobody will agree with the screen.

See, in your frame, there is some additional energy on board the ship. The fuel inside has a velocity--it's kinetic energy must be considered separately, as it disappears (is burnt and expelled) from the ship. So, the burning of the fuel leads to some additional energy in your frame.

You can try the calculations yourself--whatever fuel you use, after conserving momentum and energy, it is consistent in both frames separately.

Note that there is no need at all for the energies to be equal in different reference frames.

Assuming 100kg spaceships, will 5kJ of energy always produce 10m/s of relative velocity?

Nope, it depends on your frame. See Why does kinetic energy increase quadratically, not linearly, with speed?

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To answer your last question first - the energy depends on the square of the velocity, not the velocity itself. $$E \propto v^2$$ or $$v \propto \sqrt{E}$$

Which means when you double the velocity, you'd quadruple the energy input. For the example that you had given, to get a velocity of $20$ $m/s$ you'd have to give in an energy input of $20$ $kJ$. So clearly, when you double the energy input, you'll only see an increase in velocity proportional to $\sqrt 2$, and in your case that's $1.414 * 10 = 14.14$, which is exactly what you got.

Hopefully this clears some of your confusion up. To answer your title question - As other answers have already pointed out, both momentum and velocity are relative. In special relativity, the momentum transforms as $$p = \gamma mv$$ where $v$ is the velocity of the body in one frame of reference, and $$\gamma = \frac{1}{\sqrt{1 - u^2/c^2}}$$ where $u$ is the relative velocity between the two frames of reference. So when one observer measures a different velocity because of his frame of reference, he will also measure a different kinetic energy.

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Both momentum AND velocity must be relative to the same thing. In other words if the two cyclists are going along at the same speed, the one cyclist's momentum relative to the other is zero. If they crashed into one another, both moving at the same speed, nothing would happen.

As for the formula, you can't say that $5KJ$ produces $10m/s$ and then another $5KJ$ $4.14 m/s$. The formula is not associative.

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Kinetic energy is frame-dependant.

The reason is very simple: kinetic energy is the difference between Mc^2 (don't know how to write square here) and the rest energy M0c^2. M depends on v, which depends on the frame, so obv kinetic energy depends on the frame.

Note: I tried to give you a logical explanation other that E=1/2mv^2

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