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I was wondering. As velocity is relative, how can one calculate the kinetic energy of an object.

If for example object A has a mass of 1kg. It has a relative speed to object B of 10m/s and a relative speed to object C of 500m/s.
What would be the kinetic energy released when it is slowed down 5m/s

1/2 * 1kg * (10m/s)^2 - 1/2 * 1kg * (10m/s - 5m/s)^2 = 37.5J

1/2 * 1kg * (500m/s)^2 - 1/2 * 1kg * (500m/s - 5m/s)^2 = 2487.5J

To mee it seems like contradicotry that the velocity is relative and being squared in the formula.
Could someone explain how this works and what is the correct way of calculating?

The velocity is a dot product, that makes kinetic energy a scalar and so invariant under a transformation.

Energy is relative just as velocity

[math]\dot{\bf v}^2[/math] is invariant under rigid motion of [math]\mathbb{R}^3[/math].
Correct.
Incorrect.

You are forgetting density. 1 kg of steel and 1 kg of feathers will look and act very differently in air or water environment.

Gravity is a bullshit theory.

This guy is correct.

OP even provided proof through a thought experiment

Kinect energy is just as relative as velocity (it changes under Lorentz transformations on a special relativistic setting; under Galilean transformation otherwise), like user said.
In the thought experiment, what happens is that whatever is doing the work to decelerate the particle A will have itself a final state (we can specify its state by the energy gained) which differs from one reference frame to the other (w.r.t. B or C), hence the difference in energy

so the kinetic energy differs in every system but still the conservation of energy is respected in said systems, right?

Kinetic energy of object A is meaningless until it interacts with some second object. I mean that there is no gadget you can hold in your hand and read your own KE off the display.

Object A has a KE relative to Earth and, simultaneously, a KE relative to Mars. They are different. But CHANGES of KE are the same by everybody's measure.

Example: An astronaut floats outside the ISS, 50 meters distant. He masses 200 kg. His backpack applies a steady 100 newtons until he closes the gap and grabs a handhold. Work is force times distance. So his
delta KE is 100 nt x 50 meters. Right? 5000 joules. How fast did he hit the station? 5000 = .5 * 200 * v^2
V = 7.071 meters/sec. Average velocity is half that, so he took 50 * 2 / 7.071 = 14.142 seconds for the trip.

Someone on Earth is watching him through a telescope. The ISS moves at, say, 7000 mt/sec. The astronaut has slowed from 7007.071 mt/sec to 7000 mt/sec. His delta KE is 0.5 * 200 * (7002.071^2 - 7000^2) =
9,904,449 joules. MUCH larger. How can this be?

Well, the guy on the ground sees him traveling at an average 7003.535 mt/sec. In 14.142 seconds he moved 99,044 meters. Work is force times distance. So his delta KE is 100 nt x 99,044 meters. Which is 9,904,340 joules. The tiny error is due to my rounding off.

Is this clear? You're by no means the first to be puzzled.

>v˙2v˙2 is invariant under rigid motion of R3R3.
What does that even mean?
Brainlet here.

exactly, and you need to compare exchange of energy (and its conservation) Given a reference frame (what you mean by system I suppose). If you want to compare energies in different reference frames, then you need to do a Galilean (or Lorentz) transformation

see also
it is important to notice that by asking these questions you are making the transition from kinematics (movement alone) to dynamics (why the object is moving in the first place)

v^2 is a scalar, so it's invariant under 3D rotations, just like distance between two points

By scalar he means that V^2 doesn't have a direction.
Kinetic energies cannot be added like vectors.
This comes in handy sometimes.
A space probe can fall towards Jupiter and pick up a tremendous velocity along the way by virtue of losing potential energy during the drop. It uses its rockets to increase its velocity a little bit, X meters/sec, at perijove and then soars away from the planet, sacrificing velocity to climb out of the Jovian gravity well. Eventually it's far away from the planet.

If done properly, its velocity (with respect to Jupiter) has changed by a great deal more than X mtrs/sec during the fly-by. This is called a gravity well maneuver.

This "free lunch" is above and beyond anything it gains by stealing momentum from the motion of the planet itself.
Of course, it's not really free. It's fuel had potential energy before it approached Jupiter and that fuel is discarded there at the bottom of the gravity well.

thank you, I'm talking about scalars and 3D rotations but that may be no more clear than "invariant under rigid motion of R^3".

You've welcome.
I'm an engineer, so I think in terms of force, mass, energy, and conservation laws.

Shut the fuck up, monkey.

does this comment say anything other than "tell me to shut the fuck up"?

I see. I think your physical examples were great.

I'm a theoretical physicist, so I think in terms of conservation laws and the likes, but in a much more mathematical manner, I believe.

Pleased to meet you.

>Average velocity is half that
why?

He started motionless with respect to the ISS. Initial velocity = 0
He hit it at a final velocity of 7.071 meters/sec.
Under constant thrust, velocity varies linearly with time.
When acceleration is constant, average velocity is (initial v + final v) / 2
The time to cover a given distance is the distance divided by the average velocity.