# I am confused about the meaning, and value, of kinetic energy

This confusion has lingered in the back of my mind for years now, would be good for me to finally get a grasp on this.

Say I have an object currently at rest, and I use energy X to accelerate it to speed v. According to the standard formula, it now has a kinetic energy 1/2mv^2.

Now I use the same amount of energy X again, accelerating to 2v. But, now according to 1/2mv^2 is has *four* times the energy! So, question number one, how does this jive with the conservation of energy?

I searched a bit and found the following quote: "The law of conservation of energy does not say that when you go from one frame of reference to another frame, the energy is conserved."

So, what meaning does kinetic energy have then? It doesn't seem "innate" to the system, and worse, whenever the value changes it implies a change of frame of reference (because of changing v), thus nullifying its meaning.

Any illumination would be highly appreciated.

etotheipi
Say I have an object currently at rest, and I use energy X to accelerate it to speed v. According to the standard formula, it now has a kinetic energy 1/2mv^2.

Now I use the same amount of energy X again, accelerating to 2v. But, now according to 1/2mv^2 is has *four* times the energy! So, question number one, how does this jive with the conservation of energy?

I don't understand what this means - if you accelerate it from ##0 \text{ms}^{-1}## to ##2v##, then the kinetic energy increase is ##2mv^2##. You haven't put in ##X##, you've put in ##4X## of energy? What's the issue?

A.T.
Say I have an object currently at rest, and I use energy X to accelerate it to speed v. According to the standard formula, it now has a kinetic energy 1/2mv^2.

Now I use the same amount of energy X again, accelerating to 2v. But, now according to 1/2mv^2 is has *four* times the energy! So, question number one, how does this jive with the conservation of energy?
Depends on the details of the acceleration process, and what "use the same amount of energy" means.

I searched a bit and found the following quote: "The law of conservation of energy does not say that when you go from one frame of reference to another frame, the energy is conserved."

So, what meaning does kinetic energy have then?
Total energy is conserved over time, not across reference frames.

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It doesn't seem "innate" to the system

Maybe it doesn't seem so for you but it is (see etotheipi's answer above).

whenever the value changes it implies a change of frame of reference (because of changing v)

Changing a velocity within a frame of reference doesn't imply a change of frame of reference.

• etotheipi
PeroK
Homework Helper
Gold Member
2020 Award
Summary:: It seems kinetic energy is dependent on the frame of reference. How does this all play together with conservation of energy etc?

This confusion has lingered in the back of my mind for years now, would be good for me to finally get a grasp on this.

Say I have an object currently at rest, and I use energy X to accelerate it to speed v. According to the standard formula, it now has a kinetic energy 1/2mv^2.

Now I use the same amount of energy X again, accelerating to 2v. But, now according to 1/2mv^2 is has *four* times the energy! So, question number one, how does this jive with the conservation of energy?

I searched a bit and found the following quote: "The law of conservation of energy does not say that when you go from one frame of reference to another frame, the energy is conserved."

So, what meaning does kinetic energy have then? It doesn't seem "innate" to the system, and worse, whenever the value changes it implies a change of frame of reference (because of changing v), thus nullifying its meaning.

Any illumination would be highly appreciated.
A conserved quantity is one which is conserved over time. A quantity which is the same in all reference frames is called invariant.

Energy is conserved but not invariant.

In your example, if an object accelerates from rest to speed ##v## then it gains kinetic energy of ##\frac 1 2 mv^2## in that frame. That energy must have come from somewhere. Depending on how the acceleration is achieved you could identify where the energy came from.

If it then accelerates to a speed of ##2v## it has gained a further ##\frac 1 2 m(2v)^2 - \frac 1 2 m v^2 = \frac 3 2 mv^2##. In this case more energy must have come from somewhere else. And, if you analyse the scenario you would find where this energy came from.

• Lnewqban
Ibix
2020 Award
Summary:: It seems kinetic energy is dependent on the frame of reference. How does this all play together with conservation of energy etc?

Now I use the same amount of energy X again, accelerating to 2v. But, now according to 1/2mv^2 is has *four* times the energy! So, question number one, how does this jive with the conservation of energy?
I think you are mixing frames of reference here and that will lead you into contradiction.

Work in the frame where the object is initially at rest. Giving it energy ##X## accelerates it to ##v##, so giving it 2X accelerates it to ##\sqrt{2} v##, because ##v\propto\sqrt E##. If you work in the frame where the object is at rest after the first acceleration, then your first acceleration took away energy ##X##, reducing the speed to zero, and your second acceleration increased the speed to ##v## again in the opposite direction.

This does, indeed, seem problematic because we have different velocity changes with the same energy change. But this is fine because you aren't describing a complete system. You can't just "give an object energy and start it moving". That would violate conservation of momentum. The object has to push against something else (tires against the road, or combustion chamber against exhaust gases). If you account for the change in motion of whatever you push against, you'll find that the total energy change is frame invariant.

• Lnewqban and jbriggs444
Thanks for the replies everybody!
If I understand it correctly, kinetic energy is self-consistent provided you stay in the same frame, and that my example was flawed in the sense that I disregarded the "accelerator" which would have made the total energy stay constant when taken into account.

• Ibix and jbriggs444
kuruman
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