Why is kinetic energy 1/2 mv^2?


by CuriousBanker
Tags: 1 or 2, energy, kinetic
CuriousBanker
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May12-13, 12:44 AM
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This is probably a stupid question because the answer is likely "it just is".
I am reading Calculus: The elements by Michael Comenetz, and when talking about mQ, he goes on to say "there happens to be advantage in working with 1/2 mQ rather than mQ itself", and then gives the equation k = 1/2 mv^2, but does not explain where the 1/2 came from, he just claims there is an advantage to using it. Is there a reason for this?
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SteamKing
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May12-13, 01:03 AM
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Here is one derivation of the KE formula:

http://hyperphysics.phy-astr.gsu.edu/hbase/ke.html#c2
CuriousBanker
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May12-13, 01:16 AM
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That doesn't explain why mQ all of a sudden became 1/2mQ without explaining why....is this something I should just accept and move on?

vanhees71
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May12-13, 03:22 AM
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Why is kinetic energy 1/2 mv^2?


I don't know what mQ should be, but the energy-conservation law can be directly derived from Newton's equation of motion for a point particle in a conservative force field:
[tex]m\ddot{\vec{x}}=-\vec{\nabla} V(\vec{x}).[/tex]
Multiply this by [itex]\dot{\vec{x}}[/itex] and use the chain rule, and you get
[tex]\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right ) = -\frac{\mathrm{d}}{\mathrm{d} t} V(\vec{x}).[/tex]
Defining the total energy as
[tex]E=T+V=\frac{m}{2} \dot{\vec{x}}^2+V(\vec{x}),[/tex]
the former equation is written as the conservation law
[tex]\frac{\mathrm{d} E}{\mathrm{d} t}=0.[/tex]
This shows that it is convenient to define the kinetic energy as
[tex]T=\frac{m}{2} \dot{\vec{x}}^2[/tex]
with the factor 1/2.

Does this answer your question?
Vagn
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May12-13, 03:26 AM
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Quote Quote by CuriousBanker View Post
That doesn't explain why mQ all of a sudden became 1/2mQ without explaining why....is this something I should just accept and move on?
It comes about when considering the Work-Energy theorem and using the calculus form of Newton's second law. You can have a look at the derivation in the lower part here if the derivation by Vanhees is hard to follow.
arildno
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May12-13, 04:46 AM
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While vanhees71 has provided you with an elegant derivation that suggests why the quantity 1/2*mv^2 might be worth giving its own name, that was certainly not clear to previous generations of physicists, and Our modern concept of kinetic energy thus defined gained ascendancy in the Scientific community in the first decades of the 19th Century.

For some of the history, you can check up on the "vis viva"-concept that Leibniz advocated, "vis viva" being, in essence, m*v^2, rather than 1/2*m*v^2:
http://en.wikipedia.org/wiki/Vis_viva
-----------------
A major reason for the rather slow acceptance of/interest in the kinetic energy concept was that it was easily shown by experiments that, in general, this quantity is NOT conserved. The latter half of the 18th century gradually developed the idea that heat was a "type" of "vis viva"/(kinetic) energy, and that the energy of the system was therefore a truly fundamental quantity, on par with momentum (which had been acknowledged a fundamental place in physics since the time of Newton). Without understanding heat as a type of energy, the energy concept itself isn't very interesting.
CuriousBanker
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May12-13, 09:25 AM
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Ah, see the problem is that this comes up on page 14 of a calc book.....which is before derivatives are taught. So I guess for now I should just accept it, and then come back to it later.
Also, one more question if I may, why is q=v^2....how is it determined that v is squared?
Andrew Mason
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May12-13, 09:54 AM
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Quote Quote by CuriousBanker View Post
Ah, see the problem is that this comes up on page 14 of a calc book.....which is before derivatives are taught. So I guess for now I should just accept it, and then come back to it later.
Also, one more question if I may, why is q=v^2....how is it determined that v is squared?
This has to do with the concept of work. Kinetic energy is defined at the ability of a body to do work by virtue of its motion. The amount of work that an object moving at speed v can do is proportional to the square of its speed.

If the force it applies is constant and the body goes from speed v to 0, it is easy to work out: KE = Wv→0 = Fd = mad =m(Δv/Δt)d = m(Δv/Δt)vavg Δt = mΔv(vavg )

Since Δv = v - 0 = v and vavg = (v + 0)/2 you have:

KE = Wv→0 = mv2/2

AM
WannabeNewton
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May12-13, 10:20 AM
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This is all quite natural using the Lagrangian approach. Let us consider a system in which we just have a free particle moving with constant velocity ##\mathbf{v}## relative to an inertial frame. The homogeneity of space ad time as well as the isotropy of space, for our mechanical system (the free particle), implies that the Lagrangian can only depend on the speed ##v## i.e. ##L = L(v^{2})##.

The key tool now is the principle of Galilean relativity because if we have another frame moving with constant velocity ##\mathbf{u}## relative to our inertial frame then by the aforementioned principle, ##\mathbf{v}' = \mathbf{v} + \mathbf{u}##. However we also know that then the Lagrangian, under this Galilean boost, must transform in such a way so as to keep the equations of motion the same in both frames therefore the two Lagrangians can differ by at most a total time derivative of a smooth function of time and the configuration space coordinates. We have ##L((v')^{2}) = L(v^{2}) + 2\frac{\partial L}{\partial v^{2}}\mathbf{v}\cdot \mathbf{u}## after a first order Taylor expansion in powers of ##\mathbf{u}## which means ##2\frac{\partial L}{\partial v^{2}}\mathbf{v}\cdot \mathbf{u}## must be a total time derivative as mentioned before which can only be possible if ##\frac{\partial L}{\partial v^{2}} = \text{const.} = \alpha## i.e. ##L = \alpha v^{2}##. The constant ##\alpha## is chosen to be ##\alpha = \frac{1}{2}m## (##m## is called the mass of the particle) and the physical reason for this is given by vanhees' link (the one about ##E## being a first integral of the equations of motion).

If you want to read more, see Landau and Lifgarbagez "Mechanics" if you can get access to it.
Vanadium 50
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May12-13, 11:24 AM
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WannabeNewton, the OP says this is "before derivatives are taught". Do you believe that pointing someone at this stage to calculus of variations, Lagrangian mechanics and Landau & Lifgarbagez is likely to be helpful?
CuriousBanker
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May12-13, 11:33 AM
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I appreciate all the responses, but I guess maybe it is best to just accept it as a given, go through the rest of the book, and come back to it later right?
I am self-teaching, that is why I post here and not in a classroom. I took calc I years ago but forgot it. I am self teaching because I want to learn calculus to improve my financial trading skills but also mostly just because I want to learn as much as possible in life...I hope I am able to self-teach most things...I wish I didn't always need to get things intuitively to accept them....I guess I should just accept things as a given until it all comes together, seems to be the best way to get through a subject
WannabeNewton
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May12-13, 11:46 AM
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Quote Quote by Vanadium 50 View Post
WannabeNewton, the OP says this is "before derivatives are taught". Do you believe that pointing someone at this stage to calculus of variations, Lagrangian mechanics and Landau & Lifgarbagez is likely to be helpful?
I'm not seeing how there is any simpler way to derive the expression for kinetic energy without going into that and then adding it on to what vanhees said. How would you do it without derivatives?
Redbelly98
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May12-13, 12:41 PM
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Quote Quote by WannabeNewton View Post
How would you do it without derivatives?
Algebra-based physics courses teach that acceleration is the slope of a velocity vs. time graph, and that distance traveled is the area under a velocity vs. time graph. Students who have not taken calculus are still taught those concepts in non-calc physics.

We can use the slope and area of a v-vs.-t curve, to look at a constant force F applied to an object that is initially at rest.

The kinetic energy K of the object is equal to the work W done by the (constant) force F:
[tex]K=W= F \cdot \Delta x [/tex]
[tex]K = m a \cdot \Delta x[/tex]
[tex]K = m \cdot \frac{\Delta v}{\Delta t} \cdot (\frac{1}{2} \Delta v \cdot \Delta t)[/tex]
This is where the 1/2 comes in. Δx is the area under the v-vs.-t curve, which is a straight line going from v=0 initially to v=Δv at some time Δt. I have used the formula for the area of a triangle to express Δx in terms of this area.

Now cancel the Δt terms, and we get
[tex]K = m \cdot \Delta v \cdot \frac{1}{2} \Delta v = \frac{1}{2} m \Delta v ^2[/tex]

Hope that helps. I imagine this could also be derived using the kinematic equations for uniform acceleration.
WannabeNewton
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May12-13, 01:19 PM
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Hi Redbelly! I had interpreted the question as asking "how do we get the expression for kinetic energy without pre-defining it in terms of work etc. i.e. does it come naturally out of some other quantity without actually defining something to be the kinetic energy a priori" but if what was wanted was just to see how it comes out of the definition in terms of work done then I do apologize for the post. I don't suppose it is possible to remove it without removing the context of the consecutive posts?
DaleSpam
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May12-13, 03:59 PM
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Quote Quote by CuriousBanker View Post
This is probably a stupid question because the answer is likely "it just is".
Kinetic energy is a defined quantity, so we certainly could have chosen to define it with a factor of 1 instead of 1/2. However, we want kinetic energy to have the property that the change in KE is equal to the work done. Therefore you need the factor of 1/2 (the details of why are in the "work-energy theorem" mentioned by Vagn).
MisterX
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May12-13, 05:28 PM
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We can start with change in velocity equals acceleration times the change in time
[itex]
\Delta v = \frac{F}{m} \Delta t
[/itex]

Now multiply both sides by velocity [itex]v = \frac{\Delta x}{\Delta t}[/itex]. On the left I am leaving this as [itex]v[/itex], but on the right I am going to express it in terms of [itex]\Delta x[/itex] and [itex]\Delta t[/itex]

[itex]
v\Delta v = v\frac{F}{m} \Delta t
[/itex]
[itex]
v\Delta v = \frac{F}{m} \Delta t \frac{\Delta x}{\Delta t} = \frac{F}{m} \Delta x
[/itex]
[itex]
v\Delta v = \frac{F}{m} \Delta x
[/itex]

Now here comes the crucial step, where we get the 1/2.

[itex]v\Delta v = \frac{1}{2}\Delta (v^2)[/itex]

This is a result from calculus. The change in [itex]v^2[/itex] is equal to twice v times the change in v.

[itex]
\frac{1}{2}\Delta (v^2) = \frac{F}{m} \Delta x
[/itex]
[itex]
\Delta (\frac{1}{2}m v^2) = F \Delta x
[/itex]

Notice that the delta applies to the entire expression [itex]\frac{1}{2}m v^2[/itex]. This means the change in [itex]\frac{1}{2}m v^2[/itex] is equal to [itex]F \Delta x[/itex]. Also note that

[itex]
\Delta (\frac{1}{2}m v^2) \neq \frac{1}{2}m (\Delta v) ^2
[/itex]

Redbelly98's derivation is wrong. He used the wrong thing for [itex]\Delta x[/itex].
MisterX
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May12-13, 05:49 PM
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Here's an explanation of why
[itex]\Delta(v^2) = 2v \Delta v[/itex]

First we express the change in [itex]v^2[/itex]
[itex]\Delta(v^2) = (v + \Delta v)^2 - v^2[/itex]

[itex] = v^2 +2 v\Delta v + (\Delta v)^2 - v^2 = 2 v\Delta v + (\Delta v)^2[/itex]

Now something we do sometimes when doing calculus is to ignore anything like [itex](\Delta v)^2[/itex]. We do this because we are only interested in what happens as [itex]\Delta v[/itex] becomes very small. Of course, if [itex]\Delta v[/itex] becomes small then [itex]\Delta(v^2) [/itex] also becomes small. But the two parts, [itex]2v \Delta v[/itex] and [itex](\Delta v)^2[/itex], get small with different orders. For example, we might be interested with the rate that [itex]\Delta(v^2)[/itex] is changing in time.

[itex]\frac{\Delta (v^2)}{\Delta t} = 2 v\frac{\Delta v}{\Delta t} + \frac{(\Delta v)^2}{\Delta t}[/itex]

[itex] = 2 v\frac{\Delta v}{\Delta t} + \frac{\Delta v}{\Delta t}\Delta v [/itex]

[itex]\frac{\Delta v}{\Delta t}[/itex] is the acceleration, and even when [itex]\Delta v [/itex] and [itex]\Delta t [/itex] get very small, a = [itex]\frac{\Delta v}{\Delta t}[/itex] does not have to get small. Just think of [itex]\frac{\Delta v}{\Delta t}[/itex] as being some number. But the second term still has that extra [itex]\Delta v [/itex], so it goes away when we make [itex]\Delta v [/itex] and [itex]\Delta t [/itex] small.
jtbell
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May12-13, 06:03 PM
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Here's yet another version that assumes constant force and acceleration, so we don't have to use calculus.

We push on an object with constant force and give it constant acceleration (in a straight line) via F = ma. The object starts at position x0 with speed v0. Some time later it ends up at position x with speed v.

Define the work done on the object as W = FΔx = F(x - x0). Before people started talking about energy in the modern sense, they already knew about work in connection with simple machines such as levers, in which the output work equals the input work.

Let's see what happens if we rewrite the work equation in terms of the initial and final speeds, not positions. One of the standard kinematic equations for constant acceleration which you can find in any introductory physics textbook is

v2 = v02 + 2a(x - x0)

from which

x - x0 = (v2 - v02) / (2a)

Substitute this equation, and F = ma, into the work equation. After cancelling out a, you end up with

W = (1/2)mv2 - (1/2)mv02

Clearly the quantity (1/2)mv2 (for any arbitrary speed) is "interesting". Let's give it a name: (kinetic) energy. Now we can say that doing work on an object changes its energy by an amount equal to the work done.

It turns out you get the same result even if the force and acceleration aren't constant, but you have to use calculus to derive it.


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