Can anybody explain the formula for Kinetic Energy?

  • Thread starter fetmar
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  • #1
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I have always plugged numbers into the following forumula:

KE = 1/2m*v^2 and it works perfectly. :biggrin:

I don't understand why. :blushing: Can anybody please tell me why it is one half mass, and not all of the mass? And even more perplexing, is why the velocity needs to be squared.
 

Answers and Replies

  • #2
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Welcome to PF, fetmar.

It simply follows from calculating the work done by a force on an object. We call the work done by a force on an object as the change is kinetic energy due to the action of that force.

[tex]W = \int_{r(t_0)}^{r(t)}{\vec{F}\cdot d\vec{r}}[/tex]

Write [tex]\vec{F} = m\frac{d\vec{v}}{dt}[/tex] and [tex]d\vec{r} = \vec{v}dt[/tex]
Now evaluate the integral.

[edited on request]
 
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  • #3
daniel_i_l
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If you take away the 1/2 and use v instead of v^2 you get mv which is the momentum.
 
  • #4
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work is the force x distance.
the kinematics equation relating acceleration to distance is
a = vf^2 - vi^2/2d
F = ma
so... F = mvf^2 - mvi^2/2d
multiply this by the distance.. and the change in kinetic energy is
mvf^2-mvi^2/2

soo.. 1/2mv^2 = change in kinetic energy.
 
  • #5
russ_watters
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It isn't half the mass, the 1/2 comes into the equation when evaluating the integral neutrino posted.
 
  • #6
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If you understand that Energy is Force * Distance, then it should start to become clear. Imagine you are driving. Keep in mind that the energy required to brake is equal to the kinetic energy of the vehicle while it was in motion... If you are driving at 25 m/s and slow down with a force that causes an acceleration of -5m/s/s, you can do the math and figure out you had a braking distance of 62.5 m. Now, if you were driving at 50 m/s and were to apply the same force, not only would the TIME for you to brake double, your average velocity while slowing down would also double. Therefore the distance you brake for at double the speed, is four times that of the original. In this case, conceptually, the 1/2 is to get the average velocity. If you slow down at an even rate from 50 m/s to a stop, your average velocity was 25 m/s. I really hope this made some kind of sense.
 
  • #7
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I don't understand why. :blushing: Can anybody please tell me why it is one half mass, and not all of the mass? And even more perplexing, is why the velocity needs to be squared.
The deep answer comes from time symmetry via Noether's theorem. You can google that, but it'll only explain why velocity is squared. The factor out the front probably isn't really necessary (we could just change our units and other definitions accordingly) but then we'd still need a new constant in front of all the other energy equations (gravitational potential, electric potential, etc), in other words, it is necessary for consistency.
 
  • #8
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f.d=m.a.d=m.(v1^2-v2^2)/2d.d

=(mv1^2)/2 - (mv2^2)/2=f.d

newton(force) . distance=delta kinetic energy
 
  • #9
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all you people posting that kinematic relationship
[tex]v_f^2=v_i^2+2ad[/tex]
don't get that that is the simplified conservation of energy theorem, not that the conservation of energy theorem comes from that.
[tex]\Delta KE=\Delta U[/tex]

[tex]KE_f - KE_i = \Delta U[/tex]
[tex]\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mg\Delta h[/tex]
cancel m and multiply by 2 to clear the fraction
[tex]v_f^2 - v_i^2 = 2g \Delta h[/tex]
 
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  • #10
Andrew Mason
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I have always plugged numbers into the following forumula:

KE = 1/2m*v^2 and it works perfectly. :biggrin:

I don't understand why. :blushing: Can anybody please tell me why it is one half mass, and not all of the mass? And even more perplexing, is why the velocity needs to be squared.
The Kinetic Energy of a body represents the ability of that body to do work due to its motion. To do work is to apply a force through a distance. So ask yourself how much force x distance a body with mass m and velocity v can apply.

1. Force is the time rate of change of momentum.

2. Since speed is the time rate of change of distance, distance traveled is the average velocity x time.

3. So Force x distance is the time rate of change of momentum x average velocity x time, or (since the time cancels out): change of momentum x average velocity.

Since in going from v to 0, the average velocity is v/2 and the change in momentum is mv. So the work done is v/2 x mv = [itex]\frac{1}{2}mv^2[/itex].

AM
 
  • #11
rbj
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I have always plugged numbers into the following forumula:

KE = 1/2m*v^2 and it works perfectly. :biggrin:

I don't understand why. :blushing: Can anybody please tell me why it is one half mass, and not all of the mass? And even more perplexing, is why the velocity needs to be squared.
Fetmar,

can you deal with regular kinematics? like do you know why the distance traversed by an accelerating particle is:

[tex] d = \frac{1}{2} a t^2 \ + \ v_0 t [/tex]
?

if you know why that is true, then i can explain kinetic energy, in terms of work, to you pretty succintly.
 
  • #12
robphy
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Since in going from v to 0, the average velocity is v/2
Most likely, this implicitly assumes a constant acceleration,
as do the other responses with an acceleration "a".

neutrino's reply is more general since it doesn't require a constant acceleration.
 
  • #13
I have always plugged numbers into the following forumula:

KE = 1/2m*v^2 and it works perfectly. :biggrin:

I don't understand why. :blushing: Can anybody please tell me why it is one half mass, and not all of the mass? And even more perplexing, is why the velocity needs to be squared.
To put it simply, it's the integral of momentum.


On a different note, does the integral of (F=) ma have any physical significance?
 
  • #14
jtbell
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The integral of F with respect to position is the work done on an object by F, which equals the change in the object's kinetic energy:

[tex]\int {\vec F \cdot d \vec r} = \Delta K[/tex]

The integral of F with respect to time is the impulse delivered to an object by F, which equals the change in the object's momentum:

[tex]\int {\vec F dt} = \Delta \vec p[/tex]
 
  • #15
Andrew Mason
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Most likely, this implicitly assumes a constant acceleration,
as do the other responses with an acceleration "a".

neutrino's reply is more general since it doesn't require a constant acceleration.
There is no need to assume a constant acceleration (ie. constant force). All you need to know is that change in momentum is always average force x time interval. This is because the definition of average force is:

[tex]\bar F = \frac{\Delta p}{\Delta t}[/tex]

Hence:

[tex]\bar F \Delta d = \bar F \Delta t \frac{\Delta d}{\Delta t} = \Delta p \bar v[/tex]

AM
 
  • #16
robphy
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The key point I am trying highlight is that
Since in going from v to 0, the average velocity is v/2
is not generally true.
 
  • #17
Andrew Mason
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The key point I am trying highlight is that

Originally Posted by Andrew Mason View Post
Since in going from v to 0, the average velocity is v/2​

is not generally true.
It is if there is only deceleration. If an object of mass m travelling at speed v experiences negative acceleration of any kind (constant, not constant or arbitrary) such that it ends up at rest (speed = 0) and stays stopped, its average [of the beginning and end speed] over the entire period of deceleration will be v/2. The average is always v/2.

[Edit][PS: see below]

Andrew Mason
 
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  • #18
nrqed
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It is if there is only deceleration. If an object of mass m travelling at speed v experiences negative acceleration of any kind (constant, not constant or arbitrary) such that it ends up at rest (speed = 0) and stays stopped, its average speed over the entire period of deceleration will be v/2. The average is always v/2.

Andrew Mason
Just out of curiosity: what definition of average do you use? If one uses a time average [itex] = \frac{1}{T} \int_0^T v_x dt [/itex] it does nto seem to me that the average will always be v/2.
 
  • #19
Andrew Mason
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Just out of curiosity: what definition of average do you use? If one uses a time average [itex] = \frac{1}{T} \int_0^T v_x dt [/itex] it does nto seem to me that the average will always be v/2.
Well, you are right, generally.

But consider this when the mass is subject to only deceleration. For a reduction in speed from v0 to v1 over an arbitrarily short time interval the change in momentum is m(v1-v0) and the average speed is arbitrarily close to (v1+v0)/2. If you combine n of these intervals:

[tex]E = \sum_{j=0}^n m(v_{j+1} - v_j)(v_{j+1} + v_j)/2 = \frac{m}{2}\sum_{j=0}^n (v_j^2 - v_{j+1}^2) = = \frac{m}{2}\left((v_0^2 - v_{1}^2) + (v_1^2 - v_2^2).... + (v_{n-2}^2 - v_{n-1}^2) + (v_{n-1}^2 - v_n^2)\right)[/tex]

But the intermediate terms all cancel out and you are left with:

[tex]E = m(v_0^2-v_{n}^2)/2 = m(v_n - v_0)(v_n + v0)/2 = \Delta p(v_0+v_n)/2[/tex]

If v_n = 0 then:

[tex]W = m(v_0)(v_0/2) = \Delta p(v_0)/2[/tex]

So, to be perfectly correct, one should say that the work done is the change in momentum multiplied by the average of the beginning and end speeds ([itex]W = \Delta p (v_i + v_f)/2[/itex]).

AM
 
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  • #20
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Scientists came up with the idea of work after noticing that the quantity Fd (force times distance) is conserved. This led them to call that quantity work, W.
They then sought to manipulate the equation to see what they can extract from it and they did this: W=ΔKE=Fd=mad=
m* (Vf-V0)/t * (Vf+V0)/2 * t=(m/2) (Vf^2-V0^2)=KE(final)-KE(initial)
 
  • #21
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The key point I am trying highlight is that

is not generally true.
Well, it's true if the force is kept constant. If it isn't, the proportions are still kept the same as if it were, after all is said and done.
 
  • #22
robphy
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robphy said:
The key point I am trying highlight is that

Andrew Mason said:
Since in going from v to 0, the average velocity is v/2
is not generally true.
It is if there is only deceleration. If an object of mass m travelling at speed v experiences negative acceleration of any kind (constant, not constant or arbitrary) such that it ends up at rest (speed = 0) and stays stopped, its average [of the beginning and end speed] over the entire period of deceleration will be v/2. The average is always v/2.

[Edit][PS: see below]

Andrew Mason
Well, it's true if the force is kept constant. If it isn't, the proportions are still kept the same as if it were, after all is said and done.
Highlights mine.

Of course, the phrase "not generally true" doesn't mean NEVER TRUE.
The point of using this phrase is to suggest that it is NOT ALWAYS TRUE and relevant conditions when it is true should spelled out (especially ones that are likely to lead to misconceptions.. e.g., the above passage I first quoted).

(If it's not clear by now,
the common misconception is that "average velocity" is [always] (vi+vf)/2.
It's not always true... it is true in certain cases (as finally pointed out above)... and there are textbook situations when it isn't true.

"Average-velocity" as used in kinematics means (as nrqed points out) time-averaged-velocity:
[tex]v_{avg} \equiv \displaystyle\frac{\int v\ dt}{\int dt}= \frac{\Delta x}{\Delta t} [/tex].
Most textbooks unfortunately leave out the center expression [or at least a discretized algebraic version of it], which would probably help clear up the misconception.)
 
  • #23
I think that by "not generally true" it was just meant to suggest that there is a more general expression which always holds which does not depend on qualifying statements, like requiring that the acceleration be constant or that the force be constant, for example. When a discussion begins with a certain context and carries on until some of the joining parties loose track of these qualifiers because they are not explicitly stated every time, confusion can creep in.

Here's my question. When is it correct to use the formula, i.e. kinetic energy is one half the mass times velocity squared, and when should one revert back to the integral of the velocity with respect to the velocity, etc. in order to get the correct result?
 
  • #24
Here’s my version of the derivation of the kinetic energy formula (first post using latex, so go easy)

Kinetic energy is force times change in distance.

[tex]
KE = F \Delta x
[/tex]

Force is mass times acceleration, which is a change in velocity over time. We can write this as

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


Now the change in distance, [tex] \Delta x [/tex] can be expressed as velocity times time


[tex]
\Delta x = \bar{v} t
[/tex]


Notice that the velocity is is average velocity, where we have a constant change in velocity, or acceleration.

Average velocity can be written as the sum of the initial and final velocities divided by two.

[tex]
\bar v = \frac{v_0 + v_f}{2}
[/tex]


Now let's look at final velocity. We can write this as the initial velocity plus the change in velocity over time multiplied by time

[tex]
v_f = v_0 + \frac{\Delta v}{\Delta t} \Delta t
[/tex]

which can also be written as

[tex]
v_f = v_0 + at
[/tex]


Plugging this into the aveage velocity formula gives

[tex]
\bar v = \frac{v_0 + \left[ v_0 + \frac{\Delta v}{\Delta t} \Delta t \right]}{2}
[/tex]

which simplifies to

[tex]
\bar v = v_0 + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t
[/tex]

Now, we plug this into the change in distance formula

[tex]
\Delta x = \left[v_0 + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t \right] \Delta t
[/tex]

[tex]
= v_0 \Delta t + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t^2
[/tex]

Now, let's assume the beginning velocity is zero. Then we take the product of the above and force


[tex]

F\Delta x = \left( m\frac{\Delta v}{\Delta t} \right) \left( \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t \Delta t \right)
[/tex]


Collecting terms we arrive at


[tex]
KE = m \frac{1}{2} \Delta v \Delta v
[/tex]

Which of course is

[tex]
KE = \frac{1}{2} mv^2
[/tex]
 

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