- #1

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

I don't understand why. 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.

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- #1

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

I don't understand why. 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.

- #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]

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.

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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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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.I don't understand why. 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.

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=(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]

[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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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.

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

I don't understand why. 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.

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):

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

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Fetmar,

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

I don't understand why. 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.

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.

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Most likely, this implicitly assumes a constant acceleration,Since in going from v to 0, the average velocity is v/2

as do the other responses with an acceleration "a".

neutrino's reply is more general since it doesn't require a constant acceleration.

- #13

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To put it simply, it's the integral of momentum.

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

I don't understand why. 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.

On a different note, does the integral of (F=) ma have any physical significance?

- #14

jtbell

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[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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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: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.

[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

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is not generally true.Since in going from v to 0, the average velocity is v/2

- #17

Andrew Mason

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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 [of the beginning and end speed] over the entire period of deceleration will be v/2. The average is always v/2.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.

[Edit][PS: see below]

Andrew Mason

Last edited:

- #18

nrqed

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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.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

- #19

Andrew Mason

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Well, you are right, generally.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.

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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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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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.The key point I am trying highlight is that

is not generally true.

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robphy said:The key point I am trying highlight is that

is not generally true.Andrew Mason said:Since in going from v to 0, the average velocity is v/2

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

Highlights mine.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.

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] (v

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

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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

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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]

- #25

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Looking at the history of an idea/concept can sometimes help to get a feel for it. Wikipedia has a small bit on the history of energy: http://en.wikipedia.org/wiki/Energy and http://en.wikipedia.org/wiki/Vis_viva

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