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Kinetic Energy

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Ok, KE=1/2*m*v^2. What about when v is a vector? It seems it would take energy to change the direction of v. Assuming you are the only mass in the universe, i.e. me, all me, nothing but me then your potential energy didn''t change. So you must have added or removed kinetic energy from the system. Assuming your speed and mass didn''t change, just your direction then 1/2*m*|v|^2 is the same. So I''m confused.

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quote:
Original post by LilBudyWizer
Ok, KE=1/2*m*v^2. What about when v is a vector? It seems it would take energy to change the direction of v.

It takes a force, but no energy. Think of the moon going around earth...

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When you speak of kinetic energy, you are often thinking of it in terms of work, via the work-energy theorem, which states that:

W_net = (DELTA)K

We know that work is independent of path, provided we are talking about conservative forces (most of which are... friction is basically the only nonconservative force you ever need worry about). So because of the fact that work is independent of path, this leads to the v in 1/2mv^2 being speed rather than velocity.

In other words, the kinetic energy of an object does not depend on the direction it is heading, but instead, only on the magnitude of its velocity (well, and mass, but that''s a given ).

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Ok, I can accept that. I guess I have trouble with the idea that you could produce a force without energy. Since velocity is a contineous function in the real world then to alter the velocity it seems the force has to be applied over time. Since you have a velocity you moved during that time and thus have a force acting over a distance.

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dE = F.dr

Changing only the direction of motion means F is always directed orthogonal to dr, thus F.dr = 0.

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The solution is rather simple.....It sounds as though you are mixing scalar(speed) quantities with vector(velocity) quanties. Scalar quantities have magnitude only, whereas vector quantities have magnitude and direction. In the equation:
K=1/2mv^2
even if v was a vector quantity, once it is squared it still gives us a positive value. Thus K for the given v will always be positive. The only way K can be negative is if you are finding (delta)K when final v is less than initial v.:
(delat)K=(1/2mvf^2)-(1/2mvi^2)
This would only imply a loss on kinetic energy.

[edited by - Moorphene on March 4, 2003 11:46:48 AM]

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lol.. no idea what this post is supposed to prove

if your the only thing in the universe.. and your moving in a straight line with some velocity(v) and kinetic energy(.5mv^2).. there is no reason for your direction to change..

so why is anymore energy needed in the system?

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Just because you are the only mass doesn''t mean energy doesn''t exist. Einstein showed that the two are equivalent, so you could generate a force using some of the energy. Refering to your question about velocities, you are taking the dot product for the velocity, so v.v = mag(v)^2. Furthermore, think about this- A charged particle will move around in the path of a circle in a uniform magnetic field. No work is being done, however, the direction of the velocity is always changing. It''s kinda wierd, but it makes sense. The direction you are travelling really doesn''t change the amount of energy that you have.

Brendan

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If you are the only thing in the universe, motion is undefinable.

Motion is always measured relative to another object (i.e. in a reference frame). There is no such thing as ''absolute'' movement versus a universal co-ordinate system except in games.

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I guess I''m just confusing energy with fuel. If you are in orbit, land then return to the same orbit then you have the same potential and kinetic energy at the end as when you started. Even ignoring air friction it seems that would take a considerable amount of fuel. Half of that trip takes little fuel assuming you can survive the landing. Perhaps I should think of force over time as more closely related to fuel consumption than changes in potential/kinetic energy or work.

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