Thread: Why does v represent speed in the KE Equation?

Originally Posted by ninja9578

Your equation is incomplete

KE = (mc^2 / (sqrt(1 - (v^2/c^2))) - mc^2

WTF?

http://www.google.com/#hl=en&source=...96bd6416f4afd8

Also, c represents specifically the speed of light.

Originally Posted by ninja9578

And usually if you want to represent speed in an equation you would use ds/dt or |v|. |v| is more common.

Absolute value of velocity? That's not necessarily speed.

I am talking about just a single variable.

c is the speed of light.

KE = mv^2 only works at low speeds.
KE = (mc^2 / (sqrt(1 - (v^2/c^2))) - mc^2 works at all speeds

You need to learn a little bit more about physics.

Also two bars in vector mathematics means magnitude not absolute value. And physics is vector math.

Originally Posted by Universal Mind

It has something to do with several sub-velocities taken into account?

Oh, no. Imagine that you have a velocity V (which I called V sub-c in the image) at whatever angle. That velocity has both an x and y component, which I labeled Vb and Va respectively. You can apply the Pythagorean theorem (a^2 + b^2 = c^2) to these two components to get that resultant vector V. I only brought it up in order to demonstrate that taking the dot product of V with itself equals the same thing, and through that you can represent speed in terms of velocity in the KE equation. V dot product itself is equal to speed squared, or the magnitude of V squared.

Originally Posted by Universal Mind

WTF?

Ninja's giving you the modified version of the equation that allows it to work at relativistic speeds.

Originally Posted by ninja9578

KE = mv^2 only works at low speeds.

I am aware of the foot note, but that is the equation I am asking about. It is called the Kinetic Energy Equation. It is the subject of this thread. I am not concerned with speeds that are substantial fractions of the speed of light in this thread.

Originally Posted by ninja9578

You need to learn a little bit more about physics.

I already said that myself, dork. Stop trying to act tall.

Originally Posted by ninja9578

c Also two bars in vector mathematics means magnitude not absolute value. And physics is vector math.

In addition, absolute value is a form of magnitude, and it is used in math, which is used in physics.

Originally Posted by Invader

Oh, no. Imagine that you have a velocity V (which I called V sub-c in the image) at whatever angle. That velocity has both an x and y component, which I labeled Vb and Va respectively. You can apply the Pythagorean theorem (a^2 + b^2 = c^2) to these two components to get that resultant vector V. I only brought it up in order to demonstrate that taking the dot product of V with itself equals the same thing, and through that you can represent speed in terms of velocity in the KE equation. V dot product itself is equal to speed squared, or the magnitude of V squared.

Let's say an object moves over a 500 m path that ends 200 m north of the start in 1 second. If you calculate the kinetic energy involved, would v equal 500 m/s or 200 m/s north?

Magnitude also strips direction. Absolute value has a direction in the positive direction, magnitude doesn't. Weird little technicalities are what so frustrates high school physics students and stoners. I will get so into this thread in a little bit.

Originally Posted by Universal Mind

Let's say an object moves over a 500 m path that ends 200 m north of the start in 1 second. If you calculate the kinetic energy involved, would v equal 500 m/s or 200 m/s north?

200m north how?



In both cases the speed is going to be 500 m/s.

The average velocity (final displacement/time) in each is going to be different.
V in case 1 = 200 m/s
V in case 2 = 500 m/s
However the instantaneous velocity at the final point for each case will have the same magnitude (500m/s) but a different direction.

You'd be using the 500m/s second value for your KE equation in both of the scenarios.

Originally Posted by Invader

200m north how?



In both cases the speed is going to be 500 m/s.

The average velocity (final displacement/time) in each is going to be different.
V in case 1 = 200 m/s
V in case 2 = 500 m/s
However the instantaneous velocity at the final point for each case will have the same magnitude (500m/s) but a different direction.

You'd be using the 500m/s second value for your KE equation in both of the scenarios.

Case 1. I was talking about an ending point 200 meters north of the starting point.

Originally Posted by khh

You need the velocity. Because if you're going in a curved line, you won't be building the same energy as with a straight line.

It looks like both of you know a lot about this stuff but don't agree on the answer to my question.

Velocity is a vector and so should technically be written in bold as v or underlined as v. It has (3 in our universe) dimensions.

The v in the equation isn't bolded, so we assume it represents the scalar |v|, which is the one dimensional magnitude of v, in other words the speed.

Writing v = (v1, v2, v3) we calculate |v| as sqrt(v1^2 + v2^2 + v3^2), which as Invader says is the square root of the dot product v·v (the dot product just sums the products of the corresponding components of the vectors).

With regards to your zig-zag question; the question isn't really well defined. Neither path necessarily 'uses' more energy; you only 'use' energy when there's some resisting force against the particle which you can't 'get the energy back out of', and as you didn't specify any kind of drag in your question, both paths could use 0 energy. Imagine that what causes the zig-zag is the ball bouncing off of two parallel walls. The ball initially has kinetic energy because it has a speed, which is then transferred to elastic potential energy when the ball hits the wall (at which point the ball has speed 0 and so KE 0), and then the energy is transferred back into the kinetic energy of the ball on so on. Assuming perfact elasticity of the walls, the ball ends up with its initial energy at the end of each path.

The key point is that moving with constant velocity doesn't involve an energy change. If it did, according to the KE formula the ball would slow down even though no forces are acting on it, which violates Newton's first law.

Originally Posted by Xei

Velocity is a vector and so should technically be written in bold as v or underlined as v. It has (3 in our universe) dimensions.

The v in the equation isn't bolded, so we assume it represents the scalar |v|, which is the one dimensional magnitude of v, in other words the speed.

Writing v = (v1, v2, v3) we calculate |v| as sqrt(v1^2 + v2^2 + v3^2), which as Invader says is the square root of the dot product v·v (the dot product just sums the products of the corresponding components of the vectors).

With regards to your zig-zag question; the question isn't really well defined. Neither path necessarily 'uses' more energy; you only 'use' energy when there's some resisting force against the particle which you can't 'get the energy back out of', and as you didn't specify any kind of drag in your question, both paths could use 0 energy. Imagine that what causes the zig-zag is the ball bouncing off of two parallel walls. The ball initially has kinetic energy because it has a speed, which is then transferred to elastic potential energy when the ball hits the wall (at which point the ball has speed 0 and so KE 0), and then the energy is transferred back into the kinetic energy of the ball on so on. Assuming perfact elasticity of the walls, the ball ends up with its initial energy at the end of each path.

The key point is that moving with constant velocity doesn't involve an energy change. If it did, according to the KE formula the ball would slow down even though no forces are acting on it, which violates Newton's first law.

Hm... I believe you, but I don't think I saw v or v used to represent velocity in any of the stuff I read. I was of course talking about a zig zag path involving friction, air resistance, or other counterforce. I am not trying to get into a discussion about that zillions of potential factors that could affect motion. I am just curious about what the v in the equation represents and why v, of all variables, would be used to represent speed.

Originally Posted by ninja9578

KE = mv^2 only works at low speeds.

"Low speeds"... I wouldn't call speeds up to 0.1 * c to be "low" :p

Originally Posted by ninja9578

KE = (mc^2 / (sqrt(1 - (v^2/c^2))) - mc^2 works at all speeds

But it's hell to calculate. Basically Ke = mv^2 is the Newtonian physics that worked right up until Einstein came along with his relativity.

Originally Posted by ninja9578

Also two bars in vector mathematics means magnitude not absolute value. And physics is vector math.

It's the same basic concept, though. If you take the vector to the second power and take the square-root of that value, like you do for scalars to find the absolute value, you get the magnitude.

Originally Posted by Universal Mind

Let's say an object moves over a 500 m path that ends 200 m north of the start in 1 second. If you calculate the kinetic energy involved, would v equal 500 m/s or 200 m/s north?

Since kinetic energy doesn't have a direction, you'd need to use 500 m/s to get the right value.
edit...
wait a second, no you shouldn't. You need the velocity. Because if you're going in a curved line, you won't be building the same energy as with a straight line.