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

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.

My point about the zig zag is that you have to realise that kinetic energy and drag forces are two utterly different concepts. Once you put energy into an object so as to give it KE, travelling along a path, even if not a straight one, does not inherently alter the KE.

Textbooks can be kind of sloppy about the whole thing. Basic problems though are often set in one dimension, so v = (v1), which you may as well just write as v.

v is used to represent speed because speed is the modulus of velocity, which in one dimension you may as well just write v. They leave out the modulus signs |v| because squaring it makes it positive anyway so v^2 = |v|^2 = s^2 where s is speed (so writing 1/2m|v|^2 in one dimension is redundant as it = 1/2mv^2 anyway). Velocity is by far the more fundamental quantity in physics problems - speed is just a trivial property of velocity really - so as you progress through physics you tend to use v whenever you can, not s.

Also, in physics the convention is to use s to mean displacement, which is distance with a direction.