A problem with KE = ½mv². Help.

[IDMclean]

Ummm... Isn't Momentum one of those things that is strickly conserved? Also even for inelastic force exchange, isn't it true that every action has an equal but oppisite reaction?

 

Further for the claim of Ke = mv to be true, Energy would have to be defined as, ML/T, not ML^2/T^2. just poking holes in things.

[arkain101]

Quote:

Originally Posted by arkain101

but the transfer of momentum remains the same.

 

That is absolututely incorrect. Not only will the transfer of momentum be lesser in the shaft-having-wheel case, but also will the energy transferred be less than 500J in both cases.

 

I'm not sure if that will be worth noticing. The collision time difference will be of the order of nanoseconds.

Further, we can consider this as a inelastic collision with the lost energy not dissipated, but sent into the spinning wheel

 

First of all you cant tell how long the time of force will change, you have no idea how long the shaft is. You have no idea how heavy the wheel is, or what gear ratio it is set at. The shaft could be 50 feet long. It will push you backwards and backwards, while it slows down. The force will feel less, but the force will act over a greater distance. The Energy is equal to force x distance. In an imediate collision, the distance is almost null but the force is extreme. In the telescopic collision the same mass is moving, the force is less the time is more but the energy remains the same.

 

Here is another example. A spring 50 feet long that weighs 50kg and a bowling ball that weighs 50kg are the two moving objects.

There is a person in a cart facing an incline of 10degrees. When the bowling ball hits the cart it sends the cart 30 feet up the hill in 5 seconds. When the spring pushes the cart (we have to consider it always pushes parrallel to the direction of travel, no funky angles), the car slowly starts to move up the hill, the spring is slowing down in net velocity, but the cart is moving along and speeding up, the spring will also push the cart 30 feet up the incline. (as long as we assume no friction or energy losses). But now the (perfect spring) contains the same energy in it that it put into the cart and it shoots off back where it came from at the same speed, leaving the cart there (which has put its brakes on and is stopped).

 

The same work was done to the spring as was done to you. The equation to figure out how much work can be harnessed out of the entire system is Ke=M*V^2 .

The problem is you need to set-up a device properly to make use of this potential energy. When a bullet hits a wall, it does 1/2 M*V^2 energy to the wall, + 1/2M*V^2 Energy to itself.

[ronthepon]

Could you clarify please with some more examples?

[arkain101]

If I hit you in the face with my fist. Does my fist hit you, or do you hit my fist? We both end up with broken bones in the impact. Who hit who? Momentum is a trade off, the energy we both felt. heh thats a joke but serious.

 

heh, I will supply a very well written example soon.. Going drinkin tonight.

[ronthepon]

HA! That is well placed...

 

Actually I am not really able to get what are you pointing at. Can you state you statement in bold letters?

[arkain101]

But now the (perfect spring) contains the same energy in it that it put into the cart and it shoots off back where it came from at the same speed,

 

This is a device that is able to capture the total Ke.

 

The shaft and wheel was a device that was able to capture the total Ke.

 

You dont see many device in this world that are meant to work in this manner.

 

Example.

 

You dont see a piston in engine that can make energy when it transfers from going down on its powerstroke to up. That change in direction can turn an internal device in the piston to develop energy. But the piston is only thought of as making 1 power stroke. This would be hard to enginner so it is a bad example. But its not the engineering, its the operation of physics we are concirned with.

[ronthepon]

Well, I gotta say that I've heard of newton's third law and thats it.

 

I still don't get your point. Stupid, ain't I?

[Farsight]

All: Thanks for the feedback on this.

 

Rontheon, can I ask you to look again at your proof. You start with

 

F = ma saying Force = mass x acceleration, absolutely no problem. But then you say

 

Work = x(ma) or Work = Force x Distance, and then you use this axiom to prove itself via:

 

v² = 2ax -> x = v²/2a -> Work = ½mv²

 

There's nothing wrong your maths, it's that Work = Force x Distance. Look at it carefully. Challenge it. Ask yourself why isn't it Work = Force x Time? Now look at the sled again. Ask yourself how can be work from the spring be equal to force times distance when the distance is greater if the sled is already moving? Ask yourself whether the distance we ought to be talking about is the extension of the spring working against the load, not the distance travelled by the system through the reference frame.

[Farsight]

Tormod:

 

I was skeptical about Gary Novak's article when I first read it. There's a lot of nonsense on the Internet. But it niggled.

 

Imagine you're on that 100kg sled at rest out in the depths of space with negligible friction, air resistance, or gravity. Fire the spring gun once and you find you're travelling at 0.1 m/s. Calculate the Kinetic energy you've acquired and you find it's:

 

KE = ½mv² = ½ x 100 x .1² = 50 x .01 = .5 Joules

 

Now fire the spring gun again to find you're travelling at 0.2 m/s and work out your Kinetic energy again:

 

KE = ½mv² = ½ x 100 x .2² = 50 x .04 = 2 Joules

 

That's 2 Joules. Not 1 Joule. Your second firing gave you 1.5 Joules of added Kinetic Energy, three times as much as the first firing.

 

How can this be?

[Erasmus00]

There's nothing wrong your maths, it's that Work = Force x Distance. Look at it carefully. Challenge it. Ask yourself why isn't it Work = Force x Time?

 

Because force x time already has a name, its called impulse. You have confused conservation of momentum with conservation of energy. I suggest taking a physics 101 type course, or just picking up a book. Kleppner and Kolenkow have an excellent introduction to mechanics.

-Will

[Farsight]

I know about Impulse, Erasmus. And since this is a Physics Forum can we look at the problem please?

[CraigD]

You’re sitting on a 100kg sled on a frozen lake, with a spring gun that fires a puck backwards. The recoil accelerates the sled forwards to a velocity of 0.1 meters per second.

 

Sled and puck have equal action and reaction so you calculate the energy stored in the spring as 2 x ½mv² -> 100 x .01 = 1 Joule.

 

With the assistance of a snowcat crew running alongside with a supply of pucks you fire the spring gun a further nineteen times in quick succession, and the sled is now travelling at 2 meters per second.

 

You recalculate the spring energy as 2 x ½ mv² / 20 -> 100 x 4 / 20 = 20 Joules

…

Can anybody point out the flaw in this?

This is an interesting “trick” problem that tests one’s grasp of basic mechanics.

 

The flaw in it is this

Sled and puck have equal action and reaction so you calculate the energy stored in the spring as 2 x ½mv² -> 100 x .01 = 1 Joule

, which implies that the “law of action and reaction” states that the energy of the sled and the puck after firing the spring gun are equal and opposite. This is incorrect. Energy is a scalar – it can’t be “opposite” any vector. What is conserved is momentum, which means that, while the kinetic energy of the sled-and-puck system increases as energy stored in the spring is released into it, the momentum remains the same. It’s convenient and conventional (but not required) to start the experiment with the system momentum and energy both at zero, which I’ll do.

 

Here’s the problem, restated without the flaw, with some helpful notations and formulae:

 

You’re sitting on a 100kg sled on a frozen lake, with a spring gun that fires a puck backwards…

OK so far, except for an important piece of missing data, the mass of the puck. I’ll use an approximate value of 0.2 kg to keep the calculations simple – a regulation hockey puck is closer to 0.17 kg.

MassSled = 100 kg

MassPuck = 0.2 kg

VelocitySled = 0 m/s

VelocityPuck = 0 m/s

Momentum = 0 kg*m/s = MassSled*VelocitySled + MassPuck*VelocityPuck

Energy = 0 J = 0 kg*m^2/s^2 = MassSled*VelocitySled^2/2 + MassPuck*VelocityPuck^2/2

 

… The recoil accelerates the sled forwards to a velocity of 0.1 meters per second. Sled and puck have equal action and reaction so you calculate the energy stored in the spring as …

We have all the data we need to calculate the energy stored in the spring correctly:

VelocitySled = 0.1 m/s

Momentum= 0 kg*m/s = 100 kg*0.1 m/s + 0.2 kg*VelocityPuck

VelocityPuck= -10 kg*m/s/.2 kg = -50 m/s

Energy= 100 kg*(0.1 m/s)^2/2 + 0.2 kg*(-50 m/s)^2/2 = 250.5 J

 

A very different result than the original 1 J!

 

With the assistance of a snowcat crew running alongside with a supply of pucks you fire the spring gun a further nineteen times in quick succession …

When all this is done, here’s what the whole system looks like:

VelocitySled = 2 m/s

VelocityPuck1 = -50 m/s

VelocityPuck2 = -49.9 m/s

…

VelocityPuck20 = -48.1 m/s

 

The spring gun has been fired 20 times, releasing a total energy of 5010 J. Only 200 J are in the form of the kinetic energy of the sled. The rest is in the form of the kinetic energy of the 20 pucks.

 

:) If you calculate the total energy and momentum of the sled and 20 puck system, you’re in for a small surprise – the energy is slightly greater than 5010 J (5012.47), and the momentum greater than 0 kg*m/s (3.8)!

 

:) Where did this extra energy and momentum come from? I leave this as an exercise for the reader, with a hint: What happened to the total energy and momentum of the sled-and-increasing-number-of-pucks system each time the snowcat crew passed a new puck to the gunner on the sled?

[Farsight]

Thanks for the lengthy reply Craig. I didn't intend any trick.

 

What you're saying is that when the first puck is fired, 250 Joules go into the puck side of the system, and 0.5 Joules go into sled side of the system. So the puck is given 500 times as much energy as the sled. This is coming from the fact that the puck side of the spring weighs 1/500th as much as the sled side, so the spring extends mainly on the puck side. IMHO you're using Work = Force x Distance, as ronthepon did, to prove a restatement of this in the form KE = ½mv².

 

Now consider the situation where you and I are each on back-to back sleds. I fire my spring gun at your sled, where it is caught by a spring-loaded catcher. At what speeds do the two sleds move apart?

[arkain101]

:hihi: uh oh, I think I see my flaw, ah jeez.

[Erasmus00]

IMHO you're using Work = Force x Distance, as ronthepon did, to prove a restatement of this in the form KE = ½mv².

 

Work, much like impulse, is a DEFFINITION, not a derivation. Work is defined by

 

[math] W = \int \vec{F} \cdot d\vec{x} [/math]

 

This leads to all the other equations for energy that come up often, including elastic energy, electromagnetic energy, etc. Because chemical energy is essentially electromagnetic, this deffinition can be extended into chemistry. Because heat is essentially a measure of average kinetic energy, we can extend our ideas of energy into the realm of thermodynamics.

 

Your insistance that work needs to changed as integral over time, instead of distance, is silly. You're just swapping vocabulary.

 

In current vocabulary the integral of force over time has a name, impulse. You are advocating changing names. Even if we call impulse work, and work impulse, it doesn't actually change the physics.

-Will

[ronthepon]

All: Thanks for the feedback on this.

 

Rontheon, can I ask you to look again at your proof. You start with

 

F = ma saying Force = mass x acceleration, absolutely no problem. But then you say

 

Work = x(ma) or Work = Force x Distance, and then you use this axiom to prove itself via:

 

v² = 2ax -> x = v²/2a -> Work = ½mv²

 

There's nothing wrong your maths, it's that Work = Force x Distance. Look at it carefully. Challenge it. Ask yourself why isn't it Work = Force x Time? Now look at the sled again. Ask yourself how can be work from the spring be equal to force times distance when the distance is greater if the sled is already moving? Ask yourself whether the distance we ought to be talking about is the extension of the spring working against the load, not the distance travelled by the system through the reference frame.

I'm replying late to Popular, I know, but I just cant resist saying this.

 

Let a box be held a few meters above the ground. Force is being applied on it, but there is definitely no energy change.

 

Force into Time is positive, it should have been zero to be a correct descriptor of work. (which, by the way, refers to energy in transit by the method of macroscopic movement)

 

So Force into time Is not work.

 

But Force into distance is filling this as an accurate description of energy change by macroscopic motion, so we call W=FS the work function.

 

PS: I have seen the previous posts

[Qfwfq]

Where did this extra energy and momentum come from? I leave this as an exercise for the reader, with a hint: What happened to the total energy and momentum of the sled-and-increasing-number-of-pucks system each time the snowcat crew passed a new puck to the gunner on the sled?

Actually, you could ask what happened in between each of these times! :hihi: