energy producing experiments

[greendoor]

It doesn't make logical sense that force x time is a measure of what is accelerating the mass.

F = MA which means that A = F/M
Whichever way you look at it, it is Force that causes Acceleration.

If we want to increase in Velocity we need to Accelerate for a period of Time greater than 0 seconds - so IMO it makes perfect 'logical sense' to say "force x time is a measure of what is accelerating the mass"

Rocketry is the purest example to consider, because it is removed from the problems of air friction and let's say we are sufficiently free of gravitational attraction for that to be an issue:

Let's say we are in a stationary capsule (from our own arbitrary inertial reference plane). We have on board 4 small rockets that can create 1 unit of unbalanced Force for 1 unit of Time.

We can choose to fire these rockets one after the other - or fire them together. The same amount of fuel is expended either way. And the same amount of Force x Time is expended ...

Makes perfect sense - in this situation - that Momentum, which is equivalent to Force x Time, is the true measure of Fuel ... not Energy as we are lead to believe ...

Distance or Displacement can be a red herring, in some cases, I believe.

Don't have time to explain, but I would like to continue this ...

[greendoor]

Energy is defined as the capacity to perform Work, and Work is defined as the capacity to consume Energy. So they become self referencing to a certain extent. They are both measured in Joules.

The original definition of Work as devised by Carnot was "weight lifted through a height". Which seems like a reasonable, practical engineering thing to want to do.

Work is now defined as Force x Displacement (Distance moved).

I smell a bit of a rat. Displacement or Distance should never be confused with Height. Acting within a gravity field - which effectively functions as an Acceleration field - there is a huge difference between moving something horizontally or lifting it vertically. Which is where the Force thing comes in. It gets clouded by the fact that we are generally overcoming friction and inertia and other losses.

So Energy/Work maths is based around Distance - but I would like to point out that Distance isn't all it's cracked up to be.

In space - it takes zero Work or Energy to move vast distances. All we have to do is get something moving at any Velocity, in the right direction - and given enough Time, it will get there with no further effort. So in this context - Distance really is free.

The cost is getting our space ship up to speed in the right direction (Velocity). That required Acceleration. Which simply requires Force applied for sufficient Time.

Back on Earth, and looking for a Bessler solution - do we really care about Distance? Or Time for that matter? All we really want to do is lift a weight vertically upwards a greater Height from which it fell. If we could do that simple thing - the rest is easy. Do we care how long it takes? Not really. Time is a variable we can manipulate.

I believe what we need is Force x Time to lift our mass. And Force x Time is a variable that we can manipulate and create in abundance using an Atwoods system.

Why do I believe this could work? Because I believe we could essentially create "virtual height".

The Force of gravity acting on 1 kg is a constant - whether it falls for 1 second, or whether it falls for 1000 seconds. That Force doesn't go away when it hits the ground either. And if we opened up a trap door in the ground to a mine shaft, it could keep falling with the same Force.

The difference is that when the mass is on the ground, it stresses the earth and creates an opposing Normal Force which maybe heats the Earth, but from our perspective is a waste of motive power.

So what if we can engineer that 1 kg to fall for long periods of Time, without losing height and hitting the ground? Well this is exactly what a balanced Atwoods system does. The mass falls at a slower rate (because it is Accelerating the heavy balanced system). Virtually - the Force x Time is equivalent to the Force x Time of that same Mass free falling for the same amount of Time ...

Wubbly - I know you don't want to consider this possibility, but I really do think that IF we create massive amounts of Momentum - there remains a real possibility that we COULD use that Force x Time to launch a mass vertically upwards.

That benchmark of Momentum required return a mass in height means that we can return that mass with that amount of Momentum if we want to.

I accept that we can squander variable amounts of Momentum in returning that Mass IF we take too much Time. Which is why we have to return it fast, to avoid fight gravity for too long.

I think Bessler achieved this - and I have some ideas about why he said "a hammer receives many blows" ...

My main thought is that the Wheel must be constantly slowly Accelerating, and then very quickly Decelerating. Velocity is nothing - Acceleration is everything. Because Force causes Acceleration - and we have virtually unlimited Force available to us - we are simply robbing the Earth of a little of that Stress that would otherwise be there if the Mass was sitting on the ground ...

[nicbordeaux]

Sorry, no theories in this one, just a little amazed at how quick kids can be in understanding. That tennis ball on top of basket ball drop and rebound/transfer/launch discussed in this thread, well, my 11 year old son was really thrilled when we thumped the kitchen ceiling real hard and the wife started hollering.

So as an adult I reckon that I'm way cleverer than him and need to explain it. So I start about weight, gravity and other stuff, then ask him "so why did that tennis ball take off so fast and high ?" And the immediate answer was "the basket ball tried to bounce but it couldn't because the tennis ball was on it, so it pushed the tennis ball".

And it is as simple as that.

[pequaide]

I think almost everybody views physics in a Newtonian sense until the reprogramming starts in about the 8th grade.

For instance: Ask an 11 year old this question.

First show your 11 year old some pucks moving on a frictionless plane. Show the student(s) that the pucks move freely and that a certain amount of push gives the pucks a certain amount of motion. Then ask the student: “if you push a puck for twice as long how does that change the mount of motion that the puck has�.

I think that, with almost no exception, the student will answer: “If you push it twice as long you will get twice as much motion�. And that was also Newton’s answer.

Around the 8th grade however teachers begin to replace this common sense Newtonian Physics with the physics of Leibniz. The teacher will tell the students that instead of twice the motion the puck really has four times the quantity of motion after only being pushed twice as long.

The teacher then takes this puck with the alleged four times the motion and collides and combines it with another puck; but only two times the original motion comes back out. But never mind the teacher tells the students we will just add in two units of heat. And the indoctrination continues.

[ovyyus]

Around the 8th grade however teachers begin to replace this common sense Newtonian Physics with the physics of Leibniz. The teacher will tell the students that instead of twice the motion the puck really has four times the quantity of motion after only being pushed twice as long.

The teacher will tell the students that it is a problem of displacement not time. You shouldn't have skipped 8th grade :P

[Fletcher]

A force accelerates a body giving it momentum.

Two unequal masses given the same force horizontally [like a puck on ice] for the same time will have the same momentum [mv].

They will have very different Kinetic Energy's i.e. the capacity for mechanical work - the lighter mass with faster velocity the higher Ke proportional to the mass ratios, the velocities & the displacement.

e.g. a 1 kg mass v's a 3 kg mass - the 1 kg mass will displace 3 times as far as the 3 kg mass - the 1 kg mass will have 3 times the Ke & 3 times the velocity of the 1 kg mass.

BUT .. they have the same momentum [mv].

The same force reversed & applied to each will bring both masses to a stop - their momentums will be zero.

At any time & distance during the process of acceleration followed by equal deceleration the two masses momentum [mv] is the SAME.

BUT .. the faster one has far greater mechanical Energy to do Work [Force x Distance] i.e. m x a x D

....................................................................................................


Momentum is a more 'fundamental' property of mass than Energy - turning Force into mechanical Energy is very wasteful usually.

Nevertheless, for Work to be Done on another object displacement must occur - Ke is the measure of that capacity to do Work.

Work Done = F x D in Joules.

Leverage is a factor of mechanical Work Done [that every machine must obey], says that F1 x D1 = F2 x D2, so you cannot avoid displacement or reduce it mechanically to change the equi-potentiality of Leverage - that IS Newtonian mechanics of machines obeying Newtonian Laws !

Study 'Kinematics' [the study of movement of objects irrespective of the forces involved] & you will see that mass does not appear in the equations IINM i.e. just forces, time, velocity, & acceleration - you can rearrange the equations to get times & distances for your puck answers.

N.B. no mass or Momentum [mv] is used in kinematics, & Work Done is conservative i.e. you can't get out more than you give so that mass, acceleration, displacement relationship is always maintained.

[Fletcher]

Kinematic Equations : - founded on first principles.

Two simple cases of uniform velocity & uniform acceleration, neglecting losses.

variables :

Time = t [sec]
Distance = s [meters]
Velocity [initial] = u [meters per sec]
Velocity [final] = v [meters per sec]
Acceleration = a [meters per sec per sec]

...................................................................................................

Uniform Velocity :

If velocity is uniform at u meters per sec

Distance traveled [displacement] = Velocity x Time

s = ut

Uniform Acceleration :

Final Velocity = Initial Velocity + Increase in Velocity

v = u + at

Displacement = Average Velocity x Time

s = [v + u]/2 x t .... [but v = u + at]

s = [u + at + u]/2 x t

i.e. s = ut + 1/2at^2

Eliminating t from v = u + at we get t = [v-u]/a & putting this value into ...

s = [v + u]/2 x t we get ...

s = [[v + u][v - u]]/2a

s = [v^2 - u^2]/2a ..... or v^2 = u^2 + 2as


e.g.1. Q. A car starts from rest & attains a velocity of 90 km/hr in 10 secs - what is the average acceleration ?

Ans. Initial velocity u = 0
Final velocity v = 90 km/hr = 25 m/s
Time t = 10 sec
a = ?

v = u + at
25 = 0 + 10a
a = 25/10
a = 2.5 m/s^2

e.g.2. Q. How far will the car have traveled in 10 secs i.e the displacement ?

Ans. u = 0, v = 25 m/s, t = 10 sec, a = 2.5 m/s^2

To find s we can either use ....

s = ut + 1/2at^2 = 0 + 1/2 x 2.5 x 10^2 = 125 m

or v^2 = u^2 + 2as => 25 x 25 = 0 + 2 x 2.5 x s
:. s = [25 x 25]/[2 x 2.5] = 125 m

e.g.3. Q. A bomb is dropped from an aircraft in level flight doing 200 kts at a height of 3500 m - neglecting the effect of air resistance, how long will it be before the bomb hits the ground, & how far horizontally before target must it be dropped ?

Ans. To find time of fall we are only concerned with vertical velocity, which was zero at release.

:. u = 0, a = 9.80665 m/s^2, s = 3500 m, t = ?

s = ut + 1/2at^2
:. 3500 = 0 + 1/2 x 9.80665 x t^2
=> t^2 = [3500/9.80665] x 2 = 713
=> t = sqrt 713 = 27 sec [approx]

Since we are neglecting air friction the horizontal velocity of the bomb will throughout the fall remain the same as at moment of release i.e. the same velocity as the aircraft 200 kts - converting to meters per sec [200 x 1852]/3600 = 103 m/s [approx].

Therefore the distance the bomb will travel forward during the fall time of 27 sec is 103 x 27 = 2781 m & is the distance the bomb must be released before target.

...................................................................................................

Understand these equations with worked examples & you can work out puck on ice problems for yourself & never again get confused.

Might be a struggle for the average 8th grader but a 9th grader could do it with some help.

[path_finder]

demonstration (at 2:27) here!
http://www.youtube.com/watch?v=Vtdxarsf6BM

[pequaide]

Fletcher Quote: ‘Momentum is a more 'fundamental' property of mass than Energy - turning Force into mechanical Energy is very wasteful usually.’

Maybe momentum is a more fundamental property because it is actually conserved. Momentum is experimentally conserved in ballistic pendulums and Energy is only allegedly conserved. Kinetic energy needs a mystic friend of heat to be conserved in a ballistic pendulum and no one has ever found the heat. So at best you are only making an allegation that energy is a conserved property.

Fletcher Quote: ‘- turning Force into mechanical Energy is very wasteful usually.’

Is it wasteful or is the concept of energy conservation just plain false?

[Fletcher]

While I'm killing time for the next few minutes.

I might as well show how Kinematic equations can be used to find Energy [Capacity to do Work].

A weight of 50 newtons [50N's = 50/9.80665 = 5.01 kg, say 5 kg] is raised a height of 2 meters - it has gained additional Pe of 100 Joules i.e. [Pe = mgh = 5.01 x 9.81 x 2 = 100 Joules].

100 J was the Work Done to raise the mass to give it extra Energy of Position.

This is the Work it should be able to do returning to its original position.

N.B. Work Output = Work Input [assuming no losses] or Work Output < Work Input [with losses] - this is the Leverage Work Done equi-potentiality of F1 x D2 = F2 x D2.

Q. What is the Ke of a mass of m kg moving at v m/s^2 ?

Ans. Let us assume that gravity accelerated it uniformly at a m/s^2 from zero velocity to v m/s^2 by being pushed by a constant Force of F newtons .

The distance traveled during acceleration was s meters, then the Work Done i.e. the Ke, will be in Joules.

But v^2 = u^2 + 2as [u =0]
=> v^ = 2as
=> s = v^2/2a

N.B. => v^2 = 2as => v = sqrt [2as] = sqrt 40 = 6.32 m/s

BUT F = ma

SO Ke = Fs = ma x v^2/2a = 1/2mv^2 Joules

Thus Ke of 5 kg moving at 6.32 m/s after 2 meters

= 1/2 x 5 x 6.32^2

= 100 Joules.

[Fletcher]

You obviously don't read & understand others post very well pequiade.

Laws of Thermodynamic say that Energy is Conserved.

A body's 'Internal Energy [i.e. Pe + Ke]' is changed by Work Done on the body e.g. mechanically compressing a gas, OR adding thermal/heat Energy to INCREASE the Internal Energy - the Internal Energy raised then dissipates as heat over time.

Out of time !

[nicbordeaux]

Wubbly, thanks a load for that post from previous page 71 (oh that topics and posts were automatically numbered so we could refer to a post without quoting...). The ability shown in explaining what is usually expressed in mathematical terms only in plain language (there's a sometimes negative connation to vulgarization, thus the word is avoided) could lead one to believe that you have suffered as a teacher/professor.

Fletcher too, your input, which I can understand, is a form of fast track education. Which raises the inevitable question, French, of which I know no English equivalent : "Should you give jam to the pigs ?" Another common expression : "It is pointless to talk to fools, it only serves to educate them".

[pequaide]

"Internal Energy"? Bologna: we measure heat all the time, it is very easy with the modern technology. You should be embarrassed to suggest such a thing.

[pequaide]

I am using a tether length of almost one circumference.

The mass ratio of the wheel to missile (BB bag) is about 37 to one.

I wrap the tether clockwise and gently spin the wheel counterclockwise.

I release at 3 o’clock and the bag just misses the floor and then just misses the ceiling. The tether loop slips off the pin when the bag is headed about 45° down: but I guess that is the safest angle anyway. The bag thumps the floor about 10 feet away.

But even with gentle spins there is a distinct whirr being made by the BB bag. I know from experience that a whirr mean fairly high velocity. With spins that are just scarcely fast enough to work at all I get a whirr. And according to your Law of Conservation of Energy I should not even be able to touch the ceiling.

Once I missed my release point and hit an obstruction behind the wheel; and the 20 pound fluorocarbon fishing line broke on the obstruction. These are not moderate velocities.

I often release to early and violently thump the ceiling

These speeds are simply too high for your Law of Conservation of Energy to be correct.

[jim_mich]

Wubbly, thanks a load for that post from previous page 71 (oh that topics and posts were automatically numbered so we could refer to a post without quoting...)

The posts ARE numbered. At the top left of each post, just to the left of where it says, Posted: date and time is a little box that looks like a little tiny page of paper with its corner turned over. It's either white or yellow depending whether it is a new post or not. If you put the cursor on it then it will say "Post". It is a link to that post. I'm not sure if all browsers work like my Firefox, but I can "copy link location" using right click of my mouse and then paste wherever needed.