re: energy producing experiments

[Floor]

Back on topic ...

How much energy is delivered to an object when 1 Newton of force
is applied to it causing that object to accelerate (given that 100% of the
energy applied is transferred as acceleration) ?

This depends upon the length of time the Newton of force is applied for.

See the "Magnets Motion and Measurement part 1-9 PDF file, attached below,
for detailed explanation.

[Delburt Phend]

No it does not.

Force does not produce consistent quantities of energy.

Force produces momentum,  F = ma   a = v/t Therefore Ft = mv  it is not Ft = 1/2 mv²

If 7 N is applied to 3 kg for 4 seconds you will get;  7 N * 4 sec = 3 kg * v  = 9.33 m/sec This gives you 28 units of linear momentum and 130.66 J of energy.

If 7 N is applied to 30 kg for 4 seconds you will get;  7 N * 4 sec = 30 kg * v  = .933 m/sec This gives you 28 units of linear momentum and 13.066 J of energy.

Force does not produce energy it produces linear Newtonian momentum.

[sm0ky2]

No it does not.

Force does not produce consistent quantities of energy.

Force produces momentum,  F = ma   a = v/t Therefore Ft = mv  it is not Ft = 1/2 mv²

If 7 N is applied to 3 kg for 4 seconds you will get;  7 N * 4 sec = 3 kg * v  = 9.33 m/sec This gives you 28 units of linear momentum and 130.66 J of energy.

If 7 N is applied to 30 kg for 4 seconds you will get;  7 N * 4 sec = 30 kg * v  = .933 m/sec This gives you 28 units of linear momentum and 13.066 J of energy.

Force does not produce energy it produces linear Newtonian momentum.

It is actually an integral equation. If you want to truly understand this problem,
I suggest learning calculus and trigonometry.


FT = ma = m* (delta)v
That is the change in velocity, integrated over time


Since mass is on both sides, we take that out and we see
that the FT results in a value of acceleration.


Anything off the horizontal constraint (with respect to the gravitational field)
looks something like [FT +{<+\-}9.8m/s/s]
Which is a vectored quantity


So the vector is also integrated over time
What we see mathematically, is that both the
vector and the acceleration are changing over time.


4 seconds takes up about a full sheet of notebook paper
if you write it all out in standard academic format.


Then you will understand that every second, you are inputting
different values for acceleration and vector (change in vector changes acceleration)


The summation of which can be estimated at 1/2q^2


This is simple shorthand physics, we do this in most "public equations"
Because the average person can't perform pages of calculus for a simple problem
in the real world.


To get from one to the other is where trig comes in
A triangulation of the acceleration curve
You are correct in saying that is not the actual equation
It is an average of the change in velocity over time


The results are the same, because one works out the velocity
the other starts with the velocity and works it backwards

[sm0ky2]

Now understand that "one second" is an arbitrary value.


We can instead use a value of 1-trillionth of a second.


We see that for every sec/ 1,000,000,000,000 the acceleration and vector change
so 1 trillion sheets of paper later, we realize that the best of our equation are only an approximation.


We could divide the trillionth of a second another trillion times.


and all of this only applies to the "stationary" inertial reference of our position on the earth's surface
Which moves at 1000mph relative to the sun, which moves at 25Mmph relative to the inertial
reference of our galactic center, which moves at 1 trillion mph with respect to another arbitrary value
representing our sector of expanding space-time. After which time itself becomes  an integral quantity.

[sm0ky2]

Even the 9.8m/s/s of gravity is not entirely accurate
We can integrate gravity using tSecs (trillionths of a second)
and derive many decimals which (were it not for terminal velocity)
would become significant over long drops.


Using this level of accuracy in tests to verify the equation,
gives rise to variances in the true value of gravitational acceleration
from location to location.
Mapping the earth's gravity for the purposes of physics,
We see that even "sea level" is not the appropriate altitude in every location.
We have no consistency for our standard tests, beyond a certain accuracy.


Not to mention sea level in 1706 is meters under water today.