Another E=1/2mv^2 dilemma increasing energy required?

[poor mystic]

It's also a question about acceleration. If a weight falls a meter, it gives up kinetic energy on impact. If it falls twice as far, it gives up twice the energy.
One must carry the weight to 4 times the height if one wishes it to strike the ground at twice the speed; why take surprise that 4 times as much energy is involved?

 

[jablonsky27]

You have to take into account that in order to accelerate the mass (let's call it a block) on the train by 1 m/s relative to the ground, one must push back on the train with the same force. The impulse given to the 1 kg. block is equal and opposite to the impulse given to the train.

Let's suppose the train had a mass of 1000 kg. to make the math clear.

Initially, the KE of the train and block is:

KE = \frac{1}{2}(m_{train}+m_{block})v^2 = .5 \times 1001 \times 1^2 = 500.5 J.

By giving the 1 kg. block a forward impulse of 1 N sec. or 1 kg m/sec, the train slows down by 1/1000 m/sec. to .999 m/sec. So the 1 kg block is going 2 m/sec relative to the ground and the train is now going .999 m/sec relative to the ground. The total energy of the system relative to the ground is

KE = KE_{train} + KE_{block} = \frac{1}{2}(1000)(.999)^2 + \frac{1}{2}(1)(2)^2 = 499 + 2 = 501 J.

So the total energy did not increase by 1.5 J. It only increased by .5 J.

AM

Here you re turning my question into a conservation of energy verification. That was not what I was trying to put forward.
What I meant was this,
1. On the ground, if I get a block of mass 1 kg going at 1 m/s, I would have spent 0.5 Nm energy
2. Now, I get on a train which itself moves at 1m/s. On it, to get the same block going at 1 m/s(relative to me ie. 2 m/s w.r.t the ground) I will spend 1.5 Nm energy. This is the amount I would measure isn't it?

Isnt this difference against the principle of relativity?

 

[Andrew Mason]

Here you re turning my question into a conservation of energy verification. That was not what I was trying to put forward.
What I meant was this,
1. On the ground, if I get a block of mass 1 kg going at 1 m/s, I would have spent 0.5 Nm energy
2. Now, I get on a train which itself moves at 1m/s. On it, to get the same block going at 1 m/s(relative to me ie. 2 m/s w.r.t the ground) I will spend 1.5 Nm energy. This is the amount I would measure isn't it?

No. I showed that you would only expend .5 J., not 1.5 J. The change in total energy is only .5 Joules, not 1.5 J. (relative to the ground frame of reference). So this is all you would have to expend in order to increase the speed of the block from 1 to 2 m/sec.

Isnt this difference against the principle of relativity?

It is quite consistent with Galilean relativity. Is that what you are referring to? The measurement of the energy expended does not depend upon the motion of the observer. The measure of the energy added is the same in all inertial frames of reference.

AM

 

[Cleonis]

1. On the ground, if I get a block of mass 1 kg going at 1 m/s, I would have spent 0.5 Nm energy
2. Now, I get on a train which itself moves at 1m/s. On it, to get the same block going at 1 m/s(relative to me ie. 2 m/s w.r.t the ground)
[...]

We treat the Earth, the ground, being much heavier than the other objects in this setup, as unmoving.

The 1 kg block is subject to two stages of acceleration:
- A train with the block onboard accelerates to 1 m/s. During this stage the amount of work done upon the block is 0.5 Nm
- The block is accelerated w.r.t the train, again the work done upon the block is 0.5 Nm.

However, this evaluatoin does not give the kinetic energy of the block w.r.t the ground, which is 2 Nm.


It is in fact incorrect to think of kinetic energy as a property of an individual object. Kinetic energy is a property of the relation between two objects: Kinetic energy is a function of the relative velocity between two objects.

It may be helpful to think of kinetic energy as a form of potential energy. When two objects have a velocity relative to each other then there is a potential for energy conversion. (The most crude energy conversion process is a head-on collision, in which all energy dissipates.)

To evaluate the kinetic energy of the block w.r.t. the ground you must go back to a point in time that the block and the ground were co-moving. It's the acceleration history from that point on, (acceleration w.r.t. the ground) that counts for evaluating the block's kinetic energy w.r.t the ground.

The point is that within the scope of an evaluation the same inertial coordinate system must be used throughout. If that is done then one finds that kinetic energy slots in with the principle of relativity of inertial motion.

 

[jablonsky27]

Andrew, Cleonis,

To make sure I got this right, let me try to put into words my understanding...

Observer A is inside the train. He does 0.5 J work to increase the speed of the block by 1 m/s.

Observer B on the ground observes the energy put in by A also as 0.5 J. This breaks down as 1.5 J to increase the speed of the block from 1 m/s to 2 m/s, of which 1 J is provided by the train(since it slows down by a corresponding speed). So (1.5 - 1) J = 1 J.

Right?

 

[Andrew Mason]

Andrew, Cleonis,

To make sure I got this right, let me try to put into words my understanding...

Observer A is inside the train. He does 0.5 J work to increase the speed of the block by 1 m/s.

Observer B on the ground observes the energy put in by A also as 0.5 J. This breaks down as 1.5 J to increase the speed of the block from 1 m/s to 2 m/s, of which 1 J is provided by the train(since it slows down by a corresponding speed). So (1.5 - 1) J = 1 J.

Right?

Right.

AM

 

[jablonsky27]

Right.

AM

Thanks..

 

[Cleonis]

[...]
Still, the "leverage" description is very odd
[...]

Let me compare some forms of propulsion:

A rocket engine, accelerating a space capsule in flight. As we know, the operating principle of rocket propulsion is recoil. The acceleration of the exhaust to make it shoot out the nozzle is like the acceleration of a bullet down a gun's barrell. Rocket propulsion is recoil propulsion.

The universal rule is: to accelerate you have to push off against some other lump of mass. Usually what you push of against is much heavier than yourself - like pushing yourself off with your feet against a wall when you're swimming - but in the case of rocket propulsion: if you accelerate the exhaust to extremely high velocity then that acceleration will give you plenty of leverage.

Now the case of a jetski.
A jetski scoops up water in the front, then an internal propeller accelerates that water, and shoots it out rearward, with a high velocity relative to the jetski. Let's say 10 m/s w.r.t the jetski. The jetski propulsion gets its leverage from accelerating water rearward.

By the time the jetski itself has velocity of 10 m/s its harder to get the desired leverage. The water scooped up at the front must first be accelerated to bring it up to jetski's velocity, and then accelerated some more to expel it rearward for propulsion.
[later comment]
The 'accelerated some more' is probably not a good formulation. Anyway, it's a detail; the overall reasoning is not affected by it.
[/later comment]

To get uniform acceleration the jetski's motor will need to work harder and harder the faster the jetski goes. In other words, for uniform acceleration the energy output of the motor must rise in proportion to the current velocity.

In the case of a car the acceleration from 30 to 60 takes three times as much power as it took to accelerate from 0 to 30. During the phase where you accelerate from 30 to 60 your distance traveled is much larger. In order to have the wheels rotate sufficiently fast to propel the car you shift to a higher gear ratio, but the consequence of higher gear ratio that your mechanical advantage goes down.

 

[Andrew Mason]

In the case of a car the acceleration from 30 to 60 takes three times as much power as it took to accelerate from 0 to 30. During the phase where you accelerate from 30 to 60 your distance traveled is much larger. In order to have the wheels rotate sufficiently fast to propel the car you shift to a higher gear ratio, but the consequence of higher gear ratio that your mechanical advantage goes down.

You are trying to demonstrate in a non-mathematical way that the amount of effort required to maintain constant acceleration increases with speed. Just a suggestion here, but I would stick to the math. The amount of effort required to maintain constant acceleration increases with speed because the amount of "effort" is proportional to force x distance rather than force x time.

W = \vec{F}\cdot\vec{s}

This is not an obvious relationship. It was not until the 19th century and the advent of heat engines, that this was fully understood. This is the concept of energy - the ability to do work (applying a force over a distance). Even Newton did not appreciate its significance.

AM

 

[Cleonis]

You are trying to demonstrate in a non-mathematical way that the amount of effort required to maintain constant acceleration increases with speed. Just a suggestion here, but I would stick to the math.

The importance of formulating a theory mathematically is that it demonstrates self-consistency. (Conversely, if there is a self-contradiction then the mathematical formulation of the theory will expose that.)

However, the math does not explain.

We start with demanding self-consistency, and then we find that it follows that work done is force * distance traveled. But logical implication and explanation are not the same thing.

So I'm very curious whether I can find a visceral explanation. Maybe there is a way of presenting the physics in such a way that this non-obvious aspect of mechanics starts looking obvious after all.

I feel that visualizing aids such as 'recoil' and 'propulsion principles' and 'leverage' can play an importent part in explaining the relation between energy and work done.

 

[Lsos]

I agree. Understanding the mathematical formula is different than truly understanding the concept. Once you fully, truly understand the concept, you can derive your own mathematical formula.

 

[Andrew Mason]

The importance of formulating a theory mathematically is that it demonstrates self-consistency. (Conversely, if there is a self-contradiction then the mathematical formulation of the theory will expose that.)

However, the math does not explain.

IF you accept the PREMISE that "effort" is proportional to the magnitude of the force and the distance over which the force is applied, then it IS just a matter of math. The math explains it completely ie. the reason it takes 3 times as much energy to go from 0-30 as from 30-60 is BECAUSE the same force is applied over 3 times the distance in going from 30-60 as compared to going from 0-30.

The math does not explain that the premise is true. Experiment - physical data - shows that it is true. The math just shows you the consequences that necessarily flow from the premise.

AM

 

[Cleonis]

IF you accept the PREMISE that "effort" is proportional to the magnitude of the force and the distance over which the force is applied, then it IS just a matter of math.

I'd like to break down that premise in two independent premises.

There is the case of the suspended weights that provide a pendulum clock with energy to sustain its operation. Those weights descend in a controlled manner; they descend with uniform velocity.

Then there is the case of free fall, which you can think of as a runaway release.

Premises:
1) In a controlled release (uniform velocity) the energy output is force * distance traveled
2) In a runaway release the energy output is force * distance traveled

I regard those as independent premises, that require separate experiments for their respective verification. I think that when conservation of energy was first suspected, but not yet confirmed, premise 1) and 2) have indeed been experimented on separately.

 

[Andrew Mason]

Premises:
1) In a controlled release (uniform velocity) the energy output is force * distance traveled
2) In a runaway release the energy output is force * distance traveled

These are not premises. They flow from the definition that energy = force x distance.

If the question is "why is energy defined that way", the answer is a little more complicated. One reason energy is defined that way is because the quantity consisting of "work done", which is defined as "force x distance" and the ability to do work ("energy") are directly related. To the extent that work done on a system is not spent as heat, energy is stored in the system and the system retains the ability to do the amount of work that was done on it. This conservation rule is the basis for the first law of thermodynamics, for example. This is at the heart of the reason that we perceive that the amount of effort is related to the product of force x distance. If we exert a muscular force over a distance, we expend ourselves physically. If we apply a muscular force for a time, but not over a distance, we do not expend ourselves physically (except for a small amount of heat needed to keep the force up).

AM