What Is The Usage Of Energy In This Process?

[belovelife]

say you have a load weighing 10 tons on wheels,

 

you move the load onto a platform geared to produce electricity, it moves down one foot

 

you move the load back to the start

 

and do it again

 

what is the energy usage at each point in this process?

 

energy production

,

 

and spring to move the platform back up?

[JMJones0424]

*Please do not hesitate to correct me if I am wrong*

 

Work equals force times distance. So at 100% efficiency and zero friction, you get the same amount of energy from the falling weight as is required to lift the weight. However, there's no way to design a generator that will convert 100% of the work to energy, and there's no way to move weight with zero friction, so in the real world it will actually take more effort to raise the weight than the energy you'll generate by letting the weight drop. Ideally, you'd use energy in a form that isn't useful to you, like blowing wind, solar, running water, etc. and then use the process to convert that energy to electricity, which would be a more useful form of energy. Though I wonder if you couldn't find a better way of converting the energy, since the generator needs to rotate rather than oscillate up and down.

 

Here's the math (note, this is for straight line application of force... weight is moving up and down. If the weight were moving up an incline, you'd need coefficient of friction of the slope and angle of slope)

 

10 tons = 9078 kg

1 foot = 0.3048m

 

[math]F = mg[/math]

 

[math]F=9078kg \cdot \frac{9.807N}{kg} = 89030N[/math]

 

[math]W=Fs[/math]

 

[math]W=89030N \cdot 0.3048m=27140J[/math]

[sigurdV]

Hi there Is the "v" in "provess" a deliberate evvov or is the ouija effect showing its ugly read?

 

Or is only my ignorance shown when thinking "provess" as u s e d in the topic is not a word in english?

 

And: Mr jones has the ability of keeping mathematical formulae down to a weadable minimum. (Applauds)

[CraigD]

…

[math]W=89030N \cdot 0.3048m=27140J[/math]

Looks right to me. :thumbs_up

 

27140 Joules = 27140 Watt-seconds, so intuitively, this is enough energy to light a 100 W bulb for a bit more than 2.5 minutes, though I doubt you could get near the assumed 100% efficiency with any real machine like Belovedlife describes: slow-moving, high-force geared machines are notoriously hard to make be efficient.

 

(note, this is for straight line application of force... weight is moving up and down. If the weight were moving up an incline, you'd need coefficient of friction of the slope and angle of slope)

If you assume an ideal, frictionless system, you don’t need to consider whether the weight moves vertically, on an incline, pendulum, roller-coaster, or as a liquid flowing through a pipe.

 

There’re complicated ways to show this by calculating the force parallel to the incline and distance traveled, but it’s easier to just consider change in gravitational potential energy, which gives approximately the same formula you used starting with the definition of work.

 

This result is an approximate, not exact one, because the gravitational force on the weight decrease slightly as its height increases, but this can be ignored when the change in height is very small (such as 0.308 m) compared to its distance from the center of the Earth (about 6370000 m).

 

If the platform was very tall, you’d have to use the exact formula for GPE, and also take into account the rotation of the Earth. If the platform was really tall, taller than a geostationary orbit (about 36,000,000 m), you’d actually get work/energy raising the weight above that height, and have to apply work lower it.

 

But I digress. :) Oh, and calculations are left as an exercise to the reader. ;)

[belovelife]

and what if you add a system so only the front wheels need to be on the platform and as the shelf lowers , it pulls the weight onto the platform

[belovelife]

*Please do not hesitate to correct me if I am wrong*

 

Work equals force times distance. So at 100% efficiency and zero friction, you get the same amount of energy from the falling weight as is required to lift the weight. However, there's no way to design a generator that will convert 100% of the work to energy, and there's no way to move weight with zero friction, so in the real world it will actually take more effort to raise the weight than the energy you'll generate by letting the weight drop. Ideally, you'd use energy in a form that isn't useful to you, like blowing wind, solar, running water, etc. and then use the process to convert that energy to electricity, which would be a more useful form of energy. Though I wonder if you couldn't find a better way of converting the energy, since the generator needs to rotate rather than oscillate up and down.

 

Here's the math (note, this is for straight line application of force... weight is moving up and down. If the weight were moving up an incline, you'd need coefficient of friction of the slope and angle of slope)

 

10 tons = 9078 kg

1 foot = 0.3048m

 

[math]F = mg[/math]

 

[math]F=9078kg \cdot \frac{9.807N}{kg} = 89030N[/math]

 

[math]W=Fs[/math]

 

[math]W=89030N \cdot 0.3048m=27140J[/math]

 

what if you used a spring to move the platfrom back up, driving the generator?

[CraigD]

and what if you add a system so only the front wheels need to be on the platform and as the shelf lowers , it pulls the weight onto the platform

For this to work, the mass of the shelf [imath]M_1[/imath] times the distance it lowers [imath]-h_1[/imath] must be greater than the mass of the weight [imath]M_2[/imath] time the height It’s raised pulling it onto the platform [imath]h_2[/imath]. The difference in these, that is

 

[math]W = M_1 h_1 + M_2 h_2[/math]

 

is the work done, and also the energy stored, by the system.

 

In short, it lowers the system's potential energy.

 

what if you used a spring to move the platfrom back up, driving the generator?

If the system is 100% efficient, and has no generator or other scheme to deliver energy outside of it, it could bounce down and up on a spring forever. Since such a system doesn’t produce usable work, it’s literally useless, though.

 

Since no system involving springs is 100% efficient, a real weight on a spring converts its potential energy into forms such as heating the spring, and eventually stops bouncing.

 

In short, no sort of “tricks” improve a system like this, causing it to do more work than that calculated by the mass times height calculation in JMJones’s post #2.

[belovelife]

hmmm