Can a Falling Mass Generate 12V Using an Optimal Magnet and Coil?

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SUMMARY

The discussion centers on the feasibility of generating 12 volts from a falling mass using an optimal magnet and coil setup. Participants clarify the relationship between force, energy, and voltage, emphasizing that 1 kg corresponds to 9.81 Newtons and that energy is measured in Joules. The mechanics of a 2.5 kg weight falling 5 mm are analyzed, with the conclusion that while energy can be calculated, the voltage produced depends on the charge moved. The conversation highlights the complexities of energy conversion and the misconceptions surrounding perpetual motion and free energy devices.

PREREQUISITES
  • Understanding of Newton's laws and units of force (Newton, kgf)
  • Basic knowledge of energy units (Joules, Watts)
  • Familiarity with electromagnetic principles (magnet and coil interactions)
  • Concept of voltage and charge in electrical circuits
NEXT STEPS
  • Calculate energy released from a falling mass using gravitational potential energy formulas
  • Explore the relationship between voltage, current, and charge in electrical circuits
  • Research the principles of electromagnetic induction and Faraday's law
  • Investigate the limitations and misconceptions surrounding perpetual motion machines
USEFUL FOR

Physics students, electrical engineers, hobbyists interested in energy generation, and anyone exploring the principles of electromagnetism and energy conversion.

pete94857
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Homework Statement
Is a continuous applied force of 2.5kg over a distance of 5mm in a lateral motion equal to or more than 2.5w of mechanical energy ?
Relevant Equations
1w=1Nm=0.981kgm
1 watt equals 1 Newton metre

1 kg equals 9.81 Newton metres

2.5 kg = 24.525 Nm = 24.525 Watts of mechanical energy.
 
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This question is messed up to me, Watts are a unit of power, not energy. Are you sure you don't mean Joules?

Let me correct your units/conversions

1 Joule (J) = 1 Newton Meter (Nm)
1 watt (W) = 1 Joule/second (J/s)

also kg is a unit of mass not force.

If I were to make a correction to your question statement it would be

"Does a 2.5 Newton force applied over a distance of 5mm result in 2.5 J or more of work done?"
 
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I see, thank you.

But if 1kg is equal to 9.81 Nm what it not be , is 9.81Nm equal to 2.5 J or more ?
 
pete94857 said:
I see, thank you.

But if 1kg is equal to 9.81 Nm what it not be , is 9.81Nm equal to 2.5 J or more ?
##1kgf## (=1 kilogram force) is equal to ##9.81## Newtons and is a unit of weight $$1kgf=1kg\times g=1kg\times 9.81ms^{-2}=9.81N$$
##\text{ g is acceleration due to gravity}##

Also ##1watt=1Nms^{-1}##
 
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I'm assuming this is in a vertical movement with a reference to gravity. Therefore is applying the force laterally parallel with gravity I'm assuming the ×9.81 m/s would be cancelled out. ?

What I'm actually trying to answer is for example , if I had a coil beside a permanent magnet, the coil was connected to a cord the coil moves laterally horizontal to gravity, the cord then goes round a single pully wheel with a 2.5kg weight attached to it. It is allowed to fall 5mm , would this be enough to produce 12 v assuming the magnet and coil are optimum. Also assuming no drag or friction ?

How would I work this out ?
 
pete94857 said:
I'm assuming this is in a vertical movement with a reference to gravity. Therefore is applying the force laterally parallel with gravity I'm assuming the ×9.81 m/s would be cancelled out. ?
This is very confusing. "laterally parallel with gravity" means both horizontal and vertical. It does not clarify anything. The 9.81 that you probably mean is the acceleration of gravity. That is 9.81 m/s/s. Not 9.81 m/s.

pete94857 said:
What I'm actually trying to answer is for example , if I had a coil beside a permanent magnet, the coil was connected to a cord the coil moves laterally horizontal to gravity, the cord then goes round a single pully wheel with a 2.5kg weight attached to it. It is allowed to fall 5mm , would this be enough to produce 12 v assuming the magnet and coil are optimum. Also assuming no drag or friction ?
12 Volts is a voltage. It has dimensions of energy per unit charge. For example, joules per coulomb.

If the 2.5 kg weight falls 5 mm, that releases an amount of energy. But we cannot know how many volts correspond to that energy unless we know how much charge has been moved.

Start with the simple stuff: How much energy is released when a 2.5 kg weight falls 5 mm vertically under gravity with a strength of 9.81 m/s2 ?
 
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pete94857 said:
What I'm actually trying to answer is for example , if I had a coil beside a permanent magnet, the coil was connected to a cord the coil moves laterally horizontal to gravity, the cord then goes round a single pully wheel with a 2.5kg weight attached to it.
Your description brings the picture below to my mind. The string is pulling laterally on the coil. Pretend that the magnet is fixed on the table and "beside" instead of "inside" the coil. Am I close?

Are the ends of the coil connected to anything?


Half-Atwood with coil.png
 
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Thread closed temporarily for Moderation...
 
Unfortunately, via private communication I've discovered that the OP is working on an Over-Unity mechanism, which we do not discuss here at PF. @pete94857 -- Please read the links below to help you understand why what you are trying to do will not work.

Forbidden Topics said:
Pseudoscience, such as (but not limited to):
Perpetual motion and "free energy" discussions (see our Insights Article here)
http://en.wikipedia.org/wiki/Perpetual_motion
http://www.skepdic.com/freeenergy.html
http://www.skepdic.com/perpetual.html

https://www.physicsforums.com/insights/why-we-dont-discuss-perpetual-motion-machines-pmm/
 
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pete94857 said:
It is allowed to fall 5mm , would this be enough to produce 12 v assuming the magnet and coil are optimum. Also assuming no drag or friction ?
Depending on the details of your coil and magnet it could produce 12v or 12000v or any other voltage you want. However, a higher voltage means that either less current will flow or or the current will flow for a shorter time. The total amount of charge moved by the current, times the voltage, will be equal to the energy derived from lowering the mass.
 
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