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

In summary, the concept of generating 12V from a falling mass using an optimal magnet and coil involves converting gravitational potential energy into electrical energy through electromagnetic induction. The efficiency of this process depends on the design of the magnet and coil system, including factors like the strength of the magnet, the speed of the falling mass, and the configuration of the coil. By optimizing these parameters, it is theoretically possible to produce sufficient voltage to achieve 12V, which could be used for various applications, although practical implementation may face challenges such as energy losses and the need for a suitable load.
  • #1
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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  • #2
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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  • #3
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 ?
 
  • #4
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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  • #5
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 ?
 
  • #6
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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  • #7
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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  • #8
Thread closed temporarily for Moderation...
 
  • #9
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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  • #10
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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FAQ: Can a Falling Mass Generate 12V Using an Optimal Magnet and Coil?

1. How does a falling mass generate electricity using a magnet and coil?

A falling mass can generate electricity through electromagnetic induction. When a magnet falls through a coil of wire, the changing magnetic field induces an electric current in the coil according to Faraday's Law of Electromagnetic Induction. This current can be harnessed to produce voltage, such as 12V, depending on the design and parameters of the system.

2. What are the optimal conditions for generating 12V with a falling mass?

To generate 12V efficiently, several factors must be optimized: the strength of the magnet, the number of turns in the coil, the diameter of the coil, the speed at which the magnet falls, and the distance between the magnet and the coil. A stronger magnet and a larger number of coil turns generally increase the induced voltage. Additionally, ensuring that the magnet falls freely without obstruction maximizes its speed and the resultant voltage.

3. What types of magnets and coils are best for this application?

Neodymium magnets are often the best choice due to their strong magnetic fields. For coils, copper wire is commonly used because of its excellent conductivity. The coil should be designed with an appropriate gauge of wire and a sufficient number of turns to maximize the induced voltage. Additionally, using a core material, such as iron, can enhance the magnetic field interaction and improve efficiency.

4. What are the practical applications of generating 12V from a falling mass?

Generating 12V from a falling mass can be applied in various fields, including renewable energy systems, energy harvesting devices, and educational demonstrations. It can be used to power small electronic devices, charge batteries, or serve as a backup power source in remote locations. Additionally, it can be integrated into kinetic energy recovery systems in transportation or other mechanical systems.

5. What are the limitations of this method of electricity generation?

One major limitation is the amount of energy that can be harvested, which is dependent on the mass of the falling object, the height from which it falls, and the efficiency of the conversion process. Additionally, the system may not produce continuous power, as it relies on the periodic falling of the mass. Mechanical wear and environmental factors can also affect the longevity and reliability of the system. Finally, safety concerns must be addressed when dealing with falling objects and high-speed movements.

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