Looking for an algebraic definition of work.

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Work in physics is defined as the product of force and displacement, represented algebraically as W = Fd, where F is the force and d is the distance moved in the direction of the force. For a more rigorous definition, work involves the dot product of force and displacement vectors, especially when force varies in magnitude or direction, requiring integration. In cases of constant force acting along the same direction, the simpler formula is sufficient. The discussion highlights a basic understanding of algebra and vectors, with an interest in further learning and exploration of mathematical concepts. Engaging with physics can reignite a passion for knowledge and understanding of fundamental principles.
Diax
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Greetings and Salutations!

Currently I am not enrolled in a physics class but I am researching on my own and I was wondering if there was an algebraic definition of work and what it was if it exists. The only definition of work I know of (cursory knowledge i.e. I saw the following words written) Work is the integral of displacement over an area. Unfortunately my I'm only at a very basic algebra level.

Thanks for your time!

Peace be with you!

Diax
 
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Work is equal to Fd, where F is the force being applied and d is the distance over which the force is applied.

P.S., Welcome to Physics Forums!
 
Thank you for the prompt response!

Also thank you for the welcome! I hope to learn lots of cool stuff while I'm here. I've been out of school for a couple years now, and I've found my thirst for knowledge reawakened after some time without the soul crushing weight of public school and garbage college instructors. It's refreshing.
 
Technically the rigorous definition of work involves a dot product between the force vector and the displacement vector. And if the force is variable in either magnitude, direction or both, then an integral is involved.

But if it's just a constant force moving an object along the same direction as the force acts, then Chester's simple formula suffices.
 
Curious3141 said:
Technically the rigorous definition of work involves a dot product between the force vector and the displacement vector. And if the force is variable in either magnitude, direction or both, then an integral is involved.

But if it's just a constant force moving an object along the same direction as the force acts, then Chester's simple formula suffices.

Indeed. Like I said very basic which is to say elementary algebra here. I've never worked with a vector and all I know about the dot product is that it multiplies to zero when the angle between the vectors is orthogonic... And I know that from my research when I was trying to prove that perpendicular lines have opposite reciporical slopes. Which I did btw using the Pythagorean Theorem... I think my next post is going to be that proof. It's pretty cool how it all works out.
 
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