What is the physical significance of work?

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Discussion Overview

The discussion revolves around the concept of work in physics, specifically its definition, significance, and relationship to energy. Participants explore the mathematical formulation of work and its implications in various contexts, including dimensional analysis and unit conversions.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Homework-related

Main Points Raised

  • One participant questions the definition of work as a cross product of force and displacement, suggesting it is actually a dot product.
  • Another participant emphasizes the importance of work due to its fundamental relationship to energy, which aids in simplifying problems.
  • A participant shares a detailed cheat sheet outlining the definitions and units of force, work, energy, power, and pressure, indicating that understanding dimensional analysis is crucial for grasping these concepts.
  • The cheat sheet includes various unit conversions and relationships, such as 1 Joule being equivalent to 10^7 ergs and 1 BTU being equal to 1 Kilo-joule.

Areas of Agreement / Disagreement

There is no consensus on the definition of work, as one participant asserts it is a dot product while another initially states it is a cross product. The discussion includes multiple viewpoints on the significance of work and its relationship to energy, indicating that participants have differing perspectives on these concepts.

Contextual Notes

Participants reference various units and their relationships, but the discussion does not resolve the initial disagreement regarding the definition of work. The reliance on dimensional analysis and unit conversions suggests that understanding these concepts may depend on specific definitions and contexts.

johncena
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As the title suggests, what is work? Or, what is the physical significance of work?
My textbook define work as crossproduct of force and displacement.
But why do we need that quantity?
 
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It is the dot product of force and displacement, not the cross-product.

The reason that we need it is because of its fundamental relationship to energy which is very useful in simplifying many problems.
 


I made up this little cheat sheet that I have to refer to every time I get involved in working out work. Understanding (and remembering) the dimensional analysis (the basic MKS units of each measure, e.g., Meters per Second is speed) should help in getting the idea of each unit.

Code:
                 Getting Energy Straight

Force -- Newton -- Mass times Acceleration ( F = MA )
                    Killograms times Meters per Sec^2: (Kg x M) / S^2
                    1 Newton = 10^5 dynes
                    1 pound-force ~= 4.5 Newtons

Work --  Joule  -- Force times Distance ( W = FD )
 (aka Energy)       Newtons times Meters -- N x M:     (Kg x M^2) / S^2
                    1 Joule = 10^7 ergs
                              .74 foot-pounds
                              6.25x10^18 electron volts
                    1 BTU = 1 Kilo-joule

                    note:
                     Watt = volt x ampere
                     1 Columb -- amp-sec ~= 6.25 x 10^18 electron-second
                     Watt-seconds -- volt x coulmb
                     1 Joule = 1 Watt-second
                     1 KwHr = 3.6 Mega-joule

Power -- Watt   -- Work per Time ( P = W/S )
                    Joules per Second -- J/S:          (Kg x M^2) / S^3
                    1 HP = 550 ft-lb/s = 745.7 watts
                    1 Kw = 1.34 HP
                    1 BTU/hour = .29 watts

for extra credit:
Pressure -- Pascal -- Force per Area ( P = F/A )
                       Newtons per Meter^2 -- N/M^2:    Kg / (M x S^2)
		        1 pound/sqin (PSI) = 6.9 Kpascal
 


schip666! said:
I made up this little cheat sheet that I have to refer to every time I get involved in working out work. Understanding (and remembering) the dimensional analysis (the basic MKS units of each measure, e.g., Meters per Second is speed) should help in getting the idea of each unit.

Code:
                 Getting Energy Straight

Force -- Newton -- Mass times Acceleration ( F = MA )
                    Killograms times Meters per Sec^2: (Kg x M) / S^2
                    1 Newton = 10^5 dynes
                    1 pound-force ~= 4.5 Newtons

Work --  Joule  -- Force times Distance ( W = FD )
 (aka Energy)       Newtons times Meters -- N x M:     (Kg x M^2) / S^2
                    1 Joule = 10^7 ergs
                              .74 foot-pounds
                              6.25x10^18 electron volts
                    1 BTU = 1 Kilo-joule

                    note:
                     Watt = volt x ampere
                     1 Columb -- amp-sec ~= 6.25 x 10^18 electron-second
                     Watt-seconds -- volt x coulmb
                     1 Joule = 1 Watt-second
                     1 KwHr = 3.6 Mega-joule

Power -- Watt   -- Work per Time ( P = W/S )
                    Joules per Second -- J/S:          (Kg x M^2) / S^3
                    1 HP = 550 ft-lb/s = 745.7 watts
                    1 Kw = 1.34 HP
                    1 BTU/hour = .29 watts

for extra credit:
Pressure -- Pascal -- Force per Area ( P = F/A )
                       Newtons per Meter^2 -- N/M^2:    Kg / (M x S^2)
		        1 pound/sqin (PSI) = 6.9 Kpascal

Thanks, this might prove very helpful to me.

Ron
 

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