What is the physical significance of work?

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SUMMARY

The discussion clarifies the physical significance of work, defined as the dot product of force and displacement, contrary to the initial misconception of it being a cross product. Work is fundamentally linked to energy, represented in Joules, where 1 Joule equals Newtons times meters. The conversation emphasizes the importance of dimensional analysis in understanding work, energy, and power, detailing key relationships and conversions between units such as Newtons, Joules, Watts, and Pascals.

PREREQUISITES
  • Understanding of basic physics concepts including force, work, and energy
  • Familiarity with dimensional analysis and MKS (Meter-Kilogram-Second) units
  • Knowledge of fundamental equations such as F = MA and W = FD
  • Basic understanding of unit conversions between Joules, Watts, and other energy units
NEXT STEPS
  • Research the relationship between work and energy in thermodynamics
  • Explore advanced applications of dimensional analysis in physics problems
  • Learn about the implications of power calculations in engineering contexts
  • Study the concept of pressure and its relation to force and area in fluid mechanics
USEFUL FOR

Students of physics, educators teaching mechanics, engineers involved in energy calculations, and anyone interested in the foundational principles of work and energy in physical systems.

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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