What is the physical significance of work?

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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