Xxiv INTRODUCTION. 



23. Work and Energy. — When the point of application of a force, acting on 

 a body, moves in the direction of the force, work is done by the force, and the 

 amount is measured by the product of the force and displacement numbers. The 

 dimensional formula is therefore FL or ML^T~^. 



The work done by the force either produces a change in the velocity of the body 

 or a change of shape or configuration of the body, or both. In the first case it 

 produces a change of kinetic energy, in the second a change of potential energy. 

 The dimension formulae of energy and work, representing quantities of the same 

 kind, are identical, and the conversion factor for both is mlH~'^. 



24. Resilience. — This is the work done per unit volume of a body in distort- 

 ing it to the elastic limit or in producing rupture. The dimension formula is there- 

 fore ML^T-'^L-^ or ML-^T-^, and the conversion factor ml-H-^. 



25. Power, or Activity. — Power — or, as it is now very commonly called, ac- 

 tivity — is defined as the time rate of doing work, or if W represent work and P power 



P = — — . The dimensional formula is therefore WT~^ or ML^~^ and the con- 

 dt 



version factor mPt~^, or for problems in gravitation units more conveniently _/7/~\ 



where/ stands for the force factor. 



Examples, {a) Find the number of gram centimeters in one foot pound. 



Here the units of force are the attraction of the earth on the pound * and 

 the gram of matter, and the conversion factor is fi, where / is 453.59 and / is 

 30.48. 



Hence the number is 453.59 X 30.48 = 13825. 



(J)) Find the number of foot poundals in i 000 000 centimeter dynes. 

 Here m = i/453-59» ^= 1/30.48, and t=\; .: mPr^ = i/453-59 X 30-48^ 

 and io^mPr'= 107453.59 X 3o-48'= 2.373. 



(c) If gravity produces an acceleration of 32.2 feet per second per second, how 

 many watts are required to make one horse-power ? 



One horse-power is 550 foot pounds per second, or 550 X 32.2 = 17710 foot 

 poundals per second. One watt is 10'' ergs per second, that is, 10'^ dyne centi- 

 meters per second. The conversion factor is mPt^', where »« = 453-59, / = 30.48, 

 and /= I, and the result has to be divided by 10'', the number of dyne centime- 

 ters per second in the watt. 



Hence, ly-j 10 mPr^ 10' = 17710X 453-59 X 30.48710''= 746.3. 



(d) How many gram centimeters per second correspond to 33000 foot pounds 

 per minute ? 



The conversion factor suitable for this case isy?/'\ where/is 453.59, /is 30.48, 

 and / is 60. 



Hence, 33000//'^ =133000 X 453-59 X 30.48/60 = 7 604000 nearly. 



* It is important to remember that in problems like that here given the term " pound " or 

 " gram " refers to force and not to mass. 



