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 2 T~ 2 . 



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 



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 2 T~ 2 L- 8 or ML-'T- 2 , and the conversion factor mt~ l r 2 . 



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 __ dw^ The di mens i ona i formula is therefore WT" 1 or ML*T- 8 , and the con- 

 dt 



version factor w/ 2 /" 8 , or for problems in gravitation units more conveniently fit'*, 

 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 fl, where / is 453.59 and /is 

 30.48. 



Hence the number is 453.59 X 30.48 = 13825. 



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

 Here m = 1/453.59, /= 1/30.48, and t = i ; /. mI 2 r z = 1/453-59 X 3-48 2 , 

 and iomPr*= 107453.59 X 3Q-48 2 = 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 io 7 ergs per second, that is, io 7 dyne centi- 

 meters per second. The conversion factor is ^z/ 2 /' 8 , where * = 453-59, / = 30.48, 

 and /= i, and the result has to be divided by io 7 , the number of dyne centime' 

 ters per second in the watt. 



Hence, 177 iomf t t*/io l = 17710 X 453-59 X 3o.48 2 /io 7 =r 746.3. 



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

 per minute ? 



The conversion factor suitable for this case isflf\ where/is 453.59, /is 30.48, 

 and / is 60. 



Hence, 33000 //~ 1 = 33000 X 453-59 X 30.48/60 = 7 604 ooo 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. 



