75 kg 165 lb 555 ft Washington monument work kJ 25% energy

75 kg 165 lb 555 ft Washington monument work kJ 25% energy

 

 

How much work is done when a person weighing 75 kg (165 lb)

climbs the Washington monument. 555 ft high?

How many kilojoules must be supplied to do this muscular work,

assuming that 25% of the energy produced by the oxidation of food

in the body can be converted into muscular mechanical work.

 

 

 

1 ft = 0.3048 m 이므로,

( 참고 https://ywpop.tistory.com/20863 )

 

555 ft × (0.3048 m / 1 ft) = 169 m

 

 

 

W = FS = mgS = mgh

( 참고 https://ywpop.tistory.com/5281 )

 

= (75 kg) (9.8 m/s2) (169 m)

= (75) (9.8) (169)

= 124215 kg•m2/s2

 

 

 

1 J = 1 kg•m2/s2 이므로,

( 참고 https://ywpop.tistory.com/11834 )

 

= 124215 J

= 124 kJ

---> How much work is done

 

 

 

124 kJ / (25/100)

= 496 kJ

---> How many kilojoules must be supplied

 

 

 

답: 124 kJ, 496 kJ

 

 

 

 

[키워드] 워싱턴 기념탑 기준, Washington monument dic, 75 kg 165 lb 555 ft kJ 25%, 물리적 일 기준