Question

A person trying to lose weight (dieter) lifts a 10 kg mass through 0.5 m, 100 times. Assunme that the ppotential energy lost each time she lowers the mass is dissplated.
(a) How much work does she do against teh gravity tional force?
(b) Fat supplies $4\times {10}^{7}J$ of energy per kilogram which is concerted to mechanical energy with 20% efficiency rate. How much fat will the dieter use up? (use g=10$m/s^2$)

Answer 1

Person lifts mass( M) = 10 kg

height ( h) = 0.5 m

no of times mass lifted( n) = 100

(a) work done against gravitational force = n × change in potential energy in each time

= n × Mgh

= 100 × 10 × 9.8 × 0.5 [ g = 9.8 m/s²]

= 4.9 × 10³ joule

(b) Fat supplies energy per kilogram = 3.8 × 10⁷ Joule

A/C to question ,

20% of Fat energy convert in mechanical energy

e.g mechanical energy /Kg = 20/100 × 3.8 × 10⁷ j/Kg

= 0.76 × 10⁷ j/Kg

now,

Fat used by dieter = work done by person/mechanical energy convert by it

= 4.9× 10³/0.76 × 10⁷