A person trying to lose weight (dieter ) lifts a 10 kg mass , one thousand times , to a height of 0.5 m each time .Assume that the potential energy lost each time she lowers the mass is dissipated . (a) How much work does she do against the gravitational force ? (b) Fat supplies `3.8 xx10^(7)` J of energy per kilogram which is converted to mechanical energy with a 20 % efficiency rate . How much fat will the dieter use up ?

Explanation

(a) Mass of the weight , m = 10 kg
Height to which the person lifts the weight , h = 0.05 m
Number of times the weight is lifted , n = 1000
` :. ` Work done against gravitational force ,
= n (mgh)
` = 1000 xx10 xx 9.8 xx 0.5`
` = 49 xx10^(3) J = 49000 J `
(b) Energy equivalent of 1 kg of fat = `3.8 xx10^(7) J `
Efficiency rate = 20 %
Mechanical energy supplied by the person.s body
` 20/100 xx3.8 xx10^(7) J `
` = 1/5 xx3.8 xx10^(7) J `
Equivalent mass of fat lost by the dieter ,
` m = 1/(1/5 xx3.8 xx10^(7)) xx49 xx10^(3)`
` m = 245/(3.8) xx10^(-4) = 0.0064 kg `
= `6.4 ` g