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

Explanation

Given `m=10kg,g=9.8ms^(-2),h=0.5m,n=1000`
Energy supplied by fat is `3.8xx10^(7)J//kg`, efficiency = 20%
`therefore20%` of energy supplied by fat = `3.8xx10^(7)xx0.20=0.76xx10^(-7)Jkg^(-1)`
Work done against the gravity = n mgh = `10^(3)xx10xx9.8xx0.5=4.9xx10^(4)J`
Mass of fat used = `(4.9xx10^(4))/(0.76xx10^(7))=6.45xx10^(-3)kg`