Question

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 $3.8\times {10}^{7}J$ of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Answer 1

Mass of the weight, m = 10 kg
Height to which the person lifts the weight, h = 0.5 m
Number of times the weight is lifted, n = 1000
(a) Work done against gravitational force:
$=n\left(mgh\right)=1000\times 10\times 9.8\times 0.5=49\times {10}^{3}J=49kJ$

(b) Energy equivalent of 1 kg of fat $=3.8\times {10}^{7}J$
Efficiency rate = 20%
Mechanical energy supplied by the person's body:
$=\frac{20}{100}\times 3.8\times {10}^{7}J=\frac{1}{5}\times 3.8\times {10}^{7}J$
Equivalent mass of fat lost by the dieter:
$=\frac{1}{\frac{1}{5}\times 3.8\times {10}^7}\times 49\times {10}^3=\frac{245}{3.8}\times {10}^{-4}=6.45\times {10}^{-3}kg$