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

(a) Total work done = $mgh\times \text{no. of times}=10\times 9.8\times 0.5\times 1000=49000J$

(b) Energy equivalent of 1 kg of fat $=3.8\times {10}^{7}J$

Efficiency rate = 20%

Mechanical energy supplied by the persons body:

= $\left(20/100\right)\times 3.8\times {10}^{7}J$

= $\left(1/5\right)\times 3.8\times {10}^7$ J

Equivalent mass of fat lost by the dieter:

$=\frac{1}{\frac{1}{5}\times 3.8\times {10}^7}\times 49\times {10}^3$

$=\left(245/3.8\right)\times 10$ ${}^{-4}=6.45\times {10}^{-3}kg$