Question

A person trying to loose weight by burning fat lifts a mass of $15kg$ , five hundred times, upto a height of $2m$ . Assume that the potential energy lost each time he lowers the mass is dissipated. Fat supplies $2.8\times 10^{7}J$ of energy per $kg$ which is converted to mechanical energy with a $15\%$ efficiency rate. Consider the work is done only when the weight is lifted up. The fat he will use up is_____ $\times 10^{- 3}kg$ . Give your anwer in nearest integer. (Take $g=10ms^{- 2}$ )

Answer 1

Total work done in lifting mass
$W=\left(\right.15\times 10\times 2\left.\right)\times 500=1.5\times \left(10\right)^{5}$
If $m$ is mass of fat burnt then, energy used
$E=m\times 2.8\times 10^{7}\times \frac{15}{100}=4.2\times 10^{6}m$
Equating the two, we get
$\therefore m=\frac{1 . 5 \times 10^{5}}{4 . 2 \times 10^{6}}=35.7\times 10^{- 3}kg$