A person trying to lose weight by burning fat lifts a mass of 10 kg up to a height of 1 m 1000

step 1 In this question it is given that a person trying to lose his weight by burning a fat for this h

A person trying to lose weight by burning fat lifts a mass...

A person trying to lose weight by burning fat lifts a mass of $10 \mathrm{~kg}$ up to a height of $1 \mathrm{~m} 1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \times 10^{7} \mathrm{~J}$ of energy per $\mathrm{kg}$, which is converted into mechanical energy with a $20 \%$ efficiency rate. Take $g=9.8 \mathrm{~ms}^{-2}$ : $\quad$ [2016] (A) $6.45 \times 10^{-3} \mathrm{~kg}$ (B) $9.89 \times 10^{-3} \mathrm{~kg}$ (C) $12.89 \times 10^{-3} \mathrm{~kg}$ (D) $2.45 \times 10^{-3} \mathrm{~kg}$

A person trying to lose weight by burning fat lifts a mass of $10 \mathrm{~kg}$ up to a height of $1 \mathrm{~m} 1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \times 10^{7} \mathrm{~J}$ of energy per $\mathrm{kg}$, which is converted into mechanical energy with a $20 \%$ efficiency rate. Take $g=9.8 \mathrm{~ms}^{-2}$ : $\quad$ [2016]
(A) $6.45 \times 10^{-3} \mathrm{~kg}$
(B) $9.89 \times 10^{-3} \mathrm{~kg}$
(C) $12.89 \times 10^{-3} \mathrm{~kg}$
(D) $2.45 \times 10^{-3} \mathrm{~kg}$

Key Concepts

Efficiency of Energy Conversion

Efficiency in this context refers to the percentage of chemical energy from fat that is successfully converted into mechanical work. Due to losses (typically as heat and through other inefficiencies), only a fraction of the original energy content is used, which means that more energy (and hence more fat) must be consumed to produce the required work.

Energy Density of Fat

This is a measure of the amount of energy stored in a given mass of fat, expressed in joules per kilogram. In problems where energy input is provided by burning fat, knowing the energy density allows for the conversion between the mechanical work done and the mass of fat burned, once the system’s conversion efficiency is taken into account.

Gravitational Potential Energy

This concept involves the energy an object possesses due to its height in a gravitational field. It is quantified using the formula PE = m × g × h, where m represents the mass of the object, g is the acceleration due to gravity, and h is the height. In problems like the one provided, potential energy is calculated for each lift to determine the work done against gravity.

Work Done in Lifting

Work is the process of transferring energy via force applied over a distance. When an object is lifted, the work done is equal to the increase in its gravitational potential energy. By summing the work done over many repetitions, one can calculate the total energy input into the system.

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