Question

A person trying to lose weight by burning fat lifts a mass of $10\text{kg}$ upto a height of $1\text{m}$ for $1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up, considering work is done only when the weight is lifted up? Fat supplies $3.8\times {10}^7\text{J/kg}$ which is converted to mechanical energy at an efficiency of $20$% . Take $g=9.8{\text{ms}}^{-2}$. Consider weight to be lifted slowly.

• A. $2.45\times {10}^{-3}\text{kg}$
• B. $6.45\times {10}^{-3}\text{kg}$
• C. $9.89\times {10}^{-3}\text{kg}$
• D. $12.89\times {10}^{-3}\text{kg}$

Answer 1

The correct option is D $12.89\times {10}^{-3}\text{kg}$

Normal reaction force $N^{\prime }$ exerted by the lifter on the weight will be in upward direction. As the weight is lifted slowly, it will be in equilibrium in vertical or $y-$ direction i.e $a_y=0$

From FBD of weight:


$N^{\prime }=mg=10\times 9.8=98\text{N}$
Work done by the normal reaction in lifting the weight by $1\text{m}$ during $n=1000$ repetitions:
$W=n\times (\overrightarrow{F}.\overrightarrow{d})=nFd\cos 0^{\circ }=n{N}^{\prime }\times d\times 1=(1000\times 98\times 1)$
$\therefore W=98,000\text{J}$

Let $x\text{kg}$ fat be burnt during the process, hence energy produced ($E$) by buring of $x\text{kg}$ fat:
$E=x\text{kg}\times (3.8\times {10}^7)\text{J/kg}$
$E=3.8x\times {10}^7\text{J}$

Energy transformed into mechanical energy, which was expended by the lifter in the form of work done ($W$) is:
$\therefore W=\eta \times E...\left(i\right)$ where $\eta =20$%$=0.20$
Substituting the values in Eq. $\left(i\right)$, we get:
$98000=0.20\times 3.8x\times {10}^7$
$\Rightarrow x=\frac{98000\times 5}{3.8\times {10}^7}$

$\therefore x=\frac{490}{38}\times {10}^{-3}=12.89\times {10}^{-3}\text{kg}$