Question

A person trying to lose weight by burning fat lifts a mass of 10 kg up to a height of 1 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 $3.8\times {10}^7$ J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take $g=9.8m{s}^{-2}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let x kg of fat is burned then

$x\times 3.8\times {10}^7\times \frac{20}{100}=10\times 9.8\times 1000$

$\Rightarrow x=12.89\times {10}^{–3}kg$