Question

5. A person trying to lose weight by burning fat lifts a mass of 10 kg upto 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 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms–2 :-​

Answer 1

Answer:

A person trying to lose weight by burning fat filts a mass of 10kg upto a being of 1m1000 time . Assume that the potential energy lost each time be lower the mass is dissipated . ... Far supplies 3.8×107J of energy per kg wich is canverted to mechanical energy with x20% efficiency rate Take =9.8ms-2.

The net work done by the man will be 1000 times the work done in lifting 10kg to a height of 1m

Net work done =1000×10×9.8×1 J

Let's assume xkg of fat is burnt in doing this work. Energy balance will give the following equation,

$x× </p><p>100</p><p>20</p><p> </p><p> ×3.8×10 </p><p>7</p><p> = Net work done</p><p>x=12.8947×10 </p><p>−3</p><p> kg </p><p>$

Explanation:

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