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Question

A person trying to lose weight by burning fat lifts a mass of 10kg upto a height of 1m 1000times. 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 3.8×107J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g=9.8ms2.

A
2.45×103kg
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B
6.45×103kg
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C
9.89×103kg
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D
12.89×103kg
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Solution

The correct option is D 12.89×103kg
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×20100×3.8×107= Net work done
x=12.8947×103 kg

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