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Question
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 is 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.
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 is 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.
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Solution
The correct option is D 12.89×10−3 kg
The net work done by the man will be 1000 times the work done in lifting 10 kg to a height of 1 m.
Net work done =1000×10×9.8×1 J
=9.8×104 J
Let's assume x kg of fat is burnt in doing this work. Energy balance will give the following equation.
x×20100×3.8×107= Net work done =9.8×104 J
⇒x=12.8947×10−3 kg
The net work done by the man will be 1000 times the work done in lifting 10 kg to a height of 1 m.
Net work done =1000×10×9.8×1 J
=9.8×104 J
Let's assume x kg of fat is burnt in doing this work. Energy balance will give the following equation.
x×20100×3.8×107= Net work done =9.8×104 J
⇒x=12.8947×10−3 kg
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