Question

solve the following

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 x 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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Answer 1

Answer:

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 will be

$1000 \times 10 \times 9.8 \times 1J$

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

Net work done :

$X \times \frac{20}{100} \times 3.8 \times 10 {}^{7}$

$X = 12.89 \times 10 {}^{ - 3}$

Or refer the attachement for better understanding

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Answer 2

The answer is

$12.89 \times {10}^{ - 3} \: kg$

Refer the attachment