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 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

Answers (2)

As we discussed in

If only conservative forces act on a system, total mechnical energy remains constant -

K+U=E\left ( constant \right )

\Delta K+\Delta U=0

\Delta K=-\Delta U

-

 

 Total work done by the person in lifting the weigh = mgh

    =10 \times \9.8 \times \1 \times \1000

    =\98 \times \10^{3}J

Total mechanical energy produced by burning 1 kg fat  =(3.8 \times10^{7})\times 0.20 = 7.6 \times 10^{6}J 

Total fat burn =\frac{98 \times10^{3}}{7.6 \times 10^{6}}kg=12.89 \times 10^{-3} kg

 


Option 1)

2.45×10−3 kg

This is an incorrect option.

Option 2)

 6.45×10−3 kg

This is an incorrect option.

Option 3)

 9.89×10−3 kg

This is an incorrect option.

Option 4)

12.89×10−3 kg

 

This is the correct option.

N neha

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