m=10kg,h=1m,1000times
Work done against gravity mgh
=10×1×9.8=98J
In lifting 1000 times
=1000×98
=9.8×104J
Fat burn with 20% efficiency =3.8×107J×0.2
=7.6×106J per kg
m=7.6×1069.8×104=1.289×10−2
=12.89×10−3kg
JEE Main 2016 — Physics Mechanics
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1m 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=9.8ms−2 :
Held on 3 Apr 2016 · Verified 6 Jul 2026.
9.89×10−3kg
12.89×10−3kg
2.45×10−3kg
6.45×10−3kg
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