Work Energy Power

JEE Main level, single correct, medium

124 of 183

Question 124 of 183

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 3.8×1073.8\times {10}^{7} J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g=9.8ms−2g=9.8m{s}^{-2} :

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Answer: (B) 12.89×10−3kg12.89\times {10}^{-3}kg

m=10kg,h=1m,1000timesm=10kg,h=1m,1000times

Work done against gravity mgh

=10×1×9.8=98J=10\times 1\times 9.8=98 J

In lifting 1000 times

=1000×98=1000\times 98

=9.8×104J=9.8\times {10}^{4}J

Fat burn with 20% efficiency =3.8×107J×0.2=3.8\times {10}^{7}J\times 0.2

=7.6×106J=7.6\times {10}^{6}J per kg

m=9.8×1047.6×106=1.289×10−2m=\dfrac{9.8\times {10}^{4}}{7.6\times {10}^{6}}=1.289\times {10}^{-2}

=12.89×10−3kg=12.89\times {10}^{-3}kg

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