A person trying to lose weight by burning at 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 :

Question

A person trying to lose weight by burning at 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 : (A) 2.45 × 10-3 kg  (B) 6.45 × 10-3 kg  (C) 9.89 × 10-3 kg  (D) 12.89 × 10-3 kg

Tardigrade Admin · Accepted Answer

0.2 ×3.8 × 107 × m = 10× g ×1× 1000 m = (10 × 9.8 × 1000/0.2× 3.8× 107) = 1.289 × 10-2 kg = 12.89 × 10-3 kg

$2.45 \times 1 0^{- 3} k g$

$6.45 \times 1 0^{- 3} k g$

$9.89 \times 1 0^{- 3} k g$

$12.89 \times 1 0^{- 3} k g$

$0.2 \times 3.8 \times 1 0^{7} \times m = 10 \times g \times 1 \times 1000$
$m = \frac{10 \times 9.8 \times 1000}{0.2 \times 3.8 \times 1 0 ^{7}} = 1.289 \times 1 0^{- 2} k g = 12.89 \times 1 0^{- 3} k g$

2.45×10−3kg

6.45×10−3kg

9.89×10−3kg

12.89×10−3kg

0.2×3.8×107×m=10×g×1×1000 m=0.2×3.8×10710×9.8×1000​=1.289×10−2kg=12.89×10−3kg

$2.45 \times 1 0^{- 3} k g$

$6.45 \times 1 0^{- 3} k g$

$9.89 \times 1 0^{- 3} k g$

$12.89 \times 1 0^{- 3} k g$

$0.2 \times 3.8 \times 1 0^{7} \times m = 10 \times g \times 1 \times 1000$
$m = \frac{10 \times 9.8 \times 1000}{0.2 \times 3.8 \times 1 0 ^{7}} = 1.289 \times 1 0^{- 2} k g = 12.89 \times 1 0^{- 3} k g$