CBSE
Gujarat Board
Haryana Board
Class 10
Class 12
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 3.8×10^{7} J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g=9.8 ms^{−2}:
2.45 ×10^{−3} kg
6.45 x×10^{−3} kg
9.89 ×10^{−3} kg
12.89 ×10^{−3} kg
D.
12.89 ×10^{−3} kg
Given potential energy burnt by lifting weight
= mgh = 10 x 9.8 x 1 x 1000 = 9.8 x 10^{4} J
If mass lost by a person be m, then energy dissipated
= m x 2 x 38 x 10^{7} J /10
⇒ m = 5 x 10^{-3} x 9.8 / 3.8
= 12.89 x 10^{-3} kg
Given potential energy burnt by lifting weight
= mgh = 10 x 9.8 x 1 x 1000 = 9.8 x 10^{4} J
If mass lost by a person be m, then energy dissipated
= m x 2 x 38 x 10^{7} J /10
⇒ m = 5 x 10^{-3} x 9.8 / 3.8
= 12.89 x 10^{-3} kg
The different units of energy are :
(i) Joule
(ii) Erg
(iii) eV
(iv) KWh
(v) Calorie.