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Work, Power and Energy

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Question : 22 of 30
Marks: +1, -0
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.
(a) How much work does she do against the gravitationalforce?
(b) Fat supplies 3.8×1073.8 \times 10^7J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Solution:  
Here, m = 10 kg, h = 0.5 m, n = 1000
(a) Work done against gravitational force
W=n(mgh)W = n(mgh) =1000×(10×9.8×0.5)=49000= 1000 \times (10 \times 9.8 \times 0.5) = 49000 J
(b) Mechanical energy supplied by 1 kg of fat
=3.8×107×20100=0.76×107J/kg=3.8 \times 10^{7} \times \frac{20}{100} = 0.76 \times 10^{7} \text{J}/\text{kg}
∴ Fat is used up by the dieter =1kg0.76×107×49000= \frac{1 \text{kg}}{0.76 \times 10^{7}} \times 49000 =6.45×10−3kg=6.45 \times 10^{-3} \text{kg}
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