A person trying to lose weight (dieter) lifts a 10kg mass, one th… - Vidyadip

Question

A person trying to lose weight (dieter) lifts a $10kg$ mass, one thousand times, to a height of $0.5m$ each time. Assume that the potential energy lost each time she lowers the mass is dissipated.

How much work does she do against the gravitational force?
Fat supplies $3.8 \times 107J$ of energy per kilogram which is converted to mechanical energy with a $20\%$ efficiency rate. How much fat will the dieter use up?

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Answer

Mass of the weight, m = 10kg Height to which the person lifts the weight, h = 0.5m Number of times the weight is lifted, n = 1000 Work done against gravitational force, = n(mgh) = 1000 × 10 × 9.8 × 0.5 = 49kJ Energy equivalent of 1kg of fat = $3.8 \times 10^7J$ Efficiency rate = 20% Mechanical energy supplied by the person’s body, $=\frac{20}{100} × 3.8 × 107\text{J}$ $=\frac{1}{5} × 3.8 × 107\text{J}$ Equivalent mass of fat lost by the dieter, $=\Bigg[\frac{1}{\frac{1}{5}}× 3.8 × 10^7\Bigg]× 49 × 10^3$ $=\frac{245}{3.8}×10-4 = 6.45 × 10^{-3}\text{kg}$

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