Question

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 gravitational force ?
(b) Fat supplies 3.8 × 10\(^7\)J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Answer 1

Given Data

Parameter	Value
Mass (\(m\))	10 kg
Lifts	1000 times
Height (\(h\))	0.5 m
\(g\)	9.8 m/s²
Fat energy	\(3.8 \times 10^7\) J/kg
Efficiency	20% = 0.20

(a) Work Against Gravity

Work done = total gravitational potential energy gained:

$$W = N \cdot mgh$$ $$W = 1000 \times 10 \times 9.8 \times 0.5 = 49{,}000 \, \text{J} = 4.9 \times 10^4 \, \text{J}$$

Work done = 49 kJ

(b) Fat Consumed

Total mechanical energy required = work done / efficiency:

$$E_\text{mech} = \frac{W}{\eta} = \frac{4.9 \times 10^4}{0.20} = 2.45 \times 10^5 \, \text{J}$$

Fat mass needed:

$$m_\text{fat} = \frac{E_\text{mech}}{E_\text{fat/kg}} = \frac{2.45 \times 10^5}{3.8 \times 10^7} = 6.45 \times 10^{-3} \, \text{kg}$$ $$m_\text{fat} = 6.45 \, \text{g} \approx 6.5 \, \text{g}$$

Fat used = 6.5 grams

Key Insight

Each lift gains \(mgh = 49\) J potential energy, but lowering dissipates it as heat. Net metabolic cost is \(W/\eta = 245\) J per lift × 1000 = 245 kJ total, equivalent to ~6.5 g fat.