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 $3.8 \times 10^7 \, J$ of energy per kg which is converted to mechanical energy with a $20\%$ efficiency rate. Take $g = 9.8 \, ms^{-2}$ :

• A. $2.45 \times 10^{-3} \, kg $
• B. $6.45 \times 10^{-3} \, kg $
• C. $9.89 \times 10^{-3} \, kg $
• D. $12.89 \times 10^{-3} \, kg $

Answer 1

The Correct Option is D

To determine how much fat the person uses up while lifting a weight, we need to calculate the total work done in lifting the weight and then relate this to the amount of fat burned.

1. Calculate Work Done in Each Lift:
   The work done (\(W\)) in lifting a mass is given by the formula: \(W = m \cdot g \cdot h\), where:
   • \(m = 10\, kg\) (mass of the object)
   • \(g = 9.8 \, ms^{-2}\) (acceleration due to gravity)
   • \(h = 1\, m\) (height)
2. Total Work Done for 1000 Lifts:
   Since the weight is lifted 1000 times, the total work done is: \(W_{\text{total}} = 1000 \times 98 = 98000 \, J\)
3. Account for Efficiency:
   The mechanical energy is only 20% efficient, so the energy required from fat is given by: \(E_{\text{required}} = \frac{W_{\text{total}}}{\text{Efficiency}}\)
   \(E_{\text{required}} = \frac{98000}{0.20} = 490000 \, J\)
4. Calculate Fat Burned:
   Since fat supplies \(3.8 \times 10^7 \, J\) per kg, the amount of fat needed is: \(m_{\text{fat}} = \frac{E_{\text{required}}}{\text{Energy per kg of fat}}\)
   \(m_{\text{fat}} = \frac{490000}{3.8 \times 10^7} \approx 0.01289 \, kg\)

Thus, the amount of fat the person uses up is \(12.89 \times 10^{-3} \, kg\), which matches the correct answer: \(12.89 \times 10^{-3} \, kg\).