Work Energy

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 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms-2

  1. 6.45 × 10-3 kg
  2. 9.89 × 10-3 kg
  3. 12.89 × 10-3 kg
  4. 2.45 × 10-3 kg

Solution

Let fat used by the person is a kg.

Energy supplied by 1 kg fat = 3.8 × 107 

Energy supplied by a kg fat with 20% efficiency = 20% × a × 3.8 × 107 

Work done in lifting up = 1000 × mgh = 1000 × 10 × 9.8 × 1

20% × a × 3.8 × 107 = 9.8 × 104

a = 12.89 × 10-3

The correct option is C.