#### 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 x 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?

#### Solution 1

**Mass of the weight,**

*m*= 10 kg

Height to which the person lifts the weight, *h* = 0.5 m

Number of times the weight is lifted,* n* = 1000

∴Work done against gravitational force:

= n(*mgh*)

= 1000 × 10 × 9.8 × 0.5 = 49 kJ

(b)** **Energy equivalent of 1 kg of fat = 3.8 × 10^{7} J

Efficiency rate = 20%

Mechanical energy supplied by the person’s body: = (20/100) × 3.8 × 10^{7} J

= (1/5) × 3.8 × 10^{7} J

Equivalent mass of fat lost by the dieter:

= [ 1 / (1/5) × 3.8 × 10^{7} ]× 49 × 10^{3}

= (245 / 3.8) × 10^{-4} = 6.45 × 10^{-3} kg

#### Solution 2

Here, m = 10 kg, h = 0.5 m, n = 1000

(a) work done against gravitational force.

W = n(mgh) = 1000 x (10 x 9.8 x 0.5) = 49000J.

(b) Mechanical energy supplied by 1 kg of fat = 3.8 x 10^{7} x20/100 = 0.76 x10^{7} J/kg

∴ Fat used up by the dieter =1kg/(0.76 x 10^{7}) x 49000 = 6.45 x 10^{-3} kg