__Question 6.22__:^{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__:**a)**Mass of the weight,

*m*= 10 kg

*h*= 0.5 m

Number of times the weight is lifted,

*n*= 1000

∴Work done against gravitational force:

= n(mgh)

= 1000 X 10 X 9.8 X 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) X 3.8 X 10

^{7}J

= (1/5) X 3.8 X 10

^{7}J

Equivalent mass of fat lost by the dieter:

= [ 1 / (1/5) X 3.8 X 10

^{7}]X 49 X 10

^{3}

= (245 / 3.8) X 10

^{-4}= 6.45 X 10

^{-3}kg

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