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 × 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

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Solution

Given, the mass is 10 kg , height lifted is 0.5 m , number of times mass lifted is 1000 times and energy supplied by fat is 3.8× 10 7 J/ kg .

(a)

Let W be the work done against gravity by the dieter, h be the height of lift and m be the mass lifted, then

W=1000×( mgh )

Substitute the given values in the above expression.

W=1000×( 10×9.8×0.5 ) =49 kJ

(b)

Given, 1 kg of fat supplied 3.8× 10 7 J of energy.

Let the mechanical energy supplied by the body be E m , then

E m = 20 100 ×energy by fat = 20 100 ×3.8× 10 7 =7.6× 10 6 J

Mass of the fat lost by the body is,

mass= W E m = 1 7.6× 10 6 ×49000 =6.45× 10 −3 kg

Hence the dieter will lose up 6.45× 10 −3 kg of mass.

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