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?

Asked by Pragya Singh | 1 year ago |  162

##### Solution :-

Mass, m = 10 kg

Height to which the mass is lifted, h = 0.5 m

Number of times, 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 107 x $$\dfrac{20}{100}$$

= 0.76 x107 J/kg

Therefore, fat used up by the dieter

=$$\dfrac{1}{(0.76 \times 107)}\times 49000$$

$$\dfrac{ 49000}{(0.76 \times 10^7)}$$

= 6.45 x 10-3 kg

Answered by Pragya Singh | 1 year ago

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