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?
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 × ( m g h )
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.
Given, the mass is
(a)
Let
Substitute the given values in the above expression.
(b)
Given, 1 kg of fat supplied
Let the mechanical energy supplied by the body be
Mass of the fat lost by the body is,
Hence the dieter will lose up
