Answer: 0.57Kcal
Explanation:
Given m= 55kg, g = 9.8m/s²
The Workdone is given by:
W = mg∆h
Where ∆h is the total distance covered
∆h = 55 ×0.4m = 22m
W = 55 × 9.8 × 22
W = 11,858 J
Since 1Kilocalorie = 4,186.80Joules
W = (11,858/4186.80) Kcal
W = 2.8322Kcal
But since her efficiency is 20%, (this means the conversion ratio between her chemical energy to mechanical ; potential energy), the
The efficient Energy used for the exercise is
Ew = 0.20 × 2.8322
Ew = 0.566 ~= 0.57Kcal