Calculate how many Kilo-calories of food (CAL) consumed by a 163-lb man due to the force he applies during one repetition of a bodyweight only (single leg squat) exercise that lifts 87% of his bodyweight through a controlled full-range of motion that measures 31" in both the down and up directions of the movement? Assume that the exercise takes place under comfortable conditions, and that only about 22% of all the potential energy stored in food is available for useful work, and that 1 ft-lb = 0.000324048 Kilocalories.

the man consumed 0.539598 Kilo-calories of food

Explanation:

Given the data in the question;

first we calculate the work required to raise the given weight at the given height;

ΔW = mgh = w'h = (0.87) × h

= ( 0.87 × 163) × (31/12)ft     { 1 foot = 12 inch}

= 366.34 lb-ft

Now since 22% of all the potential energy stored in food is available for useful work;

ΔW = 0.22 × ΔPE

ΔPE = ΔW/0.22

we substitute

ΔPE = 366.34 / 0.22

= 1665.18 lb-ft

Now given that; 1 ft-lb = 0.000324048 Kilocalories

= 1665.18 × 0.000324048

= 0.539598 Kilo-calories

Therefore the man consumed 0.539598 Kilo-calories of food

568.98

explanation:

its math

volts times amps 6 x 0.5 3w