# What is the rate of energy expenditure at VO2 max of 10 mets for a 50 kg subject?

## Answers (1)

1

Here's the math:

1 MET = 3.5 ml O2/kg/min, therefore 10 METs = 35 ml/kg/min

(35 ml/kg/min ) x (50 kg) = 1,750 ml O2/min = 1.75 liters/min

A VO2 of 1 l/min yields an energy expenditure of ~5 kcal/min (it varies

depending on the relative % contribution of glucose and fatty acids to ATP production.)

At a VO2 of (1.75 liters O2/min) x (~5 kcal/l) = ~8.75 kcal/min

Keep in mind that if it's at true VO2 max. this expenditure of 8.75 kcal/min will be sustainable for only a very short time because of fatigue.

Take care.

1 MET = 3.5 ml O2/kg/min, therefore 10 METs = 35 ml/kg/min

(35 ml/kg/min ) x (50 kg) = 1,750 ml O2/min = 1.75 liters/min

A VO2 of 1 l/min yields an energy expenditure of ~5 kcal/min (it varies

depending on the relative % contribution of glucose and fatty acids to ATP production.)

At a VO2 of (1.75 liters O2/min) x (~5 kcal/l) = ~8.75 kcal/min

Keep in mind that if it's at true VO2 max. this expenditure of 8.75 kcal/min will be sustainable for only a very short time because of fatigue.

Take care.