Work and weight loss: teaching nutrition students the work equivalent of energy stored in a pound of fat (adipose).

Few nutrition students or university faculty can define the kilocalorie in terms of work (ft‐lbs or nt‐m). Much physical work is done against gravity, such as in stair‐climbing, walking uphill, and weight‐lifting. Therefore, an equation was developed that calculates energy expended during physical work against gravity based on mass lifted (M, body mass plus clothes), the acceleration of gravity (g), vertical distance (h), a value for muscular efficiency (~20–30%), the number of repetitions carried out (r), and the caloric equivalent of work done against gravity (SI units): Work = M ∗ g ∗ h ∗ r / ( 4186 ∗ muscular_efficiency ) .Near sea level, weight in pounds or kg*9.8 can be substituted for M*g. For work done near sea level, 1 kcal is equivalent to 3088 ft‐lbs (in SI units, 4186 nt‐m (J)). At a muscular efficiency of 20%, the energy stored in 1 lb of adipose (3500 kcal) is equivalent to ~2.16 x 10^6 ft‐lb (~2.93 x 10^6 nt‐m) of physical work. A person who wished to lose 1 lb of weight per week could achieve this aim by modest caloric restriction (100–200 kcal/d) combined with half an hour of work per day at ~6 METs. This amount of stair climbing (including the energy cost of descending stairs) provides a daily energy expenditure of about 3 x 10^5 ft‐lb (4.18 x 10^5 nt‐m). The equation provides very good agreement with work measured during stair climbing when energy efficiency is about 20%. Nutrition students should be taught that 1 kcal can be defined as heat or as the equivalent of 3088 ft‐lbs (or 4186 nt‐m) of work against gravity, with an adjustment for muscular efficiency. One kcal is expended when ~620 ft‐lb (840 nt‐m) of work is done. (GA Hatch project GEO00922).