The record time for a tour de france cyclist to ascend the famed 1100-m-high alpe d'huez was 37.5 min, set by marco pantani in 1997. pantani and his bike had a mass of 65 kg. assume the body works with 25 % efficiency.
a)how many calories did he expend during this climb?
b)what was his average metabolic power during the climb?

Marco Pantani set the ‘record’ when he averaged 6.63 W/kg during the 1997 Tour. So the amount of calorie marco pantani expended can be calculated by multiplying it by its mass and dividing by its efficiency

Calorie = (6.63 W/ kg) *(65 kg) *( 1 cal/s/ 4.19 W) *( 37.5 min)*( 60 s/ 1 min) / 0.25

Calorie = 925, 668 cal 

Pe=2682275.334J    ,51012j/s

Explanation:

The record time for a Tour de France cyclist to ascend the famed 1100-m-high Alpe d'Huez was 37.5 min, set by Marco Pantani in 1997. Pantani and his bike had a mass of 65 kg. Assume the body works with 25 % efficiency

Definition of terms

potential energy is energy stored due to position

power is rate of doing work

mass is the quantity of mater in a body,65kg

potential energy of the cyclist will be

Pe=m*g*h

m=65kg

g=9.81m/s^2

h=1100

the body works with 25% efficincy

.25*P=65*9.81*1100

Pe=701415/0.25

Pe=11222640J

recall

4184 Joules = 1000 calories

Pe=11222640/4.184

Pe=2682275.334 calories.  the cyclist will expend tat amount of calories during the climb

2.Power=energy/time

time= 37.5 *60=2250s

power=11222640J/2250s

power=51012j/s

His average metabolic power during the climb  is 51012j/s

need to see circuit to get time constant