Question

1) In a cycling road race consisting of a flat section and an uphill section, a particular cyclist pedals with an average power of 324 Watts for 5.50 hours on the flat section of the course, and an average power of 418 Watts for 30.0 minutes while climbing the uphill section, which has an average grade of 6%. Potentially useful information: Power = Energy/time; 1 Watt = 1 Joule/sec; 1 kcal = 4186 J (Note, 1 kcal = 1 food Calorie)

part b)

Given that the human body has an efficiency of 25.0% at converting food energy into mechanical output, how many kcal of food energy must the cyclist consume to replace the mechanical energy expended during the race?

Answer 1

Energy consumption on the flat road = 324 * 5.50 * 3600 J = 6415.2 kJ = 1532.5 kcal

Energy consumption on the climbing road = 418 * 30 *60 J = 752.4 kJ = 179.7 kcal

Total energy required = 1712.2 kcal

because only 25% energy is converting to mechanical energy so total energy by food = 1712.2 kcal *100 / 25 = 6848.8 kcal