Question

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)

(a) Calculate the average mechanical power output of the cyclist over the whole race.

(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

6.co solation (a) Total work done = 394x5.50"'+ 418 x 0.50 WTotal = 1991 watt hour Time Consumed = 5.50 +0.50 = Grou hour so Average Power out Put= 199 = 331,833 watt Pange = 1 331.833 watt. (b) Total energy consumed in Joule = 1991X3600 = 7167600 Joule of E Total = 7167600 K Calorie 4186 = 1719.279 k calorie. = 1.712779 k from calorie. so As efficiency af human body = 25% so total food Calorie to be contamed -4811712279 K food calorie. = 6.85 K food calorie 6.85 Ik food calories