A bicycle rider in the Tour de France eats a lot. If his total daily food intake were burned it wouldliberate about 8000 kcal of heat. Over the three or four weeks of the race his weight change isnegligible, less than 1%. Thus his energy input and output must balance Let’s first look at the mechanical work done by the racer. A bicycle is incredibly efficient. Theenergy lost to internal friction, even including the tires, is negligible. The expenditure against airdrag is however significant, amounting to 10 MJ per day. Each day the rider races for 6 hours.a. Compare the 8000 kcal input to the 10 MJ of work done. Something’s missing! Could the missingenergy be accounted for by the altitude change in a hard day’s racing?Regardless of how you answered (a), next suppose that on one particular day of racing there’sno net altitude change, so that we must look elsewhere to see where the missing energy went. Wehave so far neglected another part of the energy equation: the rider gives off heat. Some of thisis radiated. Some goes to warm up the air he breathes in. But by far the greatest share goessomewhere else.The rider drinks a lot of water. He doesn’t need this water for his metabolism—he is actuallycreating water when he burns food. Instead, nearly of all that liquid water leaves his body as watervapor. The thermal energy needed to vaporize water appeared in Problem 1.6 above.b. How much water would the rider have to drink in order for the energy budget to balance? Is thisreasonable?Next let’s go back to the 10 MJ of mechanical work done by the rider each day.c. The wind drag for a situation like this is a backward force of magnitude f = Bv2, were B issome constant. We measure B in a wind-tunnel to be 1.5 kg/m. If we simplify by supposing a day’sracing to be at constant speed, what is that speed? Is your answer reasonable?
A bicycle rider in the Tour de France eats a lot. If his total daily foodintake were burned it wouldliberate about 8000 kcal of heat. Over the three or four weeks of therace his weight change…

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