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?

(a) 332 W

First of all, we need to calculate the energy spent in each part of the race.

The duration of the first part is

And the average power in this part is

So the energy spent is

Similarly, the duration of the second part is

And the average power in this part is

So the energy spent is

So the total energy consumed is

And the total time taken is

Therefore, the average power output of the cyclist during the whole race is

(b) 6870 kcal

We know that the output energy needed by the cyclist to finish the race is

However, the human body converts only 25.0% of the input energy into output energy. So we can write

From which we find the food energy that the cyclist must consume:

And keeping in mind the conversion between joules and calories:

1 cal = 4.18 J

We find:

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle

your answer is, orbit around the nucleus