In the following scenario: Sloopy bikes to school every morning. Assuming she weighs 100kg (220lb) and pedals at a constant rate of 150 Watts. What is her VO2 on the ride in?It takes Sloopy (as referenced previously) 30 minutes to bike to school. About how many calories does she "burn" on the trip?

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First, restate the problem in my own words to ensure clarity: Sloopy weighs 100 kg and pedals at a constant mechanical power output of 150 watts for 30 minutes during a bike ride to school. We’re asked two things: (a) her VO2 on the ride in, and (b) about how many calories she burns on the trip. The answer choices for calories are 353 kcal, 414 kcal, 71 kcal, and 118 kcal.

Option 1: 353 kcals. This would imply a much higher energy expenditure than what a 150 W mechanical work rate for 0.5 hours typically yields when converting mechanical energy to metabolic calories with standard efficiency assumptions. If you take the basic energy calculation of work done (power × time): 150 W × 1800 s = 270,000 J ≈ 64.5 kcal of mechanical energy. Metabolic energy would normally be higher due to an efficiency factor (human leg muscles are not 100% efficient). However, even with a generous efficiency perspective, surpassing roughly 300 kcal would require either a substantially higher power output or a much longer duration. Therefore, 353 kcal seems too large for these given values.

Option 2: 414 kcals. This is even larger than 353 kcal and would imply an even greater metabolic energy expenditure than described by the simple energy-work calculation. As with the previous option, the mechanical work is 64.5 kcal; achieving ~414 kcal would necessitate a large, unlikely efficiency scenario or misconstruing units. This makes 414 kcal an unlikely match for the stated data.

Option 3: 71 kcals. This figure is very close to the straightforward calculation of mechanical energy converted to kcal if we treat 150 W sustained for 30 minutes as yielding energy on the order of 150 W × 0.5 h = 75 Wh, and convert to kcal using 1 kWh ≈ 860 kcal. 75 Wh is 0.075 kWh; 0.075 × 860 ≈ 64.5 kcal. Rounded reasonably, 71 kcal is a plausible estimate for the calories burned corresponding to the mechanical work done in this simple frame, especially when minor rounding and metabolic efficiency considerations are included. It aligns with the basic energy-in-energy-out notion for this setup.

Option 4: 118 kcals. This value sits between the rough mechanical-energy estimate (around 64–65 kcal) and the much larger figures in options 1 and 2. To justify 118 kcal, one would need to assume a higher metabolic cost than the straightforward conversion allows, such as a substantial efficiency loss or additional caloric expenditure beyond the work output. With a pure 150 W for 30 minutes scenario, 118 kcal would be a stretch beyond typical efficiency-based expectations, making this option less consistent with the given data.

Putting it together, we evaluate how each choice aligns with the core calculation: energy from work equals power times time, which is 150 W × 1800 s ≈ 270,000 J ≈ 64.5 kcal. Since calories are usually reported to the nearest whole number and rounded in educational problems, a value in the low 60s to low 70s range is most consistent. Options 353 kcal and 414 kcal are clearly far too high for the stated mechanical power and duration. Option 118 kcal is also somewhat high relative to the direct calculation, though closer than the very large values. Option 71 kcal sits closest to the direct energy-conversion estimate and is therefore the most consistent choice given the data.

In the following scenario: Sloopy bikes to school every morning. Assuming she weighs 100kg (220lb) and pedals at a constant rate of 150 Watts. What is her VO2 on the ride in?It takes Sloopy (as referenced previously) 30 minutes to bike to school. About how many calories does she "burn" on the trip?

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