Question

a 85kg teacher (including the mass of his gear) hikes up a mountain! during his hike he experiences a 9n drag (and negligible friction impeding his motion). the distance of the path and the height of the mountain are shown.
list the energy types at the initial and final time and whether work and loss (due to non - conservative forces) occur as well as the corresponding amounts of energy. include the earth as part of the system.
initial:
none = 0 j
work = j
loss? = j
final:
gpe = j
how many kilocalories did he burn if his body does work with 26% efficiency?
hint: divide the work by the efficiency (in decimal form), then convert to kilocalories.
kilocalories
kilocalories are typically just called “calories”, and an average person eats approximately 2000 kilocalories per day.

Explanation:

Step1: Calculate gravitational potential energy (final GPE)

The final gravitational potential energy is calculated using $GPE = mgh$, where $m=85\ \text{kg}$, $g=9.8\ \text{m/s}^2$, $h=720\ \text{m}$.
$$GPE = 85 \times 9.8 \times 720 = 601920\ \text{J}$$

Step2: Calculate energy loss from drag

Energy loss is work done against drag: $W_{\text{loss}} = F \times d$, where $F=9\ \text{N}$, $d=6600\ \text{m}$.
$$W_{\text{loss}} = 9 \times 6600 = 59400\ \text{J}$$

Step3: Calculate total work done by the teacher

Total work is sum of GPE and loss.
$$W_{\text{work}} = 601920 + 59400 = 661320\ \text{J}$$

Step4: Convert work to kilocalories (1 kcal = 4184 J)

First find total energy burned: $E = \frac{W_{\text{work}}}{\text{efficiency}}$, efficiency = 0.26.
$$E = \frac{661320}{0.26} = 2543538.46\ \text{J}$$
Convert to kilocalories:
$$\text{Kilocalories} = \frac{2543538.46}{4184} \approx 608$$

Answer:

initial: none = 0 J
work? work = 661320 J
loss? loss = 59400 J
final: GPE = 601920 J
608 kilocalories