Question

4. An 82 kg hiker climbs Mt. Humphrey near Flagstaff. During a two hour period, the hiker's

vertical elevation increases by 540 meters,

a. Calculate the climber's delta Eg

Answer 1

Answer:

$\Delta E = 434253.96\,J$, $\Delta P = 60.313\,W$

Explanation:

The change in energy is given by the change in gravitational potential energy:

$\Delta E = m\cdot g \cdot \Delta h$

$\Delta E = (82\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (540\,m)$

$\Delta E = 434253.96\,J$

The average rate of change in terms of time is approximately this:

$\Delta P = \frac{\Delta E}{\Delta t}$

$\Delta P = \frac{434253.96\,J}{(2\,h)\cdot (\frac{3600\,s}{1\,h} )}$

$\Delta P = 60.313\,W$