Question

The lowest point in Death Valley is $85m$below sea level. The summit of nearby Mt. Whitney has an elevation of $4420m$. What is the change in potential energy when an energetic $65kg$hiker makes it from the floor of Death Valley to the top of Mt. Whitney?

Short answer

The hiker's change in potential energy, rounded to two major digits, is$2.9\times {10}^6\mathrm{J}$.

Step-by-step solution

Introduction 

Potential energy is the energy possessed by a body as a result of its position.

$U=mgy$

Here, $m$is mass of the body, $g$is acceleration due to gravity, and $y$is height from the base.

Explanation

The change in potential energy throughout the journey from the bottom of Death Valley to the summit of Mt. Whitney is expressed as,

$\Delta U=mg\left(y_{\mathrm{f}}-y_{\mathrm{i}}\right)$

Here, $m$is mass of the hiker,$g$is acceleration due to gravity, $y_i$is height of the floor of Death Valley from the base, and $y_{\mathrm{f}}$is the height of the top of Mt. Whitney from the base.

Explanation

Because it is below sea level, the height of Death Valley's floor is deemed negative.

Substitute $65\mathrm{kg}$for $m,9.8\mathrm{m}/{\mathrm{s}}^2$for $g,-85\mathrm{m}$for$y_{\mathrm{i}}$, and $4420\mathrm{m}$for $y_{\mathrm{f}}$in the above equation $\Delta U=mg\left(y_{\mathrm{f}}-y_{\mathrm{i}}\right)$.

$\Delta U=(65\mathrm{kg})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(4420\mathrm{m}-(-85\mathrm{m}\left)\right)$

$=(65\mathrm{kg})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(4420\mathrm{m}+85\mathrm{m})$

$=2.87\times {10}^6\mathrm{J}$

The hiker's change in potential energy, rounded to two major digits, is$2.9\times {10}^6\mathrm{J}$.