Chapter 11.1, Problem 13SSC

To determine

The potential energy of the system at the top and at the bottom.

To sketch: The bar graphs for both situations.

Answer to Problem 13SSC

The potential energy of the system at the top is $PE=3.97\times {10}^4\text{ J}$ .

The potential energy of the system at the bottom is $PE=-3.53\times {10}^4\text{ J}$ .

Bar graph for both situations is shown in Figure 1.

Explanation of Solution

Given:

The mass of the rock climber is $m=90.0\text{ kg}$ .

Formula used:

The expression for the gravitational potential energy is,

$PE=mgh$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $h$ is the distance from the reference level.

Calculation:

Consider the value of acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^2$ .

The potential energy of the system at the top is,

$\begin{array}{l}PE=mg\left(h_{\text{f}}-h_{\text{i}}\right)\\ PE=\left(90.0\text{ kg}\right)\times \left(9.8\text{m}/{\text{s}}^2\right)\times \left(45.0\text{ m}-0.0\text{ m}\right)\\ PE=3.97\times {10}^4\text{ J}\end{array}$

The potential energy of the system at the bottom is,

$\begin{array}{l}PE=mg\left(h_{\text{f}}-h_{\text{i}}\right)\\ PE=\left(90.0\text{ kg}\right)\times \left(9.8\text{m}/{\text{s}}^2\right)\times \left(45.0\text{ m}-85.0\text{ m}\right)\\ PE=-3.53\times {10}^4\text{ J}\end{array}$

Sketch the bar graphs for both the situations is shown below.

Figure 1

Conclusion:

Thus, the potential energy of the system at the top is $PE=3.97\times {10}^4\text{ J}$ .

Thus, the potential energy of the system at the bottom is $PE=-3.53\times {10}^4\text{ J}$ .