Chapter 11.1, Problem 9PP

To determine

The changes in the energy of box-Earth system and change in total energy of the system.

Answer to Problem 9PP

The change in the box-Earth system’s energy to lift the box to the table is $W_1=1.1\times {10}^2\text{ J}$ .

The change in the box-Earth system’s energy to slide the box across the table is $W_2=0\text{ J}$ .

The change in the box-Earth system’s energy to lower the box to the floor is $W_3=-1.1\times {10}^2\text{ J}$ .

The system’s total energy change is $W=0\text{ J}$ .

Explanation of Solution

Given:

The mass of the box is $m=10.1\text{ kg}$ .

The height of the box from the floor is $1.1\text{ m}$ .

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 change in the box-Earth system’s energy to lift the box to the table is,

$\begin{array}{l}{W}_1=\Delta PE\\ W_1=mg\left(h_{\text{f}}-h_{\text{i}}\right)\\ W_1=\left(10.1\text{ kg}\right)\times \left(9.8\text{m}/{\text{s}}^2\right)\times \left(1.1\text{ m}-0.0\right)\\ W_1=1.1\times {10}^2\text{ J}\end{array}$

The height of the reference level does not change across the table and ignore the friction. The change in the box-Earth system’s energy to slide the box across the table is,

$\begin{array}{l}{W}_2=\Delta PE\\ W_2=mg\left(h_{\text{f}}-h_{\text{i}}\right)\\ W_2=\left(10.1\text{ kg}\right)\times \left(9.8\text{m}/{\text{s}}^2\right)\times \left(\text{1}\text{.1 m}-1.1\text{ m}\right)\\ W_2=0\text{ J}\end{array}$

The change in the box-Earth system’s energy to lower the box to the floor is,

$\begin{array}{l}{W}_3=\Delta PE\\ W_3=mg\left(h_{\text{f}}-h_{\text{i}}\right)\\ W_3=\left(10.1\text{ kg}\right)\times \left(9.8\text{m}/{\text{s}}^2\right)\times \left(\text{0}\text{.0 m}-1.1\text{ m}\right)\\ W_3=-1.1\times {10}^2\text{ J}\end{array}$

The system’s total energy change is,

$\begin{array}{l}W=W_1+W_2+W_3\\ W=\left(1.1\times {10}^2\text{ J}\right)+\left(0\text{ J}\right)+\left(-1.1\times {10}^2\text{ J}\right)\\ W=0\text{ J}\end{array}$

The system’s total energy change is $W=0\text{ J}$ .

Conclusion:

Thus, the change in the box-Earth system’s energy to lift the box to the table is $W_1=1.1\times {10}^2\text{ J}$ .

Thus, the change in the box-Earth system’s energy to slide the box across the table is $W_2=0\text{ J}$ .

Thus, the change in the box-Earth system’s energy to lower the box to the floor is $W_3=-1.1\times {10}^2\text{ J}$ .

Thus, the system’s total energy change is $W=0\text{ J}$ .