Chapter 8, Problem 17PQ

(a)

To determine

The potential energy at top and the bottom and the change in potential energy.

Answer to Problem 17PQ

The potential energy at bottom is $\underset{\_}{0}$, the potential energy at top is $\underset{\_}{717\text{J}}$ and the change in potential energy is $\underset{\_}{-717\text{J}}$.

Explanation of Solution

The diagram for the position of the child is given in figure 1.

  

  Figure 1

Write the equation of potential energy at bottom.

  $U_B=mg{y}_b$                                                                                                   (I)

Here, $U_B$ is the potential energy at bottom, $m$ is the mass, $g$ is the gravitational acceleration and $y_b$ is the bottom height.

Write the equation of potential energy at top.

  $U_T=mg{y}_T$                                                                                                   (II)

Here, $U_T$ is the potential energy at top, $m$ is the mass, $g$ is the gravitational acceleration and $y_T$ is the top height.

Write the expression for the top height.

  $y_T=h\sin \theta$                                                                                              (III)

Here, $h$ is the length and $\theta$ is the angle.

Rewrite the expression from equation (II).

  $U_T=mgh\sin \theta$                                                                                         (IV)

Write the expression for the change in potential energy.

  $\left(\Delta U\right)=U_B-U_T$                                                                                          (V)

Here, $\left(\Delta U\right)$ is the change in potential energy.

Conclusion:

Substitute $0$ for $y_b$ in equation (I) to find $U_B$.

  $\begin{array}{c}{U}_B=mg\left(0\right)\\ =0\text{J}\end{array}$

Thus, The potential energy at bottom is $\underset{\_}{0}$.

Substitute $55.0\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $8\text{ft}$ for $h$ and $33°$ for $\theta$ in equation (IV) to find $U_T$.

  $\begin{array}{c}{U}_T=\left(55.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left[\left(8\text{ft}\right)\left(\frac{0.3048\text{m}}{1\text{ft}}\right)\right]\sin \left(33.0°\right)\\ =\left(55.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(2.44\text{m}\right)\sin \left(33.0°\right)\\ =717\text{J}\end{array}$

Thus, the potential energy at top is $\underset{\_}{717\text{J}}$.

Substitute $0\text{J}$ for $U_B$, $717\text{J}$ for $U_T$ in equation (V) to find $\left(\Delta U\right)$.

  $\begin{array}{c}\left(\Delta U\right)=\left(0\text{J}\right)-\left(717\text{J}\right)\\ =-717\text{J}\end{array}$

Thus, the change in potential energy is $\underset{\_}{-717\text{J}}$.

(b)

To determine

The potential energy at top and the bottom and the change in potential energy by choosing the top of the slide as reference frame.

Answer to Problem 17PQ

The potential energy at top is $\underset{\_}{0}$, the potential energy at bottom is $\underset{\_}{-717\text{J}}$ and the change in potential energy is $\underset{\_}{-717\text{J}}$.

Explanation of Solution

Write the equation of potential energy at bottom.

  $U_B=mg{y}_b$                                                                                                         (VI)

Here, $U_B$ is the potential energy at bottom, $m$ is the mass, $g$ is the gravitational acceleration and $y_b$ is the bottom height.

Write the equation of potential energy at bottom.

  $U_T=mg{y}_T$                                                                                                       (VII)

Here, $U_T$ is the potential energy at top, $m$ is the mass, $g$ is the gravitational acceleration and $y_T$ is the top height.

Write the expression for the bottom height.

  $y_b=-h\sin \theta$                                                                                                     (VIII)

Here, $h$ is the length and $\theta$ is the angle.

Rewrite the expression from equation (I).

  $U_B=mg\left(-h\sin \theta \right)$                                                                                              (IX)

Write the expression for the change in potential energy.

  $\left(\Delta U\right)=U_B-U_T$                                                                                                   (X)

Here, $\left(\Delta U\right)$ is the change in potential energy.

Conclusion:

Substitute, $0$ for $y_b$ in equation (VII) to find $U_T$.

  $\begin{array}{c}{U}_T=mg\left(0\right)\\ =0\text{J}\end{array}$

Thus, The potential energy at top is $\underset{\_}{0}$.

Substitute, $55.0\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $8\text{ft}$ for $h$ and $33°$ for $\theta$ in equation (IX) to find $U_B$.

  $\begin{array}{c}{U}_B=\left(55.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left[-\left(8\text{ft}\right)\left(\frac{0.3048\text{m}}{1\text{ft}}\right)\right]\sin \left(33.0°\right)\\ =-\left(55.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(2.44\text{m}\right)\sin \left(33.0°\right)\\ =-717\text{J}\end{array}$

Thus, the potential energy at bottom is $\underset{\_}{-717\text{J}}$.

Substitute, $0\text{J}$ for $U_T$, $-717\text{J}$ for $U_B$ in equation (X) to find $\left(\Delta U\right)$.

  $\begin{array}{c}\left(\Delta U\right)=-\left(717\text{J}\right)-\left(0\text{J}\right)\\ =-717\text{J}\end{array}$

Thus, the change in potential energy is $\underset{\_}{-717\text{J}}$.