Chapter 9, Problem 27A

Consider these four situations.

(A) a 3-kg ball at rest atop a 5-m-tall hill

(B) a 4-kg ball at rest atop a 5-m-tall hill

(C) a 3-kg ball moving at 2 m/s atop a 5-m-tall hill

(D) a 4-kg ball moving at 2 m/s at ground level

a. Rank from greatest to least the potential energy of each ball.

b. Rank from greatest to least the kinetic energy of each ball.

c. Rank from greatest to least the total energy of each ball.

To determine

The rank of the potential energy of the ball from greatest to least.

Answer to Problem 27A

The rank of the potential energy of the ball from greatest to least is $B>A=C>D$ .

Explanation of Solution

Formula used:

The expression the potential energy as follows:

  $PE=mgh$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $h$ is the position.

Calculation:

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

For case A:

The potential energy of the ball as follows:

  $\begin{array}{l}PE=mgh\\ P{E}_{\text{A}}=\left(\text{3 kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\text{5 m}\right)\\ P{E}_{\text{A}}=147\text{J}\end{array}$

For case B:

The potential energy of the ball as follows:

  $\begin{array}{l}PE=mgh\\ P{E}_{\text{B}}=\left(\text{4 kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\text{5 m}\right)\\ P{E}_{\text{B}}=196\text{J}\end{array}$

For case C:

The potential energy of the ball as follows:

  $\begin{array}{l}PE=mgh\\ P{E}_{\text{C}}=\left(\text{3 kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\text{5 m}\right)\\ P{E}_{\text{C}}=147\text{J}\end{array}$

For case D:

The potential energy of the ball as follows:

  $\begin{array}{l}PE=mgh\\ P{E}_{\text{D}}=\left(\text{4 kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\text{0 m}\right)\\ P{E}_{\text{D}}=0\end{array}$

By comparing the values, the potential energy is $B>A=C>D$ .

Conclusion:

Thus, the rank of the potential energy of the ball from greatest to least is $B>A=C>D$ .

To determine

The rank of the kinetic energy of the ball from greatest to least.

Answer to Problem 27A

The rank of the kinetic energy of the ball from greatest to least is $D>C>A=B$ .

Explanation of Solution

Formula used:

The expression the kinetic energy as follows:

  $KE=\frac{1}{2}m{v}^2$

Here, $m$ is the mass and $v$ is the velocity.

Calculation:

For case A:

The kinetic energy of the ball as follows:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ K{E}_{\text{A}}=\frac{1}{2}\left(\text{3 kg}\right){\left(0\text{m}/\text{s}\right)}^2\\ K{E}_{\text{A}}=0\end{array}$

For case B:

The kinetic energy of the ball as follows:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ K{E}_{\text{B}}=\frac{1}{2}\left(\text{4 kg}\right){\left(0\text{m}/\text{s}\right)}^2\\ K{E}_{\text{B}}=0\end{array}$

For case C:

The kinetic energy of the ball as follows:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ K{E}_{\text{C}}=\frac{1}{2}\left(\text{3 kg}\right){\left(2\text{m}/\text{s}\right)}^2\\ K{E}_{\text{C}}=6\text{ J}\end{array}$

For case D:

The potential energy of the ball as follows:

  $\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ K{E}_{\text{D}}=\frac{1}{2}\left(\text{4 kg}\right){\left(2\text{m}/\text{s}\right)}^2\\ K{E}_{\text{D}}=8\text{ J}\end{array}$

By comparing the values, the kinetic energy is $D>C>A=B$ .

Conclusion:

Thus, the rank of the kinetic energy of the ball from greatest to least is $D>C>A=B$ .

To determine

The rank of the total energy of the ball from greatest to least.

Answer to Problem 27A

The rank of the total energy of the ball from greatest to least is $B>C>A>D$ .

Explanation of Solution

Formula used:

Show the expression the total energy as follows:

  $TE=PE+KE$

Here, $PE$ is the potential energy and $KE$ is the kinetic energy.

Calculation:

Refer part (a).

The potential energy of the ball for case A is $P{E}_{\text{A}}=147\text{J}$ .

The potential energy of the ball for case B is $P{E}_{\text{B}}=196\text{J}$ .

The potential energy of the ball for case C is $P{E}_{\text{C}}=147\text{J}$ .

The potential energy of the ball for case D is $P{E}_{\text{D}}=0$ .

Refer part (b).

The kinetic energy of the ball for case A is $K{E}_{\text{A}}=0$ .

The kinetic energy of the ball for case B is $K{E}_{\text{B}}=0$ .

The kinetic energy of the ball for case C is $K{E}_{\text{C}}=6\text{J}$ .

The kinetic energy of the ball for case D is $K{E}_{\text{D}}=8\text{ J}$ .

For case A:

The total energy of the ball as follows:

  $\begin{array}{l}TE=PE+KE\\ T{E}_{\text{A}}=147\text{ J}+0\text{ J}\\ T{E}_{\text{A}}=147\text{ J}\end{array}$

For case B:

The total energy of the ball as follows:

  $\begin{array}{l}TE=PE+KE\\ T{E}_{\text{B}}=196\text{ J}+0\text{ J}\\ T{E}_{\text{B}}=196\text{ J}\end{array}$

For case C:

The total energy of the ball as follows:

  $\begin{array}{l}TE=PE+KE\\ T{E}_{\text{C}}=147\text{ J}+6\text{ J}\\ T{E}_{\text{C}}=153\text{ J}\end{array}$

For case D:

The total energy of the ball as follows:

  $\begin{array}{l}TE=PE+KE\\ T{E}_{\text{D}}=0\text{ J}+8\text{ J}\\ T{E}_{\text{D}}=8\text{ J}\end{array}$

By comparing the values, the total energy is $B>C>A>D$ .

Conclusion:

Thus, the rank of the total energy of the ball from greatest to least is $B>C>A>D$ .