Chapter 3, Problem 3.6QP

(a)

Interpretation Introduction

Interpretation:

Kinetic energy should be calculated in the given statement by using the equation of kinetic energy

Concept Introduction:

Energy is the capacity to do work or transfer heat where work is the movement of a body using some force.  The SI unit of energy is joule ($\text{J}$).  Energy is in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.  Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.

Answer to Problem 3.6QP

The kinetic energy of a $\text{25-kg}$ mass moving at $\text{61}\text{.3 m/s}$ is $\text{4}{\text{.7 × 10}}^{\text{4}}\text{ J}$.

Explanation of Solution

To find: Determine the kinetic energy of a $\text{25-kg}$ mass moving at $\text{61}\text{.3 m/s}$

Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

Where $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.  By considering the given problem, $\text{m = 25 kg}$; $\text{u = 61}\text{.3 m/s}$.  Substitute the given values in the formula,

$\begin{array}{l}{\text{E}}_{\text{k}}\text{ = }\frac{\text{1}}{\text{2}}\text{(25 kg)(61}{\text{.3 m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}\text{ = 4}{\text{.7 × 10}}^{\text{4}}{\text{ kg×m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}\text{ = 4}{\text{.7 × 10}}^{\text{4}}\text{ J}\end{array}$

Therefore, the kinetic energy of a $\text{25-kg}$ mass moving at $\text{61}\text{.3 m/s}$ is $\text{4}{\text{.7 × 10}}^{\text{4}}\text{ J}$

(b)

Interpretation Introduction

Interpretation:

Kinetic energy should be calculated in the given statement by using the equation of kinetic energy

Concept Introduction:

Energy is the capacity to do work or transfer heat where work is the movement of a body using some force.  The SI unit of energy is joule ($\text{J}$).  Energy is in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.  Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.

Answer to Problem 3.6QP

The kinetic energy of a tennis ball weighing $\text{58}\text{.1 g}$ moving at $\text{66}\text{.2 mph}$ is $\text{29}\text{.7 J}$

Explanation of Solution

To find: Determine the kinetic energy of a tennis ball weighing $\text{58}\text{.1 g}$ moving at $\text{66}\text{.2 mph}$

Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.  By considering the given problem, $\text{m = 58}\text{.1 g}$; $\text{u = 66}\text{.2 mph}$.  Hence, ‘$\text{m}$’ in $\text{g}$ and ‘$\text{u}$’ in $\text{mph}$ should be converted into ‘$\text{m}$’ in kilograms and ‘$\text{u}$’ in meters per second.

The mass of the tennis ball in kilograms is

$\begin{array}{l}\text{m = 58}\text{.1 g × }\frac{\text{1 kg}}{{\text{1 × 10}}^{\text{3}}\text{ g}}\\ \text{m = 0}\text{.0581 kg}\end{array}$

The velocity of the tennis ball in meters per second is

$\begin{array}{l}\text{u = }\frac{\text{66}\text{.2 mi}}{\text{1 h}}\text{ × }\frac{\text{1}\text{.61 km}}{\text{1 mi}}\text{ × }\frac{{\text{1 × 10}}^{\text{3}}\text{ m}}{\text{1 km}}\text{ × }\frac{\text{1 h}}{\text{60 min}}\text{ × }\frac{\text{1 min}}{\text{60 s}}\\ \text{u = 29}\text{.6 m/s}\end{array}$

Substitute the given values in the formula,

$\begin{array}{l}{\text{E}}_{\text{k}}\text{ = }\frac{\text{1}}{\text{2}}\text{(0}\text{.0581 kg)(29}{\text{.6 m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}\text{ = 25}{\text{.5 kg×m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}\text{ = 25}\text{.5 J}\end{array}$

Therefore, the kinetic energy of a tennis ball weighing $\text{58}\text{.1 g}$ moving at $\text{66}\text{.2 mph}$ is $\text{25}\text{.5 J}$

(c)

Interpretation Introduction

Interpretation:

Kinetic energy should be calculated in the given statement by using the equation of kinetic energy

Concept Introduction:

Energy is the capacity to do work or transfer heat where work is the movement of a body using some force.  The SI unit of energy is joule ($\text{J}$).  Energy is in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.  Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.

Answer to Problem 3.6QP

The kinetic energy of a beryllium atom moving at $\text{275 m/s}$ is $\text{5}{\text{.66 × 10}}^{\text{-22}}\text{ J}$

Explanation of Solution

To find: Determine the kinetic energy of a beryllium atom moving at $\text{275 m/s}$

Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.  By considering the given problem, $\text{m }=\text{ 9}\text{.102 amu}$; $\text{u = 275 m/s}$.  Hence, ‘$\text{m}$’ in $\text{amu}$ should be converted into ‘$\text{m}$’ in kilograms.

The mass of a beryllium atom in kilograms is

$\begin{array}{l}\text{m }=\text{ 9}\text{.102 amu }\times \frac{\text{1}{\text{.661 × 10}}^{-\text{24}}\text{ g}}{\text{1 amu}}\times \frac{\text{1 kg}}{{\text{1 × 10}}^{\text{3}}\text{ g}}\\ \text{m }=\text{ 1}{\text{.4969 × 10}}^{-\text{26}}\text{ kg}\end{array}$

Substitute the given values in the formula,

$\begin{array}{l}{\text{E}}_{\text{k}}\text{ = }\frac{\text{1}}{\text{2}}\text{(1}{\text{.4969 × 10}}^{\text{-26}}{\text{ kg)(275 m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}\text{ = 5}{\text{.66 × 10}}^{\text{-22}}{\text{ kg×m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}\text{ = 5}{\text{.66 × 10}}^{\text{-22}}\text{ J}\end{array}$

Therefore, the kinetic energy of a beryllium atom moving at $\text{275 m/s}$ is $\text{5}{\text{.66 × 10}}^{\text{-22}}\text{ J}$

(d)

Interpretation Introduction

Interpretation:

Kinetic energy should be calculated in the given statement by using the equation of kinetic energy

Concept Introduction:

Energy is the capacity to do work or transfer heat where work is the movement of a body using some force.  The SI unit of energy is joule ($\text{J}$).  Energy is in the form of kinetic energy or potential energy.  Kinetic energy is the energy associated with motion.  Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.

Answer to Problem 3.6QP

The kinetic energy of a neutron moving at $\text{2}{\text{.000 × 10}}^{\text{3}}\text{ m/s}$ is $\text{3}{\text{.34 × 10}}^{\text{-21}}\text{ J}$

Explanation of Solution

To find: Determine the kinetic energy of a neutron moving at $\text{2}{\text{.000 × 10}}^{\text{3}}\text{ m/s}$ (d)

Kinetic energy (in joule) is calculated using the formula:

${\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{mu}}^{\text{2}}$

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.  By considering the given problem, $\text{m }=\text{ 1}{\text{.67493×10}}^{-\text{24}}\text{ g}$; $\text{u = 2}{\text{.000 × 10}}^{\text{3}}\text{ m/s}$.  Hence, ‘$\text{m}$’ in $\text{g}$ should be converted into ‘$\text{m}$’ in kilograms.

The mass of a neutron in kilograms is

$\begin{array}{l}\text{m }=\text{ 1}{\text{.67493 × 10}}^{-\text{24}}\text{ g }\times \frac{\text{1 kg}}{{\text{1 × 10}}^{\text{3}}\text{ g}}\\ \text{m }=\text{ 1}{\text{.67493 × 10}}^{-\text{27}}\text{ kg}\end{array}$

Substitute the given values in the formula,

$\begin{array}{l}{\text{E}}_{\text{k}}\text{ = }\frac{\text{1}}{\text{2}}\text{(1}{\text{.67493 × 10}}^{\text{-27}}\text{ kg)(2}{\text{.000 × 10}}^{\text{3}}{\text{ m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}\text{ = 3}{\text{.34 × 10}}^{\text{-21}}{\text{ kg×m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}\text{ = 3}{\text{.34 × 10}}^{\text{-21}}\text{ J}\end{array}$

Therefore, the kinetic energy of a neutron moving at $\text{2}{\text{.000 × 10}}^{\text{3}}\text{ m/s}$ is $\text{3}{\text{.34 × 10}}^{\text{-21}}\text{ J}$