Chapter 3, Problem 3.6QP

Determine the kinetic energy of (a) a 29-kg mass moving at 122 m/s, (b) a tennis ball weighing 58.5 g moving at 71.3 mph, (c) a beryllium atom moving at 355 m/s, (d) a neutron moving at 3.000 × 10³ m/s.

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.

Explanation of Solution

To find: Determine the kinetic energy of a $\text{29-kg}$ mass moving at $\text{122 m/s}$ (a)

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, $\text{m }=\text{ 29 kg}$; $\text{u }=\text{ 122 m/s}$.  Substitute the given values in the formula,

$\begin{array}{l}{\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}{\text{(29 kg)(122 m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 2}{\text{.2 × 10}}^{\text{5}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 2}{\text{.2 × 10}}^{\text{5}}\text{ J}\end{array}$

Therefore, the kinetic energy of a $\text{29-kg}$ mass moving at $\text{122 m/s}$ is $\text{2}{\text{.2 × 10}}^{\text{5}}\text{ J}$

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.

Explanation of Solution

To find: Determine the kinetic energy of a tennis ball weighing $\text{58}\text{.5 g}$ moving at $\text{71}\text{.3 mph}$

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, $\text{m }=\text{ 58}\text{.5 g}$; $\text{u }=\text{ 71}\text{.3 mph}$.  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

$\begin{array}{l}\text{m }=\text{ 58}\text{.5 g }\times \frac{\text{1 kg}}{{\text{1 × 10}}^{\text{3}}\text{ g}}\\ \text{m }=\text{ 0}\text{.0585 kg}\end{array}$

The velocity of the tennis ball in meters per second is

$\begin{array}{l}\text{u }=\frac{\text{71}\text{.3 mi}}{\text{1 h}}\times \frac{\text{1}\text{.61 km}}{\text{1 mi}}\times \frac{{\text{1 × 10}}^{\text{3}}\text{ m}}{\text{1 km}}\times \frac{\text{1 h}}{\text{60 min}}\times \frac{\text{1 min}}{\text{60 s}}\\ \text{u }=\text{ 31}\text{.89 m/s}\end{array}$

Substitute the given values in the formula,

$\begin{array}{l}{\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}\text{(0}\text{.0585 kg)(31}{\text{.89 m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 29}\text{.7 kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 29}\text{.7 J}\end{array}$

Therefore, the kinetic energy of a tennis ball weighing $\text{58}\text{.5 g}$ moving at $\text{71}\text{.3 mph}$ is $\text{29}\text{.7 J}$

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.

Explanation of Solution

To find: Determine the kinetic energy of a beryllium atom moving at $\text{355 m/s}$ (c)

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, $\text{m }=\text{ 9}\text{.102 amu}$; $\text{u }=\text{ 355 m/s}$.  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,

$\begin{array}{l}{\text{E}}_{\text{k}}\text{ = }\frac{\text{1}}{\text{2}}\text{(1}{\text{.4969 × 10}}^{-\text{26}}{\text{ kg)(355 m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}\text{ = 9}{\text{.43 × 10}}^{-\text{22}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}\text{ = 9}{\text{.43 × 10}}^{-\text{22}}\text{ J}\end{array}$

Therefore, the kinetic energy of a beryllium atom moving at $\text{355 m/s}$ is $\text{9}{\text{.43 × 10}}^{-\text{22}}\text{ J}$

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.

Explanation of Solution

To find: Determine the kinetic energy of a neutron moving at $\text{3}{\text{.000 × 10}}^{\text{3}}\text{ m/s}$ (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, $\text{m }=\text{ 1}{\text{.67493×10}}^{-\text{24}}\text{ g}$; $\text{u }=\text{ 3}{\text{.000 × 10}}^{\text{3}}\text{ m/s}$.  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,

$\begin{array}{l}{\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}\text{(1}{\text{.67493 × 10}}^{-\text{27}}\text{ kg)(3}{\text{.000 × 10}}^{\text{3}}{\text{ m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 7}{\text{.538 × 10}}^{-\text{21}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 7}{\text{.538 × 10}}^{-\text{21}}\text{ J}\end{array}$

Therefore, the kinetic energy of a neutron moving at $\text{3}{\text{.000 × 10}}^{\text{3}}\text{ m/s}$ is $\text{7}{\text{.538 × 10}}^{-\text{21}}\text{ J}$