Chapter 3, Problem 3.5QP

Determine the kinetic energy of (a) a 1.25-kg mass moving at 5.75 m/s, (b) a car weighing 3250 lb moving at 35 mph, (c) an electron moving at 475 m/s, (d) a helium atom moving at 725 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 $\text{1}\text{.25-kg}$ mass moving at $\text{5}\text{.75 m/s}$

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, $\text{m }=\text{ 1}\text{.25 kg}$; $\text{u }=\text{ 5}\text{.75 m/s}$.  Substitute the given values in the formula,

$\begin{array}{l}{\text{E}}_{\text{k}}\text{ = }\frac{\text{1}}{\text{2}}\text{(1}\text{.25 kg)(5}{\text{.75 m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 20}\text{.7 kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 20}\text{.7 J}\end{array}$

Therefore, the kinetic energy of a $\text{1}\text{.25-kg}$ mass moving at $\text{5}\text{.75 m/s}$ is $\text{20}\text{.7 J}$

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 car weighing  $\text{3250 lb}$ moving at $\text{35 mph}$

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

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

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.  By considering the given problem, $\text{m }=\text{ 3250 lb}$; $\text{u }=\text{ 35 mph}$.  Hence, ‘ $\text{m}$’ in $\text{lb}$ 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 car in kilograms is

$\begin{array}{l}\text{m }=\text{ 3250 lb }\times \frac{\text{453}\text{.6 g}}{\text{1 lb}}\times \frac{\text{1 kg}}{{\text{1 × 10}}^{\text{3}}\text{ g}}\\ \text{m }=\text{ 1474 kg}\end{array}$

The velocity of the car in meters per second is

$\begin{array}{l}\text{u }=\frac{\text{35 mi}}{\text{1 h}}\times \frac{\text{1}\text{.61 km}}{\text{1 mi}}\times \frac{{\text{1 × 10}}^{\text{3}}\text{ m}}{\text{1 km}}\times \frac{\text{1 h}}{\text{60 min}}\times \frac{\text{1 min}}{\text{60 s}}\\ \text{u }=\text{ 15}\text{.6 m/s}\end{array}$

Substitute the given values in the formula,

$\begin{array}{l}{\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}\text{(1474 kg)(15}{\text{.6 m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 1}{\text{.8 × 10}}^{\text{5}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 1}{\text{.8 × 10}}^{\text{5}}\text{ J}\end{array}$

Therefore, the kinetic energy of a car weighing $\text{3250 lb}$ moving at $\text{35 mph}$ is $\text{1}{\text{.8 × 10}}^{\text{5}}\text{ J}$

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 an electron moving at $\text{475 m/s}$

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

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

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.  By considering the given problem, $\text{m }=\text{ 9}{\text{.10938 × 10}}^{-\text{28}}\text{ g}$; $\text{u }=\text{ 475 m/s}$.  Hence, ‘ $\text{m}$’ in $\text{g}$ should be converted into ‘ $\text{m}$’ in kilograms.

The mass of an electron in kilograms is

$\begin{array}{l}\text{m }=\text{ 9}{\text{.10938 × 10}}^{-\text{28}}\text{ g × }\frac{\text{1 kg}}{{\text{1 × 10}}^{\text{3}}\text{ g}}\\ \text{m }=\text{ 9}{\text{.10938 × 10}}^{-\text{31}}\text{ kg}\end{array}$

Substitute the given values in the formula,

$\begin{array}{l}{\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}\text{(9}{\text{.10938 × 10}}^{-\text{31}}{\text{ kg)(475 m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 1}{\text{.03 × 10}}^{-\text{25}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 1}{\text{.03 × 10}}^{-\text{25}}\text{ J}\end{array}$

Therefore, the kinetic energy of an electron moving at $\text{475 m/s}$ is $\text{1}{\text{.03 × 10}}^{-\text{25}}\text{ J}$

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 helium atom moving at $\text{725 m/s}$ (d)

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

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

Where, $\text{m}$ ‒ mass in kilograms; $\text{u}$ – velocity in meters per second.  By considering the given problem, $\text{m }=\text{ 4}\text{.003 amu}$; $\text{u }=\text{ 725 m/s}$.  Hence, ‘ $\text{m}$’ in $\text{amu}$ should be converted into ‘ $\text{m}$’ in kilograms.  The factor for conversion of $\text{amu }\to \text{ g}$ is $\text{1}{\text{.661 × 10}}^{-\text{24}}\text{ g / 1 amu}$.

The mass of a helium atom in kilograms is

$\begin{array}{l}\text{m }=\text{ 4}\text{.003 amu }\times \frac{\text{1}{\text{.661 × 10}}^{-\text{24}}\text{ g}}{\text{1 amu}}\times \frac{\text{1 kg}}{{\text{1 × 10}}^{\text{3}}\text{ g}}\\ \text{m }=\text{ 6}{\text{.649 × 10}}^{-\text{27}}\text{ kg}\end{array}$

Substitute the given values in the formula,

$\begin{array}{l}{\text{E}}_{\text{k}}=\frac{\text{1}}{\text{2}}\text{(6}{\text{.649 × 10}}^{-\text{27}}{\text{ kg)(725 m/s)}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 1}{\text{.75 × 10}}^{-\text{21}}\text{ kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{E}}_{\text{k}}=\text{ 1}{\text{.75 × 10}}^{-\text{21}}\text{ J}\end{array}$

Therefore, the kinetic energy of a helium atom moving at $\text{725 m/s}$ is $\text{1}{\text{.75 × 10}}^{-\text{21}}\text{ J}$