Chapter 7 Readings: 7.2 Kinetic Energy

Foundations and Classical Definition of Kinetic Energy

Kinetic energy represents the energy possessed by an object due to its motion. The physical principle underlying this concept is that the greater the velocity of a body, the greater the effect it can exert when interacting or colliding with other bodies. This physical capacity does not depend on the directional vector of the velocity, but solely on its magnitude (speed).

Historical Evolution of the Concept

• Late Seventeenth Century: A mechanical quantity was first introduced to explain collisions between two perfectly elastic bodies. Consider an idealized scenario where a moving body undergoes a direct head-on collision with an identical body initially at rest. The first body comes to a complete halt, while the second body sets off with the initial velocity of the first. This exact phenomenon can be observed in billiards, croquet, and the motion of a Newton's Cradle.

• Energy of Motion: The underlying concept was initially tied to the forces acting upon a body and was described as the "energy of motion."

• Eighteenth Century: The term kinetic energy was formally adopted to designate this energy of motion.

The Classical Approximation

In classical mechanics, the definition of kinetic energy assumes non-relativistic conditions, meaning the speed of the particle $v$ is much less than the speed of light $c$ ($v \ll c$). When particles travel at speeds approaching the speed of light, the classical definition no longer holds, and the special theory of relativity provides an alternate expression.

Mathematical Formulation

For a single particle of mass $m$ moving at speed $v$, its kinetic energy $K$ is defined as one-half the product of its mass and the square of its speed:

$K = \frac{1}{2}mv^2$

For a system consisting of multiple constituent particles, the total kinetic energy is obtained by summing the individual kinetic energies of all particles in the system:

$K = \sum \frac{1}{2}mv^2$

Kinetic Energy in Terms of Linear Momentum

Just as Newton's second law can be formulated either as the mass multiplied by the rate of change of velocity ($\mathbf{F} = m\mathbf{a} = m\frac{d\mathbf{v}}{dt}$) or as the time rate of change of linear momentum ($\mathbf{F} = \frac{d\mathbf{p}}{dt}$), the kinetic energy of a single particle can also be expressed in terms of its linear momentum rather than its velocity.

Linear momentum $\vec{\mathbf{p}}$ is defined by:

$\vec{\mathbf{p}} = m\vec{\mathbf{v}}$

Taking the magnitude $p = mv$, solving for speed yields:

$v = \frac{p}{m}$

Substituting this expression into the kinetic energy equation gives:

$K = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}$

This formulation is especially practical when analyzing physical systems where momentum is conserved or directly measured.

Dimensional Analysis and Units of Measurement

From the algebraic expression $K = \frac{1}{2}mv^2$, the base units of kinetic energy are:

$\text{mass} \times (\text{speed})^2 = \text{kg} \cdot (\text{m/s})^2 = \text{kg}\cdot\text{m}^2/\text{s}^2$

The SI unit of force is the newton ($\text{N}$), where:

$1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$

Multiplying force by distance (meters) gives:

$\text{N} \cdot \text{m} = (\text{kg}\cdot\text{m/s}^2) \cdot \text{m} = \text{kg}\cdot\text{m}^2/\text{s}^2$

Thus, the units of kinetic energy are identical to the units of force multiplied by distance, which define work. The SI derived unit for both work and kinetic energy is the joule ($\text{J}$):

$1\,\text{J} = 1\,\text{N}\cdot\text{m} = 1\,\text{kg}\cdot\text{m}^2/\text{s}^2$

Work and kinetic energy share the same units because they represent different forms of the same unified physical property.

Quantitative Calculations: Macro to Subatomic Scales

Kinetic energy spans an immense dynamic range across the universe, necessitating specialized units for very large and very small quantities.

Example Calculations across Diverse Regimes

1. Human Scale: Running Athlete

• Parameters: Mass $m = 80\,\text{kg}$, speed $v = 10\,\text{m/s}$.

• Calculation:

$K = \frac{1}{2}(80\,\text{kg})(10\,\text{m/s})^2 = 4000\,\text{J} = 4.0\,\text{kJ}$

2. Planetary Scale: Chicxulub Asteroid Impact

The Chicxulub crater located in Yucatán represents one of the largest impact structures preserved on Earth. It was generated by an asteroid traveling at $22\,\text{km/s}$ ($22 \times 10^3\,\text{m/s}$) that released $4.2 \times 10^{23}\,\text{J}$ of kinetic energy upon impact.

• Calculation of Mass:

$m = \frac{2K}{v^2} = \frac{2(4.2 \times 10^{23}\,\text{J})}{(22 \times 10^3\,\text{m/s})^2} = 1.7 \times 10^{15}\,\text{kg}$

3. Subatomic Scale: Thermal Neutrons in Nuclear Reactors

Thermal neutrons play an essential operational role inside nuclear reactors, moving at typical speeds of approximately $2.2\,\text{km/s}$ ($2.2 \times 10^3\,\text{m/s}$) with a mass of $m = 1.68 \times 10^{-27}\,\text{kg}$.

• Calculation:

$K = \frac{1}{2}(1.68 \times 10^{-27}\,\text{kg})(2.2 \times 10^3\,\text{m/s})^2 = 4.1 \times 10^{-21}\,\text{J}$

Specialized Energy Units

• Megaton of TNT: Used to quantify explosive yields of planetary impacts and nuclear detonations:

$1\,\text{megaton} = 4.18 \times 10^{15}\,\text{J}$

  The kinetic energy released by the Chicxulub asteroid ($4.2 \times 10^{23}\,\text{J}$) is equivalent to approximately one hundred million megatons ($10^8\,\text{megatons}$) of TNT.

• Electron-Volt ($\text{eV}$): Used in subatomic, atomic, and nuclear physics:

$1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$

  The kinetic energy of the thermal neutron ($4.1 \times 10^{-21}\,\text{J}$) corresponds to approximately one fortieth of an electron-volt ($\approx 0.025\,\text{eV}$ or $\frac{1}{40}\,\text{eV}$).

Conceptual Analysis: Mass and Speed Dependence

Consider the direct relationships established by $K = \frac{1}{2}mv^2$:

Constant Kinetic Energy with Varying Mass

If two vehicles—such as a passenger car and a heavy freight truck—possess identical kinetic energies ($K_{\text{car}} = K_{\text{truck}}$), but the truck has significantly more mass ($m_{\text{truck}} > m_{\text{car}}$), their speeds must differ.

Rearranging for speed:

$v = \sqrt{\frac{2K}{m}}$

Because speed is inversely proportional to the square root of mass at fixed kinetic energy, the body with the smaller mass must possess the higher speed. Consequently, the car travels with a greater speed than the truck.

Constant Speed with Varying Mass

If the car and the truck travel at identical speeds ($v_{\text{car}} = v_{\text{truck}}$), their kinetic energies scale directly in proportion to their respective masses:

$K \propto m$

Because the truck has a larger mass, the truck possesses the greater kinetic energy.

Reference Frame Dependence of Kinetic Energy

Because velocity is fundamentally a relative quantity that depends on the observer's chosen coordinate frame, kinetic energy is also frame-dependent. Calculating the kinetic energy of an object requires defining the reference frame with respect to which velocity is evaluated.

Frame Selection

Depending on the system and the mathematical convenience of the problem, two common frames are utilized:

1. External Reference Frame: A coordinate system fixed to external surroundings where observations occur (e.g., ground, laboratory, or tracks).

2. Internal Reference Frame: A coordinate system attached to or moving along with the system of interest.

Kinetic energy changes from one reference frame to another according to the Galilean velocity addition rule in one dimension:

$\vec{\mathbf{v}}_{PT} = \vec{\mathbf{v}}_{PC} + \vec{\mathbf{v}}_{CT}$

Where:

• $P$ denotes the particle or person.

• $C$ denotes the moving carriage/car.

• $T$ denotes the track or fixed ground.

Universal Constraint on Kinetic Energy

Although the numeric magnitude of kinetic energy varies across reference frames, kinetic energy can never be negative in any reference frame. Mass $m$ is always positive ($m > 0$), and the square of any real velocity $v^2$ is strictly greater than or equal to zero ($v^2 \ge 0$). Hence, $K \ge 0$ under all circumstances.

Worked Scenario: Passenger Inside a Moving Subway Train

• Given Data:

  • Person mass: $m = 75.0\,\text{kg}$

  • Walking speed relative to subway car ($C$): $v_{PC} = 1.50\,\text{m/s}$

  • Subway car speed relative to track ($T$): $v_{CT} = 15.0\,\text{m/s}$

1. Kinetic Energy Relative to the Subway Car

Using the relative speed within the internal frame of the train carriage:

$K = \frac{1}{2}(75.0\,\text{kg})(1.50\,\text{m/s})^2 = 84.4\,\text{J}$

2. Kinetic Energy Relative to the Fixed Tracks

Because the person walks along the central aisle parallel to the train's motion, two distinct directions are possible:

$v_{PT} = v_{CT} \pm v_{PC} = (15.0 \pm 1.50)\,\text{m/s}$

• Case A: Walking Toward the Front (in direction of train travel):   $v_{PT} = 15.0\,\text{m/s} + 1.50\,\text{m/s} = 16.5\,\text{m/s}$

$K = \frac{1}{2}(75.0\,\text{kg})(16.5\,\text{m/s})^2 = 10.2\,\text{kJ}$

• Case B: Walking Toward the Back (opposite to train travel):   $v_{PT} = 15.0\,\text{m/s} - 1.50\,\text{m/s} = 13.5\,\text{m/s}$

$K = \frac{1}{2}(75.0\,\text{kg})(13.5\,\text{m/s})^2 = 6.83\,\text{kJ}$

3. Kinetic Energy in the Co-Moving Frame of the Person

In a reference frame attached directly to the walking person, their relative velocity is zero ($v_P = 0\,\text{m/s}$):

$K = \frac{1}{2}(75.0\,\text{kg})(0\,\text{m/s})^2 = 0\,\text{J}$

Scenario: Rowing Against vs. With River Currents

Suppose a boat is rowed parallel to riverbanks, and the rower's kinetic energy relative to the riverbanks is observed to be less than the kinetic energy relative to the flowing water:

$K_{\text{bank}} < K_{\text{water}} \implies v_{\text{bank}} < v_{\text{water}}$

Because the velocity relative to the bank is the vector combination of the rowing velocity relative to water and the current velocity:

$v_{\text{bank}} = |v_{\text{water}} \pm v_{\text{current}}|$

For $v_{\text{bank}}$ to be smaller than $v_{\text{water}}$, the flow of the current must oppose the boat's rowing direction ($v_{\text{bank}} = v_{\text{water}} - v_{\text{current}}$). Therefore, the boat is rowing against the current.

Classification and Internal Subdivisions of Kinetic Energy

While the kinetic energy of an individual elementary particle is a single scalar quantity, the kinetic energy of an extended or multi-particle system can be classified into distinct forms depending on the nature of the motion:

Regardless of nomenclature, every form of kinetic energy represents the exact same physical property: energy associated with physical motion.

Microscopic vs. Macroscopic Kinetic Energy Analysis

To observe the contrast between macroscopic bulk kinetic energy and internal microscopic thermal kinetic energy, consider an inflated basketball in flight.

Parameter Specifications

• Basketball Mass: $m = 624\,\text{g} = 0.624\,\text{kg}$

• Pass Flight Trajectory: Displaces $15\,\text{m}$ horizontally in $2\,\text{s}$

• Contained Air Characteristics:

  • Average molecular mass: $m_{\text{mol}} = 29\,\text{u}$

  • Unified atomic mass conversion: $1\,\text{u} = 1.66 \times 10^{-27}\,\text{kg}$

  • Average molecular random speed: $v_{\text{mol}} = 500\,\text{m/s}$ (relative to basketball)

  • Molecular population: $N = 3 \times 10^{23}\,\text{molecules}$

Comparative Calculations

1. Horizontal Translational Kinetic Energy

The average horizontal speed of the basketball relative to the court is:

$v_{\text{ball}} = \frac{15\,\text{m}}{2\,\text{s}} = 7.5\,\text{m/s}$

The macroscopic horizontal kinetic energy is:

$K_{\text{trans}} = \frac{1}{2}(0.624\,\text{kg})(7.5\,\text{m/s})^2 = 17.55\,\text{J} \approx 17.6\,\text{J}$

(Note: If the basketball were spinning in flight, its parts would also possess rotational kinetic energy in addition to this translational component).

2. Internal Thermal Kinetic Energy of Contained Gas

First, convert the average molecular mass to kilograms:

$m_{\text{mol}} = 29\,\text{u} \times (1.66 \times 10^{-27}\,\text{kg/u}) = 4.814 \times 10^{-26}\,\text{kg}$

The average translational kinetic energy of a single air molecule relative to the ball is:

$K_{\text{single}} = \frac{1}{2}(29\,\text{u})(1.66 \times 10^{-27}\,\text{kg/u})(500\,\text{m/s})^2 = 6.02 \times 10^{-21}\,\text{J}$

The total internal random kinetic energy across all $N = 3 \times 10^{23}$ molecules inside the basketball is:

$K_{\text{thermal}} = (3 \times 10^{23})(6.02 \times 10^{-21}\,\text{J}) = 1.80 \times 10^3\,\text{J} = 1.80\,\text{kJ}$

The internal thermal kinetic energy ($1800\,\text{J}$) is approximately 100 times larger than the macroscopic translational kinetic energy of the pass ($17.6\,\text{J}$).

3. Macroscopic Speed Equivalent to Internal Thermal Energy

To determine the speed $v$ at which the entire basketball would need to travel across the court to match the internal kinetic energy of its constituent air molecules ($1.80\,\text{kJ}$):

$K = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2K}{m}}$

$v = \sqrt{\frac{2(1.80 \times 10^3\,\text{J})}{0.624\,\text{kg}}} = 76.0\,\text{m/s}$

Because kinetic energy scales with the square of speed ($K \propto v^2$), an energy increase of a factor of roughly $100$ necessitates an increase in speed by a factor of $\sqrt{100} = 10$:

$7.5\,\text{m/s} \times 10 \approx 76.0\,\text{m/s}$