Comprehensive Physics Study Guide: Kinematics, Dynamics, Vectors, Work, and Energy

One-Dimensional Kinematics: Displacement, Velocity, and Acceleration

• Foundations of One-Dimensional Motion:

  • One-dimensional motion represents the simplest form of motion, confined to movement along a single straight line in either a forward or backward direction.

  • Complex motions observed throughout nature—such as the rotation of the Earth on its axis, its revolution around the sun, the motion of the solar system through the Milky Way galaxy, and the movement of the galaxy through space—are analyzed systematically in physics by isolating one-dimensional components against a defined frame of reference.

  • High-speed passenger trains (such as those operated in Japan, France, England, Germany, and South Korea with operational speeds between $200\,\text{km/h}$ and $300\,\text{km/h}$) demonstrate one-dimensional motion when traveling along a straight track.

  • Frame of Reference: A coordinate system used for specifying the precise location of objects in space and time. A frame of reference must remain fixed throughout the analysis of a specific problem and must possess a well-defined origin (starting point) from which all changes in position are measured. Any consistent frame of reference yields identical physical results, though selecting convenient reference points (such as a train station platform or a centimeter stick beneath a moving object) simplifies analysis.

• Displacement:

  • Definition: Displacement is the change in position of an object, represented as the straight-line directed segment from the object's initial position to its final position.

  • Displacement Equation:         $\Delta x = x_f - x_i$         where $\Delta x$ is the displacement, $x_f$ is the final position coordinate, and $x_i$ is the initial position coordinate. The Greek letter delta ($\Delta$) indicates a change in any physical quantity, defined universally as the final value minus the initial value.

  • One-Dimensional Vertical Displacement: When vertical motion along the y-axis is considered:         $\Delta y = y_f - y_i$

  • Displacement versus Distance: Displacement is a vector quantity describing net change in position and direction, whereas distance is a scalar measure of the total length of the path traveled. For example, a gecko moving up a tree along a measuring stick from an initial position of $y_i = 20\,\text{cm}$ up to $80\,\text{cm}$, and subsequently retreating down to $y_f = 50\,\text{cm}$, travels a total distance of $90\,\text{cm}$ ($60\,\text{cm}$ up plus $30\,\text{cm}$ down), but undergoes a displacement of only:         $\Delta y = 50\,\text{cm} - 20\,\text{cm} = +30\,\text{cm}$

  • If an object executes a round trip returning to its exact point of origin, its displacement is zero ($\Delta x = 0$), regardless of the total path distance traversed.

  • Sign Conventions:

    • Horizontal motion: Motion directed to the right or east is designated as positive ($+\Delta x$); motion directed to the left or west is designated as negative ($-\Delta x$).

    • Vertical motion: Motion directed upward or north is designated as positive ($+\Delta y$); motion directed downward or south is designated as negative ($-\Delta y$).

• Velocity and Speed:

  • Average Velocity: The total displacement divided by the time interval during which the displacement occurred.         $v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$

  • SI Units: Meters per second ($\text{m/s}$). Velocity can also be expressed in kilometers per hour ($\text{km/h}$).

  • The sign of average velocity directly corresponds to the sign of displacement because the elapsed time interval ($\Delta t$) is always positive.

  • Speed versus Velocity: Speed is a scalar quantity possessing magnitude only, defined as total distance traveled divided by the time of travel:         $\text{average speed} = \frac{\text{distance traveled}}{\text{time of travel}}$

  • Velocity is a vector quantity that possesses both a numerical magnitude (indicating speed) and a directional component. Two objects can possess identical speeds while having entirely different velocities if their directions of travel differ.

  • Average Velocity Behavior: Average velocity represents the equivalent constant velocity required to traverse a given net displacement over a designated time interval. It does not reflect momentary variations or intermediate stops during travel.

• Graphical Interpretation of Position versus Time:

  • On a graph of position versus time ($x$ vs. $t$), time is plotted on the horizontal axis and position on the vertical axis.

  • Slope as Velocity: The slope of a straight line connecting any two points on a position-time graph equals the average velocity across that time interval:         $\text{slope} = \frac{\text{rise}}{\text{run}} = \frac{\Delta x}{\Delta t} = v_{\text{avg}}$

  • A straight line with a constant positive slope indicates uniform motion with constant positive velocity.

  • A horizontal line (slope $= 0$) indicates that the object is at rest.

  • A straight line with a constant negative slope indicates uniform motion with constant negative velocity.

  • Instantaneous Velocity: The velocity of an object at a specific instant or at a precise point along its path. For an object undergoing changing velocity (represented by a curved position-time graph), instantaneous velocity is found by calculating the slope of the straight tangent line drawn to the curve at that precise point in time.

• Acceleration:

  • Definition: Acceleration is the rate at which velocity changes over time. An object accelerates if its speed changes, its direction changes, or both change simultaneously.

  • Average Acceleration Equation:         $a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}$

  • Dimensions and SI Units: Dimensions are length divided by time squared ($\text{length}/\text{time}^2$). The SI unit of acceleration is meters per second squared ($\text{m/s}^2$), which represents how many meters per second the velocity changes during each elapsed second ($(\text{m/s})/\text{s} = \text{m/s}^2$).

  • Signs of Velocity and Acceleration:

    • Positive initial velocity ($v_i > 0$) and positive acceleration ($a > 0$): Object is speeding up in the positive direction.

    • Negative initial velocity ($v_i < 0$) and negative acceleration ($a < 0$): Object is speeding up in the negative direction.

    • Positive initial velocity ($v_i > 0$) and negative acceleration ($a < 0$): Object is slowing down (decelerating) in the positive direction.

    • Negative initial velocity ($v_i < 0$) and positive acceleration ($a > 0$): Object is slowing down in the negative direction.

    • Nonzero velocity ($v_i \neq 0$) and zero acceleration ($a = 0$): Object moves at constant velocity.

• Zero velocity ($v_i = 0$) and nonzero acceleration ($a \neq 0$): Object is speeding up from rest in the direction of the acceleration. * Zero velocity ($v_i = 0$) and zero acceleration ($a = 0$): Object remains at rest.

  • Graphical Interpretation of Velocity versus Time:

    • On a velocity-time graph ($v$ vs. $t$), the slope of the curve represents the acceleration of the object ($\text{slope} = \frac{\Delta v}{\Delta t} = a$).

    • An upward-sloping line indicates positive acceleration.

    • A horizontal line indicates zero acceleration (constant velocity).

    • A downward-sloping line indicates negative acceleration.

• Kinematic Equations for Constant Acceleration:

  • For an object undergoing constant acceleration, the average velocity across an interval equals the arithmetic mean of its initial and final velocities:         $v_{\text{avg}} = \frac{v_i + v_f}{2}$

  • Equating $v_{\text{avg}} = \frac{\Delta x}{\Delta t}$ with $\frac{v_i + v_f}{2}$ yields the displacement equation without explicit acceleration:         $\Delta x = \frac{1}{2}(v_i + v_f)\Delta t$

  • Rearranging the acceleration definition gives velocity as a function of time:         $v_f = v_i + a\Delta t$

  • Substituting the expression for $v_f$ into the displacement equation yields displacement as a function of initial velocity, acceleration, and time:         $\Delta x = v_i\Delta t + \frac{1}{2}a(\Delta t)^2$

  • Eliminating $\Delta t$ between $v_f = v_i + a\Delta t$ and $\Delta x = \frac{1}{2}(v_i + v_f)\Delta t$ yields final velocity as a function of displacement:         $v_f^2 = v_i^2 + 2a\Delta x$

  • Special Case for Objects Accelerating from Rest ($v_i = 0$):         $\Delta x = \frac{1}{2}v_f\Delta t$         $v_f = a\Delta t$         $\Delta x = \frac{1}{2}a(\Delta t)^2$         $v_f^2 = 2a\Delta x$

Free Fall and Gravitational Acceleration

• The Physics of Free Fall:

  • Definition: Free fall is the motion of an object when only the force due to gravity is acting on the body, with air resistance completely disregarded.

  • In a vacuum, all objects dropped near a celestial body fall with identical, constant acceleration regardless of their mass, shape, surface area, or composition.

  • Historical Moon Demonstration: On August 2, 1971, Apollo 15 astronaut David Scott simultaneously released a hammer and a falcon feather from equal heights above the lunar surface. Because of the absence of a lunar atmosphere, both objects fell at the exact same rate and landed simultaneously, verifying that gravitational acceleration is independent of mass.

  • Free-Fall Acceleration Constant ($g$):

    • On Earth's surface, the magnitude of the acceleration due to gravity is denoted by $g$, approximately equal to $9.81\,\text{m/s}^2$ (or $32\,\text{ft/s}^2$).

    • Because gravity pulls downward toward Earth's center, in standard Cartesian coordinates where upward is positive, free-fall acceleration is directed downward:             $a_g = -g = -9.81\,\text{m/s}^2$

• Symmetry in Upward and Downward Projectile Motion:

  • When an object is thrown vertically upward with an initial upward velocity ($v_i > 0$), it experiences a constant downward acceleration of $-9.81\,\text{m/s}^2$ from the instant it leaves the hand.

  • As it ascends, its positive velocity decreases at a rate of $9.81\,\text{m/s}$ each second until it reaches its maximum height.

  • At Maximum Altitude (Peak): The object's instantaneous velocity is momentarily zero ($v = 0\,\text{m/s}$), but its acceleration remains precisely $-9.81\,\text{m/s}^2$. Acceleration is never zero at the peak.

  • As it descends, its velocity becomes negative and increases in magnitude in the downward direction at the rate of $9.81\,\text{m/s}$ each second.

  • In the absence of air resistance, the flight is completely symmetrical: the time required to rise to the peak equals the time required to fall back to the initial elevation, and the speed at any given elevation during descent equals the speed at that same elevation during ascent.

• Atmospheric Drag and Terminal Velocity:

  • When an object moves through a fluid medium such as air, the fluid produces a resistive drag force ($F_R$) opposing the motion.

  • At low speeds, $F_R$ is roughly proportional to the object's speed ($F_R \propto v$); at higher speeds, $F_R$ is roughly proportional to the square of the speed ($F_R \propto v^2$).

  • As a falling object accelerates, its speed and the opposing upward air resistance both increase until the upward drag force balances the downward gravitational force ($\sum F = 0$).

  • Terminal Velocity: The maximum, constant falling speed achieved when net acceleration drops to zero ($a = 0\,\text{m/s}^2$).

  • Terminal velocity depends on the object's mass, cross-sectional area, and shape:

    • Skydiver spread out horizontally belly-to-earth: terminal velocity is approximately $55\,\text{m/s}$ ($123\,\text{mi/h}$).

    • Skydiver curled tightly into a head-first dive or ball: terminal velocity increases up to approximately $90\,\text{m/s}$ ($200\,\text{mi/h}$).

    • Fastest terminal velocity recorded for a human skydiver: $148\,\text{m/s}$ ($331\,\text{mi/h}$) in 2015.

    • Skydiver with fully deployed parachute: terminal velocity decreases to approximately $5\,\text{m/s}$ ($11\,\text{mi/h}$).

Angular Kinematics and Rotational Motion

• Angular Position and Radian Measure:

  • Circular motion describes a point on an object rotating about a fixed axis. Linear quantities cannot describe rotational motion of extended bodies effectively because different points along the radius move through different linear distances in the same time interval.

  • All points on a rigid rotating body (except points directly on the axis of rotation) sweep through the identical angle during any given time interval.

  • Radian Definition: The angle $\theta$ subtended at the center of a circular path by an arc length $s$ equal in length to the radius $r$:         $\theta = \frac{s}{r}$

  • Because the radian is the ratio of two lengths ($\text{arc length}/\text{radius}$), units cancel; the radian ($\text{rad}$) is a dimensionless pure number.

  • For a complete circle ($360^\circ$), the arc length is the circumference ($s = 2\pi r$):         $\theta = \frac{2\pi r}{r} = 2\pi\,\text{rad} \approx 6.28\,\text{rad}$

  • Conversion Between Radians and Degrees:         $\theta\,(\text{rad}) = \frac{\pi}{180^\circ}\theta\,(\text{deg})$

• Angular Displacement, Velocity, and Acceleration:

  • Angular Displacement ($\Delta \theta$): The angle through which a point, line, or body rotates about a specified axis:         $\Delta \theta = \frac{\Delta s}{r}$         where $\Delta s$ is the change in linear arc length. By convention, counterclockwise rotation is positive ($+\Delta \theta$) and clockwise rotation is negative ($-\Delta \theta$).

  • Average Angular Velocity ($\omega_{\text{avg}}$, Greek letter omega): The ratio of angular displacement to elapsed time:         $\omega_{\text{avg}} = \frac{\Delta \theta}{\Delta t} = \frac{\theta_2 - \theta_1}{t_2 - t_1}$         The SI unit of angular velocity is radians per second ($\text{rad/s}$). Angular speed refers to the absolute magnitude of angular velocity. Rotational speed may also be given in revolutions per unit time ($1\,\text{rev} = 2\pi\,\text{rad}$).

  • Average Angular Acceleration ($\alpha_{\text{avg}}$, Greek letter alpha): The time rate of change of angular velocity:         $\alpha_{\text{avg}} = \frac{\Delta \omega}{\Delta t} = \frac{\omega_2 - \omega_1}{t_2 - t_1}$         The SI unit of angular acceleration is radians per second squared ($\text{rad/s}^2$).

  • Rigid Body Condition: In any rotating rigid structure (such as a bicycle wheel or Ferris wheel), every portion of the object must possess the identical angular velocity and angular acceleration to preserve structural shape.

• Rotational versus Linear Kinematic Analogies:

  • Angular displacement $\theta$ corresponds to linear displacement $x$.

  • Angular velocity $\omega$ corresponds to linear velocity $v$.

  • Angular acceleration $\alpha$ corresponds to linear acceleration $a$.

  • Kinematic Equations for Constant Angular Acceleration:         $\omega_f = \omega_i + \alpha\Delta t$         $\Delta \theta = \omega_i\Delta t + \frac{1}{2}\alpha(\Delta t)^2$         $\omega_f^2 = \omega_i^2 + 2\alpha\Delta \theta$         $\Delta \theta = \frac{1}{2}(\omega_i + \omega_f)\Delta t$

Vector Operations and Two-Dimensional Kinematics

• Vectors versus Scalars:

  • Scalar: A physical quantity that has magnitude but no directional orientation. Examples include speed, volume, mass, temperature, time, and distance.

  • Vector: A physical quantity that possesses both numerical magnitude and specific spatial direction. Examples include displacement, velocity, acceleration, force, and momentum.

  • Notation Conventions: In print, vector variables appear in boldface type ($\mathbf{v}$, $\mathbf{F}$, $\mathbf{a}$), whereas scalar magnitudes appear in standard italicized font ($v$, $F$, $a$). In handwritten work, vectors are represented with an arrow drawn above the symbol ($\vec{v}$).

• Graphical Vector Addition and Subtraction:

  • Vectors are represented graphically by directed arrows. The length of the arrow is scaled to be directly proportional to the vector's magnitude, and the arrowhead indicates orientation.

  • Resultant Vector ($\mathbf{R}$): A single vector representing the vector sum of two or more individual vectors.

  • Triangle (Polygon / Head-to-Tail) Method:

    • Draw the first vector to scale in the proper direction.

    • Draw the second vector starting with its tail placed precisely at the tip (head) of the first vector, maintaining its original scale and orientation.

    • Draw the resultant vector from the tail of the very first vector directly to the tip of the final vector.

    • Measure the resultant length with a ruler to find magnitude and measure the angle using a protractor.

  • Properties of Vectors:

    • Translation Invariance: Vectors may be moved parallel to themselves in a diagram without altering their physical value, provided magnitude and direction are preserved.

    • Commutative Property of Addition: Vectors can be added in any order without changing the resultant sum:             $\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$

    • Vector Subtraction: Vector subtraction is executed by adding the negative (opposite) vector:             $\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})$             The negative of a vector is defined as a vector of equal magnitude pointing in the exactly opposite direction ($180^\circ$ inversion).

    • Scalar Multiplication: Multiplying a vector by a positive scalar changes its magnitude by that factor; multiplying by a negative scalar changes its magnitude and reverses its spatial direction.

• Analytical Vector Operations:

  • Perpendicular Resultants (Pythagorean Theorem and Tangent Function):

    • Magnitude: For perpendicular vector components $\Delta x$ and $\Delta y$:             $d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$

    • Direction: Angle $\theta$ measured relative to the adjacent leg ($x$-axis):             $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\Delta y}{\Delta x}$             $\theta = \tan^{-1}\left(\frac{\Delta y}{\Delta x}\right)$

  • Resolving Vectors into Components: Any vector oriented at an angle $\theta$ to a Cartesian axis can be broken down into perpendicular scalar components parallel to the coordinate axes:         $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$         $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$

    • When $\theta$ is referenced to the horizontal axis:             $v_x = v\cos(\theta)$             $v_y = v\sin(\theta)$

  • Adding Non-Perpendicular Vectors Algebraically:

    • Step 1: Resolve each vector into its horizontal ($x$) and vertical ($y$) scalar components using sine and cosine functions.

    • Step 2: Calculate the net sum of all horizontal components to find the total horizontal displacement:             $\Delta x_{\text{tot}} = \sum \Delta x_i = \Delta x_1 + \Delta x_2 + \dots$

    • Step 3: Calculate the net sum of all vertical components to find the total vertical displacement:             $\Delta y_{\text{tot}} = \sum \Delta y_i = \Delta y_1 + \Delta y_2 + \dots$

    • Step 4: Determine the magnitude of the overall resultant vector using the Pythagorean theorem:             $d = \sqrt{(\Delta x_{\text{tot}})^2 + (\Delta y_{\text{tot}})^2}$

    • Step 5: Determine the resultant direction using the inverse tangent function:             $\theta = \tan^{-1}\left(\frac{\Delta y_{\text{tot}}}{\Delta x_{\text{tot}}}\right)$

Projectile Motion and Two-Dimensional Trajectories

• Fundamental Principles of Projectile Motion:

  • Projectile Motion: The two-dimensional motion of an object thrown, launched, or projected into the air near the Earth's surface under the sole influence of gravity.

  • Parabolic Trajectory: In the absence of air resistance, the path traced by a projectile is a parabola.

  • Independence of Motion Components: The horizontal motion and vertical motion of a projectile are completely independent of each other.

    • Horizontal velocity ($v_x$) remains constant throughout the flight because no horizontal force acts on the object ($a_x = 0\,\text{m/s}^2$).

    • Vertical motion exhibits constant downward acceleration equal to free-fall acceleration ($a_y = -g = -9.81\,\text{m/s}^2$).

  • A projectile launched horizontally from a given height strikes level ground at the exact same instant as an object dropped vertically from rest from that identical height.

• Mathematical Models of Projectile Motion:

  • Horizontally Launched Projectiles ($v_{y,i} = 0$):

    • Horizontal Equations:             $v_x = v_{x,i} = \text{constant}$             $\Delta x = v_x\Delta t$

    • Vertical Equations:             $v_{y,f} = a_y\Delta t = -g\Delta t$             $v_{y,f}^2 = 2a_y\Delta y = -2g\Delta y$             $\Delta y = \frac{1}{2}a_y(\Delta t)^2 = -\frac{1}{2}g(\Delta t)^2$

    • Flight Time:             $\Delta t = \sqrt{\frac{2\Delta y}{a_y}} = \sqrt{\frac{-2\Delta y}{g}}$

  • Projectiles Launched at an Angle $\theta$ Above the Horizontal:

    • Initial Component Resolution:             $v_{x,i} = v_i\cos(\theta)$             $v_{y,i} = v_i\sin(\theta)$

    • Horizontal Equations:             $v_x = v_i\cos(\theta) = \text{constant}$             $\Delta x = (v_i\cos(\theta))\Delta t$

    • Vertical Equations:             $v_{y,f} = v_i\sin(\theta) + a_y\Delta t$             $v_{y,f}^2 = v_i^2(\sin(\theta))^2 + 2a_y\Delta y$             $\Delta y = (v_i\sin(\theta))\Delta t + \frac{1}{2}a_y(\Delta t)^2$

  • Peak Conditions: At the apex of the parabolic flight path, the vertical velocity component is momentarily zero ($v_{y} = 0\,\text{m/s}$), while the horizontal velocity component remains $v_x = v_i\cos(\theta)$. The total speed at the peak is $v = v_x$.

  • Instantaneous Resultant Velocity: At any point along the flight path, the total velocity vector is found from its components:         $v = \sqrt{v_x^2 + v_{y,f}^2}$         $\theta = \tan^{-1}\left(\frac{v_{y,f}}{v_x}\right)$

Relative Motion and Frames of Reference

• Relativity of Velocity:

  • All velocity measurements depend entirely on the chosen frame of reference of the observer. Two observers in relative motion will measure different velocities for the exact same moving object.

  • For example, an object dropped inside an airplane cruising horizontally moves in a straight vertical line relative to an observer inside the cabin, but follows a parabolic trajectory relative to an observer stationed on the ground.

• Subscript Notation for Relative Velocity:

  • Relative velocity problems are formulated by assigning two-letter subscripts indicating the object of interest and the reference frame.

  • $\mathbf{v}_{ab}$ denotes the velocity of object $a$ with respect to reference frame $b$.

  • $\mathbf{v}_{bc}$ denotes the velocity of frame $b$ with respect to frame $c$.

  • $\mathbf{v}_{ac}$ denotes the velocity of object $a$ with respect to reference frame $c$.

  • General Vector Addition Equation for Relative Velocity:         $\mathbf{v}_{ac} = \mathbf{v}_{ab} + \mathbf{v}_{bc}$         The intermediate subscript ($b$) connects the inner terms, with the resultant retaining the initial and final subscripts.

  • Reversal Rule: Reversing reference frame subscripts reverses the vector direction (changes the algebraic sign):         $\mathbf{v}_{ba} = -\mathbf{v}_{ab}$

  • For collinear motion in one dimension, vectors simplify to signed scalar arithmetic. For perpendicular or angled relative motion (such as a boat crossing a river with a current, or a plane flying across wind currents), vector resolution and the Pythagorean theorem are applied:         $v_{ac} = \sqrt{v_{ab}^2 + v_{bc}^2}$         $\theta = \tan^{-1}\left(\frac{v_{bc}}{v_{ab}}\right)$

Classical Dynamics: Newton's Laws of Motion

• Force Concepts and Classification:

  • Definition: A force is an action exerted on an object that can alter the object's state of rest or motion. Forces describe physical interactions between an object and its surrounding environment.

  • Forces cause accelerations (changes in velocity over time), which include initiating motion from rest, bringing moving objects to rest, or altering the direction of travel.

  • SI Unit of Force: The newton ($\text{N}$), defined as the amount of net force required to impart an acceleration of $1\,\text{m/s}^2$ to a mass of $1\,\text{kg}$:         $1\,\text{N} = 1\,\text{kg} \times 1\,\text{m/s}^2 = 1\,\text{kg}\cdot\text{m/s}^2$

  • Unit Systems of Mass, Acceleration, and Force:

    • SI: Mass in kilograms ($\text{kg}$), acceleration in $\text{m/s}^2$, force in newtons ($\text{N} = \text{kg}\cdot\text{m/s}^2$).

    • cgs: Mass in grams ($\text{g}$), acceleration in centimeters per second squared ($\text{cm/s}^2$), force in dynes ($\text{dyne} = \text{g}\cdot\text{cm/s}^2$).

    • Avoirdupois (English): Mass in slugs, acceleration in feet per second squared ($\text{ft/s}^2$), force in pounds ($\text{lb} = \text{slug}\cdot\text{ft/s}^2$).

    • Conversions: $1\,\text{lb} = 4.448\,\text{N}$; $1\,\text{N} = 0.225\,\text{lb}$. (A standard $1/4\,\text{lb}$ stick of margarine exerts an approximate gravitational force of $1\,\text{N}$).

  • Contact versus Field Forces:

    • Contact Forces: Forces that result from physical contact between macroscopic bodies (e.g., pushes, pulls, normal force, tension, friction).

    • Field Forces: Forces exerted through space without physical contact. Masses generate gravitational fields, while electric charges generate electromagnetic fields. Macroscopic contact forces are fundamentally composed of microscopic electromagnetic interactions between atomic electron clouds.

• Free-Body Diagrams:

  • A force diagram displays all force vectors acting across an entire system.

  • A free-body diagram isolates a single object of interest, representing only the external forces exerted on that specific object. Forces exerted by the object on other external bodies are omitted because they do not affect the isolated body's acceleration.

  • Forces are drawn as vectors pointing away from the object's center of mass, with arrow lengths proportional to force magnitudes.

• Newton's First Law (Law of Inertia):

  • Statement: An object at rest remains at rest, and an object in motion continues in motion with constant velocity (constant speed in a straight line) unless the object experiences a net external force.

  • Inertia: The tendency of an object to resist changes in its state of motion or rest. Mass is the direct scalar measure of an object's inertia.

  • Net Force ($\sum \mathbf{F}$ or $\mathbf{F}_{\text{net}}$) The single vector sum of all external forces acting on an object:         $\mathbf{F}_{\text{net}} = \sum \mathbf{F}$

  • Equilibrium: A state in which the vector sum of all external forces acting on an object is zero ($\sum \mathbf{F} = 0$):         $\sum F_x = 0$         $\sum F_y = 0$

  • An object in equilibrium has zero acceleration ($\mathbf{a} = 0\,\text{m/s}^2$), moving at constant velocity or remaining at rest.

• Newton's Second Law:

  • Statement: The acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object's mass.

  • Equation:         $\sum \mathbf{F} = m\mathbf{a}$

  • Component form for two-dimensional dynamics:         $\sum F_x = ma_x$         $\sum F_y = ma_y$

  • The direction of the resulting acceleration is identical to the direction of the net external force.

• Newton's Third Law (Action-Reaction):

  • Statement: If two objects interact, the magnitude of the force exerted on object 1 by object 2 is equal to the magnitude of the force simultaneously exerted on object 2 by object 1, and these two forces are opposite in direction.         $\mathbf{F}_{12} = -\mathbf{F}_{21}$

  • Action and reaction forces occur simultaneously and always act on two different bodies. Consequently, action-reaction pairs never cancel each other out on a single object.

  • Newton's third law applies equally to contact forces (e.g., hammer on nail and nail on hammer) and field forces (e.g., Earth pulling downward on a falling body and the falling body simultaneously pulling upward on Earth with equal gravitational force).

Everyday Forces: Weight, Normal Force, and Friction

• Mass versus Weight:

  • Mass ($m$): An intrinsic scalar property of a body quantifying the amount of matter and the measure of its inertia. Mass is invariant across locations.

  • Weight ($F_g$): A scalar measure of the gravitational force exerted on an object by a celestial body:         $F_g = mg$

  • Because weight depends on free-fall acceleration ($g$), weight varies with celestial location and altitude. On the Moon ($g_{\text{moon}} \approx 1.6\,\text{m/s}^2$), an astronaut weighing $800\,\text{N}$ ($180\,\text{lb}$) on Earth weighs approximately $130\,\text{N}$ ($30\,\text{lb}$).

• The Normal Force ($F_n$):

  • Definition: The contact force exerted by a supporting surface on an object, acting in a direction perpendicular (normal) to the contact surface.

  • On a flat horizontal surface with no vertical applied forces, the normal force balances weight:         $F_n = mg$

  • On an inclined plane angled at $\theta$ relative to the horizontal, the normal force balances the perpendicular component of gravity:         $F_n = mg\cos(\theta)$

• The Force of Friction:

  • Friction is a contact force parallel to the contacting surfaces that opposes sliding motion.

  • At microscopic scales, friction arises from adhesive electrostatic forces and contact surface roughness at asperities.

  • Static Friction ($F_s$): The resistive force opposing the initiation of relative sliding motion between two stationary surfaces in contact. As long as the object remains at rest, $F_s = -F_{\text{applied}}$.

  • Maximum Static Friction ($F_{s,\text{max}}$): The maximum threshold force before motion begins:         $F_{s,\text{max}} = \mu_s F_n$

  • Kinetic Friction ($F_k$): The retarding force acting between surfaces sliding past each other:         $F_k = \mu_k F_n$

  • Coefficients of Friction ($\mu$): Dimensionless ratios quantifying surface interactions. Because kinetic friction is less than or equal to maximum static friction, $\mu_k \le \mu_s$.

  • Approximate Experimental Coefficients of Friction:

    • Steel on steel: $\mu_s = 0.74$, $\mu_k = 0.57$

    • Aluminum on steel: $\mu_s = 0.61$, $\mu_k = 0.47$

    • Rubber on dry concrete: $\mu_s = 1.0$, $\mu_k = 0.8$

    • Rubber on wet concrete: $\mu_k = 0.5$

    • Wood on wood: $\mu_s = 0.4$, $\mu_k = 0.2$

    • Glass on glass: $\mu_s = 0.9$, $\mu_k = 0.4$

    • Waxed wood on wet snow: $\mu_s = 0.14$, $\mu_k = 0.1$

    • Waxed wood on dry snow: $\mu_s = 0.04$

    • Metal on metal (lubricated): $\mu_s = 0.15$, $\mu_k = 0.06$

    • Ice on ice: $\mu_s = 0.1$, $\mu_k = 0.03$

    • Teflon on Teflon: $\mu_s = 0.04$, $\mu_k = 0.04$

    • Synovial joints in humans: $\mu_s = 0.01$, $\mu_k = 0.003$

• The Four Fundamental Forces of Nature:

  • Strong Nuclear Force: The strongest force in nature; binds quarks together to form protons and neutrons, and binds nucleons within atomic nuclei. Possesses an extremely short subatomic range.

  • Electromagnetic Force: Responsible for interactions between charged particles; holds atoms and molecules together; underlies all macroscopic contact forces, tension, normal force, and friction. Long range.

  • Weak Nuclear Force: Governs certain radioactive decay processes and subatomic interactions. Extremely short range.

  • Gravitational Force: The weakest fundamental force; acts over infinite range between all masses; governs planetary orbits, stars, and cosmological structures.

Work, Energy, and the Work-Kinetic Energy Theorem

• Scientific Definition of Work:

  • Definition: Work ($W$) is the product of the component of a force along the direction of displacement and the magnitude of the displacement.

  • General Work Equation:         $W = Fd\cos(\theta)$         where $F$ is the applied force magnitude, $d$ is the displacement magnitude, and $\theta$ is the angle between the force vector and the displacement vector.

  • Net Work Equation:         $W_{\text{net}} = F_{\text{net}}d\cos(\theta)$

  • Conditions for Work:

    • If $d = 0$, then $W = 0$ (e.g., pushing against an immovable wall or holding a chair stationary does zero work on the object).

    • If $\theta = 90^\circ$, then $\cos(90^\circ) = 0$, so $W = 0$ (e.g., carrying a bucket of water horizontally at constant velocity involves an upward supporting force perpendicular to displacement; no work is done on the bucket).

  • SI Unit of Work: The joule ($\text{J}$), named after James Prescott Joule:         $1\,\text{J} = 1\,\text{N}\cdot\text{m} = 1\,\text{kg}\cdot\text{m}^2/\text{s}^2$

  • Lifting an apple vertically from waist level to head level does approximately $1\,\text{J}$ of work.

  • Sign Conventions of Work:

    • Positive Work: Force component points in the same direction as displacement ($0^\circ \le \theta < 90^\circ$), speeding the object up.

    • Negative Work: Force component opposes displacement ($90^\circ < \theta \le 180^\circ$), slowing the object down (e.g., kinetic friction).

• Kinetic Energy:

  • Definition: Kinetic energy ($KE$) is the energy associated with an object's motion.

  • Kinetic Energy Equation:         $KE = \frac{1}{2}mv^2$         where $m$ is object mass and $v$ is scalar speed. Kinetic energy is a scalar quantity and is measured in joules ($\text{J}$).

• The Work-Kinetic Energy Theorem:

  • Derivation from Dynamics: For a net force $F_{\text{net}} = ma$ acting over displacement $\Delta x$:         $W_{\text{net}} = F_{\text{net}}\Delta x = ma\Delta x$         From kinematics, $a\Delta x = \frac{v_f^2 - v_i^2}{2}$. Substituting yields:         $W_{\text{net}} = m\left(\frac{v_f^2 - v_i^2}{2}\right) = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

  • Work-Kinetic Energy Theorem Statement:         $W_{\text{net}} = \Delta KE = KE_f - KE_i$         The net work done by all forces acting on an object equals the net change in the object's kinetic energy.

• Potential Energy:

  • Definition: Potential energy ($PE$) is stored energy resulting from an object's spatial position relative to its environment or internal mechanical state.

  • Gravitational Potential Energy ($PE_g$): Energy associated with an object's position relative to a gravitational field:         $PE_g = mgh$         where $h$ is height measured from an arbitrary zero reference level.

  • Elastic Potential Energy ($PE_{\text{elastic}}$): Energy stored in a deformed elastic object (e.g., stretched or compressed spring):         $PE_{\text{elastic}} = \frac{1}{2}kx^2$         where $k$ is the spring constant (force constant, in $\text{N/m}$) representing stiffness, and $x$ is the distance compressed or stretched from the relaxed (unstressed) equilibrium position.

Conservation of Energy and Power

• Conservation of Mechanical Energy:

  • Conserved Quantity: A physical property that remains constant in magnitude over time within a closed system, even if it changes form.

  • Mechanical Energy ($ME$): The sum of kinetic energy and all forms of potential energy associated with an object or group of objects:         $ME = KE + \sum PE = KE + PE_g + PE_{\text{elastic}}$

  • Nonmechanical Energy: All energy forms not associated with macroscopic motion or position (e.g., chemical, electrical, nuclear, thermal/internal energy).

  • Principle of Conservation of Mechanical Energy: In the absence of non-conservative forces such as friction or drag, the total mechanical energy of an isolated system remains constant:         $ME_i = ME_f$         $KE_i + PE_{g,i} + PE_{\text{elastic},i} = KE_f + PE_{g,f} + PE_{\text{elastic},f}$         $\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$

  • For an object descending from rest ($v_i = 0$) through height $h$ on a frictionless path of any shape:         $mgh = \frac{1}{2}mv_f^2 \implies v_f = \sqrt{2gh}$         The final speed is independent of the object's mass and independent of the incline path's shape.

  • Energy Dissipation in the Presence of Friction: When kinetic friction or fluid drag is present, mechanical energy is converted into nonmechanical thermal energy. Total energy is conserved, but mechanical energy decreases:         $W_{\text{friction}} = \Delta ME = ME_f - ME_i$

• Power:

  • Definition: Power ($P$) is the rate at which work is performed or the rate of energy transfer by any mechanism.

  • Equations for Power:         $P = \frac{W}{\Delta t}$         Substituting $W = Fd$ for a constant force along the displacement:         $P = \frac{Fd}{\Delta t} = F\left(\frac{d}{\Delta t}\right) = Fv$

  • SI Unit of Power: The watt ($\text{W}$), defined as one joule per second:         $1\,\text{W} = 1\,\text{J/s} = 1\,\text{kg}\cdot\text{m}^2/\text{s}^3$

  • English Unit: The horsepower ($\text{hp}$):         $1\,\text{hp} = 746\,\text{W}$

Modern Physics Connections: Special Relativity and Mass-Energy Equivalence

• Einstein's Postulates of Special Relativity (1905):

  • First published in Albert Einstein's landmark paper "The Electrodynamics of Moving Bodies" (1905).

  • Postulate 1: The laws of physics are identical in all inertial (non-accelerating, uniform motion) reference frames.

  • Postulate 2: The speed of light in a vacuum ($c = 3.00 \times 10^8\,\text{m/s}$) is absolute and universal, possessing the exact same value for all observers regardless of the relative motion of the source or the observer.

• Time Dilation:

  • Because the speed of light $c$ is constant, an observer on a stationary platform watching a light pulse bounce vertically inside a uniformly moving train observes the light traveling a longer diagonal path than does the passenger on the train.

  • Because the path distance is greater while speed is fixed at $c$, the stationary observer measures a longer time interval ($\Delta t'$) than the passenger's clock interval ($\Delta t$):         $\Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}$         where $v$ is relative speed and $c$ is the speed of light.

  • As $v \to c$, the denominator approaches zero and $\Delta t' \to \infty$.

  • Experimental Verifications:

    • Decay of atmospheric muons: High-speed muons produced in the upper atmosphere traveling at $0.9994c$ survive significantly longer from Earth's frame than stationary muons in laboratory environments.

    • Atomic clocks on commercial jet planes: Clocks flown around the world exhibited precise elapsed time discrepancies compared to stationary atomic clocks at the U.S. Naval Observatory, matching relativistic predictions within experimental error.

• Relativistic Addition of Velocities:

  • Classical Galilean velocity addition ($u = v + u'$) fails as velocities approach the speed of light.

  • Einstein's Relativistic Addition Equation:         $u = \frac{v + u'}{1 + \frac{vu'}{c^2}}$         where $u'$ is speed measured in moving frame, $v$ is relative frame speed, and $u$ is speed measured in the primary frame.

  • Comparison of Velocity Addition Results:

    • $v = 25\,\text{m/s}$, $u' = 15\,\text{m/s}$:             Classical: $u = 40\,\text{m/s}$; Relativistic: $u = 40\,\text{m/s}$

    • $v = 100\,000\,\text{m/s}$, $u' = 100\,000\,\text{m/s}$:             Classical: $u = 200\,000\,\text{m/s}$; Relativistic: $u = 200\,000\,\text{m/s}$

    • $v = 0.50c$, $u' = 0.50c$:             Classical: $u = 1.00c = 299\,792\,458\,\text{m/s}$; Relativistic: $u = 0.80c = 239\,833\,966\,\text{m/s}$

    • $v = 0.90c$, $u' = 0.90c$:             Classical: $u = 1.80c = 539\,626\,424\,\text{m/s}$; Relativistic: $u = 0.9945c = 298\,136\,146\,\text{m/s}$

    • $v = 0.9999c$, $u' = 0.9999c$:             Classical: $u = 1.9998c = 599\,524\,958\,\text{m/s}$; Relativistic: $u = 299\,792\,457\,\text{m/s}$

  • When a light beam is shone ($u' = c$):         $u = \frac{v + c}{1 + \frac{vc}{c^2}} = \frac{v + c}{1 + \frac{v}{c}} = \frac{c(v + c)}{c + v} = c$         No physical object or reference frame can accelerate to or exceed $c$.

• Relativistic Kinetic Energy and Rest Energy:

  • Relativistic Kinetic Energy Formula:         $KE = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} - mc^2$         At speeds where $v \ll c$, this reduces algebraically to $\frac{1}{2}mv^2$.

  • Rest Energy ($E_R$): Mass itself represents stored energy:         $E_R = mc^2$         Because the conversion factor $c^2 \approx 9.00 \times 10^{16}\,\text{m}^2/\text{s}^2$ is immense, tiny masses hold colossal energy.

  • Nuclear Reactions:

    • Nuclear Fission: Splitting atomic nuclei converts roughly $0.1\%$ ($1/1000$) of the original rest mass into kinetic and electromagnetic energy.

    • Nuclear Fusion: In the Sun, hydrogen fuses into helium, converting approximately $4.5 \times 10^6\,\text{tons}$ of solar mass into radiant energy every second. The Sun has sufficient fuel reserves to maintain fusion for approximately $5 \times 10^9\,\text{years}$.

    • Particle Accelerators: At the Stanford Linear Accelerator Center (SLAC) in California, electrons accelerate to $99.999999967\%$ of the speed of light, verifying relativistic predictions.

Structural Engineering, Biomechanics, and Historical Development

• Structural Mechanics and Engineered Systems:

  • Elevators ("Lifting Rooms"):

    • Elevators convert horizontal urban expansion into efficient vertical transport, utilizing tension lines and counterweights.

    • The counterweight balances the elevator cab weight, minimizing motor power requirements during lifting.

    • Safety Mechanisms: A centrifugal "fall feeler" (governor) spins as the room moves; if speed exceeds safe limits, flyweights swing outward, engaging teeth that trip safety grabbers. These grabbers lock directly onto vertical steel guide rails.

    • Secondary safety includes line stoppers, structural safety catches at the base, and multiple redundant hoist cables (typically $4$ to $8$ steel cables, where a single cable could support the cab).

  • Bridge Dynamics ("Tall Roads"):

    • Bridges resolve gravity into compressive and tensile structural elements.

    • Cantilever and thick spans support weight through bending resistance, but mass scales faster than material strength.

    • Suspension and cable-stayed structures distribute road deck load upward through vertical and angled steel stays to massive towers, transferring gravitational forces directly into foundational bedrock.

    • Aerodynamic design must mitigate lateral oscillating wind loads that caused historical bridge collapses.

• Kinesiology and Biomechanics:

  • Kinesiology applies vector mechanics to study physiological movement, encompassing motor control (neuroscience), biomechanics, and exercise physiology.

  • Thumb musculature produces orthogonal force vectors along two axes ($x$ and $y$); strain gauge transducers, bridge amplifiers, and analog-to-digital converters measure resultant vector forces ($F = \sqrt{F_x^2 + F_y^2}$) in real time.

  • Electrical stimulation protocols on peripheral nerves recreate coordinated muscle unit firing patterns to assist individuals with paralysis resulting from stroke or spinal trauma.

  • Astronaut Exercise Biomechanics: In microgravity, astronauts use weighted vests and spring harness systems on treadmills to recreate ground reaction forces, combating bone mineral density loss.

• Historical Timeline of Mechanics and Physics (1540–1690):

  • 1543: Andreas Vesalius (De Humani Corporis Fabrica) pioneers human anatomical science; Nicolaus Copernicus (De revolutionibus orbium coelestium) publishes the heliocentric astronomical model.

  • 1556: Akbar the Great ascends the Moghul Empire throne, fostering cultural unity in India.

  • 1564: Birth of Christopher Marlowe and William Shakespeare.

  • 1588: Defeat of the Spanish Armada by the English fleet under Queen Elizabeth I.

  • 1592: Galileo Galilei appointed professor of mathematics at the University of Padua, launching formal experimental mechanics.

  • 1603: Emergence of Kabuki theater in Japan.

  • 1605: Miguel de Cervantes publishes Don Quixote (Part I).

  • 1608–1609: Development of the optical telescope in the Netherlands; Galileo constructs his telescope; Johannes Kepler publishes Astronomia Nova, proving elliptical planetary orbits ($T^2 \propto a^3$).

  • 1632–1638: Galileo publishes the foundations of experimental mechanics and kinematics, establishing mathematical descriptions of motion.

  • 1637: René Descartes publishes Discourse on Method, establishing rationalist philosophy.

  • 1644: Qing (Manchu) Dynasty founded in China.

  • 1655: Jan Vermeer begins producing paintings depicting everyday Dutch life.

  • 1669: Nicolas Steno formulates structural principles of crystallography and stratigraphy, identifying fossils as biological remains.

  • 1678: Christiaan Huygens writes Treatise on Light, formulating wave theory and Huygens's principle ($v = f\lambda$).

  • 1687: Sir Isaac Newton publishes Philosophiae Naturalis Principia Mathematica, formulating the three universal laws of motion, classical mechanics, and universal gravitation ($\sum \mathbf{F} = m\mathbf{a}$, $F_g = G\frac{m_1 m_2}{r^2}$).

Complete Worked Sample Problems and Mathematical Exercises

• Sample Problem A (Displacement and Velocity):

  • Problem: Andrea runs during a race with an average velocity of $6.02\,\text{m/s}$ due east. Calculate Andrea's displacement after $137\,\text{s}$.

  • Given: $v_{\text{avg}} = +6.02\,\text{m/s}$, $\Delta t = 137\,\text{s}$

  • Formulation: $\Delta x = v_{\text{avg}}\Delta t$

  • Calculation: $\Delta x = (6.02\,\text{m/s})(137\,\text{s}) = 824.74\,\text{m}$

  • Result: $\Delta x = 825\,\text{m}$ due east (rounded to three significant figures).

• Sample Problem B (Average Acceleration):

  • Problem: A shuttle bus slows down with an average acceleration of $-1.8\,\text{m/s}^2$. How long does it take the bus to decelerate from $9.0\,\text{m/s}$ to a complete stop?

  • Given: $v_i = 9.0\,\text{m/s}$, $v_f = 0.0\,\text{m/s}$, $a_{\text{avg}} = -1.8\,\text{m/s}^2$

  • Formulation: $\Delta t = \frac{v_f - v_i}{a_{\text{avg}}}$

  • Calculation: $\Delta t = \frac{0.0\,\text{m/s} - 9.0\,\text{m/s}}{-1.8\,\text{m/s}^2} = \frac{-9.0\,\text{m/s}}{-1.8\,\text{m/s}^2} = 5.0\,\text{s}$

  • Result: $\Delta t = 5.0\,\text{s}$.

• Sample Problem C (Displacement with Constant Acceleration):

  • Problem: A racing car reaching $42\,\text{m/s}$ deploys its parachute and brakes, coming to a stop in $5.5\,\text{s}$. Find the distance traveled during braking.

  • Given: $v_i = 42\,\text{m/s}$, $v_f = 0.0\,\text{m/s}$, $\Delta t = 5.5\,\text{s}$

  • Formulation: $\Delta x = \frac{1}{2}(v_i + v_f)\Delta t$

  • Calculation: $\Delta x = \frac{1}{2}(42\,\text{m/s} + 0.0\,\text{m/s})(5.5\,\text{s}) = (21\,\text{m/s})(5.5\,\text{s}) = 115.5\,\text{m}$

  • Result: $\Delta x = 120\,\text{m}$ (two significant figures).

• Sample Problem D (Acceleration of a Barge):

  • Problem: A barge moving at $1.00\,\text{m/s}$ accelerates uniformly to travel $60.2\,\text{m}$ in $30.0\,\text{s}$. Determine its acceleration.

  • Given: $v_i = 1.00\,\text{m/s}$, $\Delta t = 30.0\,\text{s}$, $\Delta x = 60.2\,\text{m}$

  • Formulation: $\Delta x = v_i\Delta t + \frac{1}{2}a(\Delta t)^2 \implies a = \frac{2(\Delta x - v_i\Delta t)}{(\Delta t)^2}$

  • Calculation:         $a = \frac{2[60.2\,\text{m} - (1.00\,\text{m/s})(30.0\,\text{s})]}{(30.0\,\text{s})^2} = \frac{2(30.2\,\text{m})}{9.00 \times 10^2\,\text{s}^2} = \frac{60.4\,\text{m}}{900\,\text{s}^2} = 6.71 \times 10^{-2}\,\text{m/s}^2$

  • Result: $a = 6.71 \times 10^{-2}\,\text{m/s}^2$.

• Sample Problem E (Final Velocity After Displacement):

  • Problem: A baby stroller starts from rest, accelerating uniformly at $0.500\,\text{m/s}^2$. Find its velocity after rolling $4.75\,\text{m}$.

  • Given: $v_i = 0\,\text{m/s}$, $a = 0.500\,\text{m/s}^2$, $\Delta x = 4.75\,\text{m}$

  • Formulation: $v_f = \pm\sqrt{v_i^2 + 2a\Delta x}$

  • Calculation: $v_f = +\sqrt{0 + 2(0.500\,\text{m/s}^2)(4.75\,\text{m})} = +\sqrt{4.75\,\text{m}^2/\text{s}^2} = +2.18\,\text{m/s}$

  • Result: $v_f = +2.18\,\text{m/s}$ to the right.

• Sample Problem F (Free Fall from Gateway Arch):

  • Problem: Sally jumps off the Gateway Arch ($h = 192\,\text{m}$) in St. Louis, Missouri, landing on a safety pad in $6.4\,\text{s}$. Compute average acceleration and final impact speed.

  • Given: $v_i = 0\,\text{m/s}$, $\Delta y = -192\,\text{m}$, $\Delta t = 6.4\,\text{s}$

  • Formulation: $a = \frac{2(\Delta y - v_i\Delta t)}{(\Delta t)^2}$, $v_f = v_i + a\Delta t$

  • Calculation:         $a = \frac{2(-192\,\text{m})}{(6.4\,\text{s})^2} = \frac{-384\,\text{m}}{40.96\,\text{s}^2} = -9.375\,\text{m/s}^2 \approx -9.4\,\text{m/s}^2$         $v_f = 0 + (-9.375\,\text{m/s}^2)(6.4\,\text{s}) = -60\,\text{m/s} = -6.0 \times 10^1\,\text{m/s}$

  • Physical Interpretation: Downward acceleration is slightly less than $g = 9.81\,\text{m/s}^2$ due to air drag ($0.4\,\text{m/s}^2$ reduction).

• Sample Problem G (Vector Trigonometry on the Great Pyramid):

  • Problem: An archaeologist climbs from base to summit of the Great Pyramid of Giza ($h = 136\,\text{m}$, base width $= 2.30 \times 10^2\,\text{m}$). Climbing up the center of one face, find displacement magnitude and angle.

  • Given: $\Delta y = 136\,\text{m}$, $\Delta x = \frac{1}{2}(2.30 \times 10^2\,\text{m}) = 115\,\text{m}$

  • Formulation: $d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$, $\theta = \tan^{-1}\left(\frac{\Delta y}{\Delta x}\right)$

  • Calculation:         $d = \sqrt{(115\,\text{m})^2 + (136\,\text{m})^2} = \sqrt{13225 + 18496} = \sqrt{31721\,\text{m}^2} = 178\,\text{m}$         $\theta = \tan^{-1}\left(\frac{136\,\text{m}}{115\,\text{m}}\right) = \tan^{-1}(1.1826) = 49.8^\circ$

  • Result: Displacement is $178\,\text{m}$ at an angle of $49.8^\circ$ above horizontal.

• Sample Problem H (Resolving Interstate Highway Vector):

  • Problem: A straight stretch of Interstate 5 from Mettler to Buttonwillow, California, is $53.0\,\text{km}$ at $48.7^\circ$ north of west. Calculate northern and western displacement components.

  • Given: $d = 53.0\,\text{km}$, $\theta = 48.7^\circ$

  • Calculation:         $\Delta x = d\cos(\theta) = (53.0\,\text{km})\cos(48.7^\circ) = (53.0)(0.6600) = 35.0\,\text{km}\text{ west}$         $\Delta y = d\sin(\theta) = (53.0\,\text{km})\sin(48.7^\circ) = (53.0)(0.7513) = 39.8\,\text{km}\text{ north}$

  • Verification: $(35.0)^2 + (39.8)^2 = 1225 + 1584 = 2809 = (53.0)^2$.

• Sample Problem I (Adding Non-Perpendicular Vectors Algebraically):

  • Problem: A hiker walks $d_1 = 27.0\,\text{km}$ at $35.0^\circ$ south of east, then $d_2 = 41.0\,\text{km}$ at $65.0^\circ$ north of east. Find resultant displacement.

  • Given: $d_1 = 27.0\,\text{km}$, $\theta_1 = -35.0^\circ$; $d_2 = 41.0\,\text{km}$, $\theta_2 = 65.0^\circ$

  • Component Resolution:         $\Delta x_1 = (27.0\,\text{km})\cos(-35.0^\circ) = 22.1\,\text{km}$         $\Delta y_1 = (27.0\,\text{km})\sin(-35.0^\circ) = -15.5\,\text{km}$         $\Delta x_2 = (41.0\,\text{km})\cos(65.0^\circ) = 17.3\,\text{km}$         $\Delta y_2 = (41.0\,\text{km})\sin(65.0^\circ) = 37.2\,\text{km}$

  • Vector Sums:         $\Delta x_{\text{tot}} = 22.1\,\text{km} + 17.3\,\text{km} = 39.4\,\text{km} \approx 39\,\text{km}$         $\Delta y_{\text{tot}} = -15.5\,\text{km} + 37.2\,\text{km} = 21.7\,\text{km} \approx 22\,\text{km}$

  • Resultant Evaluation:         $d = \sqrt{(39.4\,\text{km})^2 + (21.7\,\text{km})^2} = \sqrt{1552 + 471} = \sqrt{2023} = 45\,\text{km}$         $\theta = \tan^{-1}\left(\frac{21.7\,\text{km}}{39.4\,\text{km}}\right) = \tan^{-1}(0.5508) = 29^\circ\text{ north of east}$

• Sample Problem J (Horizontal Projectile Launch from Royal Gorge Bridge):

  • Problem: A rock is kicked horizontally off the Royal Gorge Bridge ($h = 321\,\text{m}$ above Arkansas River), landing with a horizontal displacement of $45.0\,\text{m}$. Find initial launch speed.

  • Given: $\Delta y = -321\,\text{m}$, $\Delta x = 45.0\,\text{m}$, $a_y = -9.81\,\text{m/s}^2$

  • Formulation:         $\Delta t = \sqrt{\frac{2\Delta y}{a_y}} = \sqrt{\frac{2(-321\,\text{m})}{-9.81\,\text{m/s}^2}} = \sqrt{65.44\,\text{s}^2} = 8.09\,\text{s}$         $v_x = \frac{\Delta x}{\Delta t} = \frac{45.0\,\text{m}}{8.09\,\text{s}} = 5.56\,\text{m/s}$

  • Result: Launch speed $v_x = 5.56\,\text{m/s}$.

• Sample Problem K (Projectiles Launched at an Angle across Seil Sound):

  • Problem: An athlete attempts to clear the Seil Sound strait ($\Delta x = 6.00\,\text{m}$) leaping at $\theta = 35.0^\circ$. Calculate minimum launch velocity, time of flight, and maximum height reached.

  • Given: $\Delta x = 6.00\,\text{m}$, $\theta = 35.0^\circ$, $a_y = -9.81\,\text{m/s}^2$

  • Derivation:         $\Delta x = (v_i\cos(\theta))\Delta t \implies \Delta t = \frac{\Delta x}{v_i\cos(\theta)}$         At the apex ($t_{\text{peak}} = \frac{\Delta t}{2}$, $v_y = 0$):         $0 = v_i\sin(\theta) + a_y\left(\frac{\Delta t}{2}\right) \implies v_i\sin(\theta) = -\frac{a_y\Delta t}{2}$         Substituting $\Delta t$:         $v_i\sin(\theta) = \frac{-a_y\Delta x}{2v_i\cos(\theta)} \implies v_i^2 = \frac{-a_y\Delta x}{2\sin(\theta)\cos(\theta)}$

  • Calculation:         $v_i = \sqrt{\frac{-(-9.81\,\text{m/s}^2)(6.00\,\text{m})}{2\sin(35.0^\circ)\cos(35.0^\circ)}} = \sqrt{\frac{58.86}{2(0.5736)(0.8192)}} = \sqrt{\frac{58.86}{0.9397}} = \sqrt{62.64} = 7.9\,\text{m/s}$         $\Delta t = \frac{6.00\,\text{m}}{(7.91\,\text{m/s})\cos(35^\circ)} = 0.92\,\text{s}$         $y_{\text{max}} = \frac{(v_i\sin(\theta))^2}{2g} = \frac{[(7.91)\sin(35^\circ)]^2}{2(9.81)} = \frac{(4.54)^2}{19.62} = 1.05\,\text{m} \approx 1.0\,\text{m}$

• Sample Problem L (Relative Velocity across Slingsby Channel):

  • Problem: The current in Slingsby Channel flows at $v_{wb} = 30.0\,\text{km/h}$. A boat crosses perpendicular to the bank at $v_{mb} = 18.0\,\text{km/h}$. Find motorboat velocity relative to water.

  • Given: $v_{wb} = 30.0\,\text{km/h}$, $v_{mb} = 18.0\,\text{km/h}$

  • Formulation: $\mathbf{v}_{mw} = \mathbf{v}_{mb} - \mathbf{v}_{wb}$         $v_{mw} = \sqrt{v_{mb}^2 + v_{wb}^2} = \sqrt{(18.0\,\text{km/h})^2 + (30.0\,\text{km/h})^2} = \sqrt{324 + 900} = \sqrt{1224} = 35.0\,\text{km/h}$         $\theta = \tan^{-1}\left(\frac{v_{wb}}{v_{mb}}\right) = \tan^{-1}\left(\frac{30.0}{18.0}\right) = \tan^{-1}(1.667) = 59.0^\circ\text{ upstream}$

• Sample Problem M (Jaw Muscle Net Force Calculation):

  • Problem: Determine the net bite force when three equal forces of $4.33 \times 10^3\,\text{N}$ act simultaneously: $F_1$ along $-y$, $F_2$ along $+x$, and $F_3$ at $-60.0^\circ$ below the $+x$-axis.

  • Given: $F_1 = 4.33 \times 10^3\,\text{N}$ at $270^\circ$, $F_2 = 4.33 \times 10^3\,\text{N}$ at $0^\circ$, $F_3 = 4.33 \times 10^3\,\text{N}$ at $-60^\circ$

  • Components:         $F_{3x} = (4.33 \times 10^3\,\text{N})\cos(-60.0^\circ) = 2.165 \times 10^3\,\text{N}$         $F_{3y} = (4.33 \times 10^3\,\text{N})\sin(-60.0^\circ) = -3.750 \times 10^3\,\text{N}$         $\sum F_x = F_2 + F_{3x} = 4.33 \times 10^3\,\text{N} + 2.165 \times 10^3\,\text{N} = 6.495 \times 10^3\,\text{N}$         $\sum F_y = -F_1 + F_{3y} = -4.33 \times 10^3\,\text{N} + (-3.750 \times 10^3\,\text{N}) = -8.080 \times 10^3\,\text{N}$

  • Resultant Magnitude and Angle:         $F_{\text{net}} = \sqrt{(6.495 \times 10^3\,\text{N})^2 + (-8.080 \times 10^3\,\text{N})^2} = \sqrt{4.218 \times 10^7 + 6.529 \times 10^7} = 1.036 \times 10^4\,\text{N}$         $\theta = \tan^{-1}\left(\frac{-8.080 \times 10^3}{6.495 \times 10^3}\right) = \tan^{-1}(-1.244) = -51.2^\circ$

  • Result: Net force is $1.036 \times 10^4\,\text{N}$ directed at $51.2^\circ$ below the positive horizontal axis.

• Sample Problem N (Newton's Second Law on a Sliding Book):

  • Problem: A $2.2\,\text{kg}$ book is pushed across a table under a net horizontal force of $1.6\,\text{N}$ to the right. Compute acceleration.

  • Given: $m = 2.2\,\text{kg}$, $\sum F = 1.6\,\text{N}$

  • Calculation: $a = \frac{\sum F}{m} = \frac{1.6\,\text{N}}{2.2\,\text{kg}} = 0.73\,\text{m/s}^2$

  • Result: $a = 0.73\,\text{m/s}^2$ to the right.

• Sample Problem O (Kinetic Friction on Van Cargo):

  • Problem: A $20.0\,\text{kg}$ trunk is pushed across a moving van floor ($\mu_k = 0.255$). Calculate opposing kinetic friction.

  • Given: $m = 20.0\,\text{kg}$, $\mu_k = 0.255$, $g = 9.81\,\text{m/s}^2$

  • Calculation: $F_n = mg = (20.0\,\text{kg})(9.81\,\text{m/s}^2) = 196.2\,\text{N}$         $F_k = \mu_k F_n = (0.255)(196.2\,\text{N}) = 50.0\,\text{N}$

• Sample Problem P (Dynamics of Overcoming Friction with Angled Tension):

  • Problem: A rope pulls a $20.0\,\text{kg}$ crate with a force of $90.0\,\text{N}$ at $30.0^\circ$ above horizontal across a surface where $\mu_k = 0.500$. Find acceleration.

  • Given: $m = 20.0\,\text{kg}$, $F_{\text{applied}} = 90.0\,\text{N}$, $\theta = 30.0^\circ$, $\mu_k = 0.500$

  • Force Resolution:         $F_{ax} = (90.0\,\text{N})\cos(30.0^\circ) = 77.94\,\text{N}$         $F_{ay} = (90.0\,\text{N})\sin(30.0^\circ) = 45.00\,\text{N}$         $F_g = mg = (20.0\,\text{kg})(9.81\,\text{m/s}^2) = 196.2\,\text{N}$

  • Vertical Equilibrium:         $\sum F_y = F_n + F_{ay} - F_g = 0 \implies F_n = 196.2\,\text{N} - 45.00\,\text{N} = 151.2\,\text{N}$

  • Friction and Horizontal Acceleration:         $F_k = \mu_k F_n = (0.500)(151.2\,\text{N}) = 75.60\,\text{N}$         $\sum F_x = F_{ax} - F_k = 77.94\,\text{N} - 75.60\,\text{N} = 2.34\,\text{N}$         $a_x = \frac{\sum F_x}{m} = \frac{2.34\,\text{N}}{20.0\,\text{kg}} = 0.117\,\text{m/s}^2 \approx 0.12\,\text{m/s}^2\text{ to the right}$

• Sample Problem Q (Work Done in Tug-of-War):

  • Problem: A girl pulls her dog $d = 8.0\,\text{m}$ along the horizontal at $\theta = 18^\circ$ performing $190\,\text{J}$ of work. Find pulling force.

  • Given: $W = 190\,\text{J}$, $d = 8.0\,\text{m}$, $\theta = 18^\circ$

  • Calculation: $F = \frac{W}{d\cos(\theta)} = \frac{190\,\text{J}}{(8.0\,\text{m})\cos(18^\circ)} = \frac{190}{(8.0)(0.9511)} = \frac{190}{7.608} = 25\,\text{N}$

• Sample Problem R (Kinetic Energy of Silvana Cruciata):

  • Problem: Silvana Cruciata runs $18.084\,\text{km}$ in $1.000\,\text{h}$. If her kinetic energy is $694\,\text{J}$, calculate her mass.

  • Given: $\Delta x = 18084\,\text{m}$, $\Delta t = 3600\,\text{s}$, $KE = 694\,\text{J}$

  • Calculation:         $v_{\text{avg}} = \frac{18084\,\text{m}}{3600\,\text{s}} = 5.0233\,\text{m/s}$         $m = \frac{2KE}{v_{\text{avg}}^2} = \frac{2(694\,\text{J})}{(5.0233\,\text{m/s})^2} = \frac{1388}{25.234} = 55.0\,\text{kg}$

• Sample Problem S (Work-Kinetic Energy Theorem on Great Pyramid):

  • Problem: A $1.37 \times 10^4\,\text{kg}$ stone block slides from rest down the Khufu Pyramid ($h = 147\,\text{m}$, $\theta = 52.0^\circ$, $\mu_k = 0.45$). Find final kinetic energy at ground level.

  • Given: $m = 1.37 \times 10^4\,\text{kg}$, $h = 147\,\text{m}$, $\theta = 52.0^\circ$, $\mu_k = 0.45$

  • Analytical Derivation:         $d = \frac{h}{\sin(\theta)}$         $F_{\text{net}} = mg\sin(\theta) - \mu_k mg\cos(\theta) = mg(\sin(\theta) - \mu_k\cos(\theta))$         $W_{\text{net}} = F_{\text{net}}d = mg(\sin(\theta) - \mu_k\cos(\theta))\left(\frac{h}{\sin(\theta)}\right) = mgh\left(1 - \frac{\mu_k}{\tan(\theta)}\right)$

  • Calculation:         $W_{\text{net}} = (1.37 \times 10^4\,\text{kg})(9.81\,\text{m/s}^2)(147\,\text{m})\left(1 - \frac{0.45}{\tan(52.0^\circ)}\right)$         $W_{\text{net}} = (1.9756 \times 10^7\,\text{J})\left(1 - \frac{0.45}{1.280}\right) = (1.9756 \times 10^7)(1 - 0.3516) = (1.9756 \times 10^7)(0.6484) = 1.3 \times 10^7\,\text{J}$         Since $KE_i = 0$, $KE_f = W_{\text{net}} = 1.3 \times 10^7\,\text{J}$.

• Sample Problem T (Bungee Jumper Potential Energy):

  • Problem: A $70.0\,\text{kg}$ stuntman drops from a $50.0\,\text{m}$ high bridge on a bungee cord ($L_0 = 15.0\,\text{m}$, $k = 71.8\,\text{N/m}$). Cord stretches to $44.0\,\text{m}$. Find total potential energy relative to river level when he momentarily stops.

  • Given: $m = 70.0\,\text{kg}$, $h = 50.0\,\text{m} - 44.0\,\text{m} = 6.0\,\text{m}$, $x = 44.0\,\text{m} - 15.0\,\text{m} = 29.0\,\text{m}$, $k = 71.8\,\text{N/m}$

  • Gravitational Potential Energy:         $PE_g = mgh = (70.0\,\text{kg})(9.81\,\text{m/s}^2)(6.0\,\text{m}) = 4.12 \times 10^3\,\text{J}$

  • Elastic Potential Energy:         $PE_{\text{elastic}} = \frac{1}{2}kx^2 = \frac{1}{2}(71.8\,\text{N/m})(29.0\,\text{m})^2 = (35.9)(841) = 3.02 \times 10^4\,\text{J}$

  • Total Potential Energy:         $PE_{\text{tot}} = PE_g + PE_{\text{elastic}} = 4.12 \times 10^3\,\text{J} + 3.02 \times 10^4\,\text{J} = 3.43 \times 10^4\,\text{J}$

• Sample Problem U (Frictionless Playground Slide):

  • Problem: A $25.0\,\text{kg}$ child zooms down a frictionless curved slide from height $h = 3.00\,\text{m}$. Determine bottom speed.

  • Given: $h = 3.00\,\text{m}$, $v_i = 0\,\text{m/s}$, $g = 9.81\,\text{m/s}^2$

  • Conservation Equation: $mgh = \frac{1}{2}mv_f^2 \implies v_f = \sqrt{2gh}$

  • Calculation: $v_f = \sqrt{2(9.81\,\text{m/s}^2)(3.00\,\text{m})} = \sqrt{58.86\,\text{m}^2/\text{s}^2} = 7.67\,\text{m/s}$

• Sample Problem V (Curtain Hoist Motor Power Selection):

  • Problem: A theater needs to raise a $193\,\text{kg}$ curtain a distance of $7.5\,\text{m}$ in exactly $5.0\,\text{s}$ at constant speed. Available motors: $1.0\,\text{kW}$, $3.5\,\text{kW}$, and $5.5\,\text{kW}$. Select the appropriate motor.

  • Given: $m = 193\,\text{kg}$, $d = 7.5\,\text{m}$, $\Delta t = 5.0\,\text{s}$

  • Calculation:         $W = mgd = (193\,\text{kg})(9.81\,\text{m/s}^2)(7.5\,\text{m}) = 14200\,\text{J}$         $P = \frac{W}{\Delta t} = \frac{14200\,\text{J}}{5.0\,\text{s}} = 2840\,\text{W} = 2.8\,\text{kW}$

  • Conclusion: The $3.5\,\text{kW}$ motor is chosen ($1.0\,\text{kW}$ is underpowered; $5.5\,\text{kW}$ lifts too quickly).