5.1 Physical Quantities, Units & Mechanics

Physical Quantities, SI Units, Vectors & Kinematics

In physical science, every measurement consists of a numerical magnitude and an associated unit. The International System of Units (SI) provides a standardized framework used universally in physics and engineering. Physics distinguishes between base quantities, which are defined independently, and derived quantities, which are expressed as combinations of base units.

SI Base and Derived Quantities

The International System relies on seven fundamental base units. All other physical quantities in mechanics, thermodynamics, and electromagnetism are derived from these seven pillars.

Standard SI Base Units

Common Derived Quantities in Mechanics

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Vectors vs. Scalars

Physical quantities are classified based on their directional properties:

• Scalar Quantities: Described completely by magnitude alone (e.g., mass, distance, speed, work, energy, time, temperature). Scalars obey standard algebraic addition.
• Vector Quantities: Described by both magnitude and direction (e.g., displacement, velocity, acceleration, force, momentum, electric field). Vectors follow vector algebra rules, such as the head-to-tail rule and trigonometric component resolution ($F_x = F \cos\theta$, $F_y = F \sin\theta$).

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Kinematics: Describing Motion

Kinematics analyzes the motion of objects without regard to the forces causing that motion. For motion under uniform (constant) acceleration, three fundamental equations govern linear kinematics:

1. First Equation of Motion: $v_f = v_i + a t$

2. Second Equation of Motion: $S = v_i t + \frac{1}{2} a t^2$

3. Third Equation of Motion: $v_f^2 = v_i^2 + 2 a S$

Where:

• $v_i$ = initial velocity ($\text{m/s}$)
• $v_f$ = final velocity ($\text{m/s}$)
• $a$ = uniform acceleration ($\text{m/s}^2$)
• $S$ = distance/displacement ($\text{m}$)
• $t$ = time interval ($\text{s}$)

Gravitational Free Fall Note: When an object falls freely under Earth's gravity, $a$ is replaced by $g \approx 9.8\text{ m/s}^2$ (or $10\text{ m/s}^2$ on PAF Airman test problems). For upward vertical motion, acceleration is downward so $a = -g$.

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Worked Example: Kinematics Calculations

Problem: A fighter jet launched from an airfield accelerates uniformly from rest at a rate of $8\text{ m/s}^2$ along a runway of length $400\text{ m}$. Calculate:

1. The takeoff speed of the jet ($v_f$).
2. The total time ($t$) required to reach takeoff speed.

Solution:

1. Using the third kinematics equation ($v_f^2 = v_i^2 + 2aS$): Given $v_i = 0\text{ m/s}$, $a = 8\text{ m/s}^2$, $S = 400\text{ m}$. $v_f^2 = 0^2 + 2(8)(400) = 6400$ $v_f = \sqrt{6400} = 80\text{ m/s}$

2. Using the first kinematics equation ($v_f = v_i + at$): $80 = 0 + 8 t \implies t = \frac{80}{8} = 10\text{ seconds}$

Dynamics: Newton's Laws, Work, Energy & Power

While kinematics describes how objects move, dynamics explains why objects move by analyzing forces, mass, and energy interactions.

Newton's Laws of Motion

Isaac Newton formulated three fundamental laws governing classical mechanics:

1. Newton's First Law (Law of Inertia): A body remains at rest or continues to move with uniform velocity in a straight line unless acted upon by an external net force. Inertia is the natural resistance of matter to changes in its state of motion, directly measured by its mass.
2. Newton's Second Law: When a net force acts on a body, it produces an acceleration in the direction of the force. The acceleration is directly proportional to the net force and inversely proportional to the mass: $F_{\text{net}} = m a$
3. Newton's Third Law (Action-Reaction): To every action force, there is always an equal and opposite reaction force ($F_{A \to B} = -F_{B \to A}$). Action and reaction forces act simultaneously on different bodies.

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Mass, Weight, Friction & Vector Resolution

• Mass ($m$): The quantity of matter in a body ($\text{kg}$). Mass is an intrinsic scalar quantity that remains constant everywhere in the universe.
• Weight ($W$): The downward gravitational force exerted on a mass by Earth ($\text{N}$). Weight is a vector quantity given by: $W = m g$
• Friction ($f$): A force opposing relative motion between contacting surfaces. Static friction ($f_s \le \mu_s N$) acts before movement starts, while kinetic friction ($f_k = \mu_k N$) opposes moving objects. $\mu$ represents the coefficient of friction and $N$ is the normal force.

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Work, Energy Transformations & Power

Mechanical Work

Work ($W$) is performed when a force causes displacement in the direction of the force: $W = F \cdot S \cos\theta$

• When force and displacement are parallel ($\theta = 0^\circ$), $W = F \cdot S$ (maximum positive work).
• When force is perpendicular to displacement ($\theta = 90^\circ$), $W = 0$ (e.g., centripetal force doing zero work on a circular orbit).

Kinetic and Potential Energy

• Kinetic Energy ($KE$): Energy of motion possessed by a moving mass: $KE = \frac{1}{2} m v^2$
• Gravitational Potential Energy ($PE$): Energy stored due to an object's position in a gravitational field: $PE = m g h$
• Law of Conservation of Mechanical Energy: In a conservative field (ignoring air resistance), total mechanical energy remains constant: $E_{\text{total}} = KE + PE = \text{constant}$

Power

Power ($P$) is the rate at which work is done or energy is transformed: $P = \frac{W}{t} = \frac{F \cdot S}{t} = F \cdot v_{\text{avg}}$ The SI unit of power is the Watt ($\text{W} = 1\text{ Joule/second}$). Engine output is also measured in horsepower ($1\text{ hp} = 746\text{ W}$).

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Worked Example: Work, Kinetic Energy, and Power

Problem: A crane lifts a crate of mass $200\text{ kg}$ vertically through a height of $15\text{ m}$ at a constant speed in $6\text{ seconds}$. Assuming $g = 10\text{ m/s}^2$:

1. Calculate the work done by the crane.
2. Determine the power output developed by the crane motor.

Solution:

1. Since speed is constant, the upward force $F$ equals the weight $W_{\text{crate}} = mg$: $F = (200\text{ kg})(10\text{ m/s}^2) = 2000\text{ N}$ Work done $W = F \cdot h = (2000\text{ N})(15\text{ m}) = 30,000\text{ Joules}\ (30\text{ kJ})$

2. Power developed by the motor: $P = \frac{W}{t} = \frac{30,000\text{ J}}{6\text{ s}} = 5000\text{ Watts}\ (5\text{ kW})$