Comprehensive Physics Study Notes: Newton's Laws of Motion, Force, Mass, Acceleration, and Free Fall

Basic Physical Quantities and Free Fall

• Speed, Velocity, and Acceleration:

  • Motion is described using fundamental quantities including speed, velocity, and acceleration.
  • Force is the physical push or pull exerted on an object that acts as the cause of a change in motion.
  • When multiple forces act on an object in the same direction, they combine to form a net force (resultant force) that points in that direction.
  • If opposing forces acting on an object are equal in magnitude, they cancel out, resulting in a net force of zero ($F_{\text{net}} = 0\,N$).

• Free Fall Dynamics:

  • Free fall is defined as motion where gravity is the single, only force acting on an object.
  • In ideal free fall, air resistance is assumed to be negligible ($F_{\text{air}} = 0\,N$).
  • Because air resistance is negligible, all objects in free fall drop at the exact same rate regardless of their mass.
  • The weight ($W$) of an object is the gravitational force exerted on it and is calculated by:     $W = m \times g$     where $m$ is the mass of the object in kilograms ($kg$) and $g$ is the acceleration due to gravity.
  • Near the surface of the Earth, the acceleration due to gravity is approximately $g = 10\,m/s^2$.
  • Due to gravitational acceleration, a falling object's speed increases by $10\,m/s$ during each second of fall.

• Direction of Motion vs. Force of Gravity:

  • Downward Motion: When an object is dropped or thrown downward, the weight (gravitational force) acts in the direction of motion, causing its speed to increase at a constant rate of $10\,m/s$ every second.
  • Upward Motion: When an object is thrown upward, the weight acts downward (opposite to the direction of motion), causing its speed to decrease at a constant rate of $10\,m/s$ every second until it reaches zero at the peak of its trajectory.
  • In both upward and downward motions of free fall, the acceleration remains constant at $g = 10\,m/s^2$ directed downward.

Newton's First Law of Motion: Inertia

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

  • An object at rest will remain at rest, and an object in motion will continue moving at a constant speed in a straight-line path, unless acted upon by a non-zero net external force ($F_{\text{net}} \neq 0\,N$).
  • Inertia is the natural property of matter to resist any change in its state of rest or uniform motion.

• Conceptual Examples and Scenarios of Inertia:

  • Massive Ball Suspended by a String: When pulling quickly downward on a lower string attached to a heavy suspended ball, the lower string breaks because the mass of the ball exhibits inertia, resisting immediate movement and preventing the force from transferring to the upper string.
  • Automobile Braking: When driving a car and suddenly applying the brakes, passengers lurch forward because their bodies are already in motion at the speed of the car and tend to maintain that state of motion due to inertia.
  • Stone Whirled on a String: A stone swung in a circular path is held in circular motion by the string's tension supplying a centripetal force. If the string breaks, the net force drops to zero ($F = 0\,N$), and the stone immediately travels off in a straight-line path tangential to the circle at velocity $v$ due to inertia.
  • Soft Drink Can on Paper Sheet: A sheet of paper can be swiftly pulled out from underneath a soft drink can without tipping it over because the can's inertia keeps it at rest.
  • Flipping a Coin inside a Moving Airplane: When flipping a coin inside an airplane flying at a constant speed and direction, the coin lands directly back in the passenger's palm because the coin retains the horizontal velocity of the aircraft due to inertia.
  • Comparison of Mass and Inertial Resistance: An object with less mass has less weight and consequently less inertia. For instance, a lightweight hat flies off or moves further forward during a sudden stop than a heavier object because its smaller mass offers less inertial resistance to changes in motion.

Newton's Second Law of Motion: Force, Mass, and Acceleration

• Fundamental Relationship ($F_{\text{net}} = m \times a$):

  • A net force applied to an object causes an acceleration ($a$), which represents a change in velocity over time:     $a = \frac{\Delta v}{\Delta t}$
  • Acceleration is directly proportional to the applied net force:     $a \propto F_{\text{net}}$     Doubling the force exerted on a fixed mass doubles its acceleration ($2 \times F_{\text{net}} \implies 2 \times a$).
  • Acceleration is inversely proportional to the mass of the object:     $a \propto \frac{1}{m}$     An object with greater mass will experience less acceleration under the same net force. Doubling the mass under a constant force cuts the acceleration in half ($\frac{a}{2}$). Increasing mass by a factor of 5 decreases acceleration by a factor of 5.
  • Combining these yields Newton's Second Law:     $F_{\text{net}} = m \times a \quad \text{or} \quad a = \frac{F_{\text{net}}}{m}$

• SI Units of Force:

  • Force is measured in Newtons ($N$).
  • Mass is measured in kilograms ($kg$).
  • Acceleration is measured in meters per second squared ($m/s^2$).
  • Unit equivalence:     $1\,N = 1\,kg \times 1\,m/s^2 = 1\,kg\,m/s^2$

• Step-by-Step Worked Numerical Examples:

  • Example 1: Two forces acting in the same direction.
  • Given: Mass $m = 2\,kg$, Force $F_1 = 8\,N$ to the right, Force $F_2 = 4\,N$ to the right.
  • Calculate Net Force ($F_{\text{net}}$):       $F_{\text{net}} = F_1 + F_2 = 8\,N + 4\,N = 12\,N \quad (\text{to the right})$
  • Calculate Acceleration ($a$):       $a = \frac{F_{\text{net}}}{m} = \frac{12\,N}{2\,kg} = 6\,m/s^2$
  • Example 2: Three forces acting simultaneously.
  • Given: Mass $m = 4\,kg$, Force $F_1 = 10\,N$ (right), Force $F_2 = 2\,N$ (left), Force $F_3 = 5\,N$ (right).
  • Calculate Net Force ($F_{\text{net}}$):       $F_{\text{net}} = F_1 + F_3 - F_2 = 10\,N + 5\,N - 2\,N = 13\,N \quad (\text{to the right})$
  • The three distinct forces can be represented by a single equivalent net force of $13\,N$ to the right.

• Proportionality Scenarios:

  • If the applied force remains constant while the mass of a car decreases to half ($\frac{m}{2}$), the acceleration doubles ($2 \times a$).
  • If a car is pushed with twice as much net force ($2 \times F_{\text{net}}$) and the acceleration remains unchanged, the mass of the car must have doubled ($2 \times m$) to maintain that constant acceleration.

Air Resistance and Terminal Velocity

• Mechanics of Non-Free Fall Motion:

  • In a vacuum, a feather and a heavy coin drop at the exact same rate due to gravity alone ($g = 10\,m/s^2$).
  • In the presence of air, objects experience an upward friction force called air resistance ($F_{\text{air}}$) opposing downward motion.
  • Air resistance is non-constant; as an object falls faster and picks up speed ($v$), the air resistance acting on it increases ($F_{\text{air}} \uparrow$).

• Mathematical Progression to Terminal Velocity:

  • For a falling object of mass $m$, two forces act in opposite directions: downward weight ($W = m \times g$) and upward air resistance ($F_{\text{air}}$).
  • The net downward force is calculated by:     $F_{\text{net}} = W - F_{\text{air}}$
  • As the object accelerates and its speed increases:
  1. Air resistance increases ($F_{\text{air}} \uparrow$).
  2. Net force shrinks ($F_{\text{net}} \downarrow$).
  3. Acceleration decreases ($a = \frac{F_{\text{net}}}{m} \downarrow$).
  • Eventually, air resistance grows until it equals the object's weight:     $F_{\text{air}} = W$
  • At this equilibrium point:     $F_{\text{net}} = W - F_{\text{air}} = 0\,N$$a = 0\,m/s^2$
  • When acceleration drops to zero ($a = 0\,m/s^2$), the object stops speeding up and falls at a constant speed, defined as its terminal velocity (or terminal speed).

• Skydiver Dynamic Example:

  • As a skydiver falls faster and faster through the atmosphere before reaching terminal velocity:
  • Air resistance increases.
  • Net downward force decreases.
  • Acceleration decreases.

Newton's Third Law of Motion: Action and Reaction

• Law of Action-Reaction Interaction:

  • Forces always occur in equal and opposite pairs. For every action force exerted by one object on another, there is an equal magnitude and oppositely directed reaction force exerted by the second object back on the first.

• Concrete Examples of Action-Reaction Pairs:

  • Person Pushing a Wall:
  • Action Force: The hand pushing against the wall.
  • Reaction Force: The wall pushing back on the hand with an equal and opposite force.
  • Sitting on a Chair:
  • Action Force: Person's downward weight ($W$) pushing on the seat.
  • Reaction Force: The chair pushing upward against the person with an equal and opposite support force.
  • Kicking a Ball:
  • Action Force: Player's foot exerting a $1500\,N$ force against the ball.
  • Reaction Force: The ball exerting an equal $1500\,N$ reaction force back against the player's foot.
  • Firing a Cannon:
  • Action Force: The cannon exerting a forward force on the cannonball, accelerating it forward.
  • Reaction Force: The cannonball exerting an equal backward force on the cannon, causing recoil.

Comprehensive Problem Solving and Quantitative Applications

• Problem 1: Jet Acceleration during Takeoff

  • Given: A jet has mass $m = 100{,}000\,kg$. It has $4$ engines, each producing a thrust force of $F_{\text{engine}} = 50{,}000\,N$.
  • Total Force Calculation:     $F_{\text{net}} = 4 \times 50{,}000\,N = 200{,}000\,N$
  • Acceleration Calculation:     $a = \frac{F_{\text{net}}}{m} = \frac{200{,}000\,N}{100{,}000\,kg} = 2\,m/s^2$

• Problem 2: Horizontal Crate with Opposing Friction Force

  • Given: Mass of crate $m = 400\,kg$, acceleration $a = 1\,m/s^2$, friction force $F_{\text{friction}}$ is half of the crate's weight.
  • Step 1: Compute Weight ($W$):     $W = m \times g = 400\,kg \times 10\,m/s^2 = 4000\,N$
  • Step 2: Compute Friction Force ($F_{\text{friction}}$):     $F_{\text{friction}} = \frac{1}{2} \times W = \frac{1}{2} \times 4000\,N = 2000\,N \quad (\text{directed opposite to motion})$
  • Step 3: Compute Net Force ($F_{\text{net}}$):     $F_{\text{net}} = m \times a = 400\,kg \times 1\,m/s^2 = 400\,N$
  • Step 4: Solve for Applied Push Force ($F_{\text{applied}}$):     $F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}$$400\,N = F_{\text{applied}} - 2000\,N$$F_{\text{applied}} = 400\,N + 2000\,N = 2400\,N$

• Problem 3: Free Fall Time from Known Height

  • Given: Height $h = 5\,m$, gravitational acceleration $g = 10\,m/s^2$.
  • Formula:     $h = \frac{1}{2} \times g \times t^2$
  • Calculation:     $5\,m = \frac{1}{2} \times 10\,m/s^2 \times t^2$$5 = 5 \times t^2 \implies t^2 = 1 \implies t = 1\,s$

• Problem 4: Skydiver Net Force and Acceleration with Air Drag

  • Given: Skydiver mass $m = 100\,kg$, encountering air resistance $F_{\text{air}} = 400\,N$.
  • Step 1: Compute Weight ($W$):     $W = m \times g = 100\,kg \times 10\,m/s^2 = 1000\,N \quad (\text{downward})$
  • Step 2: Compute Net Force ($F_{\text{net}}$):     $F_{\text{net}} = W - F_{\text{air}} = 1000\,N - 400\,N = 600\,N \quad (\text{downward})$
  • Step 3: Compute Acceleration ($a$):     $a = \frac{F_{\text{net}}}{m} = \frac{600\,N}{100\,kg} = 6\,m/s^2$
  • Expressed in terms of $g$ ($g = 10\,m/s^2$):     $a = 0.6 \times g$

• Problem 5: Horizontal Motion of a Box with Support Force

  • Given: Box mass $m = 4\,kg$, horizontal net force $F_{\text{net}} = 12\,N$
  • Acceleration:     $a = \frac{F_{\text{net}}}{m} = \frac{12\,N}{4\,kg} = 3\,m/s^2$
  • Weight ($W$):     $W = m \times g = 4\,kg \times 10\,m/s^2 = 40\,N$
  • Normal/Support Force ($F_N$):     Because vertical motion is zero ($a_y = 0$), the support force equals the weight:     $F_N = W = 40\,N$

Practice Review Concepts and Test Solutions

• Vector vs. Scalar Quantities:

  • Vector Quantity: Possesses magnitude (size), unit, and direction. Examples: displacement, velocity, acceleration, force.
  • Scalar Quantity: Possesses magnitude and unit only, requiring simple algebraic addition. Examples: mass, distance, speed, time.
  • Mass Measurement: Kilograms ($kg$) measure mass.

• Opposing Push Forces on an Object:

  • If a girl pushes a car with a $100\,N$ force to the right and a boy pushes with a $50\,N$ force to the left:     $F_{\text{net}} = 100\,N - 50\,N = 50\,N \quad (\text{in the direction of the girl's push})$

• Constant Velocity Incline Motion:

  • When a crate slides down an inclined plane at constant velocity, its acceleration is zero ($a = 0\,m/s^2$). Thus, the net force acting on the crate is zero ($F_{\text{net}} = 0\,N$).

• Speed Calculation:

  • Given distance $d = 2\,km$ covered in time $t = 0.5\,\text{hours}$:     $\text{Speed} = \frac{\text{distance}}{\text{time}} = \frac{2\,km}{0.5\,\text{hour}} = 4\,km/h$

• Velocity vs. Acceleration Conceptual Distinctions:

  • Velocity indicates how fast an object moves and in what direction.
  • Acceleration indicates how fast velocity changes over time.
  • Velocity and acceleration are rates of one another, not the exact same physical concept.

• Free Fall Instantaneous Speed Equation:

  • For an object dropped from rest in free fall, speed after time $t$ is:     $v = g \times t$
  • After $t = 2\,s$:     $v = 10\,m/s^2 \times 2\,s = 20\,m/s$

• Weight Variance across Gravitational Fields:

  • An object's mass remains constant everywhere, but its weight is greater on Earth than on the Moon due to Earth's stronger gravitational acceleration $g$.

• Tow Truck Mass Calculation:

  • Given tow truck force $F = 4000\,N$ causing car acceleration $a = 2\,m/s^2$:     $m = \frac{F}{a} = \frac{4000\,N}{2\,m/s^2} = 2000\,kg$

• Parallel Scale Tension Distribution:

  • If two spring scales support a total weight of $1000\,N$, and one scale reads $700\,N$, the second scale must read:     $1000\,N - 700\,N = 300\,N$