Newton’s Laws of Motion

Newton’s three laws of motion, formulated by Sir Isaac Newton in his groundbreaking work *Philosophiæ Naturalis Principia Mathematica* (1687), are the bedrock of classical mechanics. These laws explain how forces influence the motion of objects, from everyday scenarios to celestial bodies. They integrate concepts like inertia, force, acceleration, and action-reaction pairs, and form the basis for deriving related formulas such as weight, momentum, work, and energy. This guide provides a detailed exploration with examples, all basic formulas, and practical applications to deepen your understanding of these principles.

First Law: Inertia

An object at rest remains at rest, and an object in motion continues in a straight line at constant velocity, unless acted upon by a net external force. This property is called inertia, which depends on an object’s mass—the greater the mass, the more inertia.

Formula: No specific equation, but inertia relates to mass \( m \) (in kg).

Example 1: Stationary Object

A 5 kg box on a frictionless table stays at rest until a force is applied.

Example 2: Moving Object

A hockey puck sliding on ice continues moving at 2 m/s until friction or a force stops it.

Second Law: F = ma (with Weight, Momentum)

The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This is expressed as:

\[ F = m \cdot a \]

Where:

  • \( F \): Force (in Newtons, N)
  • \( m \): Mass (in kg)
  • \( a \): Acceleration (in m/s²)

Related Formulas:

  • Weight: \( W = m \cdot g \) (where \( g \approx 9.8 \, \text{m/s}^2 \) is gravitational acceleration)
  • Momentum: \( p = m \cdot v \) (where \( v \) is velocity in m/s)

Example 1: Force and Acceleration

A 2 kg box is pushed with 10 N of force:

\[ F = m \cdot a \] \[ 10 = 2 \cdot a \] \[ a = \frac{10}{2} \] \[ a = 5 \, \text{m/s}^2 \]

Example 2: Weight Calculation

Find the weight of a 3 kg object on Earth (\( g = 9.8 \, \text{m/s}^2 \)):

\[ W = m \cdot g \] \[ W = 3 \cdot 9.8 \] \[ W = 29.4 \, \text{N} \]

Example 3: Momentum

Calculate the momentum of a 5 kg ball moving at 4 m/s:

\[ p = m \cdot v \] \[ p = 5 \cdot 4 \] \[ p = 20 \, \text{kg·m/s} \]

Third Law: Action-Reaction

For every action, there is an equal and opposite reaction. When one object exerts a force on another, the second exerts a force of equal magnitude but opposite direction on the first.

Formula: No direct equation, but forces are equal: \( F_{\text{action}} = -F_{\text{reaction}} \).

Example 1: Jumping

When you jump, you exert a downward force on the ground, and the ground exerts an upward force equal to your weight.

Example 2: Rocket Propulsion

A rocket pushes gas downward, and the gas pushes the rocket upward with equal force.

Applications (with Work, Kinetic Energy, Potential Energy)

Newton’s laws underpin various physical phenomena. Here are detailed examples with related formulas:

Related Formulas:

  • Work: \( W = F \cdot d \cdot \cos\theta \) (where \( d \) is displacement and \( \theta \) is the angle between force and displacement)
  • Kinetic Energy: \( KE = \frac{1}{2} m v^2 \)
  • Potential Energy (Gravitational): \( PE = m \cdot g \cdot h \) (where \( h \) is height)

Example 1: Car Acceleration

A 1000 kg car accelerates at 2 m/s². What force is applied?

\[ F = m \cdot a \] \[ F = 1000 \cdot 2 \] \[ F = 2000 \, \text{N} \]

Work done over 10 m (assuming force in direction of motion):

\[ W = F \cdot d \] \[ W = 2000 \cdot 10 \] \[ W = 20000 \, \text{J} \]

Example 2: Kinetic Energy of a Ball

Find the kinetic energy of a 2 kg ball moving at 3 m/s:

\[ KE = \frac{1}{2} m v^2 \] \[ KE = \frac{1}{2} \cdot 2 \cdot 3^2 \] \[ = \frac{1}{2} \cdot 2 \cdot 9 \] \[ = 9 \, \text{J} \]

Example 3: Potential Energy

Calculate the potential energy of a 5 kg object at 10 m height (\( g = 9.8 \, \text{m/s}^2 \)):

\[ PE = m \cdot g \cdot h \] \[ PE = 5 \cdot 9.8 \cdot 10 \] \[ = 490 \, \text{J} \]

Example 4: Elevator Motion

An 800 kg elevator accelerates upward at 1.5 m/s². What is the tension in the cable? (Weight \( W = m \cdot g \))

\[ W = m \cdot g \] \[ W = 800 \cdot 9.8 \] \[ = 7840 \, \text{N} \] \[ F_{\text{net}} = m \cdot a \] \[ F_{\text{net}} = 800 \cdot 1.5 \] \[ = 1200 \, \text{N} \] \[ T - W = F_{\text{net}} \] \[ T - 7840 = 1200 \] \[ T = 7840 + 1200 \] \[ T = 9040 \, \text{N} \]

Tension is 9040 N.

Example 5: Friction and Work

A 10 kg box is pushed 5 m with 20 N force against 5 N friction (\( \theta = 0^\circ \)):

\[ W_{\text{net}} = F \cdot d \cdot \cos\theta - F_{\text{friction}} \cdot d \] \[ = 20 \cdot 5 \cdot \cos(0^\circ) - 5 \cdot 5 \] \[ = 100 - 25 \] \[ = 75 \, \text{J} \]

Example 6: Rocket Launch

A 2000 kg rocket accelerates at 3 m/s² upward. What thrust is needed? (Ignore air resistance)

\[ F_{\text{net}} = m \cdot a \] \[ F_{\text{net}} = 2000 \cdot 3 \] \[ = 6000 \, \text{N} \] \[ W = m \cdot g \] \[ W = 2000 \cdot 9.8 \] \[ = 19600 \, \text{N} \] \[ T - W = F_{\text{net}} \] \[ T - 19600 = 6000 \] \[ T = 19600 + 6000 \] \[ T = 25600 \, \text{N} \]

Thrust is 25600 N.

Example 7: Skateboard Momentum

A 50 kg skateboarder moves at 2 m/s. What is the momentum?

\[ p = m \cdot v \] \[ p = 50 \cdot 2 \] \[ p = 100 \, \text{kg·m/s} \]