Newton's Laws Applied — Science Sage

Newton's First Law — Inertia

An object remains at rest or moves in a straight line at constant velocity unless acted upon by a net (resultant) external force.

This property is called inertia — the tendency of an object to resist changes to its state of motion. Inertia is directly proportional to mass.

Real-world examples: Seatbelts prevent passengers from continuing forward when a car stops suddenly. A tablecloth pulled quickly leaves dishes behind due to their inertia.

Newton's Second Law — Fnet = ma

When a net force acts on an object, it accelerates in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass.

F_net = ma

Symbol	Meaning	Unit
Fnet	Net (resultant) force	N
m	Mass of object	kg
a	Acceleration	m·s⁻²

Fnet = ma is a VECTOR equation. The direction of acceleration is always the same as the direction of the net force.

Newton's Third Law — Action-Reaction

For every action force, there is an equal and opposite reaction force. These forces:

Are equal in magnitude but opposite in direction.
Act on different objects — they do NOT cancel each other out.
Are the same type of force (both contact, both gravitational, etc.).

Example: When you push a wall (action), the wall pushes you back equally (reaction). The two forces act on different objects so they do not cancel.

Free Body Diagrams (FBD)

An FBD shows all forces acting on a single object as arrows from or through the object's centre. Rules:

Draw only ONE object per FBD.
Show ALL forces: weight (down), normal force (perpendicular to surface), friction (along surface opposing motion), tension (along string/rope), applied force.
Label each arrow with its symbol and direction.
Arrow length should be roughly proportional to magnitude.

Normal Force and Friction

The normal force N is perpendicular to the contact surface. On a flat surface: N = mg. On an incline at angle θ: N = mg cosθ.

Static friction prevents sliding: fs ≤ μsN. It equals the applied force up to a maximum of μsN (maximum static friction = μsN).

Kinetic friction acts while sliding: fk = μkN. Note: μk < μs always.

f_k = μ_kN f_s(max) = μ_sN

Inclined Plane Problems

Resolve the weight into components parallel and perpendicular to the slope:

W∥ = mg sinθ (down the slope)
W⊥ = mg cosθ (into the slope)
N = mg cosθ (Newton's 2nd Law perpendicular to slope)

Apply Fnet = ma along the slope: if the block slides down, take down-the-slope as positive.

Worked Example — Incline with friction: A 5 kg block on a 30° incline with μk = 0.2. Find acceleration.

N = mg cos30° = 5×10×0.866 = 43.3 N
fk = 0.2 × 43.3 = 8.66 N
W∥ = mg sin30° = 5×10×0.5 = 25 N
Fnet = 25 − 8.66 = 16.34 N (down slope)
a = Fnet/m = 16.34/5 = 3.27 m·s⁻²

Connected Objects and Tension

For two blocks connected by a string: treat the whole system for acceleration, then isolate one block for tension.

System: F_net = (m₁ + m₂) × a
Tension: T = m₂ × a (for the block with no applied force)

Lifts and Apparent Weight

The apparent weight is the normal force N from the scale in a lift:

Accelerating upward: N = m(g + a) — you feel heavier.
Accelerating downward: N = m(g − a) — you feel lighter.
Free fall: N = 0 — weightlessness.
Constant velocity: N = mg — normal weight.

IEB Extension — Atwood Machine with Friction

In an ideal Atwood machine, two masses hang over a massless, frictionless pulley. IEB may introduce a pulley with mass (moment of inertia), or friction in the pulley bearing. Qualitatively: a massive pulley slows the acceleration because energy is used to spin the pulley. The tension on each side of a massive pulley is different (unlike an ideal pulley where tensions are equal for a stationary pulley problem).

IEB Extension — Non-Inertial Reference Frames

An inertial frame is one that is not accelerating. Newton's Laws hold exactly in inertial frames. In a non-inertial (accelerating) frame (e.g., a rotating carousel), a fictitious force appears — the centrifugal force. It is not a real force; it arises because the observer is accelerating. The Coriolis effect (deflection of moving objects on Earth) is another fictitious force in Earth's rotating frame.

Force and Motion Simulator

Surface Type

Parameters

Mass (kg)5 kg

Applied F (N)60 N

μk0.20

Normal N

—N

Friction fk

—N

Fnet

—N

Accel.

—m·s⁻²

Question 1 of 8

A 4 kg block is pushed with a net force of 20 N. Its acceleration is:

Question 2 of 8

A book rests on a table. The reaction force to the weight of the book (Earth pulls book down) is:

Question 3 of 8

A 10 kg block is on a 37° incline. The component of weight parallel to the slope is (sin 37° ≈ 0.60, g = 10 m·s⁻²):

Question 4 of 8

A person of mass 60 kg stands in a lift accelerating upward at 2 m·s⁻². Their apparent weight is (g = 10 m·s⁻²):

Question 5 of 8

Two blocks (3 kg and 5 kg) are connected by a string on a frictionless surface. A 40 N force is applied to the 5 kg block. The tension in the string is:

Question 6 of 8

The coefficient of kinetic friction between a block and surface is 0.3. The block has a mass of 4 kg (g = 10 m·s⁻²). The kinetic friction force is:

Question 7 of 8

A 5 kg block is released from rest on a 37° incline with friction. μk = 0.25 (g = 10 m·s⁻², sin37°≈0.60, cos37°≈0.80). What is the block's acceleration down the slope?

Question 8 of 8

Block A (5 kg) rests on a rough horizontal table (μk = 0.20) and is connected by a light string over a frictionless pulley to Block B (5 kg), which hangs vertically over the edge. Using g = 10 m·s⁻², find the acceleration of the system.

IEB Extended Questions

IEB Question 1

An Atwood machine has masses 3 kg and 5 kg. If the pulley has significant mass (rotational inertia), compared to an ideal massless pulley, the acceleration of the system will be:

IEB Question 2

A person feels pushed outward while riding a roundabout (merry-go-round). This "centrifugal force" is best described as:

Draw a free body diagram for every problem where forces are involved. Show all formulae, substitutions, and units clearly. Take g = 10 m·s⁻².

Question 1 — FBD Drawing (4 marks)

A 6 kg block is being pushed across a rough horizontal surface by a force of 30 N at 20° below the horizontal. Draw a fully labelled free body diagram showing all forces acting on the block. Name each force and indicate its direction.

Question 2 — Inclined Plane (8 marks)

A 12 kg crate slides down a 40° incline. The coefficient of kinetic friction between the crate and incline is 0.25. (a) Draw a free body diagram for the crate. (b) Calculate the normal force. (c) Calculate the kinetic friction force. (d) Find the net force on the crate and its acceleration. State the direction of acceleration.

Question 3 — Connected Objects (7 marks)

Block A (8 kg) rests on a frictionless horizontal surface and is connected by a light string over a frictionless pulley to Block B (3 kg) which hangs vertically. (a) Draw a FBD for each block separately. (b) Write Newton's Second Law equations for each block. (c) Solve for the acceleration of the system and the tension in the string.

Question 4 — Lift Apparent Weight (5 marks)

A 70 kg person stands on a bathroom scale in a lift. Calculate the reading on the scale (apparent weight) when the lift is: (a) stationary, (b) accelerating upward at 3 m·s⁻², (c) accelerating downward at 3 m·s⁻², (d) in free fall. Comment on what each reading means physically.

Question 5 — Friction Coefficient from Stopping Distance (6 marks)

A car of mass 1200 kg travelling at 25 m·s⁻¹ brakes to a stop over a distance of 80 m on a horizontal road. Assuming the only horizontal force is kinetic friction: (a) Use kinematics to find the deceleration. (b) Apply Newton's Second Law to find the friction force. (c) Calculate the coefficient of kinetic friction between the tyres and the road.

Question 6 — Reading Braking Data from a Table (6 marks)

A different 1000 kg car has its velocity recorded every second after the brakes are applied on a horizontal road:

Time (s)	0	1	2	3	4	5
Velocity (m·s⁻¹)	24	20	16	12	8	4

(a) Using the table (not a formula), show that the car's deceleration is constant, and state its value.
(b) Apply Newton's Second Law to calculate the net (friction) force acting on the car.
(c) Assuming the only horizontal force is kinetic friction, calculate the coefficient of kinetic friction between the tyres and the road.
(d) If the same braking data continued, at what time would the car reach 0 m·s⁻¹? Use the pattern in the table, not a new formula, to justify your answer.