Laws of Motion — vidhyapath 2027

Concept Core

Three laws. One framework. All of classical mechanics.

Newton's First Law — Law of Inertia

A body at rest stays at rest, and a body in motion stays in motion with constant velocity, unless acted upon by a net external force. This defines inertia and the inertial reference frame.

Net force = 0 ⟺ acceleration = 0 ⟺ constant velocity

Newton's Second Law — Law of Force

The net force on a body equals the rate of change of momentum. For constant mass: F = ma. Force and acceleration are always in the same direction.

F_net = ma (constant mass) | F = dp/dt (general)

III

Newton's Third Law — Action & Reaction

For every action there is an equal and opposite reaction. Forces act on different objects — they never cancel each other.

F_AB = −F_BA (forces on different bodies)

Free Body Diagram (FBD) — The Master Tool

        Isolate ONE object — draw it as a box or dot
Draw every force acting ON it: Weight (↓), Normal (⊥ surface), Tension (along string), Friction (opposing motion)
Choose coordinates: x along motion, y perpendicular
Apply ΣFₓ = maₓ and ΣFᵧ = maᵧ separately
For connected bodies: string inextensible → same acceleration magnitude

      

Types of Forces

Weight: W = mg downward through centre of mass

Normal (N): Perpendicular to surface, away from it

Tension (T): Along string, away from body (pulls, never pushes)

Friction (f): Opposing relative motion or tendency

Friction — Static vs Kinetic

Static: f_s ≤ μ_s N (adjustable up to max)

Kinetic: f_k = μ_k N (constant when sliding)

Key: μ_s > μ_k always. Body starts sliding when applied force exceeds f_s(max). Then friction drops to f_k.

Inclined Plane Setup

For mass m on incline at angle θ:

N = mg cosθ
F_along = mg sinθ
a = g(sinθ − μcosθ)

Atwood Machine

Two masses m₁ > m₂ over frictionless pulley:

a = (m₁−m₂)g/(m₁+m₂)
T = 2m₁m₂g/(m₁+m₂)

Apparent Weight in Lift

Lift going up (a ↑): W_app = m(g+a) — feels heavier

Lift going down (a ↓): W_app = m(g−a) — feels lighter

Free fall (a = g): W_app = 0 — weightlessness

Formula Vault

Complete formula set for Laws of Motion — exam-ready.

Newton's 2nd Law

F = ma

F in N, m in kg, a in m/s²

Weight

W = mg

g ≈ 10 m/s² in EAPCET

Impulse

J = FΔt = Δp

Change in momentum

Static Friction Max

f_s(max) = μ_s N

Before slipping; f_s ≤ μ_s N

Kinetic Friction

f_k = μ_k N

During sliding; μ_k < μ_s

Normal on Incline

N = mg cosθ

θ from horizontal

Incline Acceleration

a = g(sinθ − μcosθ)

Frictionless: a = g sinθ

Atwood Acceleration

a = (m₁−m₂)g/(m₁+m₂)

m₁ > m₂

Atwood Tension

T = 2m₁m₂g/(m₁+m₂)

Same throughout string

Lift Apparent Weight ↑

W_app = m(g + a)

a = lift's upward acceleration

Lift Apparent Weight ↓

W_app = m(g − a)

Zero when a = g

Connected Bodies (system)

a = F_net / M_total

Ignore internal tensions

Worked Examples

5 problems — easy to EAPCET-level to conceptual trap. Click to expand.

EasyFind acceleration: 5 kg block, 20 N force, 5 N friction▾

A 5 kg block on a horizontal surface is pulled by 20 N. Kinetic friction = 5 N. Find acceleration.

Net horizontal force: F_net = 20 − 5 = 15 N

F = ma: 15 = 5 × a → a = 3 m/s²

✓ Acceleration = 3 m/s²

MediumAtwood machine: m₁ = 8 kg, m₂ = 2 kg — find a and T▾

Atwood machine with m₁ = 8 kg, m₂ = 2 kg, frictionless pulley. Find a and T. (g = 10 m/s²)

For m₁ (going down): 80 − T = 8a ...(i)

For m₂ (going up): T − 20 = 2a ...(ii)

Add: 60 = 10a → a = 6 m/s²; T = 20 + 12 = 32 N

✓ a = 6 m/s², T = 32 N

Medium60 kg person in lift decelerating at 4 m/s² — apparent weight?▾

A 60 kg person is in a lift that is moving upward but decelerating at 4 m/s². What does the scale read? (g = 10 m/s²)

Decelerating while going up = accelerating downward. So a_effective = −4 (downward).

W_app = m(g − a) = 60(10 − 4) = 60 × 6 = 360 N

The person feels lighter (360 N vs actual 600 N). Scale reads 36 kg equivalent.

✓ Scale reads 360 N

EAPCET Level3 blocks (4+2+1 kg) pulled by 21 N — find T₂ between B and C▾

Blocks A(4 kg), B(2 kg), C(1 kg) on a frictionless surface connected in series. 21 N pulls A. Find tension T₂ between B and C. (g = 10)

Total mass = 7 kg. Common acceleration: a = 21/7 = 3 m/s²

Isolate block C: only T₂ acts on it → T₂ = m_C × a = 1 × 3 = 3 N

✓ T₂ = 3 N (T₁ between A&B = 3 × 3 = 9 N)

Trap QuestionWhy don't action-reaction forces cancel? Horse pulls cart...▾

Horse pulls cart with force F. Cart pulls horse back with F (3rd Law). Why does the system move? ⚠️ Classic EAPCET conceptual trap.

The trap: "Equal and opposite → cancel → no motion." WRONG. Cancellation requires forces on the same body.

Horse-on-cart and cart-on-horse act on different bodies. They never cancel.

For the horse: ground pushes horse forward (reaction to horse's foot push). If this exceeds cart's pull-back force, horse+cart system accelerates.

✓ Action-reaction pairs act on different bodies — apply Newton's 2nd Law to each body separately.

Mistake DNA

4 errors distilled from EAPCET distractor analysis — know these and never lose marks.

⚖️

Confusing Mass (kg) and Weight (N)

Substituting weight (in N) into F = ma where mass (in kg) is required.

❌ Wrong

F = Weight × a
= 50 N × 3 = 150 ✗

✓ Correct

F = mass × a
= 5 kg × 3 = 15 N ✓

In F = ma, m is in kg. Weight = mg is force in N. They differ by factor g = 10.

📐

Writing N = mg on an Inclined Plane

Normal force equals mg only on a horizontal surface. On an incline, N = mg cosθ.

❌ Wrong

On 30° incline:
N = mg = 100 N ✗

✓ Correct

N = mg cos30°
= 100 × 0.866 = 86.6 N ✓

Normal is perpendicular to the surface. Only the component of mg perpendicular to the incline balances it.

🔗

Including Tension in System-Level Force Balance

Tension is internal to the system — it cancels when treating multiple connected bodies together.

❌ Wrong

F_net = F − T + T − f
Writes extra tension terms ✗

✓ Correct

F_net = F − f only
T cancels as internal ✓
a = F_net/M_total

For a SYSTEM: only external forces matter. Find acceleration from system equation, then find T by isolating one body.

🪁

Forgetting Pseudo Force in Accelerating Frame

In lift/train problems, treating the accelerating frame as inertial gives wrong apparent weight.

❌ Wrong

Lift up at a=3 m/s²:
W_app = mg = 600 N ✗
(ignores acceleration)

✓ Correct

W_app = m(g+a)
= 60×13 = 780 N ✓
Person feels heavier ✓

In an accelerating frame: add pseudo force −ma₀ opposite to acceleration, or use W_app = m(g ± a) directly.

Chapter Intelligence

Exam trends, scoring strategy, and connections to other chapters.

EAPCET Topic Weightage (2019–2024)

Newton's 2nd Law numericals

Friction problems

Connected bodies/strings

Apparent weight / lifts

Atwood machine

3rd Law conceptual

High-Yield PYQ Patterns

Block on incline — find a Atwood tension calculation Apparent weight in lift μ from deceleration data 3 blocks in series — find T Maximum static friction Pseudo force in elevator

Exam Strategy — 4 Marks in 6 Minutes

Always draw the FBD first. 30 seconds drawing saves 2 minutes of algebraic confusion.
For connected bodies: find common acceleration from the full system first, then isolate one body for tension.
Memorise: N = mgcosθ and F_along = mgsinθ. These appear in ~40% of EAPCET mechanics questions.
If you blank on the Atwood formula, rederive using two equations in 45 seconds — faster than guessing.
This chapter links directly to Work-Energy (W = F·d uses forces from FBD) and Circular Motion (centripetal force = net force toward center).
For "which block moves first" questions: compare applied force to f_s(max) = μ_s N. If applied force > f_s(max), it moves.