NEET Physics — Laws of Motion Complete Chapter Guide

Chapter Overview & Weightage

Laws of Motion is the backbone of classical mechanics. Every force problem you will ever face in physics — from pulleys to circular motion to electrostatics — traces back to Newton’s three laws. NEET consistently tests this chapter with 2-3 questions.

Laws of Motion carries 4-5% weightage in NEET. Friction-based problems and free body diagrams (FBDs) dominate. Circular motion questions also appear from this chapter.

Year	NEET Q Count	Key Topics Tested
2025	2	Friction on incline, pulley system
2024	3	Circular motion, pseudo force
2023	2	Connected bodies, friction
2022	2	Lift problems, banking of roads
2021	3	Pulley, friction, centripetal force

graph TD
    A[Laws of Motion] --> B[Newton's Laws]
    A --> C[Friction]
    A --> D[Circular Motion]
    B --> E[FBD and Equilibrium]
    B --> F[Connected Bodies]
    B --> G[Lift Problems]
    C --> H[Static Friction]
    C --> I[Kinetic Friction]
    C --> J[Inclined Plane]
    D --> K[Banking of Roads]
    D --> L[Conical Pendulum]

Key Concepts You Must Know

Tier 1 (Always asked)

Free body diagrams — this is THE skill for this chapter
Friction: static vs kinetic, angle of friction, friction on incline
Newton’s second law applied to connected bodies (pulleys, strings)
Circular motion — centripetal acceleration, banking of roads

Tier 2 (Frequently asked)

Pseudo force in non-inertial frames (lift, accelerating cart)
Constraint relations for pulleys and strings
Limiting friction and minimum force to move a block on a rough surface

Tier 3 (Occasional)

Conical pendulum
Friction with variable applied force angle
Death-well (vertical circular motion with friction)

Important Formulas

\vec{F}_{net} = m\vec{a}

For a system of connected bodies with total mass $M$ and net external force $F_{ext}$ :

a_{system} = \frac{F_{ext}}{M_{total}}

Then find individual tensions by isolating each body.

Static friction: $f_s \leq \mu_s N$ (self-adjusting up to the limit)

Kinetic friction: $f_k = \mu_k N$ (constant once sliding starts)

On an inclined plane at angle $\theta$ :

Normal force: $N = mg\cos\theta$
Component along incline: $mg\sin\theta$
Condition to start sliding: $\tan\theta \geq \mu_s$

Centripetal acceleration: $a_c = \dfrac{v^2}{r} = \omega^2 r$

Banking angle (no friction): $\tan\theta = \dfrac{v^2}{rg}$

Maximum speed on a banked road with friction:

v_{max} = \sqrt{rg\left(\frac{\mu + \tan\theta}{1 - \mu\tan\theta}\right)}

For any problem with connected bodies (pulleys, strings), first draw the FBD of each body separately, then write $F = ma$ for each, and solve simultaneously. Never try to do it “in your head” — the FBD is not optional, it IS the method.

Solved Previous Year Questions

PYQ 1 — NEET 2024

Problem: A block of mass 5 kg on a rough horizontal surface is pulled by a force of 20 N at an angle of 30° above the horizontal. If $\mu_k = 0.2$ , find the acceleration. (Take $g = 10$ m/s $^2$ )

Solution:

Resolve the applied force:

Horizontal: $F_x = 20\cos 30° = 10\sqrt{3}$ N
Vertical: $F_y = 20\sin 30° = 10$ N

Normal force (the upward pull reduces it):

N = mg - F_y = 50 - 10 = 40 \text{ N}

Friction: $f_k = \mu_k N = 0.2 \times 40 = 8$ N

Net horizontal force: $F_{net} = 10\sqrt{3} - 8 = 17.32 - 8 = 9.32$ N

a = \frac{F_{net}}{m} = \frac{9.32}{5} = \mathbf{1.86 \text{ m/s}^2}

The classic error: students forget that the vertical component of the applied force changes the normal force. When you pull at an angle above horizontal, $N$ decreases (which reduces friction). When you push at an angle below horizontal, $N$ increases. This small detail changes the answer significantly.

PYQ 2 — NEET 2023

Problem: Two blocks of masses 3 kg and 5 kg are connected by a light string over a frictionless pulley (Atwood’s machine). Find the acceleration and the tension in the string.

Solution:

For an Atwood machine, the system acceleration:

a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(5 - 3) \times 10}{3 + 5} = \frac{20}{8} = \mathbf{2.5 \text{ m/s}^2}

Tension:

T = \frac{2m_1 m_2 g}{m_1 + m_2} = \frac{2 \times 3 \times 5 \times 10}{8} = \mathbf{37.5 \text{ N}}

For any Atwood machine: $a = \dfrac{(m_2 - m_1)g}{m_1 + m_2}$ and $T = \dfrac{2m_1 m_2 g}{m_1 + m_2}$ . These formulas save time, but derive them once so you understand where they come from.

PYQ 3 — NEET 2022

Problem: A car is moving on a banked road of radius 100 m with banking angle 30°. Find the speed for which no friction is needed. (Take $g = 10$ m/s $^2$ )

Solution:

For no friction, the centripetal force is entirely provided by the horizontal component of the normal force:

\tan\theta = \frac{v^2}{rg}

v^2 = rg\tan\theta = 100 \times 10 \times \tan 30° = 1000 \times \frac{1}{\sqrt{3}}

v = \sqrt{\frac{1000}{\sqrt{3}}} = \sqrt{577.35} \approx \mathbf{24 \text{ m/s}}

Difficulty Distribution

Difficulty	% of Questions	What to Expect
Easy	35%	Direct FBD, simple friction, lift problems
Medium	50%	Pulleys, banking, friction on incline with applied force
Hard	15%	Pseudo forces, constraint relations, variable friction

Expert Strategy

Week 1: FBD mastery. Every problem in this chapter starts and ends with a correct free body diagram. Practise drawing FBDs for blocks on inclines, pulley systems, and bodies in circular motion. Identify all forces: weight, normal, friction, tension, applied force.

Week 2: Friction problems. Master the three cases — block on flat surface, block on incline (ascending and descending), and block being pulled/pushed at an angle. For each, write the expression for normal force first.

Week 3: Circular motion and banking. These are high-scoring because the formulas are direct. Practise banked road (with and without friction), conical pendulum, and vertical circular motion problems.

When stuck on a connected-body problem, use the “system approach” first to find acceleration (treat all bodies as one system), then isolate individual bodies to find internal forces like tension. This two-step method works for 90% of NEET-level problems.

Common Traps

Trap 1 — Treating static friction as always equal to $\mu_s N$ . Static friction is self-adjusting. It equals $\mu_s N$ ONLY at the point of slipping. For a block at rest on a gentle incline, friction equals $mg\sin\theta$ , which may be much less than $\mu_s N$ .

Trap 2 — Forgetting the reaction to tension. In a pulley problem, the string pulls the heavier block upward AND the lighter block upward. The tension is the same throughout a massless string over a frictionless pulley. But the acceleration directions are opposite for the two blocks.

Trap 3 — Using $mg$ as centripetal force in banking problems. On a banked road, $mg$ acts vertically downward. It is the horizontal component of the normal force (not $mg$ ) that provides centripetal force. Resolving forces correctly is the whole challenge here.

Trap 4 — Pseudo force direction errors. Pseudo force acts opposite to the acceleration of the non-inertial frame. In a lift accelerating upward, the pseudo force is downward (adding to weight). Students sometimes reverse this.

Trap 5 — String constraint confusion. When one end of a string moves by $x$ , the other end moves by $x$ too (inextensible string). But in a pulley system with multiple strings, the constraint relation depends on the configuration. Draw the string path carefully and differentiate string length with respect to time.