Direct answer

Newton's laws connect force and motion.

Newton's laws connect force and motion. JEE questions use free-body diagrams, pulleys, connected blocks, friction and circular motion dynamics.

Key ideas

Laws of Motion and Friction: what you must understand

  • Draw a free-body diagram for each body before writing equations
  • Static friction adjusts up to μₛN; kinetic friction is μₖN
  • Pseudo force −ma acts in a non-inertial frame
  • In circular motion the net force towards the centre is mv²/r
Formula sheet

Must-know formulas for Laws of Motion and Friction

  • F_net = ma = dp/dt
  • f_s ≤ μₛN, f_k = μₖN
  • Banked road (no friction): tanθ = v² / rg
  • Impulse J = FΔt = Δp
Print

One-page formula sheet

Printable Laws of Motion and Friction formula sheet: formulas, key ideas and traps on one page. Save it as a PDF for offline revision.

Avoid

Common mistakes students make

  • Assuming friction always equals μN
  • Forgetting the normal force changes on an incline or with an applied angled force
  • Treating centripetal force as an extra force on the FBD
For the exam

How this topic is asked in JEE

Appears frequently. Connected-body and friction questions are standard. Always check first whether the block moves at all.

Prerequisites: Kinematics (Motion in 1D and 2D).

NCERT map

Where Laws of Motion and Friction is in NCERT

Chapter names follow the current rationalised NCERT textbooks. Read these sections before solving PYQs.

ClassBookChapterSections JEE asks from
Class 11Physics Part ILaws of MotionNewton's First Law of Motion; Newton's Second Law of Motion; Newton's Third Law of Motion; Conservation of Momentum; Friction
JEE Advanced level

A JEE Advanced-level example on Laws of Motion and Friction

Question. A block A of mass 4 kg rests on a rough horizontal table with coefficient of friction μ = 0.25. It is connected by a light inextensible string passing over a smooth pulley at the edge of the table to another block B of mass 6 kg hanging freely. The system is released from rest. Find the acceleration of the system and the tension in the string. (Take g = 10 m/s²)

Show solution
  1. Draw free body diagrams: Block A has tension T pulling it forward and friction f = μm_A g = 0.25 × 4 × 10 = 10 N opposing motion. Block B has weight 60 N downward and tension T upward.
  2. For block A (horizontal motion): T − f = m_A × a, so T − 10 = 4a
  3. For block B (vertical motion, taking downward positive): m_B g − T = m_B × a, so 60 − T = 6a
  4. Add both equations: (T − 10) + (60 − T) = 4a + 6a gives 50 = 10a, so a = 5 m/s²
  5. Substitute back: T = 10 + 4 × 5 = 30 N

Answer: Acceleration = 5 m/s², Tension = 30 N

This example was solved twice independently and both solutions agree.

Free lectures

Best free lectures for Laws of Motion and Friction

Full-chapter free lectures picked from YouTube search results by views, from established JEE channels (mostly in Hinglish). Use the subtopic list to jump straight to the part you need (timestamps come from each video's own chapter list). Watch one, then solve PYQs the same day.

FAQs

Frequently asked questions

Is Laws of Motion and Friction important for JEE Main 2027?

Appears frequently. It is part of the official 'Laws of Motion' unit in the JEE Main 2027 syllabus.

What are the most important formulas in Laws of Motion and Friction?

F_net = ma = dp/dt; f_s ≤ μₛN, f_k = μₖN; Banked road (no friction): tanθ = v² / rg.

How should I study Laws of Motion and Friction for JEE?

Understand the key ideas from NCERT or one full lecture, learn the formula sheet, then solve 25–30 past JEE questions on the topic. Connected-body and friction questions are standard. Always check first whether the block moves at all.

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