Why this chapter is worth more than its question count
Laws of Motion is where the free-body diagram is taught, and the free-body diagram is the tool you use for the rest of mechanics. Work and energy problems begin by identifying forces. Rotational motion adds torques to the same diagram. Even fluid and electrostatics questions involving equilibrium reuse it unchanged. A student who is fluent here is faster in four other chapters; one who is not pays the cost repeatedly.
It is also a chapter where questions are typically multi-step, so errors compound. A sign error in the first equation does not produce a slightly wrong answer - it produces a completely different one, which will usually be among the options.
How JEE actually asks Laws of Motion
As with every chapter, NTA publishes no official weightage, so treat any figure you find as a coaching estimate derived from past papers, and confirm the syllabus and pattern for your session in the NTA information bulletin.
The stable pattern in past papers is that questions are situational rather than formula-driven. You are given a physical arrangement - blocks on a wedge, a pulley system, a body in a lift, a vehicle on a banked road - and asked for an acceleration, a tension or a minimum coefficient of friction. Very few questions can be answered by substituting into a single equation, which is why practising varied arrangements matters more here than drilling any particular formula.
Key concepts, compressed
- ✓Newton's first law defines the inertial frame: a body with no net force keeps its velocity. This is what makes the third and fourth points below necessary.
- ✓Newton's second law is F = ma applied per body, along a chosen axis. Choose the axis along the acceleration wherever possible.
- ✓Newton's third law pairs are equal, opposite and act on different bodies. The weight of a block and the normal force on it are not a pair - both act on the block.
- ✓Static friction adjusts to prevent relative motion, up to a maximum of μₛN. Kinetic friction has the fixed value μₖN and opposes relative sliding.
- ✓Tension is the same throughout a string only when the string is massless and the pulley is massless and frictionless.
- ✓Uniform circular motion has constant speed and non-zero acceleration, directed toward the centre with magnitude v²/r.
Formulas you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Newton's second law | F(net) = ma | applied per body, per axis |
| Maximum static friction | f(max) = μₛN | friction is ≤ this until slipping |
| Kinetic friction | f = μₖN | fixed value while sliding |
| Normal force on an incline | N = mg cosθ | not mg |
| Acceleration on a frictionless incline | a = g sinθ | independent of mass |
| Apparent weight in a lift | N = m(g ± a) | + when accelerating up |
| Centripetal acceleration | a(c) = v²/r | directed toward the centre |
| Banking angle (frictionless) | tanθ = v²/rg | independent of mass |
| Atwood machine | a = (m₁ − m₂)g/(m₁ + m₂) | massless string and pulley |
| Impulse | J = FΔt = Δp | equals change in momentum |
The five mistakes that cost the most marks
- ✓Writing friction as μN when the body is not sliding. Static friction takes whatever value is needed to prevent motion, up to μₛN - so on a stationary block the friction usually is not μₛN at all.
- ✓Assuming the normal force equals mg. It equals mg only on a horizontal surface with no vertical acceleration and no other vertical force.
- ✓Drawing centripetal force on the free-body diagram as an extra arrow. It is the resultant of the real forces, not one of them - drawing it separately double-counts.
- ✓Forgetting the pseudo force in an accelerating frame, or applying it in the wrong direction. It is −ma, so it points opposite to the frame's acceleration.
- ✓Treating tension as uniform across a pulley that has mass. Equal tension on both sides requires the pulley to be massless and frictionless.
Practice set 1: Newton's laws and free-body diagrams
1. A force of 20 N acts on a 5 kg block on a frictionless horizontal surface. What is its acceleration?
4 m/s². Forces on the block: weight 50 N down, normal force 50 N up, applied force 20 N horizontal. The vertical forces cancel, so the net force is 20 N horizontally and a = F/m = 20/5 = 4 m/s². Trivial as stated, but writing out the vertical cancellation is the habit that makes the harder questions work.
2. A book rests on a table. Which force is the Newton's third-law reaction to the book's weight?
The gravitational force the book exerts on the Earth. Third-law pairs act on different bodies, and weight is the Earth pulling the book, so its pair must be the book pulling the Earth. The normal force from the table is not the answer - it acts on the same body as the weight, and it happens to be equal only because the book is in equilibrium.
3. A 60 kg person stands in a lift accelerating upward at 2 m/s². What does the floor push up with?
720 N. Taking upward as positive: N − mg = ma, so N = m(g + a) = 60 × (10 + 2) = 720 N. The person's actual weight is unchanged at 600 N; what increases is the normal force, which is what a weighing scale reads. That distinction between weight and apparent weight is the point of the question.
4. The cable of the same lift snaps and it falls freely. What does the scale now read?
Zero. In free fall a = −g, so N = m(g − g) = 0. The person and the lift accelerate downward together at the same rate, so the floor exerts no force. This is weightlessness, and note that gravity is still acting at full strength - the absence is of the contact force, not of gravity.
5. Two blocks of 3 kg and 2 kg are in contact on a frictionless surface. A 10 N force is applied to the 3 kg block, pushing both. What is the acceleration, and what is the contact force between them?
a = 2 m/s² and the contact force is 4 N. Treat the pair as one 5 kg system first: a = 10/5 = 2 m/s². Then isolate the 2 kg block - the only horizontal force on it is the contact force, so F = 2 × 2 = 4 N. The system-then-isolate sequence is the standard method and is worth doing in that order every time.
6. In an Atwood machine, masses of 5 kg and 3 kg hang from a light string over a frictionless pulley. Find the acceleration and the tension.
a = 2.5 m/s² and T = 37.5 N. Using a = (m₁ − m₂)g/(m₁ + m₂) = (5 − 3) × 10 / 8 = 20/8 = 2.5 m/s². For the tension, T = 2m₁m₂g/(m₁ + m₂) = 2 × 5 × 3 × 10 / 8 = 300/8 = 37.5 N. Check it against the lighter block directly: T − 30 = 3 × 2.5, so T = 37.5 N. Note the tension lies between the two weights, 30 N and 50 N, as it must.
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Practice set 2: friction
7. A 10 kg block sits on a horizontal surface with μₛ = 0.2. A horizontal force of 15 N is applied. What is the friction force?
15 N, and the block does not move. The maximum static friction is μₛN = 0.2 × 100 = 20 N. Since the applied 15 N is less than that, static friction exactly balances it at 15 N and the block stays at rest. Answering 20 N is the standard error - that is the maximum available, not the value in play.
8. The applied force on the same block is increased to 30 N, and μₖ = 0.15. What is its acceleration?
1.5 m/s². The applied force now exceeds the 20 N maximum static friction, so the block slides and kinetic friction takes over at μₖN = 0.15 × 100 = 15 N. Net force = 30 − 15 = 15 N, so a = 15/10 = 1.5 m/s². Notice that friction drops the moment sliding begins, which is why a stuck object lurches forward once it breaks free.
9. A block is placed on a frictionless incline of 30°. What is its acceleration down the slope?
5 m/s². Resolve the weight along and perpendicular to the incline. Along the slope the component is mg sinθ, so a = g sinθ = 10 × sin30° = 10 × 0.5 = 5 m/s². The mass cancels, which is why every object slides down a frictionless incline at the same rate - the same reason all bodies fall equally fast.
10. What is the minimum coefficient of static friction needed to keep a block at rest on a 30° incline?
About 0.58. At the point of slipping, mg sinθ = μₛ mg cosθ, so μₛ = tanθ = tan30° ≈ 0.577. The mass cancels here too. Note the practical consequence: a coefficient of 0.5 would not be enough, so a block with μₛ = 0.5 on a 30° slope slides.
11. The angle of an incline is slowly increased until a block just begins to slide, at 45°. What is μₛ?
1.0. That angle is the angle of repose, and at it μₛ = tanθ = tan45° = 1. This is a standard experimental method for measuring the coefficient of static friction, and questions often describe the procedure without naming it.
12. Does friction always oppose motion?
No - it opposes relative motion between surfaces, which is not the same thing. When you walk, friction on your shoe acts forward and is what propels you. When a block rests on an accelerating truck bed, friction on the block acts forward and is the only force accelerating it. Questions on this point are common precisely because the loose version of the rule is so widely taught.
Practice set 3: circular motion
13. A car travels at 10 m/s around a circular track of radius 100 m. What is its centripetal acceleration?
1 m/s². a(c) = v²/r = 100/100 = 1 m/s², directed toward the centre. The car's speed is constant, but its velocity is not, because the direction changes continuously - which is precisely why there is an acceleration despite constant speed.
14. What is the correct banking angle for a curve of radius 20 m at a design speed of 10 m/s, ignoring friction?
About 26.6°, since tanθ = v²/rg = 100/(20 × 10) = 0.5, giving θ = arctan(0.5) ≈ 26.6°. The mass does not appear, so the same banking works for a motorcycle and a lorry. Above the design speed the vehicle needs friction as well, which is what the follow-up question usually introduces.
15. What is the minimum speed at the top of a vertical circular loop for an object on the inside of the track?
v = √(gr). At the minimum speed the normal force falls to zero and gravity alone supplies the centripetal force: mg = mv²/r, so v² = gr. Below that speed the object leaves the track. The mass cancels again - a recurring feature of this chapter worth noticing, because it lets you discard options that depend on mass.
16. In uniform circular motion, what is the work done by the centripetal force over one full revolution?
Zero. The centripetal force is always perpendicular to the velocity, so it does no work at any instant, and the displacement over a full revolution is zero in any case. This is why the speed stays constant despite a force acting continuously - the force changes direction, not speed.
17. A stone tied to a string is whirled in a horizontal circle. The string breaks. Which way does the stone travel?
Along the tangent at the point of release, in a straight line - then as a projectile under gravity. With the tension gone there is no inward force, so Newton's first law applies and the stone continues with the velocity it had, which was tangential. It does not fly radially outward, which is the intuitive but wrong answer the options will offer.
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Practice set 4: connected bodies, frames and momentum
18. A 2 kg block hangs from a spring balance attached to the ceiling of a lift accelerating downward at 2 m/s². What does the balance read?
16 N. Taking downward as the direction of acceleration: mg − T = ma, so T = m(g − a) = 2 × (10 − 2) = 16 N. The reading is less than the 20 N at rest. Note the symmetry with question 3 - upward acceleration increases the reading, downward decreases it, and free fall takes it to zero.
19. A ball of mass 0.2 kg strikes a wall at 10 m/s and rebounds at 10 m/s. What is the impulse on the ball?
4 N·s, directed away from the wall. Impulse equals the change in momentum. Taking the initial direction as positive, the initial momentum is 0.2 × 10 = 2 kg·m/s and the final is 0.2 × (−10) = −2 kg·m/s. The change is −2 − 2 = −4 kg·m/s, so the magnitude is 4 N·s. Answering zero comes from treating the speeds as unchanged and forgetting that momentum is a vector.
20. Why is a longer follow-through, as when catching a cricket ball, less painful?
Because impulse is fixed by the required change in momentum, and J = FΔt. Increasing the contact time Δt reduces the average force F proportionally. The ball must be brought to rest either way, so the total impulse is the same - only the peak force changes. Airbags, crumple zones and landing mats all use the same principle.
21. A block is at rest on the floor of a truck. The truck accelerates forward at 2 m/s² and the block does not slide. What force accelerates the block, and in which direction?
Static friction, acting forward. The block needs a forward net force of ma to accelerate with the truck, and friction is the only horizontal force available. This is the clearest case of friction acting in the direction of motion rather than against it - and it also sets up the follow-up: the minimum μₛ required is a/g = 0.2.
22. A person inside an accelerating car feels pushed backward into the seat. Is there a real backward force?
No. In the ground frame, the seat pushes the person forward and there is no backward force at all - the sensation is inertia, the body's tendency to keep its original velocity. In the car's non-inertial frame you may model it as a pseudo force of −ma acting backward, but that force has no third-law pair and no physical origin. Questions test whether you can tell a real force from a frame artefact.
23. Two blocks of 4 kg and 6 kg are connected by a light string on a frictionless surface, and a 20 N force pulls the 6 kg block. Find the acceleration and the tension.
a = 2 m/s² and T = 8 N. For the system: a = 20/10 = 2 m/s². For the 4 kg block, the string tension is its only horizontal force, so T = 4 × 2 = 8 N. Check with the 6 kg block: 20 − 8 = 12 = 6 × 2. Always run that second check - it costs seconds and catches sign errors reliably.
Practice set 5: mixed and conceptual
24. A 1 kg block is pressed against a vertical wall by a horizontal force of 50 N. If μₛ = 0.3, will it slide?
No. The normal force here is horizontal and equal to the applied 50 N, so the maximum static friction is μₛN = 0.3 × 50 = 15 N, acting vertically. The weight is only 10 N, which is less than 15 N, so friction holds it. The instructive point is that the normal force is the applied force, not mg - pressing harder increases friction on a vertical wall.
25. Why does a rocket accelerate in the vacuum of space, with nothing to push against?
By Newton's third law together with conservation of momentum. The rocket expels exhaust gases backward and the gases push the rocket forward with an equal and opposite force. No external medium is required - the interaction is between the rocket and its own propellant, which is why rockets work better in vacuum than in atmosphere.
26. Two forces of 3 N and 4 N act on a 1 kg body at right angles. What is its acceleration?
5 m/s². Forces add as vectors, so the resultant is √(3² + 4²) = 5 N, and a = 5/1 = 5 m/s² in the direction of that resultant. Adding the magnitudes to get 7 N is the error being tested, and 7 will be among the options.
27. A body moves at constant velocity. What can you conclude about the forces on it?
That the net force is zero - not that no forces act. A car moving at a steady speed has thrust, drag, weight and normal force acting, all balancing. Constant velocity means zero acceleration, which by the second law means zero resultant, and questions phrase this as 'no force acts' specifically to see whether you accept it.
28. A block on a frictionless incline is connected over a pulley to a hanging block of equal mass. Which way does the system move, for an incline of 30°?
The hanging block descends. The force pulling the system one way is the hanging weight mg; the other way it is the incline component mg sin30° = 0.5mg. Since mg exceeds 0.5mg, the hanging mass wins, and the acceleration is (mg − mg sin30°)/(2m) = g(1 − 0.5)/2 = 2.5 m/s². Comparing the two driving terms before calculating tells you the direction immediately.
29. Why is it easier to pull a lawn roller than to push it at the same angle?
Because pulling has an upward vertical component that reduces the normal force, while pushing has a downward one that increases it. Since friction is μN, the reduced normal force when pulling means less friction to overcome. The horizontal components are identical, so the entire difference comes through N.
30. A lift moves upward at a constant velocity of 3 m/s. What does a 50 kg passenger's weighing scale read?
500 N. Constant velocity means zero acceleration, so N = mg = 50 × 10 = 500 N - the same as at rest. Only acceleration changes the apparent weight, not velocity. The specified 3 m/s is a distractor, and questions in this chapter routinely supply a velocity where only acceleration matters.
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How to study this chapter efficiently
- ✓Draw a separate free-body diagram for each body, every time, even when the arrangement looks simple. Most students who find this chapter hard are trying to hold three sets of forces in their head simultaneously.
- ✓Decide first whether the body is static or sliding. That choice determines whether friction is an unknown bounded by μₛN or a known μₖN, and getting it wrong makes the rest of the working irrelevant.
- ✓Solve the system as a whole for the acceleration, then isolate one body for the internal force. Doing it in that order turns most connected-body questions into two short steps.
- ✓Check your answer against a limiting case. Does the tension lie between the two weights? Does the acceleration vanish when the masses are equal? These checks take seconds and catch most sign errors.
- ✓Work in the ground frame by default. Pseudo forces are useful but are a common source of error under time pressure, and almost every JEE question can be done without them.
- ✓Practise arrangements rather than formulas - wedges, lifts, pulleys, banked roads, vertical circles - because the exam varies the arrangement, not the physics.
Turn this into active practice
This chapter rewards being tested more than almost any other, because its errors are systematic. If you consistently mishandle the normal force on an incline, you will keep doing so until something forces you to notice - and rereading the theory does not.
The JEE Laws of Motion quiz on QUFF generates fresh questions on this chapter, scores them instantly and explains each answer, so a repeated mistake becomes visible quickly. Do a timed set, note which of the five mistakes above each error belongs to, and repeat the set a day later to check that the correction held.