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Newton's Laws
Foundations and Limits of Classical Mechanics
Newton's empirical laws accurately describe macroscopic motion but are limited to speeds much lower than the speed of light.
$$v \ll c$$
The validity range for Newtonian mechanics
$v$=Velocity of the object(m/s)
$c$=Speed of light(≈ 3 × 10⁸ m/s)
$v \approx c$
→Requires relativistic mechanics (Einstein)
Relativistic Limit: As objects approach $c$, mass effectively increases and time dilates, breaking classical assumptions.
Newton's First Law: The Law of Inertia
A body maintains its state of rest or uniform motion unless compelled to change by an unbalanced external force.
$$\sum F = 0 \implies a = 0$$
Equilibrium condition where motion state remains constant
$\sum F$=Vector sum of all external forces(N)
$a$=Acceleration(m/s²)
Static Equilibrium
→Object remains at rest ($v = 0$)
Dynamic Equilibrium
→Object maintains constant velocity ($v = $ constant)
Inertia: The inherent resistance of matter to changes in its motion; quantified by mass.
Inertial Frame of Reference: A coordinate system where this law holds true — must not be accelerating or rotating.
Constant Velocity ≠ Zero Force: A moving object with zero net force continues at the same speed and direction indefinitely.
Newton's Second Law: Quantifying Dynamics
Acceleration is directly proportional to the net force and inversely proportional to mass.
$$F = ma$$
The fundamental equation of motion connecting force and change in motion
$F$=Net external force(N (kg·m/s²))
$m$=Inertial mass(kg)
$a$=Acceleration(m/s²)
$F = 0$
→No acceleration — reduces to Newton's First Law
$a = g$ (free fall)
→$F = mg$ — the object's weight
Vector Direction: The acceleration vector always points in the exact direction of the resultant force vector.
Unit Definition: One Newton (N) is the force required to accelerate 1 kg by 1 m/s².
Proportionality Shortcut: If force and mass both change, find $a' = \frac{F'}{m'}$ directly — compare ratios instead of recalculating from scratch.
Weight vs Mass: Weight $W = mg$ is the gravitational force; mass is the intrinsic property. On the Moon ($g_{moon} \approx 1.6$ m/s²), weight decreases but mass stays the same.
Dimensional Check: $[F] = [m][a] = \text{kg} \cdot \text{m/s}^2 = [MLT^{-2}]$. If your answer doesn't have force dimensions, something went wrong.
Newton's Third Law: Interaction Pairs
Every interaction involves a pair of forces that are equal in magnitude and opposite in direction, acting on different bodies.
$$F_{AB} = -F_{BA}$$
Action and reaction forces act simultaneously on different bodies
$F_{AB}$=Force exerted by object A on object B(N)
$F_{BA}$=Force exerted by object B on object A(N)
Common misconception
→These forces NEVER cancel because they act on different objects.
Book on table
→Book pushes table down (weight); table pushes book up (normal force). The action-reaction pair is book→Earth and Earth→book, not book↔table.
Different Bodies: Action and reaction always act on two separate objects — never on the same body.
Simultaneity: The reaction occurs at the exact same instant as the action; one cannot exist without the other.
Friction and Normal Force
The normal force is the perpendicular contact force exerted by a surface on an object resting on it, adjusting to prevent penetration.
$$N = mg\cos\theta$$
Normal force on an inclined plane at angle θ to the horizontal
$N$=Normal (perpendicular) force from the surface(N)
$m$=Mass of the object(kg)
$g$=Gravitational acceleration(≈ 9.8 m/s²)
$\theta$=Angle of incline from horizontal(degrees or radians)
$\theta = 0°$ (flat surface)
→$N = mg$ — normal force equals weight
$\theta = 90°$ (vertical)
→$N = 0$ — no normal force, object is in free fall along surface
On flat surfaces: Normal force equals the object's weight ($N = mg$) only when no other vertical forces act.
Not always equal to weight: If someone pushes down on the object, $N > mg$. If a rope pulls upward, $N < mg$.
Perpendicular only: Normal force is always ⊥ to the surface, regardless of other forces.
Friction opposes relative motion (or attempted motion) between surfaces in contact. It depends on the normal force and the nature of the surfaces.
$$f = \mu N$$
Friction force is proportional to the normal force
$f$=Frictional force(N)
$\mu$=Coefficient of friction (static $\mu_s$ or kinetic $\mu_k$)(dimensionless)
$N$=Normal force(N)
Object at rest, small applied force
→Static friction matches applied force exactly ($f_s \leq \mu_s N$)
Object sliding
→Kinetic friction is constant at $f_k = \mu_k N$
Static vs Kinetic: Static friction adjusts up to a maximum $\mu_s N$; kinetic friction is a fixed value $\mu_k N$ once sliding begins.
Direction: Friction always acts opposite to the direction of motion (or attempted motion).
Limiting Case: $f_{s,max} = \mu_s N$ is the threshold — any applied force exceeding this initiates motion.
Friction on Inclines: On an incline, the component of gravity along the surface is $mg\sin\theta$. The block slides when $mg\sin\theta > \mu_s mg\cos\theta$, giving the critical angle $\theta_c = \tan^{-1}(\mu_s)$.
What Moves a Car?: The engine spins the tyres, which push backward on the road. By Newton's Third Law, the road pushes the tyres forward — this static friction is the force that accelerates the car.
Tension in Connected Systems
Tension is the pulling force transmitted through a string, rope, or cable when it is pulled tight by forces acting at each end.
Massless String: In idealized problems, the string has negligible mass, so tension is the same throughout its length.
Inextensible String: The string doesn't stretch, so connected objects share the same acceleration magnitude.
Direction: Tension always pulls toward the center of the string — it can only pull, never push.
When two or more objects are connected by a string, they form a system that can be analyzed as a whole or as individual free bodies.
$$a = \frac{F_{net}}{m_{total}}$$
System acceleration for connected bodies
$a$=Common acceleration of the system(m/s²)
$F_{net}$=Total unbalanced force on the entire system(N)
$m_{total}$=Combined mass of all connected objects(kg)
System Approach: First find the acceleration of the whole system using total net force and total mass.
Individual FBD: Then isolate one object and apply $F = ma$ to find tension in the connecting string.
Atwood Machine: Two masses on a pulley — $a = \frac{(m_1 - m_2)g}{m_1 + m_2}$ and $T = \frac{2m_1 m_2 g}{m_1 + m_2}$.
Limiting Case ($m_1 = m_2$): If both masses are equal, $a = 0$ and $T = mg$ — the system is in equilibrium and the tension simply supports the weight.
Limiting Case ($m_2 \to 0$): The heavier mass is essentially in free fall: $a \to g$ and $T \to 0$.
Kinematic Analysis via Velocity-Time Graphs
A velocity-time graph provides a complete history of motion, where slope identifies acceleration and area determines displacement.
$$s = \int v \, dt$$
The area under a v-t curve represents total distance
$s$=Total distance/displacement(m)
$v$=Velocity (function of time)(m/s)
$dt$=Infinitesimal time interval(s)
$a = 0$ (constant velocity)
→$s = v \times t$ — area forms a rectangle
Uniform acceleration from rest
→$s = \frac{1}{2}vt$ — area forms a triangle
Uniform acceleration from $v_i$
→$s = \frac{1}{2}(v_i + v_f)t$ — area forms a trapezoid
Gradient = Acceleration: The slope at any point equals the instantaneous acceleration.
Area = Distance: The geometric area between the curve and time axis gives the distance covered.
Negative Velocity: Area below the time axis represents displacement in the negative direction.
Application Strategy
Worked Examples
Problem Statement
An object of mass 5 kg has its velocity increase from 10 m/s to 30 m/s in 4 seconds under constant force. Find the force.
Newton's Second LawAcceleration from kinematics
Strategic Toolkit
$$F = ma = m \cdot \frac{v - u}{t}$$
Step 1
Find Acceleration
Use the kinematic definition of acceleration.
$a = \frac{v - u}{t} = \frac{30 - 10}{4} = 5\text{ m/s}^2$
Step 2
Apply Newton's Second Law
Multiply mass by acceleration.
$F = ma = 5 \times 5 = 25\text{ N}$
Problem Statement
Calculate the total distance covered by a car:
1. Accelerates from 0 to 20 m/s in 5 s.
2. Maintains 20 m/s for 10 s.
3. Decelerates to 0 m/s in 4 s.
Area of Triangle = 0.5 × b × hArea of Rectangle = w × h
Strategic Toolkit
$$s = A_1 + A_2 + A_3$$
Step 1
Acceleration Phase (Triangle)
Area of the first triangle (0–5 s).
$A_1 = \frac{1}{2} \times 5 \times 20 = 50\text{ m}$
Step 2
Constant Velocity Phase (Rectangle)
Area of the rectangle (5–15 s).
$A_2 = 10 \times 20 = 200\text{ m}$
Step 3
Deceleration Phase (Triangle)
Area of the final triangle (15–19 s).
$A_3 = \frac{1}{2} \times 4 \times 20 = 40\text{ m}$
Step 4
Total Distance
Sum all areas.
$s = 50 + 200 + 40 = 290\text{ m}$
Problem Statement
Two blocks of mass 3 kg and 5 kg are connected by a light string on a frictionless surface. A force of 24 N is applied to the 5 kg block. Find the acceleration and the tension in the string.
System approachIndividual free body diagramNewton's Second Law
Strategic Toolkit
$$a = \frac{F_{net}}{m_{total}}, \quad T = m_1 \cdot a$$
Step 1
System Acceleration
Treat both blocks as one system.
$a = \frac{F}{m_1 + m_2} = \frac{24}{3 + 5} = 3\text{ m/s}^2$
Step 2
Tension (FBD of 3 kg block)
The only horizontal force on the 3 kg block is the tension pulling it forward.
$T = m_1 \cdot a = 3 \times 3 = 9\text{ N}$
Step 3
Verify (FBD of 5 kg block)
The 5 kg block has applied force forward and tension backward.
$F - T = m_2 \cdot a \Rightarrow 24 - 9 = 15 = 5 \times 3$ ✓
Problem Statement
A 10 kg block is pushed with 60 N of horizontal force on a surface with $\mu_k = 0.3$. Find the acceleration.
Normal forceKinetic frictionNet force
Strategic Toolkit
$$a = \frac{F_{applied} - f_k}{m}$$
Step 1
Find Normal Force
On a flat surface with only horizontal applied force.
$N = mg = 10 \times 9.8 = 98\text{ N}$
Step 2
Find Friction Force
Use kinetic friction formula.
$f_k = \mu_k N = 0.3 \times 98 = 29.4\text{ N}$
Step 3
Find Net Force and Acceleration
Subtract friction from applied force, then divide by mass.
$a = \frac{60 - 29.4}{10} = \frac{30.6}{10} = 3.06\text{ m/s}^2$
Problem Statement
Two masses of 7 kg and 3 kg hang on either side of a frictionless pulley connected by a light string. Find the acceleration and the tension in the string. (Take $g = 10\text{ m/s}^2$)
Atwood machineConnected systemsNewton's Second Law
Strategic Toolkit
$$a = \frac{(m_1 - m_2)g}{m_1 + m_2}, \quad T = \frac{2m_1 m_2 g}{m_1 + m_2}$$
Step 1
Find System Acceleration
The net force is the difference in weights; total mass is the sum.
$a = \frac{(7 - 3) \times 10}{7 + 3} = \frac{40}{10} = 4\text{ m/s}^2$
Step 2
Find Tension
Use the Atwood tension formula (or isolate the lighter mass: $T - m_2 g = m_2 a$).
$T = m_2(g + a) = 3(10 + 4) = 42\text{ N}$
Step 3
Verify with Heavier Mass
For the 7 kg mass: $m_1 g - T = m_1 a$.
$70 - 42 = 28 = 7 \times 4$ ✓
Problem Statement
A 4 kg block slides down a 30° incline with $\mu_k = 0.2$. Find its acceleration. (Take $g = 10\text{ m/s}^2$)
Inclined planeComponent resolutionKinetic frictionNormal force on incline
Strategic Toolkit
$$a = g(\sin\theta - \mu_k \cos\theta)$$
Step 1
Resolve Weight Components
Find the gravitational components along and perpendicular to the incline.
$mg\sin 30° = 4 \times 10 \times 0.5 = 20\text{ N (along)}, \quad mg\cos 30° = 4 \times 10 \times 0.866 = 34.64\text{ N (perpendicular)}$
Step 2
Find Friction Force
Normal force equals the perpendicular component; friction opposes motion down the incline.
$f_k = \mu_k N = 0.2 \times 34.64 = 6.93\text{ N}$
Step 3
Find Net Force and Acceleration
Net force along incline = gravity component minus friction.
$a = \frac{20 - 6.93}{4} = \frac{13.07}{4} = 3.27\text{ m/s}^2$
Memorize formulas
Newton's Second Law
The fundamental link between force and acceleration.
Formula
F = ma
Weight
Gravitational force on an object near Earth's surface.
Formula
W = mg
Third Law Reciprocity
Action and reaction are always equal and opposite.
Formula
$F_{12} = -F_{21}$
Friction Force
Force opposing motion, proportional to normal force.
Formula
$f = \mu N$
Atwood Machine Acceleration
Acceleration of two masses connected over a pulley.
Formula
$a = \frac{(m_1 - m_2)g}{m_1 + m_2}$
Atwood Machine Tension
Tension in the string of an Atwood machine.
Formula
$T = \frac{2m_1 m_2 g}{m_1 + m_2}$
Normal Force on Incline
Perpendicular contact force on an inclined plane.
Formula
$N = mg\cos\theta$
Acceleration on Rough Incline
Net acceleration of a block sliding down a rough incline.
Formula
$a = g(\sin\theta - \mu_k \cos\theta)$
Review
1
Two objects A and B have masses $m$ and $3m$ respectively. The same force $F$ is applied to each. If A covers a distance $d$ in time $t$ starting from rest, the distance covered by B in the same time is:
Ans
$d/3$
Review
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Compare the accelerations
With the same force applied, acceleration is inversely proportional to mass. So the 3m object has one-third the acceleration of the m object.
$a_A = \frac{F}{m}, \quad a_B = \frac{F}{3m} = \frac{a_A}{3}$
Use distance from rest
Starting from rest, distance in a fixed time is proportional to acceleration through $d = \tfrac{1}{2}at^2$. Therefore the heavier object covers one-third the distance.
$d_B = \frac{1}{2}a_B t^2 = \frac{1}{3}\left(\frac{1}{2}a_A t^2\right) = \frac{d}{3}$
2
A person of mass 60 kg stands in a lift accelerating upward at 2 m/s². The reading on a weighing scale under the person is: (Take $g = 10\text{ m/s}^2$)
Ans
720 N
Review
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Write Newton's second law vertically
The scale reads the normal force. Since the lift accelerates upward, the net upward force must be $ma$.
$N - mg = ma$
Solve for the scale reading
Rearrange to get $N = m(g+a)$ and substitute the values.
$N = 60(10+2) = 720\text{ N}$
3
A 5 kg block rests on a horizontal surface ($\mu_s = 0.4$, $\mu_k = 0.3$). A horizontal force of 15 N is applied. The friction force acting on the block is: (Take $g = 10\text{ m/s}^2$)
Ans
15 N (static friction, no motion)
Review
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Check the limiting static friction
Static friction can adjust up to its maximum value. First find the maximum possible static friction on the block.
$f_{s,\max} = \mu_s mg = 0.4 \times 5 \times 10 = 20\text{ N}$
Match the applied force
Because the applied force is smaller than the maximum static friction, the block does not move and static friction equals the applied force.
$f_s = 15\text{ N}$
4
Three blocks of masses 2 kg, 3 kg, and 5 kg are connected by light strings in a line on a frictionless surface. A force of 20 N is applied to the 5 kg block. The tension in the string between the 2 kg and 3 kg blocks is:
Ans
4 N
Review
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Find the common acceleration
All three blocks move together on a frictionless surface, so the entire 10 kg system accelerates under the 20 N force.
$a = \frac{20}{2+3+5} = 2\text{ m/s}^2$
Isolate the 2 kg block
The string between the 2 kg and 3 kg blocks only needs to accelerate the 2 kg block, so tension equals mass times acceleration for that block.
$T = 2 \times 2 = 4\text{ N}$
5
A car of mass 1000 kg is moving at constant velocity on a road where the friction force is 500 N. The engine provides a driving force of:
Ans
500 N
Review
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Use constant-velocity motion
Constant velocity means zero acceleration, so the net force on the car must be zero.
$\Sigma F = 0$
Balance the resistive force
The driving force must exactly cancel friction for the resultant force to vanish.
$F_{\text{drive}} = 500\text{ N}$
6
A velocity-time graph shows a straight line from (0, 0) to (4, 20) followed by a horizontal line from (4, 20) to (10, 20). The total distance covered in 10 seconds is:
Ans
160 m
Review
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Find the area under the rising segment
Distance on a velocity-time graph is the area under the curve. The first section is a triangle.
$\tfrac{1}{2} \times 4 \times 20 = 40\text{ m}$
Add the constant-velocity section
The second section is a rectangle with width 6 s and height 20 m/s.
$6 \times 20 = 120\text{ m}, \quad 40 + 120 = 160\text{ m}$
7
In an Atwood machine, two masses $m_1 = 5$ kg and $m_2 = 3$ kg are connected by a string over a frictionless pulley. The acceleration of the system is: (Take $g = 10\text{ m/s}^2$)
Ans
2.5 m/s²
Review
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Apply the Atwood machine formula
The net driving force is the weight difference, and the total inertia is the sum of the two masses.
$a = \frac{(m_1-m_2)g}{m_1+m_2}$
Substitute the values
Use the given masses and $g$ to evaluate the acceleration.
$a = \frac{(5-3)\times 10}{5+3} = \frac{20}{8} = 2.5\text{ m/s}^2$
8
A 4 kg block is sliding on a rough horizontal surface with $\mu_k = 0.3$. No external push is applied. The deceleration of the block is: (Take $g = 10\text{ m/s}^2$)
Ans
3 m/s²
Review
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Find the kinetic friction force
Since the block is sliding, kinetic friction is the only horizontal force and equals $\mu_k mg$.
$f_k = 0.3 \times 4 \times 10 = 12\text{ N}$
Convert friction to deceleration
The acceleration magnitude is friction divided by mass, so the block slows down at a constant rate.
$a = \frac{12}{4} = 3\text{ m/s}^2$
9
A book lies on a table. The normal force exerted by the table on the book is the reaction to:
Ans
The book pressing down on the table (not the book's weight)
Review
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Identify the interaction pair
Newton's third-law pairs act on different bodies and come from the same interaction. The table's normal force on the book pairs with the book's push on the table.
table on book \leftrightarrow book on table
Exclude the weight option
The book's weight is due to Earth acting on the book, so its reaction is the book's gravitational pull on Earth, not the normal force.
normal force is not the reaction to weight
10
An object of mass 10 kg is accelerating at 3 m/s² on a rough surface. If the coefficient of kinetic friction is 0.2, what is the applied force? (Take $g = 10\text{ m/s}^2$)
Ans
50 N
Review
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Compute the friction force
On a horizontal surface, kinetic friction is $\mu_k mg$.
$f_k = 0.2 \times 10 \times 10 = 20\text{ N}$
Apply Newton's second law
The applied force must overcome friction and still produce the stated acceleration.
$F_{\text{applied}} - 20 = 10 \times 3 \Rightarrow F_{\text{applied}} = 50\text{ N}$
11
A block slides down a smooth incline of angle 30°. The acceleration of the block is: (Take $g = 10\text{ m/s}^2$)
Ans
5 m/s²
Review
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Use the component along the incline
On a smooth incline, only the component of gravity parallel to the plane causes acceleration.
$a = g\sin\theta$
Substitute the angle
For a 30° incline, $\sin 30^\circ = 0.5$.
$a = 10 \times 0.5 = 5\text{ m/s}^2$
12
In an Atwood machine with masses 4 kg and 6 kg, the heavier mass descends 5 m from rest. The speed of the masses at that point is: (Take $g = 10\text{ m/s}^2$)
Ans
$\sqrt{20}$ m/s
Review
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Find the system acceleration
Use the Atwood formula with the heavier mass moving downward.
$a = \frac{(6-4)\times 10}{6+4} = 2\text{ m/s}^2$
Use the kinematics relation
Starting from rest, relate speed to displacement using $v^2 = u^2 + 2as$.
$v^2 = 2 \times 2 \times 5 = 20 \Rightarrow v = \sqrt{20}\text{ m/s}$
13
Two blocks (3 kg and 5 kg) are in contact on a frictionless surface. A force of 40 N pushes the 3 kg block into the 5 kg block. The contact force between the blocks is:
Ans
25 N
Review
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Find the acceleration of the pair
Treat both blocks as one system to get the common acceleration under the 40 N push.
$a = \frac{40}{3+5} = 5\text{ m/s}^2$
Use the block being pushed indirectly
The contact force is the force that accelerates the 5 kg block alone, so it equals mass times acceleration for that block.
$F_c = 5 \times 5 = 25\text{ N}$
Question 1 of 13
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1
Two objects A and B have masses $m$ and $3m$ respectively. The same force $F$ is applied to each. If A covers a distance $d$ in time $t$ starting from rest, the distance covered by B in the same time is: