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🎯⭐ INTERACTIVE LESSON

Newton's Laws with Calculus

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Newton's Laws with Calculus - Complete Interactive Lesson

Part 1: Newton's First Law

⚛️ Newton's First Law

Part 1 of 7 — Newton's First Law

An object remains at rest or in uniform motion unless acted upon by a net external force.

Key Ideas

  • Inertia: the tendency of objects to resist changes in motion
  • Mass is a measure of inertia
  • An object in equilibrium has ∑F⃗=0\sum \vec{F} = 0
  • This law defines inertial reference frames

Worked Example

A 5 kg block sits on a frictionless table with two horizontal forces: 20 N to the right and 20 N to the left. Describe its motion.

Step 1: ∑F=20−20=0\sum F = 20 - 20 = 0 N

Step 2: Net force is zero, so the block is in equilibrium.

Answer: The block remains at rest (or continues at constant velocity if already moving). ✅

Concept Check 🎯

Newton's First Law Practice 🧮

  1. A book sits on a table with no net force. What is its acceleration (inm/s2)(in m/s^{2})?

  2. A ball rolls at 10 m/s on a frictionless surface. What is its speed (m/s) after 5 seconds?

  3. Two forces act on an object: 15 N right and 10 N left. What is the net force magnitude (in N)?

Concept Check 🔍

Practice

#ProblemHint
1If net force is zero, what is the acceleration?Newton's first law
2What property measures resistance to acceleration?Starts with "i"
3Can an object move at 10 m/s with zero net force?Think about constant velocity

Challenge Question 📋

Part 2: Newton's Second Law with Calculus

⚛️ Newton's Second Law with Calculus

Part 2 of 7 — Newton's Second Law with Calculus

F⃗net=ma⃗=mdv⃗dt=dp⃗dt\vec{F}_{\text{net}} = m\vec{a} = m\frac{d\vec{v}}{dt} = \frac{d\vec{p}}{dt}

When force varies with time or position, use calculus:

  • v(t)=v0+∫0tF(t′)m dt′v(t) = v_0 + \int_0^t \frac{F(t')}{m}\,dt'
  • x(t)=x0+∫0tv(t′) dt′x(t) = x_0 + \int_0^t v(t')\,dt'

Worked Example

A 2 kg object has F(t)=6tF(t) = 6t N. Find velocity at t=3t = 3 s if v(0)=0v(0) = 0.

a(t)=Fm=6t2=3ta(t) = \frac{F}{m} = \frac{6t}{2} = 3t

v(3)=∫033t dt=3t22∣03=272=13.5v(3) = \int_0^3 3t\,dt = \frac{3t^2}{2}\Big|_0^3 = \frac{27}{2} = 13.5 m/s ✅

Concept Check 🎯

Newton's Second Law with Calculus 🧮

  1. A 4 kg object has a net force of 20 N. What is its acceleration (m/s2)(m/s^{2})?

  2. A 2 kg mass has F(t)=6tF(t) = 6t N. What is the velocity (m/s) at t=3t = 3 s if starting from rest? (Hint: integrate a(t)=3ta(t) = 3t from 0 to 3). Give answer as a decimal.

  3. A constant force of 3 N acts on a 1 kg mass starting from rest. What is the velocity (m/s) at t=3t = 3 s?

Concept Check 🔍

Practice

#ConceptFormula
1F=maF = maConstant force
2v=∫a dtv = \int a\,dtVariable force
3x=∫v dtx = \int v\,dtPosition from velocity

Challenge Question 📋

Part 3: Newton's Third Law

⚛️ Newton's Third Law

Part 3 of 7 — Newton's Third Law

For every action force, there is an equal and opposite reaction force.

F⃗A on B=−F⃗B on A\vec{F}_{A \text{ on } B} = -\vec{F}_{B \text{ on } A}

  • Action-reaction pairs act on different objects
  • They are equal in magnitude, opposite in direction
  • They are the same type of force

Worked Example

A person pushes a wall with 50 N. What force does the wall exert on the person?

By Newton's Third Law: The wall exerts 50 N on the person in the opposite direction. ✅

These are an action-reaction pair acting on different objects (person and wall).

Concept Check 🎯

Newton's Third Law 🧮

  1. You push a desk with 50 N. How many newtons does the desk push back on you?

  2. A swimmer pushes water backward with 100 N. What force (N) does the water exert on the swimmer?

  3. Object A exerts 20 N on Object B. What force (N) does Object B exert on Object A?

Concept Check 🔍

Practice

#Action ForceReaction Force
1Earth pulls apple downApple pulls Earth up
2Bat hits ballBall hits bat
3Foot pushes ground backGround pushes foot forward

Challenge Question 📋

Part 4: Friction Forces

⚛️ Friction Forces

Part 4 of 7 — Friction Forces

  • Static friction: fs≤μsNf_s \leq \mu_s N (prevents motion)
  • Kinetic friction: fk=μkNf_k = \mu_k N (opposes sliding motion)

Where NN is the normal force and μ\mu is the coefficient of friction.

For an object on an incline at angle θ\theta:

  • N=mgcos⁡θN = mg\cos\theta
  • Friction along the incline (kinetic, or maximum static): f=μmgcos⁡θf = \mu mg\cos\theta

Worked Example

A 10 kg block on a horizontal surface has μk=0.3\mu_k = 0.3. Find the kinetic friction force.

N=mg=10×10=100N = mg = 10 \times 10 = 100 N

fk=μkN=0.3×100=30f_k = \mu_k N = 0.3 \times 100 = 30 N ✅

Concept Check 🎯

Friction Forces 🧮

  1. A 10 kg block on a flat surface has μk=0.3\mu_k = 0.3. Kinetic friction force (N)? (Use g=10g = 10 m/s2m/s^{2})

  2. A 5 kg block requires 20 N to start sliding on a flat surface. What is μs\mu_s? (Use g=10g = 10 m/s2m/s^{2})

  3. A 10 kg block is on a surface with μk=0.5\mu_k = 0.5. What horizontal force (N) is needed to move it at constant velocity? (Use g=10g = 10 m/s2m/s^{2})

Concept Check 🔍

Practice

#ScenarioKey Formula
1Flat surfacef=μmgf = \mu mg
2Inclined planef=μmgcos⁡θf = \mu mg\cos\theta
3Threshold of slidingfs=μsNf_s = \mu_s N

Challenge Question 📋

Part 5: Circular Motion

⚛️ Circular Motion

Part 5 of 7 — Circular Motion

For uniform circular motion:

ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r

Fc=mv2rF_c = \frac{mv^2}{r}

where ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T} is the angular velocity.

Centripetal force is not a new force — it is the net inward force (tension, gravity, friction, normal force, etc.).

Worked Example

A 2 kg ball on a 0.5 m string moves at 4 m/s in a horizontal circle. Find the centripetal force.

Fc=mv2r=2(16)0.5=64F_c = \frac{mv^2}{r} = \frac{2(16)}{0.5} = 64 N ✅

Concept Check 🎯

Circular Motion 🧮

  1. A car goes around a curve of radius 50 m at 20 m/s. What is the centripetal acceleration (m/s2)(m/s^{2})?

  2. An object moves in a circle of radius 2 m with centripetal acceleration 8 m/s2m/s^{2}. What is the speed (m/s)?

  3. A 2 kg ball on a 0.5 m string moves at 4 m/s. What is the tension/centripetal force (N)?

Concept Check 🔍

Practice

#ConceptFormula
1Centripetal accelerationac=v2/ra_c = v^2/r
2Centripetal forceFc=mv2/rF_c = mv^2/r
3Period and frequencyT=2πr/vT = 2\pi r / v

Challenge Question 📋

Part 6: Problem-Solving Workshop

⚛️ Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

Strategy for Newton's Law Problems

  1. Draw a free-body diagram (FBD)
  2. Choose coordinate axes (often along incline)
  3. Write ∑Fx=max\sum F_x = ma_x and ∑Fy=may\sum F_y = ma_y
  4. Solve the system of equations
  5. Check units and reasonableness

Worked Example

A 5 kg block is pushed with 40 N along a surface with μk=0.2\mu_k = 0.2. Find the acceleration. (g=10g = 10 m/s2m/s^{2})

N=mg=50N = mg = 50 N, fk=0.2×50=10f_k = 0.2 \times 50 = 10 N

∑F=40−10=30\sum F = 40 - 10 = 30 N

a=30/5=6a = 30/5 = 6 m/s2m/s^{2} ✅

Concept Check 🎯

Problem-Solving Workshop 🧮

  1. A 5 kg block is pushed with 40 N on a surface (μk=0.2\mu_k = 0.2, g=10g = 10 m/s2m/s^{2}). What is the acceleration (m/s2)(m/s^{2})?

  2. Net force on a 3 kg object is 12 N. What is the acceleration (m/s2)(m/s^{2})?

  3. A 10 kg object accelerates at 2 m/s2m/s^{2} while a 15 N friction force opposes its motion. What horizontal applied force (N) acts on it?

Concept Check 🔍

Practice

#Problem TypeKey Steps
1Horizontal surface with frictionFBD → Fnet=Fapp−fkF_{net} = F_{app} - f_k
2Incline problemsComponents along and perpendicular
3Connected objectsShared acceleration, separate FBDs

Challenge Question 📋

Part 7: Review & Applications

⚛️ Review & Applications

Part 7 of 7 — Review & Applications

Key Formulas

  • ∑F=ma=mdvdt\sum F = ma = m\frac{dv}{dt}
  • fs≤μsNf_s \leq \mu_s N, fk=μkNf_k = \mu_k N
  • Fc=mv2/rF_c = mv^2/r
  • Action-reaction: FAB=−FBAF_{AB} = -F_{BA}

Worked Example

An Atwood machine has masses 3 kg and 5 kg. Find the acceleration. (g=10g = 10 m/s2m/s^{2})

a=(m2−m1)gm1+m2=(5−3)(10)3+5=208=2.5a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(5-3)(10)}{3+5} = \frac{20}{8} = 2.5 m/s2m/s^{2} ✅

Concept Check 🎯

Review & Applications 🧮

  1. Atwood machine: masses 3 kg and 5 kg. Acceleration (m/s2)(m/s^{2})? (g=10g = 10 m/s2m/s^{2})

  2. A 10 kg object hangs from a rope. What is the tension (N)? (g=10g = 10 m/s2m/s^{2})

  3. A 4 kg block is pushed with 30 N. Friction is 10 N. What is the net force (N)?

Concept Check 🔍

Practice

#TopicKey Formula
1Newton's Laws overviewF=maF = ma
2Frictionf=μNf = \mu N
3Circular motionFc=mv2/rF_c = mv^2/r

Challenge Question 📋