problem solving about second law of motion

Gurumuda Networks

Newton’s second law of motion – problems and solutions

Solved problems in Newton’s laws of motion – Newton’s second law of motion

1. A 1 kg object accelerated at a constant 5 m/s 2 . Estimate the net force needed to accelerate the object.

Mass (m) = 1 kg

Acceleration (a) = 5 m/s 2

Wanted : net force (∑F)

We use Newton’s second law to get the net force.

∑ F = (1 kg)(5 m/s 2 ) = 5 kg m/s 2 = 5 Newton

2. Mass of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object’s acceleration….

Net force (∑F) = 2 Newton

Wanted : The magnitude and direction of the acceleration (a)

a = 2 m/s 2

The direction of the acceleration = the direction of the net force (∑F)

3. Object’s mass = 2 kg, F 1 = 5 Newton, F 2 = 3 Newton. The magnitude and direction of the acceleration is…

Mass (m) = 2 kg

F 1 = 5 Newton

F 2 = 3 Newton

net force :

∑ F = F 1 – F 2 = 5 – 3 = 2 Newton

The magnitude of the acceleration :

a = 1 m/s 2

Direction of the acceleration = direction of the net force = direction of F 1

4. Object’s mass = 2 kg, F 1 = 10 Newton, F 2 = 1 Newton. The magnitude and direction of the acceleration is…

Newton's second law of motion – problems and solutions 3

F 2 = 1 Newton

F 1 = 10 Newton

F 1x = F 1 cos 60 o = (10)(0.5) = 5 Newton

Net force :

∑ F = F 1x – F 2 = 5 – 1 = 4 Newton

Direction of the acceleration = direction of the net force = direction of F 1x

5. F 1 = 10 Newton, F 2 = 1 Newton, m 1 = 1 kg, m 2 = 2 kg. The magnitude and direction of the acceleration is…

Mass 1 (m 1 ) = 1 kg

Mass 2 (m 2 ) = 2 kg

The net force :

∑ F = F 1 – F 2 = 10 – 1 = 9 Newton

a = ∑F / (m 1 + m 2 )

a = 9 / (1 + 2)

a = 3 m/s 2

The direction of the acceleration = the direction of the net force = direction of F 1

A 40-kg block accelerated by a force of 200 N. Acceleration of the block is 3 m/ s 2 . Determine the magnitude of friction force experienced by the block.

Newton's second law of motion – problems and solutions 7

Mass (m) = 40 kg

Force (F) = 200 N

Acceleration (a) = 3 m/s 2

Wanted: Friction force (F g )

The equation of Newton’s second law of motion

∑ F = net force, m = mass, a = acceleration

The direction of force F rightward, the direction of friction force leftward (the direction of friction force is opposite with the direction of object’s motion).

Choose rightward as positive and leftward as negative.

F – F g = m a

200 – F g = (40)(3)

200 – F g = 120

F g = 200 – 120

F g = 80 Newton

The correct answer is D.

7. Block A with a mass of 100-gram place above block B with a mass of 300 gram, and then block b pushed with a force of 5 N vertically upward. Determine the normal force exerted by block B on block A.

Newton's second law of motion – problems and solutions 2

Force (F) = 5 Newton

Mass of block A (m A ) = 100 gram = 0.1 kg

Mass of block B (m B ) = 300 gram = 0.3 kg

Acceleration of gravity (g) = 10 m/s 2

Weight of block A (w A ) = (0.1 kg)(10 m/s 2 ) = 1 kg m/s 2 = 1 Newton

Weight of block B (w B ) = (0.3 kg)(10 m/s 2 ) = 3 kg m/s 2 = 3 Newton

Wanted : Normal force exerted by block B to block A

F = push force (act on block B)

w A = weight of block A (act on block A)

w B = weight of block B (act on block B)

N A = normal force exerted by block B on block A (Act on block A)

N A ’ = normal force exerted by block A on block B (Act on block B)

Apply Newton’s second law of motion on both blocks :

F – w A – w B + N A – N A ’ = (m A + m B ) a

N A and N A ’ are action-reaction forces that have the same magnitude but opposite in direction so eliminated from the equation.

F – w A – w B = (m A + m B ) a

5 – 1 – 3 = (0.1 + 0.3) a

5 – 4 = (0.4) a

1 = (0.4) a

a = 1 / 0.4

a = 2.5 m/s 2

Apply Newton’s second law of motion on block A :

N A – w A = m A a

N A – 1 = (0.1)(2.5)

N A – 1 = 0.25

N A = 1 + 0.25

N A = 1.25 Newton

The correct answer is B.

8. An object with weight of 4 N supported by a cord and pulley. A force of 2 N acts on the block and one end of the cord pulled by a force of 9 N. Determine the net force acts on object X.

Newton's second law of motion – problems and solutions 4

B. 4 N downward

C. 9 N upward

D. 9 N downward

Weight of X (w X ) = 4 Newton

Pull force (F x ) = 2 Newton

Tension force (F T ) = 9 Newton

Wanted: Net force acts on object X

Vertically upward forces that act on object X :

The tension force has the same magnitude in all part of the cord. So the tension force is 9 N.

Vertically downward forces that act on object X :

There are two forces that act on object X and both forces are vertically downward, the horizontal component of weight w x and the horizontal component of force F x .

Net force act on the object X :

F T – w X – F x = 9 – 4 – 2 = 9 – 6 = 3

The net force act on the object X is 3 Newton, vertically upward.

The correct answer is A.

9. An object initially at rest on a smooth horizontal surface. A force of 16 N acts on the object so the object accelerated at 2 m/s 2 . If the same object at rest on a rough horizontal surface so the friction force acts on the object is 2 N, then determine the acceleration of the object if the same force of 16 N acts on the object.

A. 1.75 m/s 2

B. 1.50 m/s 2

C. 1.00 m/s 2

D. 0.88 m/s 2

Force (F) = 16 Newton = 16 kg m/s 2

Acceleration (a) = 2 m/s 2

Friction force (F fric ) = 2 Newton = 2 kg m/s 2

Wanted : Object’s acceleration ?

Smooth horizontal surface (no friction force) :

Newton's second law of motion – problems and solutions 5

Mass of object is 8 kilogram.

Rough horizontal surface (there is a friction force) :

Newton's second law of motion – problems and solutions 6

F – F fric = m a

16 – 2 = 8 a

a = 1.75 m/s 2

Object’s acceleration is 1.75 m/s 2 .

10. Tom and Andrew push an object on the smooth floor. Tom push the object with a force of 5.70 N. If the mass of the object is 2.00 kg and acceleration experienced by the object is 2.00 ms -2 , then determine the magnitude and direction of force act by Tom.

A. 1.70 N and its direction is opposite with force acted by Andre.w

B. 1.70 N and its direction same as force acted by Andrew

C. 2.30 N and its direction is opposite with force acted by Andrew.

D. 2.30 N and its direction same as force acted by Andrew.

Push force acted by Andrew (F 1 ) = 5.70 Newton

Mass of object (m) = 2.00 kg

Acceleration (a) = 2.00 m/s 2

Wanted : Magnitude and direction of force acted by Tom (F 2 ) ?

Apply Newton’s second law of motion :

F 1 + F 2 = m a

5.70 + F 2 = (2)(2)

5.70 + F 2 = 4

F 2 = 4 – 5.70

F 2 = – 1.7 Newton

Minus sign indicated that (F 2 ) is opposite with push force act by Andrew (F 1 ).

11. If the mass of the block is the same, which figure shows the smallest acceleration?

Newton's first law and Newton's second law 2

Net force A :

ΣF = 4 N + 2 N – 3 N = 6 N – 3 N = 3 Newton, leftward

Net force B :

ΣF = 2 N + 3 N – 4 N = 5 N – 4 N = 1 Newton, rightward

Net force C :

ΣF = 4 N + 3 N – 2 N = 7 N – 2 N = 5 Newton, rightward

Net force D :

ΣF = 3 N + 4 N + 2 N = 9 Newton, rightward

The equation of Newton’s second law :

a = acceleration, ΣF = net force, m = mass

Based on the above formula, the acceleration (a) is directly proportional to the net force (ΣF) and inversely proportional to mass (m). If the mass of an object is the same, the greater the resultant force, the greater the acceleration or the smaller the resultant force, the smaller the acceleration. Based on the above calculation, the smallest net force is 1 Newton so the acceleration is also smallest.

12. Some forces act on an object with a mass of 20 kg, as shown in the figure below.

Determine the object’s acceleration.

Mass of object (m) = 20 kg

Net force (ΣF) = 25 N + 30 N – 15 N = 40 N

Wanted: Acceleration of an object

Object’s acceleration calculated using the equation of Newton’s second law :

a = ΣF / m = 40 N / 20 kg = 2 N/kg = 2 m/s 2

13. Which statements below describes Newton’s third law?

(1) Passengers pushed forward when the bus braked suddenly

(2) B ooks on paper are not falling when the paper is pulled quickly

(3) When playing skateboard when the foot pushes the ground back then the skateboard will slide forward

(4) O ars pushed backward, boats moving forward

(1) Newton’s first law

(2) Newton’s first law

(3) Newton’s third law

(4) Newton’s third law

[wpdm_package id=’470′]

Mass and weight
Normal force
Newton’s second law of motion
Friction force
Motion on the horizontal surface without friction force
The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
Motion on the inclined plane without friction force
Motion on the rough inclined plane with the friction force
Motion in an elevator
The motion of bodies connected by cord and pulley
Two bodies with the same magnitude of accelerations
Rounding a flat curve – dynamics of circular motion
Rounding a banked curve – dynamics of circular motion
Uniform motion in a horizontal circle
Centripetal force in uniform circular motion

Discover more from Physics

Subscribe now to keep reading and get access to the full archive.

Type your email…

TPC and eLearning
Read Watch Interact
What's NEW at TPC?
Practice Review Test
Teacher-Tools
Subscription Selection
Seat Calculator
Ad Free Account
Edit Profile Settings
Classes (Version 2)
Student Progress Edit
Task Properties
Export Student Progress
Task, Activities, and Scores
Metric Conversions Questions
Metric System Questions
Metric Estimation Questions
Significant Digits Questions
Proportional Reasoning
Acceleration
Distance-Displacement
Dots and Graphs
Graph That Motion
Match That Graph
Name That Motion
Motion Diagrams
Pos'n Time Graphs Numerical
Pos'n Time Graphs Conceptual
Up And Down - Questions
Balanced vs. Unbalanced Forces
Change of State
Force and Motion
Mass and Weight
Match That Free-Body Diagram
Net Force (and Acceleration) Ranking Tasks
Newton's Second Law
Normal Force Card Sort
Recognizing Forces
Air Resistance and Skydiving
Solve It! with Newton's Second Law
Which One Doesn't Belong?
Component Addition Questions
Head-to-Tail Vector Addition
Projectile Mathematics
Trajectory - Angle Launched Projectiles
Trajectory - Horizontally Launched Projectiles
Vector Addition
Vector Direction
Which One Doesn't Belong? Projectile Motion
Forces in 2-Dimensions
Being Impulsive About Momentum
Explosions - Law Breakers
Hit and Stick Collisions - Law Breakers
Case Studies: Impulse and Force
Impulse-Momentum Change Table
Keeping Track of Momentum - Hit and Stick
Keeping Track of Momentum - Hit and Bounce
What's Up (and Down) with KE and PE?
Energy Conservation Questions
Energy Dissipation Questions
Energy Ranking Tasks
LOL Charts (a.k.a., Energy Bar Charts)
Match That Bar Chart
Words and Charts Questions
Name That Energy
Stepping Up with PE and KE Questions
Case Studies - Circular Motion
Circular Logic
Forces and Free-Body Diagrams in Circular Motion
Gravitational Field Strength
Universal Gravitation
Angular Position and Displacement
Linear and Angular Velocity
Angular Acceleration
Rotational Inertia
Balanced vs. Unbalanced Torques
Getting a Handle on Torque
Torque-ing About Rotation
Properties of Matter
Fluid Pressure
Buoyant Force
Sinking, Floating, and Hanging
Pascal's Principle
Flow Velocity
Bernoulli's Principle
Balloon Interactions
Charge and Charging
Charge Interactions
Charging by Induction
Conductors and Insulators
Coulombs Law
Electric Field
Electric Field Intensity
Polarization
Case Studies: Electric Power
Know Your Potential
Light Bulb Anatomy
I = ∆V/R Equations as a Guide to Thinking
Parallel Circuits - ∆V = I•R Calculations
Resistance Ranking Tasks
Series Circuits - ∆V = I•R Calculations
Series vs. Parallel Circuits
Equivalent Resistance
Period and Frequency of a Pendulum
Pendulum Motion: Velocity and Force
Energy of a Pendulum
Period and Frequency of a Mass on a Spring
Horizontal Springs: Velocity and Force
Vertical Springs: Velocity and Force
Energy of a Mass on a Spring
Decibel Scale
Frequency and Period
Closed-End Air Columns
Name That Harmonic: Strings
Rocking the Boat
Wave Basics
Matching Pairs: Wave Characteristics
Wave Interference
Waves - Case Studies
Color Addition and Subtraction
Color Filters
If This, Then That: Color Subtraction
Light Intensity
Color Pigments
Converging Lenses
Curved Mirror Images
Law of Reflection
Refraction and Lenses
Total Internal Reflection
Who Can See Who?
Formulas and Atom Counting
Atomic Models
Bond Polarity
Entropy Questions
Cell Voltage Questions
Heat of Formation Questions
Reduction Potential Questions
Oxidation States Questions
Measuring the Quantity of Heat
Hess's Law
Oxidation-Reduction Questions
Galvanic Cells Questions
Thermal Stoichiometry
Molecular Polarity
Quantum Mechanics
Balancing Chemical Equations
Bronsted-Lowry Model of Acids and Bases
Classification of Matter
Collision Model of Reaction Rates
Density Ranking Tasks
Dissociation Reactions
Complete Electron Configurations
Elemental Measures
Enthalpy Change Questions
Equilibrium Concept
Equilibrium Constant Expression
Equilibrium Calculations - Questions
Equilibrium ICE Table
Ionic Bonding
Lewis Electron Dot Structures
Limiting Reactants
Line Spectra Questions
Mass Stoichiometry
Measurement and Numbers
Metals, Nonmetals, and Metalloids
Metric Estimations
Metric System
Molarity Ranking Tasks
Mole Conversions
Name That Element
Names to Formulas
Names to Formulas 2
Nuclear Decay
Particles, Words, and Formulas
Periodic Trends
Precipitation Reactions and Net Ionic Equations
Pressure Concepts
Pressure-Temperature Gas Law
Pressure-Volume Gas Law
Chemical Reaction Types
Significant Digits and Measurement
States Of Matter Exercise
Stoichiometry Law Breakers
Stoichiometry - Math Relationships
Subatomic Particles
Spontaneity and Driving Forces
Gibbs Free Energy
Volume-Temperature Gas Law
Acid-Base Properties
Energy and Chemical Reactions
Chemical and Physical Properties
Valence Shell Electron Pair Repulsion Theory
Writing Balanced Chemical Equations
Mission CG1
Mission CG10
Mission CG2
Mission CG3
Mission CG4
Mission CG5
Mission CG6
Mission CG7
Mission CG8
Mission CG9
Mission EC1
Mission EC10
Mission EC11
Mission EC12
Mission EC2
Mission EC3
Mission EC4
Mission EC5
Mission EC6
Mission EC7
Mission EC8
Mission EC9
Mission RL1
Mission RL2
Mission RL3
Mission RL4
Mission RL5
Mission RL6
Mission KG7
Mission RL8
Mission KG9
Mission RL10
Mission RL11
Mission RM1
Mission RM2
Mission RM3
Mission RM4
Mission RM5
Mission RM6
Mission RM8
Mission RM10
Mission LC1
Mission RM11
Mission LC2
Mission LC3
Mission LC4
Mission LC5
Mission LC6
Mission LC8
Mission SM1
Mission SM2
Mission SM3
Mission SM4
Mission SM5
Mission SM6
Mission SM8
Mission SM10
Mission KG10
Mission SM11
Mission KG2
Mission KG3
Mission KG4
Mission KG5
Mission KG6
Mission KG8
Mission KG11
Mission F2D1
Mission F2D2
Mission F2D3
Mission F2D4
Mission F2D5
Mission F2D6
Mission KC1
Mission KC2
Mission KC3
Mission KC4
Mission KC5
Mission KC6
Mission KC7
Mission KC8
Mission AAA
Mission SM9
Mission LC7
Mission LC9
Mission NL1
Mission NL2
Mission NL3
Mission NL4
Mission NL5
Mission NL6
Mission NL7
Mission NL8
Mission NL9
Mission NL10
Mission NL11
Mission NL12
Mission MC1
Mission MC10
Mission MC2
Mission MC3
Mission MC4
Mission MC5
Mission MC6
Mission MC7
Mission MC8
Mission MC9
Mission RM7
Mission RM9
Mission RL7
Mission RL9
Mission SM7
Mission SE1
Mission SE10
Mission SE11
Mission SE12
Mission SE2
Mission SE3
Mission SE4
Mission SE5
Mission SE6
Mission SE7
Mission SE8
Mission SE9
Mission VP1
Mission VP10
Mission VP2
Mission VP3
Mission VP4
Mission VP5
Mission VP6
Mission VP7
Mission VP8
Mission VP9
Mission WM1
Mission WM2
Mission WM3
Mission WM4
Mission WM5
Mission WM6
Mission WM7
Mission WM8
Mission WE1
Mission WE10
Mission WE2
Mission WE3
Mission WE4
Mission WE5
Mission WE6
Mission WE7
Mission WE8
Mission WE9
Vector Walk Interactive
Name That Motion Interactive
Kinematic Graphing 1 Concept Checker
Kinematic Graphing 2 Concept Checker
Graph That Motion Interactive
Two Stage Rocket Interactive
Rocket Sled Concept Checker
Force Concept Checker
Free-Body Diagrams Concept Checker
Free-Body Diagrams The Sequel Concept Checker
Skydiving Concept Checker
Elevator Ride Concept Checker
Vector Addition Concept Checker
Vector Walk in Two Dimensions Interactive
Name That Vector Interactive
River Boat Simulator Concept Checker
Projectile Simulator 2 Concept Checker
Projectile Simulator 3 Concept Checker
Hit the Target Interactive
Turd the Target 1 Interactive
Turd the Target 2 Interactive
Balance It Interactive
Go For The Gold Interactive
Egg Drop Concept Checker
Fish Catch Concept Checker
Exploding Carts Concept Checker
Collision Carts - Inelastic Collisions Concept Checker
Its All Uphill Concept Checker
Stopping Distance Concept Checker
Chart That Motion Interactive
Roller Coaster Model Concept Checker
Uniform Circular Motion Concept Checker
Horizontal Circle Simulation Concept Checker
Vertical Circle Simulation Concept Checker
Race Track Concept Checker
Gravitational Fields Concept Checker
Orbital Motion Concept Checker
Angular Acceleration Concept Checker
Balance Beam Concept Checker
Torque Balancer Concept Checker
Aluminum Can Polarization Concept Checker
Charging Concept Checker
Name That Charge Simulation
Coulomb's Law Concept Checker
Electric Field Lines Concept Checker
Put the Charge in the Goal Concept Checker
Circuit Builder Concept Checker (Series Circuits)
Circuit Builder Concept Checker (Parallel Circuits)
Circuit Builder Concept Checker (∆V-I-R)
Circuit Builder Concept Checker (Voltage Drop)
Equivalent Resistance Interactive
Pendulum Motion Simulation Concept Checker
Mass on a Spring Simulation Concept Checker
Particle Wave Simulation Concept Checker
Boundary Behavior Simulation Concept Checker
Slinky Wave Simulator Concept Checker
Simple Wave Simulator Concept Checker
Wave Addition Simulation Concept Checker
Standing Wave Maker Simulation Concept Checker
Color Addition Concept Checker
Painting With CMY Concept Checker
Stage Lighting Concept Checker
Filtering Away Concept Checker
InterferencePatterns Concept Checker
Young's Experiment Interactive
Plane Mirror Images Interactive
Who Can See Who Concept Checker
Optics Bench (Mirrors) Concept Checker
Name That Image (Mirrors) Interactive
Refraction Concept Checker
Total Internal Reflection Concept Checker
Optics Bench (Lenses) Concept Checker
Kinematics Preview
Velocity Time Graphs Preview
Moving Cart on an Inclined Plane Preview
Stopping Distance Preview
Cart, Bricks, and Bands Preview
Fan Cart Study Preview
Friction Preview
Coffee Filter Lab Preview
Friction, Speed, and Stopping Distance Preview
Up and Down Preview
Projectile Range Preview
Ballistics Preview
Juggling Preview
Marshmallow Launcher Preview
Air Bag Safety Preview
Colliding Carts Preview
Collisions Preview
Engineering Safer Helmets Preview
Push the Plow Preview
Its All Uphill Preview
Energy on an Incline Preview
Modeling Roller Coasters Preview
Hot Wheels Stopping Distance Preview
Ball Bat Collision Preview
Energy in Fields Preview
Weightlessness Training Preview
Roller Coaster Loops Preview
Universal Gravitation Preview
Keplers Laws Preview
Kepler's Third Law Preview
Charge Interactions Preview
Sticky Tape Experiments Preview
Wire Gauge Preview
Voltage, Current, and Resistance Preview
Light Bulb Resistance Preview
Series and Parallel Circuits Preview
Thermal Equilibrium Preview
Linear Expansion Preview
Heating Curves Preview
Electricity and Magnetism - Part 1 Preview
Electricity and Magnetism - Part 2 Preview
Vibrating Mass on a Spring Preview
Period of a Pendulum Preview
Wave Speed Preview
Slinky-Experiments Preview
Standing Waves in a Rope Preview
Sound as a Pressure Wave Preview
DeciBel Scale Preview
DeciBels, Phons, and Sones Preview
Sound of Music Preview
Shedding Light on Light Bulbs Preview
Models of Light Preview
Electromagnetic Radiation Preview
Electromagnetic Spectrum Preview
EM Wave Communication Preview
Digitized Data Preview
Light Intensity Preview
Concave Mirrors Preview
Object Image Relations Preview
Snells Law Preview
Reflection vs. Transmission Preview
Magnification Lab Preview
Reactivity Preview
Ions and the Periodic Table Preview
Periodic Trends Preview
Intermolecular Forces Preview
Melting Points and Boiling Points Preview
Reaction Rates Preview
Ammonia Factory Preview
Stoichiometry Preview
Gaining Teacher Access
Tasks and Classes
Tasks - Classic
Subscription
Subscription Locator
1-D Kinematics
Newton's Laws
Vectors - Motion and Forces in Two Dimensions
Momentum and Its Conservation
Work and Energy
Circular Motion and Satellite Motion
Thermal Physics
Static Electricity
Electric Circuits
Vibrations and Waves
Sound Waves and Music
Light and Color
Reflection and Mirrors
About the Physics Interactives
Task Tracker
Usage Policy
Newtons Laws
Vectors and Projectiles
Forces in 2D
Momentum and Collisions
Circular and Satellite Motion
Balance and Rotation
Electromagnetism
Waves and Sound
Forces in Two Dimensions
Work, Energy, and Power
Circular Motion and Gravitation
Sound Waves
1-Dimensional Kinematics
Circular, Satellite, and Rotational Motion
Einstein's Theory of Special Relativity
Waves, Sound and Light
QuickTime Movies
About the Concept Builders
Pricing For Schools
Directions for Version 2
Measurement and Units
Relationships and Graphs
Rotation and Balance
Vibrational Motion
Reflection and Refraction
Teacher Accounts
Task Tracker Directions
Kinematic Concepts
Kinematic Graphing
Wave Motion
Sound and Music
About CalcPad
1D Kinematics
Vectors and Forces in 2D
Simple Harmonic Motion
Rotational Kinematics
Rotation and Torque
Rotational Dynamics
Electric Fields, Potential, and Capacitance
Transient RC Circuits
Light Waves
Units and Measurement
Stoichiometry
Molarity and Solutions
Thermal Chemistry
Acids and Bases
Kinetics and Equilibrium
Solution Equilibria
Oxidation-Reduction
Nuclear Chemistry
NGSS Alignments
1D-Kinematics
Projectiles
Circular Motion
Magnetism and Electromagnetism
Graphing Practice
About the ACT
ACT Preparation
For Teachers
Other Resources
Newton's Laws of Motion
Work and Energy Packet
Static Electricity Review
Solutions Guide
Solutions Guide Digital Download
Motion in One Dimension
Work, Energy and Power
Frequently Asked Questions
Purchasing the Download
Purchasing the CD
Purchasing the Digital Download
About the NGSS Corner
NGSS Search
Force and Motion DCIs - High School
Energy DCIs - High School
Wave Applications DCIs - High School
Force and Motion PEs - High School
Energy PEs - High School
Wave Applications PEs - High School
Crosscutting Concepts
The Practices
Physics Topics
NGSS Corner: Activity List
NGSS Corner: Infographics
About the Toolkits
Position-Velocity-Acceleration
Position-Time Graphs
Velocity-Time Graphs
Newton's First Law
Newton's Second Law
Newton's Third Law
Terminal Velocity
Projectile Motion
Forces in 2 Dimensions
Impulse and Momentum Change
Momentum Conservation
Work-Energy Fundamentals
Work-Energy Relationship
Roller Coaster Physics
Satellite Motion
Electric Fields
Circuit Concepts
Series Circuits
Parallel Circuits
Describing-Waves
Wave Behavior Toolkit
Standing Wave Patterns
Resonating Air Columns
Wave Model of Light
Plane Mirrors
Curved Mirrors
Teacher Guide
Using Lab Notebooks
Current Electricity
Light Waves and Color
Reflection and Ray Model of Light
Refraction and Ray Model of Light
Classes (Legacy Version)
Teacher Resources
Subscriptions

problem solving about second law of motion

Newton's Laws
Einstein's Theory of Special Relativity
About Concept Checkers
School Pricing
Newton's Laws of Motion
Newton's First Law
Newton's Third Law

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Join our Telegram Channel for Free PDF Download

Problems Based on Newton’s Second Law of Motion

Last modified on: 1 year ago
Reading Time: 16 Minutes

NEWTON’S LAWS OF MOTION

Newton’s First Law of Motion – Any object remains in the state of rest or in uniform motion along a straight line, until it is compelled to change the state by applying external force.
Newton’s Second Law of Motion – The rate of change of momentum is directly proportional to the force applied in the direction of force.
Newton’s Third Law of Motion – For every action there is an equal and opposite reaction.

NEWTON’S SECOND LAW OF MOTION

Newton’s second Law of Motion states that The rate of change of momentum is directly proportional to the force applied in the direction of force.

For example; when acceleration is applied on a moving vehicle, the momentum of the vehicle increases and the increase is in the direction of motion because the force is being applied in the direction of motion. On the other hand, when brake is applied on the moving vehicle, the momentum of the vehicle decreases and the decrease is in the opposite direction of motion because the force is being applied in the opposite direction of motion.

Force = mass × acceleration

Newton’s Second Law of Motion gives the relation between force, mass and acceleration of an object.

According to the relation obtained above, Newton’s Second Law can be modified as follows: The product of mass and acceleration is the force acting on the object. The SI unit of Force: Newton (N)

From the above relation it is clear that Acceleration increases with increase in force and vice versa. Acceleration decreases with increase in mass and vice versa. That’s why a small vehicle requires less force to attain more acceleration while a heavy vehicle requires more force to get the same acceleration

Q.1. What is acceleration produced by a force of 12 Newton exerted on an object of mass 3 kg?

Q.2. What force would be needed to produce an acceleration of 4m/s 2 on a ball of mass 6 kg?

Q.3. A force of 4 N acts on a body of mass 2 kg for 4 s. Assuming that the body to be initially at rest, find (i) its velocity when the force stops acting (ii) the distance covered in 10 s after the force starts acting.

Q.4. A feather of mass 10 g is dropped from a height. It is found to fall down with a constant velocity. What is the net force acting on it?

Q.5. A force of 5 N gives a mass m1, an acceleration of 8 m/s 2 , and a mass m 2 , an acceleration of 24 m/s 2 . What acceleration would give if both the masses are tied together?

Q.6. Calculate the force required to impart a car a velocity of 30m/s in 10 seconds. The mass of the car is 1500 kg.

Q.7. A force of 5 N gives a mass m1, an acceleration of 10 m/s 2 , and a mass m 2 , an acceleration of 20 m/s 2 . What acceleration would it give if both the masses were tied together?

Q.8. A 150 g ball traveling at 30m/s strikes the palm of a players hand and is stopped in 0.06 sec. Calculate the force exerted by the ball on the hand.

Q.9. A motorcycle is moving with a velocity of 90km/h and it takes 5 seconds to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcycle if its mass along with the rider is 200kg.

Q.10. What is the momentum of a man of mass 75kg when he walks with a velocity of 2m/s?

Q.11. What would be the force required to produce an acceleration of 2m/s 2 in a body of mass 12 kg ?What would be the acceleration it the force were doubled?

Q.12. A man pushes a box of mass 50 kg with a force of 80N. What will be the acceleration of the box? What would be the acceleration if the mass were doubled?

Q.13. A certain force exerted for 1.2 second raises the speed of an object from 1.8m/s to 4.2 m/s. Later, the same force is applied for 2 second. How much does the speed change in 2 second?

Q.14. A constant force acts on an object of mass 5 kg for duration of 2 second. It increases the object’s velocity from 3cm/s to 7m/s. Find the magnitude of the applied force. Now if the force were applied for a duration of 5 seconds, what would be the final velocity of the object?

Q.15. A motorcar is moving with a velocity of 108km/h and it takes 4 seconds to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar it its mass along with the passengers is 1000 kg.

Q.16. A hockey ball of mass 200g traveling from west to east at 10m/s is struck by a hockey stick. As a result, then ball gets turned back and now has a speed of 5m/s. If the ball and hockey stick were in contact for 0.2 s, calculate (i) initial and final momentum of the ball (ii) rate of change of momentum of the ball (iii) the force exerted by hockey stick on the ball.

Q.17. A stone of mass 500 g is thrown with a velocity of 20m/s across the frozen surface of a lake. It comes to rest after traveling a distance of 0.1 km. Calculate force of friction between the stone and frozen surface of lake.

Q.18. A body starts from rest and rolls down a hill with a constant acceleration. If its travels 400 m in 20 seconds, calculate the force acting on the body if its mass is 10kg.

Q.19. For how long should a force of 100 N act on a body of mass 20 kg so that it acquires a velocity of 100 m/s?

Q.20. A body of mass 1 kg is kept at rest. A constant force of 6.0 N starts acting on it . Find the time taken by the body to move through a distance of 12m.

Extra Questions For Class 9 Science Chapter 1 Matter in Our Surroundings

Last modified on:3 years agoReading Time:29MinutesExtra Questions For Class 9 Science Chapter 1 Matter in Our Surroundings Very Short Answer Type Questions (One Mark Each) Q.1. Degree Celsius and Kelvin are two units of measuring temperature. Which of these are SI and non-SI units ? Answer Ans. Degree Celsius – Non-SI. Kelvin – SI. Q.2. What…

Extra Questions For Class 9 Science Chapter 2 Is Matter Around Us Pure

Last modified on:3 years agoReading Time:12MinutesExtra Questions For Class 9 Science Chapter 2 Is Matter Around Us Pure Very Short Answer Type Questions (One Mark Each) Q.1. Suggest separation techniques one would need to employ to separate the mixtures :(a) Mercury and water(b) Potassium Chloride and ammonium chloride. Answer Ans. (a) Using separating funnel. (b) Using…

Extra Questions For Class 9 Science Chapter 5 fundamental unit of life

Last modified on:12 months agoReading Time:17MinutesExtra Questions For Class 9 Science Chapter 5 Fundamental Unit of Life Very Short Answer Type Questions (One Mark Each) Q.1. Name the scientist who discovered cell. Answer Ans. Robert Hooke discovered cell in 1665. Q.2. Who discovered nucleus ? Answer Ans. Robert Brown discovered nucleus. Q.3. Who coined the term…

Download CBSE Books

Exam Special Series:

Sample Question Paper for CBSE Class 10 Science (for 2024)
Sample Question Paper for CBSE Class 10 Maths (for 2024)
CBSE Most Repeated Questions for Class 10 Science Board Exams
CBSE Important Diagram Based Questions Class 10 Physics Board Exams
CBSE Important Numericals Class 10 Physics Board Exams
CBSE Practical Based Questions for Class 10 Science Board Exams
CBSE Important “Differentiate Between” Based Questions Class 10 Social Science
Sample Question Papers for CBSE Class 12 Physics (for 2024)
Sample Question Papers for CBSE Class 12 Chemistry (for 2024)
Sample Question Papers for CBSE Class 12 Maths (for 2024)
Sample Question Papers for CBSE Class 12 Biology (for 2024)
CBSE Important Diagrams & Graphs Asked in Board Exams Class 12 Physics
Master Organic Conversions CBSE Class 12 Chemistry Board Exams
CBSE Important Numericals Class 12 Physics Board Exams
CBSE Important Definitions Class 12 Physics Board Exams
CBSE Important Laws & Principles Class 12 Physics Board Exams
10 Years CBSE Class 12 Chemistry Previous Year-Wise Solved Papers (2023-2024)
10 Years CBSE Class 12 Physics Previous Year-Wise Solved Papers (2023-2024)
10 Years CBSE Class 12 Maths Previous Year-Wise Solved Papers (2023-2024)
10 Years CBSE Class 12 Biology Previous Year-Wise Solved Papers (2023-2024)
ICSE Important Numericals Class 10 Physics BOARD Exams (215 Numericals)
ICSE Important Figure Based Questions Class 10 Physics BOARD Exams (230 Questions)
ICSE Mole Concept and Stoichiometry Numericals Class 10 Chemistry (65 Numericals)
ICSE Reasoning Based Questions Class 10 Chemistry BOARD Exams (150 Qs)
ICSE Important Functions and Locations Based Questions Class 10 Biology
ICSE Reasoning Based Questions Class 10 Biology BOARD Exams (100 Qs)

✨ Join our Online JEE Test Series for 499/- Only (Web + App) for 1 Year

✨ Join our Online NEET Test Series for 499/- Only for 1 Year

Discover more from Gurukul of Excellence

Subscribe now to keep reading and get access to the full archive.

Type your email…

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Applications of Newton’s Laws

Solving Problems with Newton’s Laws

Learning objectives.

By the end of the section, you will be able to:

Apply problem-solving techniques to solve for quantities in more complex systems of forces
Use concepts from kinematics to solve problems using Newton’s laws of motion
Solve more complex equilibrium problems
Solve more complex acceleration problems
Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Identify the physical principles involved by listing the givens and the quantities to be calculated.
Sketch the situation, using arrows to represent all forces.
Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in (Figure) (a). Then, as in (Figure) (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

$\stackrel{\to }{T}$

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See (Figure) (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. (Figure) (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in (Figure) (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

$\sum {F}_{x}=m{a}_{x},\phantom{\rule{0.5em}{0ex}}\sum {F}_{y}=m{a}_{y}.$

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Different Tensions at Different Angles Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in (Figure) . Find the tension in each wire, neglecting the masses of the wires.

A sketch of a traffic light suspended from two wires supported by two poles is shown. (b) Some forces are shown in this system. Tension T sub one pulling the top of the left-hand pole is shown by the vector arrow along the left wire from the top of the pole, and an equal but opposite tension T sub one is shown by the arrow pointing up along the left-hand wire where it is attached to the light; the left-hand wire makes a thirty-degree angle with the horizontal. Tension T sub two is shown by a vector arrow pointing downward from the top of the right-hand pole along the right-hand wire, and an equal but opposite tension T sub two is shown by the arrow pointing up along the right-hand wire, which makes a forty-five degree angle with the horizontal. The traffic light is suspended at the lower end of the wires, and its weight W is shown by a vector arrow acting downward. (c) The traffic light is the system of interest, indicated by circling the traffic light. Tension T sub one starting from the traffic light is shown by an arrow along the wire making an angle of thirty degrees with the horizontal. Tension T sub two starting from the traffic light is shown by an arrow along the wire making an angle of forty-five degrees with the horizontal. The weight W is shown by a vector arrow pointing downward from the traffic light. A free-body diagram is shown with three forces acting on a point. Weight W acts downward; T sub one and T sub two act at an angle with the vertical. A coordinate system is shown, with positive x to the right and positive y upward. (d) Forces are shown with their components. T sub one is decomposed into T sub one y pointing vertically upward and T sub one x pointing along the negative x direction. The angle between T sub one and T sub one x is thirty degrees. T sub two is decomposed into T sub two y pointing vertically upward and T sub two x pointing along the positive x direction. The angle between T sub two and T sub two x is forty five degrees. Weight W is shown by a vector arrow acting downward. (e) The net vertical force is zero, so the vector equation is T sub one y plus T sub two y equals W. T sub one y and T sub two y are shown on a free body diagram as equal length arrows pointing up. W is shown as a downward pointing arrow whose length is twice as long as each of the T sub one y and T sub two y arrows. The net horizontal force is zero, so vector T sub one x is equal to minus vector T sub two x. T sub two x is shown by an arrow pointing toward the right, and T sub one x is shown by an arrow pointing toward the left.

Solution First consider the horizontal or x -axis:

${F}_{\text{net}\phantom{\rule{0.2em}{0ex}}x}={T}_{2x}-{T}_{1x}=0.$

Thus, as you might expect,

${T}_{1x}={T}_{2x}.$

This gives us the following relationship:

${T}_{1}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}30\text{°}={T}_{2}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}45\text{°}.$

Now consider the force components along the vertical or y -axis:

${F}_{\text{net}\phantom{\rule{0.2em}{0ex}}y}={T}_{1y}+{T}_{2y}-w=0.$

This implies

${T}_{1y}+{T}_{2y}=w.$

Substituting the expressions for the vertical components gives

${T}_{1}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}30\text{°}+{T}_{2}\text{sin}\phantom{\rule{0.2em}{0ex}}45\text{°}=w.$

which yields

$1.366{T}_{1}=\left(15.0\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right).$

Significance Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion .

Particle Acceleration

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

$2.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}$

The angle is given by

$\theta ={\text{tan}}^{-1}\left(\frac{{F}_{2}}{{F}_{1}}\right)={\text{tan}}^{-1}\left(\frac{3.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}}{2.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}}\right)=53.1\text{°}.$

However, Newton’s second law states that

${F}_{\text{net}}=ma.$

Substituting known values gives

${F}_{\text{D}}=\left(4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}\right)-\left(5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(7.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)=7.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}$

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

$1.20\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2},$

which gives

${F}_{\text{s}}=735\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}$

Significance The scale reading in (Figure) (a) is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

${F}_{\text{net}}=ma=0={F}_{\text{s}}-w$

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In (Figure) (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

$1.20\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}.$

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

${m}_{1}$

Strategy We draw a free-body diagram for each mass separately, as shown in (Figure) . Then we analyze each one to find the required unknowns. The forces on block 1 are the gravitational force, the contact force of the surface, and the tension in the string. Block 2 is subjected to the gravitational force and the string tension. Newton’s second law applies to each, so we write two vector equations:

$\stackrel{\to }{T}+{\stackrel{\to }{w}}_{1}+\stackrel{\to }{N}={m}_{1}{\stackrel{\to }{a}}_{1}$

Solution The component equations follow from the vector equations above. We see that block 1 has the vertical forces balanced, so we ignore them and write an equation relating the x -components. There are no horizontal forces on block 2, so only the y -equation is written. We obtain these results:

$\begin{array}{cccc}\mathbf{\text{Block 1}}\hfill & & & \mathbf{\text{Block 2}}\hfill \\ \sum {F}_{x}=m{a}_{x}\hfill & & & \sum {F}_{y}=m{a}_{y}\hfill \\ {T}_{x}={m}_{1}{a}_{1x}\hfill & & & {T}_{y}-{m}_{2}g={m}_{2}{a}_{2y}.\hfill \end{array}$

Solving for a :

$a=\frac{{m}_{2}-{m}_{1}}{{m}_{1}+{m}_{2}}g=\frac{4\phantom{\rule{0.2em}{0ex}}\text{kg}-2\phantom{\rule{0.2em}{0ex}}\text{kg}}{4\phantom{\rule{0.2em}{0ex}}\text{kg}+2\phantom{\rule{0.2em}{0ex}}\text{kg}}\left(9.8\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)=3.27\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}.$

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

What Force Must a Soccer Player Exert to Reach Top Speed? A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

Strategy To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

$\text{Δ}v=8.00\phantom{\rule{0.2em}{0ex}}\text{m/s}$

Substituting the known values yields

$a=\frac{8.00\phantom{\rule{0.2em}{0ex}}\text{m/s}}{2.50\phantom{\rule{0.2em}{0ex}}\text{s}}=3.20\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}.$

Substituting the known values of m and a gives

${F}_{\text{net}}=\left(70.0\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(3.20\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)=224\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}$

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Significance This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

Check Your Understanding The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

$5.00\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{m/s}$

Solution We have

$a=\frac{\text{Δ}v}{\text{Δ}t}=\frac{\left(6.00\stackrel{^}{i}+12.00\stackrel{^}{j}\text{m/s}\right)-\left(5.00\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)}{2.00\phantom{\rule{0.2em}{0ex}}\text{s}}=3.00\stackrel{^}{i}+3.50\stackrel{^}{j}{\text{m/s}}^{2}$

The magnitude of the force is now easily found:

$F=\sqrt{{\left(4.50\phantom{\rule{0.2em}{0ex}}\text{N}\right)}^{2}+{\left(5.25\phantom{\rule{0.2em}{0ex}}\text{N}\right)}^{2}}=6.91\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}$

Check Your Understanding Find the direction of the resultant for the 1.50-kg model helicopter.

49.4 degrees

$F=\left(820.0t\right)\phantom{\rule{0.2em}{0ex}}\text{N,}$

Significance Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving (Figure) (a), whereas only the mass of the truck (since it supplied the force) was of use in (Figure) (b).

$v=\frac{ds}{dt}$

Strategy The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

${y}_{0}=0$

We replace ds with dy because we are dealing with the vertical direction,

$ady=vdv,\text{ }\phantom{\rule{0.5em}{0ex}}\left(-0.00100{v}^{2}-9.80\right)dy=vdv.$

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

$h=114\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$

Significance Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t ), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

Check Your Understanding If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

Newton’s laws of motion can be applied in numerous situations to solve motion problems.

${F}_{\text{net}}=ma$

The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating vertically, the normal force is less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force is always less than the full weight of the object.
Some problems contain several physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics to solve these problems.

Conceptual Questions

To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at g . Why do they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

The scale is in free fall along with the astronauts, so the reading on the scale would be 0. There is no difference in the apparent weightlessness; in the aircraft and in orbit, free fall is occurring.

$\stackrel{\to }{F}$

a. 170 N; b. 170 N

Find the tension in each of the three cables supporting the traffic light if it weighs 2.00 × 10 2 N.

A sketch of a traffic light suspended by a cable that is in turn suspended from two other cables is shown. Tension T sub 3 is the tension in the cable connecting the traffic light to the upper cables. Tension T sub one is the tension in the upper cable pulling up and to the left, making a 41 degree angle with the horizontal. Tension T sub two is the tension pulling up and to the right, making a 63 degree angle with the horizontal. Force vector w equal to 200 Newtons pulls vertically downward on the traffic light.

Two muscles in the back of the leg pull upward on the Achilles tendon, as shown below. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?

An Achilles tendon is shown in the figure with two forces exerted on it by the lateral and medial heads of the gastrocnemius muscle. F sub one, equal to two hundred Newtons, is shown as a vector making an angle twenty degrees to the right of vertical, and F sub two, equal to two hundred Newtons, is shown making an angle of twenty degrees left of vertical.

376 N pointing up (along the dashed line in the figure); the force is used to raise the heel of the foot.

After a mishap, a 76.0-kg circus performer clings to a trapeze, which is being pulled to the side by another circus artist, as shown here. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.

A circus performer hanging from a trapeze is being pulled to the right by another performer using a rope. Her weight is shown by a vector w acting vertically downward. The trapeze rope exerts a tension, T sub one, up and to the left, making an angle of fifteen degrees with the vertical. The second performer pulls with tension T sub two, making an angle of ten degrees above the positive x direction.

A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow the first dolphin if it was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)

When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?

$2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{kg}$

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110.0 kg.

A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110.0 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

$2.40\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\text{;}$

a. 10 kg; b. 90 N; c. 98 N; d. 0

$1.2\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$

The device shown below is the Atwood’s machine considered in (Figure) . Assuming that the masses of the string and the frictionless pulley are negligible, (a) find an equation for the acceleration of the two blocks; (b) find an equation for the tension in the string; and (c) find both the acceleration and tension when block 1 has mass 2.00 kg and block 2 has mass 4.00 kg.

An Atwood machine consisting of masses suspended on either side of a pulley by a string passing over the pulley is shown. Mass m sub 1 is on the left and mass m sub 2 is on the right.

Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. The table and the pulley are frictionless. (a) Find the acceleration of the system. (b) Find the tension in the rope. (c) Find the speed with which the hanging mass hits the floor if it starts from rest and is initially located 1.0 m from the floor.

Block m sub 1 is on a horizontal table. It is connected to a string that passes over a pulley at the edge of the table. The string then hangs straight down and connects to block m sub 2, which is not in contact with the table. Block m sub 1 has acceleration a sub 1 directed to the right. Block m sub 2 has acceleration a sub 2 directed downward.

Shown below are two carts connected by a cord that passes over a small frictionless pulley. Each cart rolls freely with negligible friction. Calculate the acceleration of the carts and the tension in the cord.

Two carts connected by a string passing over a pulley are on either side of a double inclined plane. The string passes over a pulley attached to the top of the double incline. On the left, the incline makes an angle of 37 degrees with the horizontal and the cart on that side has mass 10 kilograms. On the right, the incline makes an angle of 53 degrees with the horizontal and the cart on that side has mass 15 kilograms.

Introduction To Motion
Newtons Second Law Of Motion And Momentum

Newton's Second Law Of Motion

Newton’s second law of motion, unlike the first law of motion, pertains to the behaviour of objects for which all existing forces are unbalanced. The second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. This article discusses Newton’s second law in detail.

What is Newton’s Second Law of Motion?

Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration.

Sir Issac newton

Defining Newton’s Second Law of Motion

Newton’s second law states that the acceleration of an object depends upon two variables – the net force acting on the object and the mass of the object. The acceleration of the body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body. This means that as the force acting upon an object is increased, the acceleration of the object is increased. Likewise, as the mass of an object is increased, the acceleration of the object is decreased.

Acceleration is directly proportional to the net force and inversely proportional to the mass.

Newton’s second law can be formally stated as,

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

This statement is expressed in equation form as,

Deriving Newton’s Second Law

For Changing Mass

Let us assume that we have a car at a point (0) defined by location X 0 and time t 0 . The car has a mass m 0 and travels with a velocity v 0 . After being subjected to a force F, the car moves to point 1 which is defined by location X 1 and time t 1 . The mass and velocity of the car change during the travel to values m 1 and v 1 . Newton’s second law helps us determine the new values of m 1 and v 1 if we know the value of the acting force.

Taking the difference between point 1 and point 0, we get an equation for the force acting on the car as follows:

Let us assume the mass to be constant. This assumption is good for a car because the only change in mass would be the fuel burned between point “1” and point “0”. The weight of the fuel is probably small relative to the rest of the car, especially if we only look at small changes in time. Meanwhile, if we were discussing the flight of a bottle rocket, then the mass does not remain constant, and we can only look at changes in momentum.

For Constant Mass

For a constant mass, Newton’s second law can be equated as follows:

Application of Second Law

Newton’s second law is applied to identify the amount of force needed to make an object move or make it stop. Following are a few examples that we have listed to help you understand this point:

Kicking a ball

When we kick a ball, we exert force in a specific direction. The stronger the ball is kicked, the stronger the force we put on it and the further away it will travel.

Pushing a cart

It is easier to push an empty cart in a supermarket than a loaded one, and more mass requires more acceleration.

Two people walking

Among the two people walking, if one is heavier than the other, the one weighing heavier will walk slower because the acceleration of the person weighing lighter is greater.

Get a glimpse of Newton’s second law of motion being taught in BYJU’S classes.

Newton’s Second Law Solved Examples

If there is a block of mass 2kg, and a force of 20 N is acting on it in the positive x-direction, and a force of 30 N in the negative x-direction, then what would be its acceleration?

We first have to calculate the net force acting on it to calculate its acceleration.

The negative acceleration indicates that the block is slowing and its acceleration vector is moving in an opposite direction directed opposite to the direction of motion.

How much horizontal net force is required to accelerate a 1000 kg car at 4 m/s 2 ? Solution: Newton’s 2nd Law relates an object’s mass, the net force on it, and its acceleration: Therefore, we can find the force as follows: F net = ma Substituting the values, we get 1000 kg × 4 m/s 2 = 4000 N Therefore, the horizontal net force is required to accelerate a 1000 kg car at 4 m/s -2 is 4000 N.

Newton’s second law is applied in daily life to a great extent. For instance, in Formula One racing, the engineers try to keep the mass of cars as low as possible. Low mass will imply more acceleration, and the more the acceleration, the chances to win the race are higher.

Frequently Asked Questions – FAQs

How does newton’s second law of motion apply to rockets.

According to Newton’s second law of motion, we know that force is a product of mass and acceleration. When a force is applied to the rocket, the force is termed as thrust. The greater the thrust, the greater will be the acceleration. Acceleration is also dependent on the rocket’s mass, and the lighter the rocket faster is the acceleration.

How does Newton’s second law apply to a car crash?

According to the definition of Newton’s second law of motion, force is the dot product of mass and acceleration. The force in a car crash is dependent either on the mass or the acceleration of the car. As the acceleration or mass of the car increases, the force with which a car crash takes place will also increase.

What is the other name for Newton’s second law?

What are some daily life examples of newton’s second law of motion.

Acceleration of the rocket is due to the force applied, known as thrust, and is an example of Newton’s second law of motion.
Another example of Newton’s second law is when an object falls from a certain height, the acceleration increases because of the gravitational force.

Write the formula for Newton’s second law of motion?

State newton’s second law of motion, for a constant mass, how is newton’s second law equated, define net force., state true or false:net force is the vector sum of all forces acting on a body., watch the video and learn more about newton’s laws of motion.

Stay tuned to BYJU’S and KEEP FALLING IN LOVE WITH LEARNING!!

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

Visit BYJU’S for all Physics related queries and study materials

Your result is as below

Request OTP on Voice Call

Register with BYJU'S & Download Free PDFs

Share full article

For more audio journalism and storytelling, download New York Times Audio , a new iOS app available for news subscribers.

April 10, 2024 • 22:49 Trump’s About-Face on Abortion
April 9, 2024 • 30:48 How Tesla Planted the Seeds for Its Own Potential Downfall
April 8, 2024 • 30:28 The Eclipse Chaser
April 7, 2024 The Sunday Read: ‘What Deathbed Visions Teach Us About Living’
April 5, 2024 • 29:11 An Engineering Experiment to Cool the Earth
April 4, 2024 • 32:37 Israel’s Deadly Airstrike on the World Central Kitchen
April 3, 2024 • 27:42 The Accidental Tax Cutter in Chief
April 2, 2024 • 29:32 Kids Are Missing School at an Alarming Rate
April 1, 2024 • 36:14 Ronna McDaniel, TV News and the Trump Problem
March 29, 2024 • 48:42 Hamas Took Her, and Still Has Her Husband
March 28, 2024 • 33:40 The Newest Tech Start-Up Billionaire? Donald Trump.
March 27, 2024 • 28:06 Democrats’ Plan to Save the Republican House Speaker

How Tesla Planted the Seeds for Its Own Potential Downfall

Elon musk’s factory in china saved his company and made him ultrarich. now, it may backfire..

Hosted by Katrin Bennhold

Featuring Mara Hvistendahl

Produced by Rikki Novetsky and Mooj Zadie

With Rachelle Bonja

Edited by Lisa Chow and Alexandra Leigh Young

Original music by Marion Lozano , Diane Wong , Elisheba Ittoop and Sophia Lanman

Engineered by Chris Wood

Listen and follow The Daily Apple Podcasts | Spotify | Amazon Music

When Elon Musk set up Tesla’s factory in China, he made a bet that brought him cheap parts and capable workers — a bet that made him ultrarich and saved his company.

Mara Hvistendahl, an investigative reporter for The Times, explains why, now, that lifeline may have given China the tools to beat Tesla at its own game.

On today’s episode

Mara Hvistendahl , an investigative reporter for The New York Times.

A car is illuminated in purple light on a stage. To the side, Elon Musk is standing behind a lectern.

Background reading

A pivot to China saved Elon Musk. It also bound him to Beijing .

Mr. Musk helped create the Chinese electric vehicle industry. But he is now facing challenges there as well as scrutiny in the West over his reliance on China.

There are a lot of ways to listen to The Daily. Here’s how.

We aim to make transcripts available the next workday after an episode’s publication. You can find them at the top of the page.

Fact-checking by Susan Lee .

The Daily is made by Rachel Quester, Lynsea Garrison, Clare Toeniskoetter, Paige Cowett, Michael Simon Johnson, Brad Fisher, Chris Wood, Jessica Cheung, Stella Tan, Alexandra Leigh Young, Lisa Chow, Eric Krupke, Marc Georges, Luke Vander Ploeg, M.J. Davis Lin, Dan Powell, Sydney Harper, Mike Benoist, Liz O. Baylen, Asthaa Chaturvedi, Rachelle Bonja, Diana Nguyen, Marion Lozano, Corey Schreppel, Rob Szypko, Elisheba Ittoop, Mooj Zadie, Patricia Willens, Rowan Niemisto, Jody Becker, Rikki Novetsky, John Ketchum, Nina Feldman, Will Reid, Carlos Prieto, Ben Calhoun, Susan Lee, Lexie Diao, Mary Wilson, Alex Stern, Dan Farrell, Sophia Lanman, Shannon Lin, Diane Wong, Devon Taylor, Alyssa Moxley, Summer Thomad, Olivia Natt, Daniel Ramirez and Brendan Klinkenberg.

Our theme music is by Jim Brunberg and Ben Landsverk of Wonderly. Special thanks to Sam Dolnick, Paula Szuchman, Lisa Tobin, Larissa Anderson, Julia Simon, Sofia Milan, Mahima Chablani, Elizabeth Davis-Moorer, Jeffrey Miranda, Renan Borelli, Maddy Masiello, Isabella Anderson and Nina Lassam.

Katrin Bennhold is the Berlin bureau chief. A former Nieman fellow at Harvard University, she previously reported from London and Paris, covering a range of topics from the rise of populism to gender. More about Katrin Bennhold

Mara Hvistendahl is an investigative reporter for The Times focused on Asia. More about Mara Hvistendahl

school Campus Bookshelves
menu_book Bookshelves
perm_media Learning Objects
login Login
how_to_reg Request Instructor Account
hub Instructor Commons
Download Page (PDF)
Download Full Book (PDF)
Periodic Table
Physics Constants
Scientific Calculator
Reference & Cite
Tools expand_more
Readability

selected template will load here

This action is not available.

6.3: Solving Problems with Newton's Laws (Part 2)

Last updated
Save as PDF
Page ID 5883

Newton’s Laws of Motion and Kinematics

Example 6.6: What Force Must a Soccer Player Exert to Reach Top Speed?

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is $\Delta$v = 8.00 m/s . We are given the elapsed time, so $\Delta$t = 2.50 s. The unknown is acceleration, which can be found from its definition: $$a = \frac{\Delta v}{\Delta t} \ldotp$$Substituting the known values yields $$a = \frac{8.00\; m/s}{2.50\; s} = 3.20\; m/s^{2} \ldotp$$
Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, $$F_{net} = ma \ldotp$$Substituting the known values of m and a gives $$F_{net} = (70.0\; kg)(3.20\; m/s^{2}) = 224\; N \ldotp$$

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Significance

This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

Exercise 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7: What Force Acts on a Model Helicopter?

A 1.50-kg model helicopter has a velocity of 5.00 $\hat{j}$ m/s at t = 0. It is accelerated at a constant rate for two seconds (2.00 s) after which it has a velocity of (6.00 $\hat{i}$ + 12.00 $\hat{j}$) m/s. What is the magnitude of the resultant force acting on the helicopter during this time interval?

We can easily set up a coordinate system in which the x-axis ($\hat{i}$ direction) is horizontal, and the y-axis ($\hat{j}$ direction) is vertical. We know that $\Delta$t = 2.00s and $\Delta$v = (6.00 $\hat{i}$ + 12.00 $\hat{j}$ m/s) − (5.00 $\hat{j}$ m/s). From this, we can calculate the acceleration by the definition; we can then apply Newton’s second law.

\[a = \frac{\Delta v}{\Delta t} = \frac{(6.00 \hat{i} + 12.00 \hat{j}\; m/s) - (5.00 \hat{j}\; m/s)}{2.00\; s} = 3.00 \hat{i} + 3.50 \hat{j}\; m/s^{2}$$ $$\sum \vec{F} = m \vec{a} = (1.50\; kg)(3.00 \hat{i} + 3.50 \hat{j}\; m/s^{2}) = 4.50 \hat{i} + 5.25 \hat{j}\; N \ldotp\]

The magnitude of the force is now easily found:

\[F = \sqrt{(4.50\; N)^{2} + (5.25\; N)^{2}} = 6.91\; N \ldotp\]

The original problem was stated in terms of $\hat{i}$ − $\hat{j}$ vector components, so we used vector methods. Compare this example with the previous example.

Exercise 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8: Baggage Tractor

Figure $\PageIndex{7}$(a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by F = (820.0t) N, find the speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?

Figure (a) shows a baggage tractor driving to the left and pulling two luggage carts. The external forces on the system are shown. The forces on the tractor are F sub tractor, horizontally to the left, N sub tractor vertically up, and w sub tractor vertically down. The forces on the cart immediately behind the tractor, cart A, are N sub A vertically up, and w sub A vertically down. The forces on cart B, the one behind cart A, are N sub B vertically up, and w sub B vertically down. Figure (b) shows the free body diagram of the tractor, consisting of F sub tractor, horizontally to the left, N sub tractor vertically up, w sub tractor vertically down, and T horizontally to the right.

A free-body diagram shows the driving force of the tractor, which gives the system its acceleration. We only need to consider motion in the horizontal direction. The vertical forces balance each other and it is not necessary to consider them. For part b, we make use of a free-body diagram of the tractor alone to determine the force between it and cart A. This exposes the coupling force $\vec{T}$, which is our objective.

$$\sum F_{x} = m_{system} a_{x}\; and\; \sum F_{x} = 820.0t,$$so $$820.0t = (650.0 + 250.0 + 150.0)a$$ $$a = 0.7809t \ldotp$$Since acceleration is a function of time, we can determine the velocity of the tractor by using a = $\frac{dv}{dt}$ with the initial condition that v 0 = 0 at t = 0. We integrate from t = 0 to t = 3: $$\begin{split}dv & = adt \\ \int_{0}^{3} dv & = \int_{0}^{3.00} adt = \int_{0}^{3.00} 0.7809tdt \\ v & = 0.3905t^{2} \big]_{0}^{3.00} = 3.51\; m/s \ldotp \end{split}$$
Refer to the free-body diagram in Figure $\PageIndex{7}$(b) $$\begin{split} \sum F_{x} & = m_{tractor} a_{x} \\ 820.0t - T & = m_{tractor} (0.7805)t \\ (820.0)(3.00) - T & = (650.0)(0.7805)(3.00) \\ T & = 938\; N \ldotp \end{split}$$

Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving Figure $\PageIndex{7}$(a), whereas only the mass of the truck (since it supplied the force) was of use in Figure $\PageIndex{7}$(b).

Recall that v = $\frac{ds}{dt}$ and a = $\frac{dv}{dt}$. If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have dt = $\frac{ds}{v}$ and dt = $\frac{dv}{a}$. Now, equating these expressions, we have $\frac{ds}{v}$ = $\frac{dv}{a}$. We can rearrange this to obtain a ds = v dv.

Example 6.9: Motion of a Projectile Fired Vertically

A 10.0-kg mortar shell is fired vertically upward from the ground, with an initial velocity of 50.0 m/s (see Figure $\PageIndex{8}$). Determine the maximum height it will travel if atmospheric resistance is measured as F D = (0.0100 v 2 ) N, where v is the speed at any instant.

(a) A photograph of a soldier firing a mortar shell straight up. (b) A free body diagram of the mortar shell shows forces F sub D and w, both pointing vertically down. Force w is larger than force F sub D.

The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

Initially, y 0 = 0 and v 0 = 50.0 m/s. At the maximum height y = h, v = 0. The free-body diagram shows F D to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

\[\begin{split} \sum F_{y} & = ma_{y} \\ -F_{D} - w & = ma_{y} \\ -0.0100 v^{2} - 98.0 & = 10.0 a \\ a & = -0.00100 v^{2} - 9.80 \ldotp \end{split}\]

The acceleration depends on v and is therefore variable. Since a = f(v), we can relate a to v using the rearrangement described above,

\[a ds = v dv \ldotp\]

We replace ds with dy because we are dealing with the vertical direction,

\[\begin{split} ady & = vdv \\ (−0.00100v^{2} − 9.80)dy & = vdv \ldotp \end{split}\]

We now separate the variables (v’s and dv’s on one side; dy on the other):

\[\begin{split} \int_{0}^{h} dy & = \int_{50.0}^{0} \frac{vdv}{(-0.00100 v^{2} - 9.80)} \\ & = - \int_{50.0}^{0} \frac{vdv}{(-0.00100 v^{2} + 9.80)} \\ & = (-5 \times 10^{3}) \ln(0.00100v^{2} + 9.80) \Big|_{50.0}^{0} \ldotp \end{split}\]

Thus, h = 114 m.

Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

Exercise 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

10.7 Newton’s Second Law for Rotation

Learning objectives.

By the end of this section, you will be able to:

Calculate the torques on rotating systems about a fixed axis to find the angular acceleration
Explain how changes in the moment of inertia of a rotating system affect angular acceleration with a fixed applied torque

In this section, we put together all the pieces learned so far in this chapter to analyze the dynamics of rotating rigid bodies. We have analyzed motion with kinematics and rotational kinetic energy but have not yet connected these ideas with force and/or torque. In this section, we introduce the rotational equivalent to Newton’s second law of motion and apply it to rigid bodies with fixed-axis rotation.

Newton’s Second Law for Rotation

We have thus far found many counterparts to the translational terms used throughout this text, most recently, torque, the rotational analog to force. This raises the question: Is there an analogous equation to Newton’s second law, Σ F → = m a → , Σ F → = m a → , which involves torque and rotational motion? To investigate this, we start with Newton’s second law for a single particle rotating around an axis and executing circular motion. Let’s exert a force F → F → on a point mass m that is at a distance r from a pivot point ( Figure 10.37 ). The particle is constrained to move in a circular path with fixed radius and the force is tangent to the circle. We apply Newton’s second law to determine the magnitude of the acceleration a = F / m a = F / m in the direction of F → F → . Recall that the magnitude of the tangential acceleration is proportional to the magnitude of the angular acceleration by a = r α a = r α . Substituting this expression into Newton’s second law, we obtain

Multiply both sides of this equation by r ,

Note that the left side of this equation is the torque about the axis of rotation, where r is the lever arm and F is the force, perpendicular to r . Recall that the moment of inertia for a point particle is I = m r 2 I = m r 2 . The torque applied perpendicularly to the point mass in Figure 10.37 is therefore

The torque on the particle is equal to the moment of inertia about the rotation axis times the angular acceleration . We can generalize this equation to a rigid body rotating about a fixed axis.

If more than one torque acts on a rigid body about a fixed axis, then the sum of the torques equals the moment of inertia times the angular acceleration:

The term I α I α is a scalar quantity and can be positive or negative (counterclockwise or clockwise) depending upon the sign of the net torque. Remember the convention that counterclockwise angular acceleration is positive. Thus, if a rigid body is rotating clockwise and experiences a positive torque (counterclockwise), the angular acceleration is positive.

Equation 10.25 is Newton’s second law for rotation and tells us how to relate torque, moment of inertia, and rotational kinematics. This is called the equation for rotational dynamics . With this equation, we can solve a whole class of problems involving force and rotation. It makes sense that the relationship for how much force it takes to rotate a body would include the moment of inertia, since that is the quantity that tells us how easy or hard it is to change the rotational motion of an object.

Deriving Newton’s Second Law for Rotation in Vector Form

As before, when we found the angular acceleration, we may also find the torque vector. The second law Σ F → = m a → Σ F → = m a → tells us the relationship between net force and how to change the translational motion of an object. We have a vector rotational equivalent of this equation, which can be found by using Equation 10.7 and Figure 10.8 . Equation 10.7 relates the angular acceleration to the position and tangential acceleration vectors:

We form the cross product of this equation with r → r → and use a cross product identity (note that r → · α → = 0 r → · α → = 0 ):

We now form the cross product of Newton’s second law with the position vector r → , r → ,

Identifying the first term on the left as the sum of the torques, and m r 2 m r 2 as the moment of inertia, we arrive at Newton’s second law of rotation in vector form:

This equation is exactly Equation 10.25 but with the torque and angular acceleration as vectors. An important point is that the torque vector is in the same direction as the angular acceleration.

Applying the Rotational Dynamics Equation

Before we apply the rotational dynamics equation to some everyday situations, let’s review a general problem-solving strategy for use with this category of problems.

Problem-Solving Strategy

Rotational dynamics.

Examine the situation to determine that torque and mass are involved in the rotation. Draw a careful sketch of the situation.
Determine the system of interest.
Draw a free-body diagram. That is, draw and label all external forces acting on the system of interest.
Identify the pivot point. If the object is in equilibrium, it must be in equilibrium for all possible pivot points––chose the one that simplifies your work the most.
Apply ∑ i τ i = I α ∑ i τ i = I α , the rotational equivalent of Newton’s second law, to solve the problem. Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation.
As always, check the solution to see if it is reasonable.

Example 10.16

Calculating the effect of mass distribution on a merry-go-round.

The moment of inertia of a solid disk about this axis is given in Figure 10.20 to be 1 2 M R 2 . 1 2 M R 2 . We have M = 50.0 kg M = 50.0 kg and R = 1.50 m R = 1.50 m , so I = ( 0.500 ) ( 50.0 kg ) ( 1.50 m ) 2 = 56.25 kg-m 2 . I = ( 0.500 ) ( 50.0 kg ) ( 1.50 m ) 2 = 56.25 kg-m 2 . To find the net torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that τ = r F sin θ = ( 1.50 m ) ( 250.0 N ) = 375.0 N-m . τ = r F sin θ = ( 1.50 m ) ( 250.0 N ) = 375.0 N-m . Now, after we substitute the known values, we find the angular acceleration to be α = τ I = 375.0 N-m 56.25 kg-m 2 = 6.67 rad s 2 . α = τ I = 375.0 N-m 56.25 kg-m 2 = 6.67 rad s 2 .
We expect the angular acceleration for the system to be less in this part because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia I , we first find the child’s moment of inertia I c I c by approximating the child as a point mass at a distance of 1.25 m from the axis. Then I c = m R 2 = ( 18.0 kg ) ( 1.25 m ) 2 = 28.13 kg-m 2 . I c = m R 2 = ( 18.0 kg ) ( 1.25 m ) 2 = 28.13 kg-m 2 . The total moment of inertia is the sum of the moments of inertia of the merry-go-round and the child (about the same axis): I = 28.13 kg-m 2 + 56.25 kg-m 2 = 84.38 kg-m 2 . I = 28.13 kg-m 2 + 56.25 kg-m 2 = 84.38 kg-m 2 . Substituting known values into the equation for α gives α = τ I = 375.0 N-m 84.38 kg-m 2 = 4 .44 rad s 2 . α = τ I = 375.0 N-m 84.38 kg-m 2 = 4 .44 rad s 2 .

Significance

Check your understanding 10.7.

The fan blades on a jet engine have a moment of inertia 30.0 kg-m 2 30.0 kg-m 2 . In 10 s, they rotate counterclockwise from rest up to a rotation rate of 20 rev/s. (a) What torque must be applied to the blades to achieve this angular acceleration? (b) What is the torque required to bring the fan blades rotating at 20 rev/s to a rest in 20 s?

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction

Authors: William Moebs, Samuel J. Ling, Jeff Sanny
Publisher/website: OpenStax
Book title: University Physics Volume 1
Publication date: Sep 19, 2016
Location: Houston, Texas
Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
Section URL: https://openstax.org/books/university-physics-volume-1/pages/10-7-newtons-second-law-for-rotation

© Jan 19, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

IMAGES

Newton’s Second Law: Statement, Examples, and Equation
4.3 Newton’s Second Law of Motion: Concept of a System
Newton's Second Law of Motion Explained, Examples, Word Problems
Newton Second Law Of Motion Worksheets
Newton's Second Law example problem
Newtons 2nd law of motion

VIDEO

Newton's Second Law of Motion
second order linear differential equation with variable coefficients (part 1)
Newton's Second Law of Motion
More applications of Newton's second law of motion
Newton's Second Law of Motion (Lecture Series)
Newton's second law of motion |⚡3d animation

COMMENTS

Newton's second law of motion
Solution : We use Newton's second law to get the net force. ∑ F = m a. ∑ F = (1 kg) (5 m/s2) = 5 kg m/s2 = 5 Newton. See also Electromagnetic induction Induced EMF - Problems and Solutions. 2. Mass of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object's acceleration…. Known :
6.2: Solving Problems with Newton's Laws (Part 1)
It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton's second law in components along the different directions. Then, you have the following equations: ∑Fx = max, ∑Fy = may. (6.2.1) (6.2.1) ∑ F x = m a x, ∑ F y = m a y.
4.3 Newton's Second Law of Motion
Now, let's rearrange Newton's second law to solve for acceleration. We get. a = Fnet m or a = ΣF m. a = F net m or a = Σ F m. 4.3. In this form, we can see that acceleration is directly proportional to force, which we write as. a ∝ Fnet, a ∝ F net, 4.4. where the symbol ∝ ∝ means proportional to.
6.1 Solving Problems with Newton's Laws
Determine the system of interest. The result is a free-body diagram that is essential to solving the problem. Apply Newton's second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line. Check the solution to see whether it is reasonable.
Newton's second law review (article)
Newton's second law of motion. Newton's second law says that the acceleration and net external force are directly proportional, and there is an inversely proportional relationship between acceleration and mass. For example, a large force on a tiny object gives it a huge acceleration, but a small force on a huge object gives it very little ...
Newton's Second Law of Motion
Newton's second law describes the affect of net force and mass upon the acceleration of an object. Often expressed as the equation a = Fnet/m (or rearranged to Fnet=m*a), the equation is probably the most important equation in all of Mechanics. It is used to predict how an object will accelerated (magnitude and direction) in the presence of an unbalanced force.
Finding force
- [Instructor] Let's solve two problems on Newton's Second Law. Here's the first one. A heavy couch of 60 kilogram was accelerated from rest to 20 meters per second in five seconds on a frictionless surface. Find the force acting on it. Okay, let's see what we're dealing with. We have a couch which is accelerated from rest.
4.3 Newton's Second Law of Motion: Concept of a System
4.6 Problem-Solving Strategies; 4.7 Further Applications of Newton's Laws of Motion; 4.8 Extended Topic: The Four Basic Forces—An Introduction; Glossary; ... Newton's second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton's second law ...
PDF Newton's Second Law of Motion Problems Worksheet
In the second set of problems below, you will be given the force applied to an object and the acceleration of that object, and then will need to solve for mass, using the equation m = F/a. In other words, you will need to divide the force by the acceleration to calculate the mass. Show your work in the space provided. Be
Newtons Second Law Problems
The Solve It! (with Newton's Second Law) Concept Builder provides learners plenty of practice using the F net = m•a equation to analyze situations involving unbalanced forces and accelerations. Much more than the usual Concept Builder, this activity demands that learners solve numerical problems.
2.4: Newton's Second Law of Motion- Force and Acceleration
In equation form, Newton's second law of motion is a = Fnet m a = F net m, often written in the more familiar form: Fnet = ma F net = m a. The weight w w of an object is defined as the force of gravity acting on an object of mass mm. Given acceleration due to gravity g g, the magnitude of weight is: w = mg w = m g.
Newton's second law: Solving for force, mass, and acceleration
Newton's second law of motion. More on Newton's second law. Understand: force, mass, and acceleration. Newton's second law review. Newton's second law: Solving for force, mass, and acceleration . Finding force - Newton's second law (Solved example) Science > Mechanics (Essentials) - Class 11th > ... Problem. A stunt woman of ...
PDF Practice Problems for Newton's Second Law of Motion
The relationship is stated by Newton's second law of motion, Force=Mass x Acceleration. -or-. F=ma. where F is the force, m is the mass, and a is the acceleration. The units are Newtons (N) for force, kilograms (kg) for mass, and meters per second squared (m/s2) for acceleration. The other forms of the equation can be used to solve for mass or ...
Newtons Second Law and Problem Solving
Newton's Second Law and Problem Solving (PDF) The Curriculum Corner contains a complete ready-to-use curriculum for the high school physics classroom. This collection of pages comprise worksheets in PDF format that developmentally target key concepts and mathematics commonly covered in a high school physics curriculum.
Problems Based on Newton's Second Law of Motion
Q.8. A 150 g ball traveling at 30m/s strikes the palm of a players hand and is stopped in 0.06 sec. Calculate the force exerted by the ball on the hand. Q.9. A motorcycle is moving with a velocity of 90km/h and it takes 5 seconds to stop after the brakes are applied.
What is Newton's second law? (article)
We know objects can only accelerate if there are forces on the object. Newton's second law tells us exactly how much an object will accelerate for a given net force. a = Σ F m. To be clear, a is the acceleration of the object, Σ F is the net force on the object, and m is the mass of the object.
Solving Problems with Newton's Laws
Determine the system of interest. The result is a free-body diagram that is essential to solving the problem. Apply Newton's second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line. Check the solution to see whether it is reasonable.
Introduction to Dynamics: Newton's Laws of Motion
4.2 Newton's First Law of Motion: Inertia; 4.3 Newton's Second Law of Motion: Concept of a System; 4.4 Newton's Third Law of Motion: Symmetry in Forces; 4.5 Normal, Tension, and Other Examples of Forces; 4.6 Problem-Solving Strategies; 4.7 Further Applications of Newton's Laws of Motion; 4.8 Extended Topic: The Four Basic Forces—An ...
Newton's Second Law Of Motion
Solution: Newton's 2nd Law relates an object's mass, the net force on it, and its acceleration: Therefore, we can find the force as follows: Fnet = ma. Substituting the values, we get. 1000 kg × 4 m/s 2 = 4000 N. Therefore, the horizontal net force is required to accelerate a 1000 kg car at 4 m/s -2 is 4000 N.
17.1: Solving Problems with Newton's Laws (Part 1)
Problem-Solving Strategy: Applying Newton's Laws of Motion. Identify the physical principles involved by listing the givens and the quantities to be calculated. Sketch the situation, using arrows to represent all forces. Determine the system of interest, and draw a free body diagram. Apply Newton's second law to solve the problem.
4.6: Problem-Solving Strategies
Problem-Solving Strategy for Newton's Laws of Motion. Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure(a).
How Tesla Planted the Seeds for Its Own Potential Downfall
When Elon Musk set up Tesla's factory in China, he made a bet that brought him cheap parts and capable workers — a bet that made him ultrarich and saved his company.
6.3: Solving Problems with Newton's Laws (Part 2)
Newton's laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters. ... the knowns, and the unknowns involved in the problem. The second step is to solve ...
10.7 Newton's Second Law for Rotation
Identifying the first term on the left as the sum of the torques, and mr2 m r 2 as the moment of inertia, we arrive at Newton's second law of rotation in vector form: τnet = Στ = Iα . τ net = Σ τ → = I α →. 10.26. This equation is exactly Equation 10.25 but with the torque and angular acceleration as vectors.

Newton’s second law of motion – problems and solutions

Share this:

Discover more from Physics

More Questions

Extra Questions For Class 9 Science Chapter 1 Matter in Our Surroundings

Extra Questions For Class 9 Science Chapter 2 Is Matter Around Us Pure

Extra Questions For Class 9 Science Chapter 5 fundamental unit of life

Download CBSE Books

Leave a Reply Cancel reply

Discover more from Gurukul of Excellence

Solving Problems with Newton’s Laws

Problem-Solving Strategies

Particle Equilibrium

Particle Acceleration

Newton’s Laws of Motion and Kinematics

Conceptual Questions

Newton's Second Law Of Motion

What is Newton’s Second Law of Motion?

Defining Newton’s Second Law of Motion

Deriving Newton’s Second Law

For Changing Mass

For Constant Mass

Application of Second Law

Kicking a ball

Pushing a cart

Two people walking

Newton’s Second Law Solved Examples

Top 10 Most Important and Expected Questions on Laws of Motion.

Frequently Asked Questions – FAQs

How does Newton’s second law apply to a car crash?

What is the other name for Newton’s second law?

Write the formula for Newton’s second law of motion?

Leave a Comment Cancel reply

Register with BYJU'S & Download Free PDFs

How Tesla Planted the Seeds for Its Own Potential Downfall

Listen and follow The Daily Apple Podcasts | Spotify | Amazon Music

On today’s episode

Background reading

6.3: Solving Problems with Newton's Laws (Part 2)

Newton’s Laws of Motion and Kinematics

Example 6.6: What Force Must a Soccer Player Exert to Reach Top Speed?

Exercise 6.4

Example 6.7: What Force Acts on a Model Helicopter?

Exercise 6.5

Example 6.8: Baggage Tractor

Example 6.9: Motion of a Projectile Fired Vertically

Exercise 6.6

10.7 Newton’s Second Law for Rotation

Newton’s Second Law for Rotation

Deriving Newton’s Second Law for Rotation in Vector Form

Applying the Rotational Dynamics Equation

Problem-Solving Strategy

Example 10.16

Significance

IMAGES

VIDEO

COMMENTS