problem solving on law of acceleration

Newton’s second law of motion – problems and solutions

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

Wanted : net force (∑F)

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

Mass (m) = 2 kg

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

Force (F) = 200 N

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

200 – F g = 120

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

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

5 – 4 = (0.4) a

N A – w A = m A a

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.

Weight of X (w X ) = 4 Newton

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

The correct answer is A.

Determine the object’s acceleration.

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

4.3 Newton's Second Law of Motion

Section learning objectives.

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

Describe Newton’s second law, both verbally and mathematically
Use Newton’s second law to solve problems

Teacher Support

The learning objectives in this section will help students master the following standards:

(D) calculate the effect of forces on objects, including the law of inertia, the relationship between force and acceleration, and the nature of force pairs between objects.

Before beginning this section, review forces, acceleration, acceleration due to gravity (g), friction, inertia, and Newton’s first law.

Section Key Terms

freefall

Newton’s second law of motion

weight

Describing Newton’s Second Law of Motion

[BL] [OL] Review the concepts of inertia and Newton’s first law. Explain that, according to Newton’s first law, a change in motion is caused by an external force. For instance, a ball that is pitched changes its speed and direction when it is hit by a bat.

[BL] [OL] [AL] Write the equation for Newton’s second law and show how it can be solved for all three variables, F , m , and a . Explain the practical implications for each case. Ask students how the other two variables would behave if one quantity is held constant.

Misconception Alert

Students might confuse the terms equal and proportional .

Newton’s first law considered bodies at rest or bodies in motion at a constant velocity . The other state of motion to consider is when an object is moving with a changing velocity, which means a change in the speed and/or the direction of motion. This type of motion is addressed by Newton’s second law of motion , which states how force causes changes in motion. Newton’s second law of motion is used to calculate what happens in situations involving forces and motion, and it shows the mathematical relationship between force, mass , and acceleration . Mathematically, the second law is most often written as

where F net (or ∑ F ) is the net external force , m is the mass of the system, and a is the acceleration. Note that F net and ∑ F are the same because the net external force is the sum of all the external forces acting on the system.

First, what do we mean by a change in motion ? A change in motion is simply a change in velocity: the speed of an object can become slower or faster, the direction in which the object is moving can change, or both of these variables may change. A change in velocity means, by definition, that an acceleration has occurred. Newton’s first law says that only a nonzero net external force can cause a change in motion, so a net external force must cause an acceleration. Note that acceleration can refer to slowing down or to speeding up. Acceleration can also refer to a change in the direction of motion with no change in speed, because acceleration is the change in velocity divided by the time it takes for that change to occur, and velocity is defined by speed and direction.

From the equation F net = m a , F net = m a , we see that force is directly proportional to both mass and acceleration, which makes sense. To accelerate two objects from rest to the same velocity, you would expect more force to be required to accelerate the more massive object. Likewise, for two objects of the same mass, applying a greater force to one would accelerate it to a greater velocity.

Now, let’s rearrange Newton’s second law to solve for acceleration. We get

In this form, we can see that acceleration is directly proportional to force, which we write as

where the symbol ∝ ∝ means proportional to .

This proportionality mathematically states what we just said in words: acceleration is directly proportional to the net external force. When two variables are directly proportional to each other, then if one variable doubles, the other variable must double. Likewise, if one variable is reduced by half, the other variable must also be reduced by half. In general, when one variable is multiplied by a number, the other variable is also multiplied by the same number. It seems reasonable that the acceleration of a system should be directly proportional to and in the same direction as the net external force acting on the system. An object experiences greater acceleration when acted on by a greater force.

It is also clear from the equation a = F net / m a = F net / m that acceleration is inversely proportional to mass, which we write as

Inversely proportional means that if one variable is multiplied by a number, the other variable must be divided by the same number. Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. This relationship is illustrated in Figure 4.5 , which shows that a given net external force applied to a basketball produces a much greater acceleration than when applied to a car.

Applying Newton’s Second Law

[BL] Review how to convert between units.

[OL] [AL] Ask students to give examples of Newton’s second law.

Students might confuse weight, which is a force, and g , which is the acceleration due to gravity.

[BL] [OL] [AL] Ask students if they think an astronaut weighs the same on the moon as they do on Earth. Talk about the difference between mass and weight.

Before putting Newton’s second law into action, it is important to consider units. The equation F net = m a F net = m a is used to define the units of force in terms of the three basic units of mass, length, and time (recall that acceleration has units of length divided by time squared). The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of 1 m/s 2 . That is, because F net = m a , F net = m a , we have

One of the most important applications of Newton’s second law is to calculate weight (also known as the gravitational force ), which is usually represented mathematically as W . When people talk about gravity , they don’t always realize that it is an acceleration. When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that the net external force acting on an object is responsible for the acceleration of the object. If air resistance is negligible, the net external force on a falling object is only the gravitational force (i.e., the weight of the object).

Weight can be represented by a vector because it has a direction. Down is defined as the direction in which gravity pulls, so weight is normally considered a downward force. By using Newton’s second law, we can figure out the equation for weight.

Consider an object with mass m falling toward Earth. It experiences only the force of gravity (i.e., the gravitational force or weight), which is represented by W. Newton’s second law states that F net = m a . F net = m a . Because the only force acting on the object is the gravitational force, we have F net = W . F net = W . We know that the acceleration of an object due to gravity is g , so we have a = g . a = g . Substituting these two expressions into Newton’s second law gives

This is the equation for weight—the gravitational force on a mass m . On Earth, g = 9.80 m/s 2 , g = 9.80 m/s 2 , so the weight (disregarding for now the direction of the weight) of a 1.0-kg object on Earth is

Although most of the world uses newtons as the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb.

Recall that although gravity acts downward, it can be assigned a positive or negative value, depending on what the positive direction is in your chosen coordinate system. Be sure to take this into consideration when solving problems with weight. When the downward direction is taken to be negative, as is often the case, acceleration due to gravity becomes –g = −9.8 m/s 2 and the weight force is –mg.

When the net external force on an object is its weight, we say that it is in freefall . In this case, the only force acting on the object is the force of gravity. On the surface of Earth, when objects fall downward toward Earth, they are never truly in freefall because there is always some upward force due to air resistance that acts on the object (and there is also the buoyancy force of air, which is similar to the buoyancy force in water that keeps boats afloat).

Gravity varies slightly over the surface of Earth, so the weight of an object depends very slightly on its location on Earth. Weight varies dramatically away from Earth’s surface. On the moon, for example, the acceleration due to gravity is only 1.67 m/s 2 . Because weight depends on the force of gravity, a 1.0-kg mass weighs 9.8 N on Earth and only about 1.7 N on the moon.

It is important to remember that weight and mass are very different, although they are closely related. Mass is the quantity of matter in an object (how much stuff there is, or how hard it is to accelerate it) and does not vary, but weight is the gravitational force on an object and is proportional to the force of gravity. It is easy to confuse the two, because our experience is confined to Earth, and the weight of an object is essentially the same no matter where you are on Earth. Adding to the confusion, the terms mass and weight are often used interchangeably in everyday language; for example, our medical records often show our weight in kilograms, but never in the correct unit of newtons.

Mass and Weight

Explain that even though a scale gives a mass, it actually measures weight. Scales are calibrated to show the correct mass on Earth. They would give different results on the moon, because the force of gravity is weaker on the moon.

In this activity, you will use a scale to investigate mass and weight.

1 bathroom scale
What do bathroom scales measure?
When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight—similar to rubber bands expanding when pulled.
The springs provide a measure of your weight (provided you are not accelerating). This is a force in newtons (or pounds). In most countries, the measurement is now divided by 9.80 to give a reading in kilograms, which is a of mass. The scale detects weight but is calibrated to display mass.
If you went to the moon and stood on your scale, would it detect the same mass as it did on Earth?

Grasp Check

While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why?

The reading increases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction of your weight.
The reading increases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction opposite to your weight.
The reading decreases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction of your weight.
The reading decreases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction opposite to your weight.

Tips For Success

Only net external force impacts the acceleration of an object. If more than one force acts on an object and you calculate the acceleration by using only one of these forces, you will not get the correct acceleration for that object.

Watch Physics

Newton’s second law of motion.

This video reviews Newton’s second law of motion and how net external force and acceleration relate to one another and to mass. It also covers units of force, mass, and acceleration, and reviews a worked-out example.

True or False —If you want to reduce the acceleration of an object to half its original value, then you would need to reduce the net external force by half.

Worked Example

What acceleration can a person produce when pushing a lawn mower.

Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N parallel to the ground. The mass of the mower is 240 kg. What is its acceleration?

Because F net and m are given, the acceleration can be calculated directly from Newton’s second law: F net = m a .

Solving Newton’s second law for the acceleration, we find that the magnitude of the acceleration, a , is a = F net m . a = F net m . Entering the given values for net external force and mass gives

Inserting the units kg ⋅ m/s 2 kg ⋅ m/s 2 for N yields

The acceleration is in the same direction as the net external force, which is parallel to the ground and to the right. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion, because we are given that the net external force is in the direction in which the person pushes. Also, the vertical forces must cancel if there is no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is reasonable for a person pushing a mower; the mower’s speed must increase by 0.21 m/s every second, which is possible. The time during which the mower accelerates would not be very long because the person’s top speed would soon be reached. At this point, the person could push a little less hard, because he only has to overcome friction.

What Rocket Thrust Accelerates This Sled?

Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on humans at high accelerations. Rocket sleds consisted of a platform mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust, T , for the four-rocket propulsion system shown below. The sled’s initial acceleration is 49 m/s 2 , 49 m/s 2 , the mass of the system is 2,100 kg, and the force of friction opposing the motion is 650 N.

The system of interest is the rocket sled. Although forces act vertically on the system, they must cancel because the system does not accelerate vertically. This leaves us with only horizontal forces to consider. We’ll assign the direction to the right as the positive direction. See the free-body diagram in Figure 4.8.

We start with Newton’s second law and look for ways to find the thrust T of the engines. Because all forces and acceleration are along a line, we need only consider the magnitudes of these quantities in the calculations. We begin with

where F net F net is the net external force in the horizontal direction. We can see from Figure 4.8 that the engine thrusts are in the same direction (which we call the positive direction), whereas friction opposes the thrust. In equation form, the net external force is

Newton’s second law tells us that F net = m a , so we get

After a little algebra, we solve for the total thrust 4 T :

which means that the individual thrust is

Inserting the known values yields

The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and to test the apparatus designed to protect fighter pilots in emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 g . (Recall that g, the acceleration due to gravity, is 9 .80 m/s 2 . 9 .80 m/s 2 . An acceleration of 45 g is 45 × 9 .80 m/s 2 , 45 × 9 .80 m/s 2 , which is approximately 440 m/s 2 . 440 m/s 2 . ) Living subjects are no longer used, and land speeds of 10,000 km/h have now been obtained with rocket sleds. In this example, as in the preceding example, the system of interest is clear. We will see in later examples that choosing the system of interest is crucial—and that the choice is not always obvious.

Practice Problems

If 1 N is equal to 0.225 lb, how many pounds is 5 N of force?

How much force needs to be applied to a 5-kg object for it to accelerate at 20 m/s 2 ?

Check Your Understanding

Use the questions in Check Your Understanding to assess whether students have achieved the section learning objectives. If students are struggling with a specific objective, the Check Your Understanding assessment will help identify which is causing the problem and direct students to the relevant content.

What is the mathematical statement for Newton’s second law of motion?

F = m a F = m a

Newton’s second law describes the relationship between which quantities?

Force, mass, and time
Force, mass, and displacement
Force, mass, and velocity
Force, mass, and acceleration
Acceleration is the rate at which displacement changes.
Acceleration is the rate at which force changes.
Acceleration is the rate at which velocity changes.
Acceleration is the rate at which mass changes.

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 Texas Education Agency (TEA). The original material is available at: https://www.texasgateway.org/book/tea-physics . Changes were made to the original material, including updates to art, structure, and other content updates.

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

Authors: Paul Peter Urone, Roger Hinrichs
Publisher/website: OpenStax
Book title: Physics
Publication date: Mar 26, 2020
Location: Houston, Texas
Book URL: https://openstax.org/books/physics/pages/1-introduction
Section URL: https://openstax.org/books/physics/pages/4-3-newtons-second-law-of-motion

© Jun 7, 2024 Texas Education Agency (TEA). 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.

StickMan Physics

Animated Physics Lessons

F=ma Practice Problems

F=ma problem set.

Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations.

In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems.

1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right?

a = 33.33 m/s 2 Right

2. What is the acceleration of the 25 kg box that has 50 N of force applied to the right?

a=2.0 m/s 2 Right

3. What is the acceleration of the 3 kg box that has 25 N of force applied to the right and 55 N left?

a = 10.0 m/s 2 Left

4. What is the acceleration of the 5 kg box that has a 25 N force and 50 N force applied both right?

a = 15.0 m/s 2 Right

5. What is the acceleration of the 25 kg box that has a 100 N force north and 50 N force east applied?

a= 4.47 m/s 2

5b. What direction would this box accelerate?

63.43° North of East

6. Does a 795 kg Lorinser speedy, 6300 kg elephant, or 8.6 kg wagon have more inertia and why?

6300 kg Elephant

The more mass the more inertia

7. How much force is required to accelerate a 795 kg Lorinser Speedy by 15 m/s 2 ?

F = 11925 N

8. How much force is required to accelerate an 8.6 kg wagon by 15 m/s 2 ?

F = 129 N

9. How much does a 6300 kg elephant accelerate when you apply 500 N of force?

a = 0.0794 m/s 2

10. What is the mass of an object if it takes a net force of 40 N to accelerate at a rate of 0.88 m/s 2 ?

m = 45.45 kg

11. How much force is required to accelerate a 0.142 kg baseball to 44.7 m/s during a pitchers 1.5 meter delivery?

F = 94.58 N

12. A 0.050 kg golf ball leaves the tee at a speed of 75.0 m/s. The club is in contact with the ball for 0.020 s. What is the net force of the club on the ball?

F = 187.5 N

13. A 90.0 kg astronaut receives a 30.0 N force from her jetpack. How much faster is she be moving after 2.00 seconds?

0.667 m/s faster

14. A 795 kg car starts from rest and travels 41 m in 3.0 s. How much force did the car engine provide?

F = 7242 N

15. Joe and his sailboat have a combined weight of 450 kg. How far has Joe sailed when he started at 5 m/s and a gust of wind provided 600 Newtons of force for 4 seconds?

x = 30.64 m

16. Tom pulls a 45 kilogram wagon with a force of 200 Newtons at a 15° angle to the horizontal from rest. How much faster will the wagon be moving after 2 seconds?

v f = 8.58 m/s

Back to the Newtons Second Law Lesson
Continue to Mass and Weight
Back to the Main Forces Page
Back to the Stickman Physics Home Page
Equation Sheet

Terms and Conditions - Privacy Policy

HIGH SCHOOL

ACT Tutoring
SAT Tutoring
PSAT Tutoring
ASPIRE Tutoring
SHSAT Tutoring
STAAR Tutoring

GRADUATE SCHOOL

MCAT Tutoring
GRE Tutoring
LSAT Tutoring
GMAT Tutoring
AIMS Tutoring
HSPT Tutoring
ISAT Tutoring
SSAT Tutoring

Search 50+ Tests

Loading Page

math tutoring

Elementary Math
Pre-Calculus
Trigonometry

science tutoring

Foreign languages.

Mandarin Chinese

elementary tutoring

Computer Science

Search 350+ Subjects

Video Overview
Tutor Selection Process
Online Tutoring
Mobile Tutoring
Instant Tutoring
How We Operate
Our Guarantee
Impact of Tutoring
Reviews & Testimonials
About Varsity Tutors

AP Physics 1 : Newton's Second Law

Study concepts, example questions & explanations for ap physics 1, all ap physics 1 resources, example questions, example question #1 : newton's second law.

$problem solving on law of acceleration$

Use Newton's second law to solve this problem.

$problem solving on law of acceleration$

When the elevator is not moving, we get

$problem solving on law of acceleration$

However, when the elevator is accelerating downward, the man appears to be lighter since the elevator is negating some of the force from gravity. Written as an equation, we have:

$problem solving on law of acceleration$

Putting in our values, we get:

$problem solving on law of acceleration$

A skydiver of mass 70kg has jumped out of a plane two miles above the surface of the earth. After 20 seconds, he has reached terminal velocity, meaning he is no longer accelerating. What is the force of the air on the skydiver's body?

$problem solving on law of acceleration$

This question is testing your understanding of terminal velocity and Newton's second law. Since the skydiver is at terminal velocity, the force of the air is equal to the force of gravity, resulting in zero net force and thus no acceleration. We just need to calculate the force of gravity on the skydiver to find the force of the air:

$problem solving on law of acceleration$

There are two forces in play in this scenario. The first is gravity, and the second is air resistance. Since they are opposing each other, we can write:

$problem solving on law of acceleration$

Substituting in Newton's second law, we get:

$problem solving on law of acceleration$

Rearranging for the force of air resistance, we get:

$problem solving on law of acceleration$

Plugging in our values from the problem statement:

$problem solving on law of acceleration$

Example Question #3 : Newton's Second Law

A diver of 50kg jumps from a platform 20m high into a pool. If the diver decelerates at a constant rate to zero velocity in 0.8 seconds after hitting the water, what is the force that the water exerts on the diver?

$problem solving on law of acceleration$

We can use the equation for conservation of energy to calculate the velocity of the diver as he hits the water:

$problem solving on law of acceleration$

Cancel out initial kinetic and final potential energies, and plug in our expressions:

$problem solving on law of acceleration$

Cancel out mass and rearranging for final velocity:

$problem solving on law of acceleration$

Plug in our values:

$problem solving on law of acceleration$

We know that the diver then decelerates from this velocity to zero in 0.8 seconds, so we can calculate the acceleration:

$problem solving on law of acceleration$

Then use Newton's second law to calculate the force on the diver:

$problem solving on law of acceleration$

Example Question #2 : Newton's Second Law

$problem solving on law of acceleration$

The truck experiences the greater force and the greater acceleration

The truck experiences the greater force and the car experiences the greater acceleration

The car and the truck experience equal force and the car experiences greater acceleration

The car experiences the greater force and the greater acceleration

Both the car and the truck experience equal force and acceleration

The car and the truck experience equal and opposite forces, but since the car has a smaller mass it will experience greater acceleration than the truck according to the equation F = ma.

$problem solving on law of acceleration$

A greater mass will decrease the acceleration.

$problem solving on law of acceleration$

Since we are neglecting air resistance, there are two forces in play: gravity and friction. Therefore, we can use Newton's second law to write the following:

$problem solving on law of acceleration$

Substituting in an expression for the force of gravity and rearranging for the force of friction, we get:

$problem solving on law of acceleration$

Example Question #6 : Newton's Second Law

$problem solving on law of acceleration$

Use Newton's second law.

$problem solving on law of acceleration$

A constant force of 30N acts on a a 10kg box as shown in the diagram. If the box is originally at rest, what will be its velocity after 5s?

$problem solving on law of acceleration$

The box has a constant force acting on it pulling it towards the left. Therefore we can write this as:

$problem solving on law of acceleration$

Since the force is constant, this means that is is causing the box to move with a constant acceleration that we can calculate using Newton's second law of motion:

$problem solving on law of acceleration$

Now that we know the acceleration, we can calculate the final velocity after 5 seconds:

$problem solving on law of acceleration$

So we have that

$problem solving on law of acceleration$

Note that the negative sign indicates that the box is moving to the left.

Example Question #7 : Newton's Second Law

$problem solving on law of acceleration$

The force applied by the weaker person can be calculated using Newton's second law, which states:

$problem solving on law of acceleration$

The net force is equal to the product of the mass of the object and the acceleration of the object. We were given the mass and acceleration of the object, but only the ratio of the applied forces:

$problem solving on law of acceleration$

Is it possible to have a non-zero number of forces acting on an object (of non-zero mass), yet the object doesn't acclerate?

Newton's second law states that the net force, or the vector sum of all the forces acting on an object, equals the mass times the acceleration. So, it is possible to have forces act on an object without acceleration if the forces are oriented such that they vector sum to zero. An example would be a person sitting in a chair. Gravity and the normal force both act on the person. However, these forces are equal in magnitude and opposite in direction. So the person doesn't accelerate.

Report an issue with this question

If you've found an issue with this question, please let us know. With the help of the community we can continue to improve our educational resources.

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC 101 S. Hanley Rd, Suite 300 St. Louis, MO 63105

Or fill out the form below:

Contact Information

Complaint details.

Email address:
Your name:
Feedback:

Please wait while your request is being verified...

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.

Newton's Second Law of Motion - Basic Problem

Please do provide a detailed step-by-step solution and a final answer.

A truck with a mass of 3,500 kg, including the passengers, has an engine that produces a net horizontal force of 525 N. Assuming that there is no other forces involve in the motion, find the following:

a.) acceleration of the truck

b.) starting from rest, how long will it take for the truck to reach the velocity of 11.5 m/s?

2. An object with a mass m1 accelerates at 8.0 m/s2 when a force F is applied. A second object with a mass m2 accelerates at 4.0 m/s2 when the same force F is applied to it.

a.) Find the value of the ratio m2/m1.

b.) Find the acceleration of the combined mass under the action of force F.

1 Expert Answer

John B. answered • 10/20/21

Licensed Physics Instructor with Industry Experience

a) Since the engine is the only force acting, the net force is 525 N.

a=0.15 m/s 2

b) Using v=v 0 +at

11.5=0+(0.15)t

t=11.5/0.15

t=76.7 seconds

For the first mass F=m1(8)

For the second mass F=m2(4)

a) Therefore m2/m1 =(F/4)/(F/8) = 8/4 = 2

b) The combined mass is now m2 + m1 = F/8 + F/4 = 3F/8

Using F = (Combined Mass) x a

a= F/(3F/8)

a= 8/3 m/s 2 or 2.67 m/s 2

Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem. Most questions answered within 4 hours.

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

I have a physics question

Answers · 10

physics buoyant

Answers · 6

how many days are in 2 million seconds?

Answers · 5

how does surface area affect rate of active transport?

find an online tutor

Physics tutors
AP Physics tutors
AP Physics C tutors
Science tutors
Fluid Mechanics tutors
Thermodynamics tutors
Statics tutors
ACT Science tutors

related lessons

Center of Mass
Need help with something else? Try one of our lessons.

Acceleration Lesson: Physics Basics, Formulas, and Applications

Lesson overview, introduction to acceleration lesson, what is acceleration, what is the history of the concept of acceleration, what is the formula for acceleration, what are the types of acceleration, how are velocity and acceleration related, what are real-world applications of acceleration, how does acceleration affect relate to forces and motion, how do you calculate acceleration.

Understanding motion is fundamental to grasping the principles that govern our physical world, from the flight of a bird to the launch of a rocket. In this acceleration lesson, we will understand the concept of acceleration, a key element that reveals how objects change their motion over time. Acceleration is not just about speed-it's about how quickly things speed up, slow down, or change direction.

As we learn about this concept, we'll understand how it applies to various real-world scenarios, from the thrill of roller coasters to the precision of space missions. This lesson on acceleration will guide you through the history, types, and mathematical aspects of acceleration, and will show you how this seemingly simple concept plays a crucial role in everything from sports to engineering.

Acceleration is a fundamental concept in physics that describes how the velocity of an object changes over time. Unlike speed, which only tells us how fast an object is moving, acceleration provides a deeper understanding of motion by indicating how quickly an object's speed is increasing or decreasing, and in what direction. This makes acceleration a vector quantity, meaning it has both magnitude ( how much ) and direction ( which way ).

Acceleration occurs whenever there is a change in an object's velocity, whether that change involves speeding up, slowing down, or altering direction. For example, when a car starts from a complete stop and begins to move forward, it is accelerating. Similarly, when it comes to a stop at a traffic light, it is also experiencing acceleration-specifically, negative acceleration or deceleration. Even when a car rounds a bend at a constant speed, it is accelerating because the direction of its velocity is changing.

The concept of acceleration has a rich history that spans centuries, evolving from early philosophical ideas to the rigorous scientific framework we use today. Understanding acceleration as we do now-an object's change in velocity over time-required contributions from several pioneering thinkers who challenged prevailing notions of motion and laid the groundwork for classical mechanics.

Early Philosophical Ideas In ancient Greece, philosophers like Aristotle proposed early theories of motion. Aristotle believed that objects required a continuous force to maintain motion and that heavier objects fell faster than lighter ones. However, his views did not distinguish between speed and acceleration, and his ideas were largely qualitative without rigorous experimental validation. Aristotle's theories dominated Western thought for nearly two millennia, until the Renaissance era when new scientific methods began to emerge.
Galileo Galilei's Revolutionary Work The first significant shift in understanding acceleration came in the late 16th and early 17th centuries with Galileo Galilei. Often referred to as the "father of modern science," Galileo challenged Aristotle's theories by introducing a mathematical and experimental approach to studying motion. Through his experiments with inclined planes and observations of falling bodies, Galileo discovered that all objects accelerate at the same rate in the absence of air resistance, regardless of their mass. This was a groundbreaking realization that contradicted the long-held Aristotelian belief. Galileo introduced the concept of uniformly accelerated motion. He formulated laws that described how velocity changes with time under constant acceleration, laying the groundwork for future studies.
Isaac Newton and the Formalization of Acceleration Building on Galileo's work, Sir Isaac Newton further developed the concept of acceleration in the 17th century within the framework of his three laws of motion. Newton's Second Law of Motion, F=ma ( net force equals mass times acceleration ), quantitatively defined acceleration as the result of a net an unbalanced force acting on a mass. This law provided a clear and mathematical definition of acceleration and connected it to the concept of force, transforming acceleration from a purely observational phenomenon to a fundamental principle in physics. Newton's work in " Philosophiæ Naturalis Principia Mathematica " (1687) established the laws of motion and universal gravitation, which became the foundation of classical mechanics. The understanding of acceleration was now solidly grounded in mathematics and could be applied to explain a wide range of natural and artificial phenomena, from falling apples to planetary orbits.
19th and 20th Century Developments The study of acceleration continued to evolve with the advancements in mathematics and physics. During the 19th century, the development of calculus allowed scientists to describe acceleration more precisely. This was particularly important for understanding non-uniform acceleration, where an object's rate of acceleration changes over time. Calculus-based physics provided the tools to analyze complex systems, such as those involving rotational motion and varying forces. In the 20th century, Albert Einstein's Theory of Relativity introduced new ways of thinking about acceleration, especially when approaching the speed of light. Einstein's work showed that acceleration and gravity are interconnected, leading to a deeper understanding of how mass, energy, and spacetime interact.
Modern Interpretations and Applications Today, the concept of acceleration is fundamental in various fields, from classical mechanics to modern engineering, astrophysics, and even artificial intelligence. In engineering, understanding acceleration is critical for designing safe vehicles, bridges, and buildings. In astrophysics, acceleration is key to understanding the motion of celestial bodies and the expansion of the universe. Concepts like angular acceleration, centripetal acceleration, and gravitational acceleration continue to drive advancements in technology and deepen our understanding of the natural world.

The formula for acceleration is a fundamental equation in physics that quantifies how an object's velocity changes over time. Acceleration is defined as the rate of change of velocity with respect to time. Because it involves both a change in speed and direction, acceleration is a vector quantity, meaning it has both magnitude and direction. Understanding the formula for acceleration is crucial for solving a variety of problems related to motion, from the simplest to the most complex scenarios.

Basic Formula for Acceleration

The most commonly used formula for constant (unchanging) acceleration is:

Acceleration (a)= Change in Velocity (Δv)/Time Taken (Δt)

Δv=v f −v i Is the change in velocity, calculated by subtracting the initial velocity (v i ) from the final velocity (v f ).
Δt Is the time interval during which the change in velocity occurs.

This formula tells us how quickly an object's velocity changes over a specific period of time. The unit of acceleration is meters per second squared (m/s²), indicating how much the velocity changes per second, every second.

Components of the Formula

Initial Velocity (v i ) This is the velocity of the object at the start of the time period. It can be zero if the object starts from rest or has any positive or negative value, depending on the direction and speed.
Final Velocity (v f ) This is the velocity of the object at the end of the time period. It reflects the new speed and direction after acceleration has taken place.
Change in Velocity (Δv) The difference between the final velocity and the initial velocity represents how much the velocity has increased or decreased over the time interval. A positive change in velocity indicates speeding up, while a negative change indicates slowing down (deceleration).
Time Interval (Δt) This is the duration over which the change in velocity occurs. Time must be measured in seconds for the standard unit of acceleration to be meters per second squared.

Take These Quizzes

How much do you know about Acceleration?

Acceleration, as a measure of how an object's velocity changes over time, can manifest in various forms depending on the nature of the motion. Understanding the different types of acceleration is crucial in physics because it provides insights into how forces affect objects in different scenarios. The types of acceleration can be broadly categorized based on the uniformity of the motion, the direction of motion, and specific cases such as circular motion. Here's a detailed look at the primary types of acceleration

Fig: Graphical Representation of the Various Types of Acceleration

1. Uniform Acceleration

Uniform acceleration occurs when an object's velocity changes at a constant rate over time. This means that the same amount of velocity is added or subtracted for each unit of time. In uniform acceleration, the acceleration value remains constant, and the motion follows a predictable pattern easy to understand.

Examples A classic example of uniform acceleration is an object in free fall under the influence of gravity (neglecting air resistance). Here, the object accelerates uniformly at approximately 9.8 m/s 2 near the Earth's surface. Another example is a car accelerating at a constant rate from a standstill to a certain speed.
Key Characteristics In uniform acceleration, the velocity-time graph is a straight line, and the slope of this line represents the constant acceleration. Equations of motion for uniformly accelerated bodies, such as v=u+at and s=ut+1/2at 2 , are commonly used to describe and predict this type of motion.

2. Non-Uniform Acceleration

Non-uniform acceleration, also known as variable acceleration, occurs when an object's velocity changes at a non-constant rate. In this case, the acceleration can vary over time, meaning that the amount of velocity added or subtracted per unit of time is not constant.

Examples A car navigating through city traffic, where it speeds up, slows down, and turns at varying rates, experiences non-uniform acceleration. Similarly, a roller coaster moving along a track with different slopes and curves accelerates non-uniformly as it speeds up and slows down due to changes in elevation and direction.
Key Characteristics In non-uniform acceleration, the velocity-time graph is not a straight line; it can be curved or take on various shapes depending on the nature of the acceleration. Calculus is often required to analyze non-uniform acceleration, particularly when dealing with instantaneous acceleration, which is the acceleration at a specific moment in time.

3. Centripetal Acceleration

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is directed toward the center of the circle around which the object is moving. Even if the object maintains a constant speed, the continuous change in direction means that it is constantly accelerating toward the center.

Examples Centripetal acceleration is observed in a car turning around a curve, a satellite orbiting a planet, or a ball attached to a string being swung in a circular motion. Even if the speed remains constant, the change in direction creates an inward acceleration.
Key Characteristics Centripetal acceleration (a c ) is calculated using the formula

a c = v 2 /r

where v is the object's tangential velocity and r is the radius of the circular path. The force causing centripetal acceleration, called centripetal force, is essential for keeping the object in circular motion and is provided by different sources depending on the scenario (e.g., tension in a string, gravitational force for planets, friction for a car).

4. Tangential Acceleration

Tangential acceleration is the component of acceleration that is tangent to the circular path of an object in rotational motion. It is associated with the change in the magnitude of the velocity (speed) of an object moving along a curved path, rather than a change in direction.

Examples A car speeding up or slowing down while moving along a circular track demonstrates tangential acceleration. Another example is a spinning disc where the speed of rotation increases or decreases; points on the edge of the disc experience tangential acceleration.
Key Characteristics Tangential acceleration is distinct from centripetal acceleration in that it affects the speed of the object moving in a circular path, not the direction. It is calculated using:

a t = dv/dt

where dv is the change in velocity and dt is the change in time. Tangential acceleration is often analyzed in conjunction with centripetal acceleration when dealing with objects in non-uniform circular motion.

5. Angular Acceleration

Angular acceleration occurs when an object's angular velocity changes with time. This type of acceleration is significant when dealing with rotational motion, where the object rotates around a fixed axis.

Examples A figure skater spinning faster by pulling in their arms or a merry-go-round that gradually speeds up or slows down exhibits angular acceleration. The wheels of a car that are accelerating rotationally while the car speeds up also experience angular acceleration.
Key Characteristics Angular acceleration (α) is the rate of change of angular velocity (ω) over time:

Angular acceleration is crucial for understanding dynamics in rotational systems and is measured in radians per second squared (rad/s 2 ).

6. Deceleration (Negative Acceleration)

Deceleration, or negative acceleration, occurs when the acceleration of an object is in the opposite direction to its velocity, causing the object to slow down. While technically still a form of acceleration, deceleration specifically refers to a reduction in speed.

Examples A car coming to a stop at a red light or a ball rolling to a halt on a flat surface experiences deceleration. A spacecraft using reverse thrusters to slow down as it approaches a docking station is also decelerating.
Key Characteristics In deceleration, the velocity decreases over time, and the acceleration vector points in the opposite direction to the velocity vector. The magnitude of deceleration can be calculated similarly to other types of linear acceleration, but it results in a decrease in the object's speed.

Velocity and acceleration are two fundamental concepts in physics that describe the motion of objects. While velocity tells us how fast an object is moving and in which direction, acceleration describes how the velocity of that object changes over time. Understanding the relationship between velocity and acceleration is crucial for analyzing various types of motion, from everyday occurrences like driving a car to more complex scenarios such as the motion of planets.

Understanding Velocity

Velocity is a vector quantity that specifies both the speed and direction of an object's motion. It is defined as the rate of change of an object's position with respect to time. For example, if a car is moving eastward at 60 km/h, its velocity is 60 km/h east. Velocity can change in two ways: either the speed changes while the direction remains constant or the direction changes while the speed remains constant.

Understanding Acceleration

Acceleration is also a vector quantity and is defined as the rate at which an object's velocity changes with time. When an object speeds up, slows down, or changes direction, it is experiencing acceleration. Acceleration can be positive (when an object speeds up in the direction of motion) or negative (when it slows down, often called deceleration).

The Relationship Between Velocity and Acceleration

Direct Relationship Acceleration is directly related to changes in velocity. When acceleration is constant and positively aligns with the velocity, the velocity speed of an object increases linearly over time. For instance, when a car accelerates steadily from a stop, its velocity increases uniformly.
Change in Velocity Over Time Acceleration determines how quickly the velocity changes. If an object has a constant acceleration, its velocity changes at a steady rate. Conversely, if the acceleration is zero, the velocity remains constant.
Instantaneous vs. Average Values Both velocity and acceleration can be described as instantaneous or average values. Instantaneous velocity and acceleration refer to their values at a specific moment in time, while average velocity and acceleration refer to their values over a period of time.
Direction Matters Since both velocity and acceleration are vector quantities, their directions are crucial. When velocity and acceleration vectors point in the same direction, the object's speed increases. When they point in opposite directions, the object's speed decreases. For example, a car moving forward and accelerating in the forward direction will increase its speed, whereas a car moving forward but accelerating backward will slow down.
Graphical Representation Velocity and acceleration are often represented on graphs to visualize their relationship. A velocity-time graph can show how velocity changes over time, where the slope represents acceleration. For motion along a 1D line, if the slope is positive, the object is accelerating; if it is negative, the object is decelerating. A flat slope indicates constant velocity with zero acceleration.

Fig: Graphical Representation of the Velocity-Acceleration Relationship

Equations of Motion In physics, the relationship between velocity and acceleration is mathematically expressed through the equations of motion. For linear motion with constant acceleration, the relationship is given by:
v is the final velocity at time t,
u is the initial velocity,
a is the acceleration,
t is the time.

This equation demonstrates that the final a resulting velocity of an object depends on its initial velocity, the acceleration, and the time over which the acceleration is applied.

Importance of the Relationship Understanding the relationship between velocity and acceleration is essential for predicting and analyzing the motion of objects. This knowledge helps in a variety of fields, from engineering and automotive safety to space exploration and sports science. It allows for the precise control of moving objects, the design of vehicles and machines, and the development of simulations and models to predict natural phenomena.

Acceleration is a fundamental concept not only in physics but also in numerous real-world applications across various fields. Understanding how acceleration works helps us design safer vehicles, optimize athletic performance, launch spacecraft, and much more. Here's a detailed look at some of the practical applications of acceleration

1. Automotive Engineering and Safety

One of the most common applications of acceleration is in the field of automotive engineering. Engineers need to understand acceleration to design vehicles that can accelerate and decelerate safely and efficiently. Acceleration is crucial in scenarios like

Vehicle Performance Acceleration determines how quickly a car can reach a certain speed, which is a key factor in vehicle performance. Engineers design engines, transmissions, and braking systems that can provide optimal acceleration and deceleration rates for different types of vehicles.
Crash Safety In crash tests, understanding negative acceleration (deceleration) is critical. Engineers analyze how quickly a vehicle decelerates upon impact to ensure that safety features such as airbags and seatbelts minimize injuries. The faster the deceleration, the greater the force experienced by occupants, so safety systems are designed to manage these forces.
Anti-lock Braking Systems (ABS) ABS technology uses acceleration sensors to prevent wheels from locking up during braking. By rapidly modulating brake pressure, ABS ensures that the tires maintain traction with the road surface, allowing drivers to maintain steering control while decelerating.

2. Space Exploration and Astronomy

Acceleration plays a vital role in space exploration, where it is crucial for launching rockets, maneuvering spacecraft, and maintaining orbits.

Rocket Launches Rockets must achieve a certain acceleration to escape Earth's gravitational pull. This involves overcoming the force of gravity and atmospheric drag to reach escape velocity. Engineers carefully calculate the required acceleration to ensure a successful launch.
Orbital Mechanics Once in space, spacecraft need to adjust their velocity and acceleration to enter stable orbits around celestial bodies. Understanding acceleration helps in plotting trajectories, planning orbital insertions, and making course corrections using thrusters.
Planetary Missions Space missions, such as those to Mars or the Moon, rely on precise calculations of acceleration and deceleration to safely land rovers and other equipment on planetary surfaces. These calculations also account for gravitational forces and the need to decelerate as the spacecraft approaches the target.

3. Sports and Human Performance

Athletes and trainers use the concept of acceleration to improve performance in various sports.

Track and Field Sprinters need to maximize their acceleration out of the starting blocks to gain an early advantage in races. Coaches analyze acceleration phases to optimize training regimens, focusing on explosive power and stride length.
Cycling and Swimming In sports like cycling and swimming, athletes strive to maintain consistent acceleration while reducing drag and resistance. Understanding how acceleration affects speed and endurance helps athletes refine their techniques and enhance overall performance.
Biomechanics and Injury Prevention Analyzing acceleration forces on joints and muscles can help in designing better training programs and equipment, thereby reducing the risk of injury.

4. Robotics and Automation

Acceleration is a key factor in the design and operation of robots and automated systems.

Robotic Arms In manufacturing, robotic arms must accelerate and decelerate precisely to handle objects, assemble parts, or perform welding tasks. Understanding acceleration ensures smooth operation, accuracy, and efficiency in high-speed manufacturing environments.
Drones and Autonomous Vehicles Drones and self-driving cars use sensors to detect acceleration and adjust their movement accordingly. These systems rely on algorithms that calculate acceleration to navigate safely, avoid obstacles, and maintain control in varying conditions.

5. Aviation and Aerospace

Understanding acceleration is critical for designing aircraft that can safely take off, fly, and land.

Takeoff and Landing Pilots must accelerate aircraft to a specific speed (takeoff speed) to become airborne. Similarly, during landing, they decelerate the aircraft while managing the descent rate. Engineers design aircraft systems that can handle these acceleration and deceleration forces safely.
Flight Dynamics Pilots and flight engineers use concepts of centripetal and angular acceleration when executing maneuvers such as turns, climbs, and descents. Understanding these accelerations helps maintain stability and passenger comfort.

6. Everyday Applications

Acceleration is part of everyday life, even in activities that may seem mundane.

Elevators Elevators accelerate and decelerate to provide smooth transitions between floors. Engineers calculate the optimal rates of acceleration to ensure comfort and safety for passengers.
Amusement Park Rides Roller coasters and other rides rely on calculated acceleration to provide thrilling experiences. Designers must balance acceleration and deceleration to ensure both excitement and safety.

Acceleration plays a pivotal role in understanding forces and motion in physics. It is directly linked to Newton's Second Law of Motion, which states that the net force acting on an object is equal to the mass of that object multiplied by its acceleration. This relationship forms the basis for understanding how objects move and interact with forces in both everyday situations and complex physical systems. Acceleration affects forces and motion by determining how objects speed up, slow down, or change direction under the influence of various forces. Here is a detailed exploration of how acceleration affects forces and motion

1. Newton's Second Law of Motion and Acceleration

Newton's Second Law of Motion provides a clear mathematical relationship between force, mass, and acceleration

F is the net force acting on an object,
m is the mass of the object,
a is the acceleration of the object.

This law indicates that for a given mass, the force applied to an object is directly proportional to the acceleration it experiences. Therefore, if you apply a greater force to an object of the same mass, its acceleration will increase. Conversely, for a constant force, an object with a larger mass will have a smaller acceleration.

2. The Role of Acceleration in Different Types of Motion

Acceleration affects various types of motion differently. Here are some examples of how acceleration influences different motion scenarios

Linear Motion In straight-line or linear motion, acceleration determines how quickly an object speeds up or slows down along a path. For instance, when a car accelerates down a straight road, it gains speed in the direction of its motion. If the driver brakes, the car experiences a negative acceleration (deceleration), reducing its speed. The force applied by the engine or brakes causes these changes in velocity.
Rotational Motion For objects in rotational motion, such as spinning wheels or planets orbiting a star, angular acceleration describes how quickly their angular velocity changes. In this context, a torque (rotational force) causes angular acceleration, influencing how fast or slow an object rotates. The relationship is expressed by the formula:
τ is the torque applied,
I is the moment of inertia (resistance to rotational acceleration),
α is the angular acceleration.

A higher torque results in greater angular acceleration, causing the object to spin faster.

Projectile Motion In projectile motion, acceleration due to gravity (9.8 m/s 2 on Earth) affects the vertical component of an object's velocity. As a projectile is launched, it accelerates downward due to gravity, while its horizontal velocity remains constant (assuming no air resistance). This results in a curved trajectory known as a parabolic path. The force of gravity constantly accelerates the projectile downwards, affecting its overall motion.

3. The Influence of Acceleration on Forces

Acceleration also affects the forces experienced by objects in different contexts

Centripetal Force in Circular Motion When an object moves in a circular path, it experiences centripetal acceleration directed towards the center of the circle. This acceleration requires a centripetal force, given by:

F c = mv 2 / r

v is the tangential velocity,
r is the radius of the circular path.

The centripetal force is necessary to keep the object moving in a circular path. If this force is removed, the object will move in a straight line tangent to the circle due to inertia.

Frictional Forces and Acceleration Friction opposes motion, and its effect depends on the acceleration of an object. When an object accelerates, frictional forces can increase. For example, a car accelerating on a road experiences friction between its tires and the surface, which affects how quickly it can speed up or slow down. Engineers design tire treads and braking systems to manage friction effectively, optimizing acceleration and deceleration.

4. Real-World Applications of Acceleration and Forces

Acceleration is crucial in various real-world applications where forces must be managed and controlled

Vehicle Dynamics In automotive design, acceleration is directly linked to the forces that must be managed to ensure stability and safety. The force exerted by an engine determines how quickly a car can accelerate while braking systems control deceleration forces to prevent skidding or loss of control.
Aerospace Engineering In aerospace engineering, understanding acceleration is vital for designing aircraft and spacecraft that can withstand the forces experienced during takeoff, flight, and landing. Pilots must manage acceleration forces during maneuvers to maintain control and passenger comfort. In space travel, thrust from rockets provides the necessary force to achieve the required acceleration for escaping Earth's gravity.
Athletic Performance Athletes utilize acceleration in sports to maximize performance. Sprinters focus on explosive acceleration from the starting blocks, while soccer players use quick accelerations to change direction and evade opponents. Understanding the forces involved in acceleration helps athletes train effectively and prevent injuries.

5. Acceleration in Gravitational Fields and Free Fall

In gravitational fields, acceleration due to gravity plays a significant role in determining the forces on objects:

Free Fall When an object is in free fall, it accelerates downward due to gravity. The force exerted by gravity on the object is proportional to its mass and acceleration:

where g is the acceleration due to gravity. All objects in free fall experience the same gravitational acceleration, regardless of their mass, assuming no air resistance. This force affects how quickly objects accelerate towards the ground.

Gravitational Acceleration in Planetary Orbits Acceleration also affects how planets and satellites orbit celestial bodies. The gravitational force provides the centripetal force necessary for orbital motion, causing acceleration that keeps the bodies in orbit. A balance between gravitational pull and orbital velocity determines the stability of these orbits.

6. The Interplay Between Inertia and Acceleration

Inertia is the property of an object to resist changes in its state of motion. Acceleration is the change in an object's velocity, and any force that causes acceleration must overcome the object's inertia. Heavier objects with greater mass require more force to accelerate at the same rate as lighter objects. This relationship is evident in scenarios such as moving furniture or launching rockets, where significant forces are needed to overcome the inertia of large masses.

Calculating acceleration involves finding how quickly an object's velocity changes over time. Acceleration is defined as the rate of change of velocity per unit of time. The formula for calculating constant acceleration is

Acceleration (a)= Change in Velocity (Δv)/ Time Taken (Δt)

The unit of acceleration is meters per second squared (m/s²). Here are five examples to help understand how to calculate acceleration:

Example 1 Calculating Acceleration for a Car

A car accelerates from rest to a speed of 20 m/s in 5 seconds. What is the acceleration of the car?

Initial velocity (vi) = 0 m/s (starts from rest)
Final velocity (vf) = 20 m/s
Time taken (Δt) = 5 s

Acceleration (a) = (vf - vi) / Δt a = (20 - 0) / 5 a = 20 / 5 a = 4 m/s²

Answer The acceleration of the car is 4 m/s².

Example 2 Deceleration of a Bicycle

A cyclist is traveling at 15 m/s and comes to a complete stop in 3 seconds. What is the acceleration (deceleration) of the bicycle?

Initial velocity (vi) = 15 m/s
Final velocity (vf) = 0 m/s (stops)
Time taken (Δt) = 3 s

Acceleration (a) = (vf - vi) / Δt a = (0 - 15) / 3 a = -15 / 3 a = -5 m/s²

Answer The deceleration of the bicycle is -5 m/s² (negative indicates slowing down).

Example 3 Calculating Acceleration for an Object in Free Fall

An object is dropped from a height, and after 2 seconds, it reaches a velocity of 19.6 m/s. What is the acceleration due to gravity?

Initial velocity (vi) = 0 m/s (dropped from rest)
Final velocity (vf) = 19.6 m/s
Time taken (Δt) = 2 s

Acceleration (a) = (vf - vi) / Δt a = (19.6 - 0) / 2 a = 19.6 / 2 a = 9.8 m/s²

Answer The acceleration due to gravity is 9.8 m/s².

Example 4 Acceleration of a Train

A train increases its speed from 10 m/s to 30 m/s in 10 seconds. What is the acceleration of the train?

Initial velocity (vi) = 10 m/s
Final velocity (vf) = 30 m/s
Time taken (Δt) = 10 s

Acceleration (a) = (vf - vi) / Δt a = (30 - 10) / 10 a = 20 / 10 a = 2 m/s²

Answer The acceleration of the train is 2 m/s².

Example 5 Calculating the Negative Acceleration (Deceleration) of a Car

A car moving at 25 m/s slows down to 5 m/s in 4 seconds. What is the acceleration of the car?

Initial velocity (vi) = 25 m/s
Final velocity (vf) = 5 m/s
Time taken (Δt) = 4 s

Acceleration (a) = (vf - vi) / Δt a = (5 - 25) / 4 a = -20 / 4 a = -5 m/s²

Answer The deceleration of the car is -5 m/s² (negative indicates slowing down).

These examples illustrate how to calculate acceleration in various scenarios, whether speeding up, slowing down or in free fall under gravity.

What’s up with Acceleration? Fun Trivia Quiz

In this lesson on acceleration, we've learned how this fundamental concept helps us understand changes in an object's motion over time. From its historical development through the work of Galileo and Newton to its practical applications in fields like automotive safety, space exploration, and sports, acceleration plays a vital role in both science and everyday life.

By understanding the different types of acceleration and learning how to calculate it, we gain insights into how forces affect motion. This knowledge is essential for solving real-world problems, designing advanced technologies, and continuing to expand our understanding of the physical world. Acceleration is more than a physics concept; it's a powerful tool that shapes our interactions with the world around us.

Featured Quizzes

Create a Quiz

Wait! Here's an interesting quiz for you.

TPC and eLearning
What's NEW at TPC?
Read Watch Interact
Practice Review Test
Teacher-Tools
Request a Demo
Get A Quote
Subscription Selection
Seat Calculator
Ad Free Account
Edit Profile Settings
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?
Lab Equipment
Lab Procedures
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
Intermolecular Forces Questions
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
Chemical Reactions Preview
Intermolecular Forces Preview
Melting Points and Boiling Points Preview
Bond Energy and Reactions Preview
Reaction Rates Preview
Ammonia Factory Preview
Stoichiometry Preview
Nuclear Chemistry Preview
Gaining Teacher Access
Task Tracker Directions
Conceptual Physics Course
On-Level Physics Course
Honors Physics Course
Chemistry Concept Builders
All Chemistry Resources
Users Voice
Tasks and Classes
Webinars and Trainings
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
Measurement and Calculations
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
Atomic Physics
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
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
Newton's Laws of Motion
Work and Energy Packet
Static Electricity Review
NGSS Alignments
1D-Kinematics
Projectiles
Circular Motion
Magnetism and Electromagnetism
Graphing Practice
About the ACT
ACT Preparation
For Teachers
Other Resources
Solutions Guide
Solutions Guide Digital Download
Motion in One Dimension
Work, Energy and Power
Chemistry of Matter
Measurement and the Metric System
Names and Formulas
Algebra Based On-Level Physics
Honors Physics
Conceptual Physics
Other Tools
Frequently Asked Questions
Purchasing the Download
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
Teacher Resources
Subscriptions

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
Sample Problems and Solutions
Kinematic Equations Introduction
Solving Problems with Kinematic Equations
Kinematic Equations and Free Fall
Kinematic Equations and Kinematic Graphs

Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep)

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

t = 32.8 s

v = 0 m/s

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

t = 5.21 s

v = 0 m/s

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

t = 2.6 s

v = 0 m/s

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

v = 18.5 m/s

v = 46.1 m/s

t = 2.47 s

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

v = 0 m/s

d = -1.40 m

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

v = 0 m/s

v = 444 m/s

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

v = 0 m/s

v = 7.10 m/s

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

v = 0 m/s

v = 65 m/s

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

v = 22.4 m/s

v = 0 m/s

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

a = -9.8 m/s

v = 0 m/s

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

v = 0 m/s

v = 521 m/s

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

(NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use: v f = v i + a*t

0 m/s = v i + (-9.8 m/s 2 )*(3.13 s)

0 m/s = v i - 30.7 m/s

v i = 30.7 m/s (30.674 m/s)

Now use: v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = (30.7 m/s) 2 + 2*(-9.8 m/s 2 )*(d)

0 m 2 /s 2 = (940 m 2 /s 2 ) + (-19.6 m/s 2 )*d

-940 m 2 /s 2 = (-19.6 m/s 2 )*d

(-940 m 2 /s 2 )/(-19.6 m/s 2 ) = d

Return to Problem 13

v = 0 m/s

d = -370 m

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

v = 367 m/s

v = 0 m/s

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

t = 3.41 s

v = 0 m/s

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

a = -3.90 m/s

v = 0 m/s

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

v = 0 m/s

v = 88.3 m/s

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

v = 0 m/s

v = m/s

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

IMAGES

Solving problems for acceleration
Acceleration Practice Problems with solutions
Solving Problems Calculating the Average Acceleration of an Object
Acceleration Formula Problems by Mr Fry's Physical Science
How To Solve Acceleration Problems
Laws of Motion Problems with Solutions Six

VIDEO

Week 6: Problem Solving: Law of GroundwaterMovement, Darcy's Law and Application
Problem Solving Law Of Motion
What is ISAACS/ILAC?
Physics. Problem solving. 01_10
Physics 71 Day 5 [Part 2/8]
🚀Conquer Life's Obstacles With 3 Legendary Laws!

COMMENTS

PDF Newton's Second Law of Motion Problems Worksheet
Practice solving problems involving force, mass and acceleration using Newton's second law equation. Fill in the blanks with the correct answers and show your work for each problem.
Newton's second law of motion
Learn how to apply Newton's second law of motion to solve problems involving force, acceleration, mass and friction. See examples of solved problems with explanations and diagrams.
PDF 10 Worksheet Practice Problems for Newton's 2 law
This web page contains a Microsoft Word document with 10 practice problems based on Newton's second law of motion, which relates force, mass, and acceleration. The problems involve solving for force, mass, or acceleration using the equation F=ma and giving the units in newtons, kilograms, and meters per second squared.
6.1 Solving Problems with Newton's Laws
Learn how to apply Newton's laws of motion to solve problems involving acceleration, force, and momentum. This web page is part of a free textbook on university physics volume 1, licensed under a Creative Commons Attribution 4.0 International License.
Newton's Law Problem Sets
Practice problems on Newton's first law of motion, the law of inertia, and the law of acceleration. Solve problems involving force, mass, weight, and acceleration using the calculator pad and audio guided solutions.
4.3 Newton's Second Law of Motion
The web page you requested is not available due to a technical issue. It is part of a free textbook on physics that covers Newton's second law of motion and change in velocity.
Newton's Law Problem Sets
Find 20 ready-to-use problem sets on Newton's laws of motion, with solutions and explanations. These problems involve forces, accelerations, mass, weight, friction, air resistance, tension, and two-body problems.
Newton's Second Law of Motion
Learn how the net force and mass determine the acceleration of an object according to Newton's second law. See examples, interactive diagrams, and practice problems on finding acceleration and force values.
F=ma Practice Problems
Solve problems involving net force, acceleration, and inertia using Newton's second law (F=ma) and 1D motion equations. See answers and explanations for 16 practice problems with different scenarios and forces.
AP Physics 1 : Newton's Second Law
Learn how to apply Newton's second law to solve problems involving forces, acceleration, and weight. See example questions with explanations and solutions for AP Physics 1 exam preparation.
Newton Second Law of Motion Example Problems with Answers
Learn how to calculate net force, mass and acceleration of an object using Newton's 2nd law of motion formula. See examples of problems with solutions for 6th grade physics students.
Solving Problems with Newton's Laws
Learn how to apply problem-solving strategies and Newton's laws of motion to solve complex problems involving forces, equilibrium, and acceleration. See examples of free-body diagrams, kinematic equations, and calculus for different situations.
Newton's Second Law of Motion
Learn how to apply Newton's second law of motion to calculate the force, mass, and acceleration of an object. Watch a 19-minute tutorial with examples, diagrams, and worksheets.
Newton's Laws of Motion: Law of Acceleration
Learn about the second law of motion, the law of acceleration, and how it relates to force and mass. Watch a video created by The Learning Bees, a channel for grade 8 science students in the ...
Solving Fnet = m•a Problems
Learn how to use Newton's Second Law equation (Fnet = m•a) to calculate acceleration or individual force values from given information. Watch a video tutorial with examples, transcript, and interactive practice exercises.
Newton's Second Law of Motion
a.) acceleration of the truck. b.) starting from rest, how long will it take for the truck to reach the velocity of 11.5 m/s? 2. An object with a mass m1 accelerates at 8.0 m/s2 when a force F is applied. A second object with a mass m2 accelerates at 4.0 m/s2 when the same force F is applied to it. a.) Find the value of the ratio m2/m1. b.)
PHYSICS: GRADE 8: Law of Acceleration/Force
In this video, I solved some basic applications of the formulas derived according to the second law of motion. But what is the Second Law of motion?Newton's ...
Acceleration Lesson: Physics Basics, Formulas, and Applications
Newton's Second Law of Motion and Acceleration. Newton's Second Law of Motion provides a clear mathematical relationship between force, mass, and acceleration. F=ma ... This knowledge is essential for solving real-world problems, designing advanced technologies, and continuing to expand our understanding of the physical world. Acceleration is ...
Kinematic Equations: Sample Problems and Solutions
Practice using the four kinematic equations to solve problems involving uniformly accelerated motion. See answers and solutions for various scenarios, such as free fall, car acceleration, rocket-powered sleds, and more.

Newton’s second law of motion – problems and solutions

Share this:

4.3 Newton's Second Law of Motion

Teacher Support

Section Key Terms

Describing Newton’s Second Law of Motion

Misconception Alert

Applying Newton’s Second Law

Mass and Weight

Grasp Check

Tips For Success

Watch Physics

Worked Example

What Rocket Thrust Accelerates This Sled?

Practice Problems

Check Your Understanding

F=ma Practice Problems

HIGH SCHOOL

GRADUATE SCHOOL

Search 50+ Tests

math tutoring

science tutoring

elementary tutoring

Search 350+ Subjects

AP Physics 1 : Newton's Second Law

Example Question #3 : Newton's Second Law

Example Question #2 : Newton's Second Law

Example Question #6 : Newton's Second Law

Example Question #7 : Newton's Second Law

Report an issue with this question

DMCA Complaint

Contact Information

Please wait while your request is being verified...

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 - Basic Problem

1 Expert Answer

Still looking for help? Get the right answer, fast.

RELATED TOPICS

I have a physics question

physics buoyant

how many days are in 2 million seconds?

how does surface area affect rate of active transport?

find an online tutor

related lessons

Acceleration Lesson: Physics Basics, Formulas, and Applications

Basic Formula for Acceleration

Components of the Formula

1. Uniform Acceleration

2. Non-Uniform Acceleration

3. Centripetal Acceleration

4. Tangential Acceleration

5. Angular Acceleration

6. Deceleration (Negative Acceleration)

Check Your Understanding

Solutions to Above Problems

IMAGES

VIDEO

COMMENTS