Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 1 : Fundamentals of Force and Newton's Laws

Study concepts, example questions & explanations for AP Physics 1

Create An Account

All AP Physics 1 Resources

7 Diagnostic Tests 170 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #871 : Ap Physics 1

A pitcher throws a fastball at $100\ \frac{\textup{mi}}{hr}$ and the catcher brings the ball to a complete rest over a distance of $0.150\textup{ m}$ . What force did the catcher exert on the baseball to bring it to rest? $\textup{m}_{\textup{baseball}}=0.145\textup{ kg}$ .

$1\textup{ in}=2.54\textup{ cm}$

Possible Answers:

$1730\textup{ N}$

$1830\textup{ N}$

$1930\textup{ N}$

$1630\textup{ N}$

Correct answer:

$1930\textup{ N}$

Explanation:

First, convert to metric units.

$\frac{100mi}{hr}=\frac{5280ft}{1mi}=\frac{12in}{1ft}=\frac{2.54cm}{1in}=\frac{1m}{100cm}=\frac{1hr}{60min}=\frac{1min}{60s}=44.7\frac{m}{s}$

Now, use the equation for velocity.

$v=\frac{\Delta x}{\Delta t}$

Plug in and solve for the final time.

$44.7\frac{m}{s}=\frac{0.150m-0m}{t_{f}-0s}$

$t_{f}=0.0034s$

Now use the equation for acceleration.

$a=\frac{\Delta v}{\Delta t}$

Plug in and solve for the acceleration.

$a=\frac{0\frac{m}{s}-44.7\frac{m}{s}}{0.0034s-0s}$

$a=13320.6\frac{m}{s^{2}}$

Use Newton's second law to solve for the force.

F=ma

Plug in and solve for the force that the catcher exerts on the ball.

$F=0.145Kg(13320.6\frac{m}{s^{2}})$

F=1930N

The catcher will exert an amazing 1930N of force to stop the fastball in such a short distance.

Report an Error

Example Question #32 : Newton's Second Law

Cody pulls his $65\textup{ kg}$ brother on the fresh snow during Christmas break. In order for his brother to have fun he must pull him at a constant speed of $6.0\ \frac{\textup{m}}{\textup{s}}$ . The coefficient of kinetic friction between the sled bottom and the fresh powder is 0.60 . What horizontal pulling force must Cody apply to give his brother a fun ride? Assume the sled has no mass and neglect air resistance.

Possible Answers:

$360\textup{ N}$

$380\textup{ N}$

$340\textup{ N}$

$320\textup{ N}$

Correct answer:

$380\textup{ N}$

Explanation:

The weight is being compensated by the normal force.

$F_{n}=w$

Weight is equal to mass times the acceleration due to gravity.

$F_{n}=mg$

$F_{n}=65Kg(9.8\frac{m}{s^{2}})$

$F_{n}=637N$

The sled does not have to accelerate so the kinetic frictional force is equal to the pulling force .

$P=F_{k}$

The equation for kinetic frictional force is inserted below.

$P=\mu _{k}F_{n}$

Plug in and solve for the pulling force.

P=0.60(637N)

P=380N

Cody must apply at least 380N of horizontal pulling force.

Report an Error

Example Question #31 : Newton's Second Law

In the military a paratrooper and his/her gear can weigh in excess of $1100\textup{ N}$ . While skydiving, a paratrooper can experience an upward acceleration of $9.75\ \frac{\textup{m}}{\textup{s}^{2}}$ when he/she opens the parachute. What force does the paratrooper experience at this time?

Possible Answers:

$550\textup{ N}$

$5050\textup{ N}$

$1090\textup{ N}$

$2190\textup{ N}$

Correct answer:

$2190\textup{ N}$

Explanation:

The weight of the paratrooper and his or her gear is 1100N

w=mg

Plug in and solve for the mass.

$1100N=m(9.8\frac{m}{s^{2}})$

m=112.2Kg

The force due to air resistance causes an acceleration of $9.75\frac{m}{s^{2}}$ .

$F_{air}=ma$

Plug in and solve for the force due to air resistance.

$F_{air}=112.2Kg(9.75\frac{m}{s^{2}})$

$F_{air}=1090N$

The paratrooper is accelerating upward so we know that the force due to air resistance is greater than the force of gravity.

$F_{net}=w+F_{air}$

Plug in the known values and solve for the combined net force.

$F_{net}=1100N+1090N$ $F_{net}=2190N$

Report an Error

Example Question #34 : Newton's Second Law

A new start up shoe company claims to make the lightest basketball shoe that still produces tremendous grip to allow players to move smoothly and freely throughout the game. In a laboratory test done on typical hard wood floor revealed that the $0.27\textup{ kg}$ shoe when swung at $15\ \frac{\textup{m}}{\textup{s}}$ takes $14\textup{ m}$ to come to a complete stop. Neglecting air resistance, what is the coefficient of kinetic friction between the shoe bottoms and the hard wood floor?

Possible Answers:

0.75

0.65

0.55

0.45

Correct answer:

0.45

Explanation:

Use the equation for velocity.

$v=\frac{\Delta v}{\Delta t}$

Plug in and solve for the final time.

$15\frac{m}{s}=\frac{14m-0m}{t_{f}-0s}$

$t_{f}=0.93s$

Now use the equation for acceleration.

$a=\frac{\Delta v}{\Delta t}$

Plug in and solve for the acceleration.

$a=\frac{15\frac{m}{s}-0\frac{m}{s}}{0.93s-0s}$

$a=-16.1\frac{m}{s^{2}}$

Now use Newton's second law.

$F_{k}=ma$

Plug in and solve for the kinetic frictional force.

$F_{k}=0.27Kg(-16.1\frac{m}{s^{2}})$

$F_{k}=4.34N$

Now solve for the normal force.

$F_{n}=w$

$F_{n}=mg$

Plug in and solve for the normal force.

$F_{n}=0.27Kg(9.8\frac{m}{s^{2}})$

$F_{n}=9.65N$

Now solve for the coefficient of kinetic friction.

$F_{k}=\mu _{k}(F_{n})$

Plug in and solve for the coefficient.

$4.34N=\mu _{k}(9.65N)$

$\mu _{k}=0.45$

The coefficient of kinetic friction between the floor and the bottom of these shoes was determined to be 0.45 .

Report an Error

Example Question #31 : Newton's Second Law

John has just finished chopping a bunch of firewood and has to get it in a pile by his house. He elects to use a sled to pull the wood across the gravel road to his house. The road and the sled maintain a coefficient of static friction determined to be 0.35 . John is able to cram $375\textup{ kg}$ of wood into the sled. After chopping all the wood he is tired but convinced he can move the sled. If John can pull with a max force of $645\textup{ N}$ can he pull the sled to his house and if not how many newtons of force is he short?

Possible Answers:

No, he would need $645\textup{ N}$ more.

No, he would need $300\textup{ N}$ more.

No, he would need $500\textup{ N}$ more.

Yes, the $645\textup{ N}$ will suffice to move the sled up to his house.

Correct answer:

No, he would need $645\textup{ N}$ more.

Explanation:

The sled and wood has a mass of 375Kg

w=mg

Plug ina and solve.

$w=375Kg(9.8\frac{m}{s^{2}})$

w=3675N

$w=F_{n}$

Use the equation for static friction.

$F_{s}=\mu _{s}(F_{n})$

Plug in and solve for the static frictional force.

$F_{s}=0.35(3675N)$

$F_{s}=1286.25N$

John will not be able to move the sled and would have to double his efforts to move it.

Report an Error

Example Question #845 : Newtonian Mechanics

During time period , a rocket ship deep in space of mass 2.55*10^5kg travels from <0,10>km to <5,5>km . During time period , the rocket fires. During time period , the rocket travels from <-15,0>km to <-30,-5>km .

Time periods , , and all took 10seconds.

Determine the average force during time period .

Possible Answers:

$7.7*10^7\ \frac{\textup{m}}{\textup{s}}$

$1.2*10^7\ \frac{\textup{m}}{\textup{s}}$

None of these

$2.3*10^7\ \frac{\textup{m}}{\textup{s}}$

$5.1*10^7\ \frac{\textup{m}}{\textup{s}}$

Correct answer:

$5.1*10^7\ \frac{\textup{m}}{\textup{s}}$

Explanation:

Finding initial momentum:

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<5000-0,5000-10000>}{10}$

$\overrightarrow{P_i}=<1.275*10^8,-1.275*10^8>\frac{kg*m}{s}$

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Converting to and plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<-30000+15000,-5000-0>}{10}$

$\overrightarrow{P}_f=<-3.825*10^8,-1.275*10^8>\frac{kg*m}{s}$

Using

$Impulse=\Delta\overrightarrow{P}=\overrightarrow{P_f}-\overrightarrow{P_i}$ Plugging in values:

$\overrightarrow{\Delta P} =<-3.825*10^8,-1.275*10^8>-<1.275*10^8,-1.275*10^8>$

$\Delta\overrightarrow{P}=<5.1*10^8,0>\frac{kg*m}{s}$

Using:

$\Delta \overrightarrow{P}=\overrightarrow{F}*\Delta t$

Solving for $\overrightarrow{F}$ and plugging in values

$\overrightarrow{F}=\frac{<5.1*10^8,0>\frac{kg*m}{s}}{10s}$

$\overrightarrow{F}=5.1*10^7\ \frac{\textup{m}}{\textup{s}}$

Report an Error

Example Question #272 : Forces

Which of the following scenarios is an example of no net force being involved?

Possible Answers:

A box is lowered at a constant speed

A person on a ferris wheel moves at a constant speed

A mass attached to a pendulum moves back and forth with constant frequency

A car gradually slows down as it approaches a red light

An apple falls from a tree

Correct answer:

A box is lowered at a constant speed

Explanation:

In this question, we're being asked to identify a situation in which no net force is involved.

In order for there to be no net force on an object, then that object must not be accelerating. Hence, we need to look at each answer choice to see if there is any acceleration involved.

For a car that is gradually slowing down, a net force will be involved. This is because the car is undergoing a negative acceleration; it is accelerating in the direction directly opposite to its direction of motion.

In the case of an apple falling from a tree, a net force is also involved. This is because as the apple falls, it is being subjected to the acceleration due to gravity.

A mass attached to a pendulum will also experience a net force. This is because the pendulum is constantly going back and forth, meaning that the direction of acceleration is constantly going back and forth.

Since the motion of a person riding a ferris wheel is circular, there is a continuously changing velocity. Even if speed is kept constant, the direction of motion is always changing, thus the rider experiences a net acceleration and thus a net force.

Finally, a box being lowered at a constant rate will not experience a net force. This is because the box has a constant velocity, which means that no acceleration is involved.

Report an Error

Example Question #847 : Newtonian Mechanics

A quarterback throws a football a horizontal distance of 50 yd to a wide receiver. The ball was airborne for 5 s . The ball had an initial speed of 50 mph . The ball has a mass of .43kg .

$1mph=.447\frac{m}{s}$

1 yd=.914 m

If the quarterback’s throw took .75 s , determine the average force of the throw.

Possible Answers:

<5.5,11.5>N

<9.8,9.8>N

<2.6,3.7>N

<9.5,5.5>N

None of these

Correct answer:

<5.5,11.5>N

Explanation:

Determining horizontal component of velocity:

d_x=v_x*t

$v_x=\frac{d_x}{t}$

$v_x=\frac{45.7}{5}$

$v_x=9.14\frac{m}{s}$

Using

v_x^2+v_y^2=v^2

Solving for v_y

$v_y=\sqrt{v^2-v_x^2}$

Combining equations

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

Converting mph to mps

$50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}$

Converting yards to meters

$50yards*\frac{.914 meters}{1 yard}=45.7 meters$

Plugging in values:

$v_y=\sqrt{v^2-(\frac{d_x}{t})^2}$

$v_y=\sqrt{22.35^2-(\frac{45.7}{5})^2}$

$v_y=20.4\frac{m}{s}$

$\overrightarrow{v}=<9.14,20.4>\frac{m}{s}$

Using

v_f=v_i+a*t

$<9.14,20.4>\frac{m}{s}=<0,0>+<a_x,a_y>*.75$

$<a_x,a_y>=<12.8,26.7>\frac{m}{s}$

Using

F=ma

$F=.43kg*<12.8,26.7>\frac{m}{s}$

F=<5.5,11.5>N

Report an Error

Example Question #54 : Fundamentals Of Force And Newton's Laws

A truck is pulling a trailer and is accelerating at $1\frac{m}{s^2}$ . The truck has mass 1500kg and the trailer has mass 2000kg . Determine the force the truck is applying on the trailer. Ignore friction.

Possible Answers:

None of these

1000N

1500N

3000N

2000N

Correct answer:

2000N

Explanation:

The mass of the truck is irrelevant to this problem.

The trailer has a mass of 2000kg , and is accelerating at $1\frac{m}{s^2}$ , thus, using

F=ma

F=2000*1

F=2000N

Report an Error

Example Question #881 : Ap Physics 1

A truck is pulling a trailer and is accelerating at $1\frac{m}{s^2}$ . The truck has mass 1500kg and the trailer has mass 2000kg . Determine the force and direction the trailer is applying on the truck.

Possible Answers:

2000N , forwards

2000N , backwards

1500N , forwards

3000N , backwards

1500N , backwards

Correct answer:

2000N , backwards

Explanation:

The mass of the truck is irrelevant to this problem.

The trailer has a mass of 2000kg , and is accelerating at $1\frac{m}{s^2}$ .

F=ma

F=2000*1

F=2000N

This is the force the truck is applying to the trailer to accelerate it forward. Based on Newton's Laws, the trailer is applying the same force, in the opposite direction.

Report an Error

View AP Physics 1 Tutors

Dossevi
Certified Tutor

cole Ste Genevive Versailles, Bachelor of Science, Mathematics. Institut National Polytechnique de Grenoble, Master of Scienc...

View AP Physics 1 Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View AP Physics 1 Tutors

Austin
Certified Tutor

University of Arizona, Bachelor of Science, Physiology.

All AP Physics 1 Resources

7 Diagnostic Tests 170 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Reading Tutors in San Diego, Physics Tutors in Atlanta, Reading Tutors in Seattle, Calculus Tutors in San Francisco-Bay Area, Spanish Tutors in Boston, Spanish Tutors in San Francisco-Bay Area, Biology Tutors in Atlanta, GRE Tutors in Denver, French Tutors in New York City, English Tutors in Washington DC

Popular Courses & Classes

MCAT Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in Los Angeles, MCAT Courses & Classes in San Diego, SAT Courses & Classes in Boston, GRE Courses & Classes in Seattle, ISEE Courses & Classes in Denver, GMAT Courses & Classes in Chicago, GRE Courses & Classes in Atlanta, SAT Courses & Classes in Seattle, LSAT Courses & Classes in Denver

Popular Test Prep

SAT Test Prep in Dallas Fort Worth, GMAT Test Prep in Seattle, ACT Test Prep in Phoenix, ISEE Test Prep in Chicago, MCAT Test Prep in Boston, ACT Test Prep in New York City, SAT Test Prep in Boston, LSAT Test Prep in Los Angeles, ACT Test Prep in Los Angeles, SSAT Test Prep in Dallas Fort Worth

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:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Physics 1 : Fundamentals of Force and Newton's Laws

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #871 : Ap Physics 1

Example Question #32 : Newton's Second Law

Example Question #31 : Newton's Second Law

Example Question #34 : Newton's Second Law

Example Question #31 : Newton's Second Law

Example Question #845 : Newtonian Mechanics

Example Question #272 : Forces

Example Question #847 : Newtonian Mechanics

Example Question #54 : Fundamentals Of Force And Newton's Laws

Example Question #881 : Ap Physics 1

All AP Physics 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: