Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics C: Mechanics : Energy

Study concepts, example questions & explanations for AP Physics C: Mechanics

Create An Account

All AP Physics C: Mechanics Resources

2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Work, Energy, And Power

A 0.8kg ball is dropped from rest from a cliff that is 150m high. Use conservation of energy to find the vertical velocity of the ball right before it hits the bottom of the cliff.

Possible Answers:

$40.0\frac{m}{s}$

$73.4\frac{m}{s}$

$32.8\frac{m}{s}$

$54.2\frac{m}{s}$

$92.3\frac{m}{s}$

Correct answer:

$54.2\frac{m}{s}$

Explanation:

The conservation of energy equation is K_1+U_1=K_2+U_2 .

The ball starts from rest so . It starts at a height of 150m, so . When the ball reaches the bottom, height is zero and thus, and $K_2=\frac{1}{2}mv^2$ . The conservation of energy equation can be adjusted below.

U_1=K_2

$mgh=\frac{1}{2}mv^2$

Solve for v.

$v=\sqrt{2gh}$

$v=\sqrt{2(9.8\ \frac{m}{s^2})(150\ m)}$

$v=54.2\ \frac{m}{s}$

Report an Error

Example Question #4 : Work, Energy, And Power

Starting from rest, a skateboarder travels down a 25^o incline that's 22m long. Using conservation of energy, calculate the skateboarder's speed when he reaches the bottom. Ignore friction.

Possible Answers:

$6.3\frac{m}{s}$

$32.8\frac{m}{s}$

$2.9\frac{m}{s}$

$21.8\frac{m}{s}$

$13.5 \frac{m}{s}$

Correct answer:

$13.5 \frac{m}{s}$

Explanation:

Conservation of energy states that K_1+U_1=K_2+U_2 .

The skateboarder starts from rest; thus, and . At the bottom of the incline, $K_2=\frac{1}{2}mv^2$ and .

U_1=K_2

$mgh=\frac{1}{2}mv^2$

Solve for v.

$v=\sqrt{2gh}$

Using trigonometry, $h=22\sin(25)$ .

$v=\sqrt{2(9.8\ \frac{m}{s^2})(22)(\sin(25))}$

$v=13.5\ \frac{m}{s}$

Report an Error

Example Question #1 : Energy

A $\small 5.0kg$ bowling ball is dropped from $\small 5.0m$ above the ground. What will its velocity be when it is $\small 1.0m$ above the ground?

Possible Answers:

$10.8\frac{m}{s}$

$8.9\frac{m}{s}$

$9.9\frac{m}{s}$

$78.4\frac{m}{s}$

$98.0\frac{m}{s}$

Correct answer:

$8.9\frac{m}{s}$

Explanation:

Relevant equations:

K_i+U_i=K_f+U_f

$K = \frac{1}{2}mv^2$

U = mgh

Determine initial kinetic and potential energies when the ball is dropped.

K_i = 0

$U_i = (5kg)(9.8\frac{m}{s^2})(5m)=245J$

Determine final kinetic and potential energies, when the ball has fallen to $\small 1.0m$ above the ground.

$K_f=\frac{1}{2}(5kg)v_f^2$

$U_f = (5kg)(9.8\frac{m}{s^2})(1m)=49J$

Use conservation of energy to equate initial and final energy sums.

$0 + 245J = \frac{1}{2}(5kg)v_f^2+49J$

Solve for the final velocity.

$0 + 245J = \frac{1}{2}(5kg)v_f^2+49J$

$v_f^2=\frac{245J-49J}{\frac{1}{2}(5kg)}=78.4\frac{m^2}{s^2}$

$v_f=8.85 \frac{m}{s} \approx 8.9\frac{m}{s}$

Report an Error

Example Question #1 : Understanding Conservation Of Energy

A solid metal object with mass of $25\: kg$ is dropped from rest at the surface of a lake that is $20\:m$ deep. The water exerts a drag force on the object as it sinks. If the total work done by the drag force is - $3800\: J$ , what is the speed of the object when it hits the sand at the bottom of the lake?

Possible Answers:

$9.38\:m/s$

$19.8\:m/s$

$16.0\:m/s$

$11.7\:m/s$

$14.4\:m/s$

Correct answer:

$9.38\:m/s$

Explanation:

This is a conservation of energy problem. First we have to find the work done by gravity. This can be found using:

$W= F\cdot d, F = mg \\ W = 25\:kg \cdot 9.8\:m/s^2 \cdot 20\:m = 4900\:J$

It is given to us that the work done by the drag force is $-3800\:J$ which means that work is done in the opposite direction. We take the net work by adding the two works together we get, $1100\:J$ of net work done on the block.

Since this is a conservation of energy problem, we set the net work equal to the kinetic energy equation:

$W = \frac{1}{2}mv^2$

is the mass of the block and we are trying to solve for .

$v = \sqrt{\frac{2 \cdot 1100}{25}} = 9.38\:m/s$

Report an Error

Example Question #3 : Energy

If a roller coaster car is traveling at $15\: m/s$ when it is $24\: m$ above the ground, how fast is it going when $10\: m$ above the ground?

Possible Answers:

$18.3\:m/s$

$19.5\:m/s$

$21.5\:m/s$

$22.3\:m/s$

$20.5\:m/s$

Correct answer:

$22.3\:m/s$

Explanation:

This is a classic conservation of energy problem. We know that potential energy and kinetic energy both have to conserve. So we use the following equation:

$\frac{1}{2}mv_o^2+mgh_1 = \frac{1}{2}mv_f^2 + mgh_2$

What this equations says is, the sum of kinetic and potential energy is same at varying heights and velocities.

We can simplify this equation by cancelling out all the m terms.

$\frac{1}{2}v_o^2+gh_1 = \frac{1}{2}v_f^2 + gh_2$

We know all the terms except for v_f , which is the final speed we are trying to solve for h_1 , which is $24\:m$ , h_2 is $10\:m$ and v_o is $15\:m/s$ .

If we plug in all the numbers and solve for v_f , we get $22.3\:m/s$ .

Report an Error

Example Question #151 : Mechanics Exam

A train car with a mass of 2400 kg starts from rest at the top of a 150 meter-high hill. What will its velocity be when it reaches the bottom of the hill, assuming that the bottom of the hill is the reference level.

Possible Answers:

$51.65 \frac{m}{s}$

$55.0 \frac{m}{s^2}$

$54.25 \frac{m}{s}$

$0 \frac{m}{s}$

$51.65 \frac{m}{s^2}$

Correct answer:

$54.25 \frac{m}{s}$

Explanation:

The law of conservation of energy states:

PE_i + KE_i = PE_f + KE_f

If the car starts at rest, then the initial kinetic energy = 0 J.

If the car ends at the reference height, the final potential energy = 0 J.

Subsituting these values, the equation becomes:

PE_i + 0 J = 0 J + KE_f

The initial potential energy can be determined by:

$PE_i = mgh_i \\ PE_i = 2400 kg \cdot 9.81 \frac{m}{s^2} \cdot 150 m\\ PE_i = 3531600 \frac{kg \cdot m^2}{s^2}\\ PE_i = 3531600 J$

The final kinetic energy equation is:

$KE_f = \frac{1}{2}mv_f^2\\ KE_f = \frac{1}{2}(2400 kg)v_f^2$

Substituting the initial potential energy and final kinetic energy into our modified conservation of energy equation, we get:

$3531600 J = \frac{1}{2}(2400 kg)v_f^2\\ 3531600 \frac{kg\cdot m^2}{s^2} = 1200 kg \cdot v_f^2\\ \frac{3531600 \frac{kg\cdot m^2}{s^2}}{1200 kg}=v_f^2\\ 2943 \frac{m^2}{s^2}=v_f^2\\ \sqrt{2943 \frac{m^2}{s^2}}=\sqrt{v_f^2}\\ 54.25 \frac{m}{s}=v_f$

Report an Error

Example Question #1 : Kinetic Energy

An object has a mass of 5kg and has a position described by the given function:

x(t)=2x^2-5x+12

What is the object's kinetic energy after two seconds?

Possible Answers:

7.5J

27.5J

45.2J

22.5J

Correct answer:

22.5J

Explanation:

Kinetic energy is defined by the equation:

$KE=\frac{1}{2}mv^2$

Taking the derivative of the position function allows us to obtain the velocity function:

x'(t)=v(t)=4x-5

We can now determine the velocity after two seconds:

v(2)=4(2)-5=3

Now that we know our velocity, we can solve for the kinetic energy.

$KE=\frac{1}{2}(5kg)(3\frac{m}{s})^2=22.5J$

Report an Error

Example Question #4 : Energy

An object starts from rest and accelerates at a rate of $2.5\frac{m}{s^2}$ . If the object has a mass of 10kg, what is its kinetic energy after three seconds?

Possible Answers:

226.5J

562.64J

281.25J

75J

Correct answer:

281.25J

Explanation:

Kinetic energy is given by the equation:

$KE=\frac{1}{2}mv^2$

We can find the velocity using the given acceleration and time:

$v=v_o+a\Delta t$

$v=0\frac{m}{s}+(2.5\frac{m}{s^2})(3s)=7.5\frac{m}{s}$

Use this velocity to find the kinetic energy after three seconds:

$KE=\frac{1}{2}(10kg)(7.5\frac{m}{s})^2=281.25J$

Report an Error

Example Question #4 : Energy

A 120kg box has a kinetic energy of 2300J. What is its velocity?

Possible Answers:

$6.2\frac{m}{s}$

$7.2\frac{m}{s}$

$38.3\frac{m}{s}$

$40.1\frac{m}{s}$

$19.4\frac{m}{s}$

Correct answer:

$6.2\frac{m}{s}$

Explanation:

The formula for kinetic energy of an object is:

$KE=\frac{1}{2}mv^2$

The problem gives us the mass and the kinetic energy, and asks for the velocity, so we can rearrange the equation:

$v=\sqrt{\frac{2KE}{m}}$

Use our given values for kinetic energy and mass to solve:

$v=\sqrt{\frac{2(2300J)}{120kg}}=6.2\frac{m}{s}$

Report an Error

Example Question #1 : Potential Energy

Calculate how much potential energy a vertical standing spring gains if someone puts a 4kg box on top of it. Take the spring constant to be 120N/s.

Possible Answers:

12.2 J

3.0J

6.5 J

8.9 J

8.0J

Correct answer:

6.5 J

Explanation:

First, find out how much the spring compresses when the 4kg box is put on top of it. To find this, consider what forces are acting on the box when it is resting on the spring. The forces acting on it are the upward spring force, , and the downward gravitational force,; thus, the net force acting on the box is given by the equation below.

$F_{net}=F_s-F_g=-kx-mg=0$

We have chosen the upward direction to be positive and the downward direction to be negative. The net force is equal to zero because the box is at rest. Solve for x.

$x=-\frac{mg}{k}$

$x=-\frac{(4\ kg)(-9.8\ \frac{m}{s^2})}{120\ \frac{N}{m}}$

$x=0.33\ \text{m}$

Then use this value to find the potential energy of the spring.

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}(120\ \frac{N}{m})(0.33\ m)^2$

$U=6.5\ J$

So when the box is placed on top of the spring, the spring gains a potential energy of 6.5J.

Report an Error

View Tutors

Constance
Certified Tutor

Loyola University-Chicago, Bachelors, Psychology. Sam Houston State University, Masters, Counseling Psychology.

View Tutors

Anne
Certified Tutor

University of Maryland-Baltimore, Doctor of Science, Accounting.

View Tutors

Satweka
Certified Tutor

The University of Texas at Dallas, Master of Science, Biomedical Engineering.

All AP Physics C: Mechanics Resources

2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Dallas Fort Worth, Computer Science Tutors in Denver, Math Tutors in Seattle, Chemistry Tutors in Denver, GMAT Tutors in San Diego, Statistics Tutors in Phoenix, GRE Tutors in San Diego, MCAT Tutors in Philadelphia, Biology Tutors in Phoenix, Calculus Tutors in Denver

Popular Courses & Classes

GMAT Courses & Classes in Miami, GRE Courses & Classes in Boston, ACT Courses & Classes in Miami, ISEE Courses & Classes in Philadelphia, ISEE Courses & Classes in Chicago, ACT Courses & Classes in Houston, ISEE Courses & Classes in Denver, MCAT Courses & Classes in Philadelphia, SSAT Courses & Classes in Miami, LSAT Courses & Classes in Houston

Popular Test Prep

GRE Test Prep in Seattle, MCAT Test Prep in Miami, GRE Test Prep in Dallas Fort Worth, ACT Test Prep in Houston, ACT Test Prep in San Diego, GMAT Test Prep in Washington DC, ISEE Test Prep in New York City, ISEE Test Prep in Washington DC, SSAT Test Prep in Washington DC, SSAT Test Prep in Denver

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 C: Mechanics : Energy

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #3 : Work, Energy, And Power

Example Question #4 : Work, Energy, And Power

Example Question #1 : Energy

Example Question #1 : Understanding Conservation Of Energy

Example Question #3 : Energy

Example Question #151 : Mechanics Exam

Example Question #1 : Kinetic Energy

Example Question #4 : Energy

Example Question #4 : Energy

Example Question #1 : Potential Energy

All AP Physics C: Mechanics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: