Call Now to Set Up Tutoring:

(888) 888-0446

High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

Create An Account

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Harmonic Motion

A spring has a spring constant of $16\frac{N}{m}$ .

If a force of 20.5N is used to stretch out the spring, what is the total displacement of the spring?

Possible Answers:

1.28m

0.78m

3.33m

328m

Correct answer:

1.28m

Explanation:

For this problem, use Hooke's law:

$F=k\Delta x$

In this formula, is the spring constant, $\Delta x$ is the compression of the spring, and is the necessary force. We are given the spring constant and the force, allowing us to solve for the displacement.

Plug in our given values and solve.

$F=k\Delta x$

$20.5N=16\frac{N}{m}\Delta x$

$\frac{20.5N}{16\frac{N}{m}}=\Delta x$

$1.28m=\Delta x$

Note that both the force and the displacement are positive because the stretching force will pull in the positive direction. If the spring were compressed, the change in distance would have been negative.

Report an Error

Example Question #212 : Motion And Mechanics

A vertical spring with a spring constant of $11.1\frac{N}{m}$ is stationary. An 8kg mass is attached to the end of the spring. What is the maximum displacement that the spring will stretch?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

-2.82m

-17.92m

-1.11m

-14.13m

-0.45m

Correct answer:

-14.13m

Explanation:

The best way to solve this problem is by using energy. Notice that the spring on its own is stationary. That means its initial total energy at that moment is zero. When the mass is attached, the spring stretches out, giving it spring potential energy ( $PE_{spring}$ ).

Where does that energy come from? The only place it can come from is the addition of the mass. Since the system is vertical, this mass will have gravitational potential energy.

Use the law of conservation of energy to set these two energies equal to each other:

$PE_{mass}=PE_{spring}$

$mg\Delta y=\frac{1}{2}k\Delta y^2$

We are trying to solve for displacement, and now we have an equation in terms of our variable.

Start by diving both sides by $\Delta y$ to get rid of the $\Delta y^2$ on the right side of the equation.

$\frac{mg\Delta y}{\Delta y}=\frac{\frac{1}{2}k\Delta y^2}{\Delta y}$

$mg=\frac{1}{2}k\Delta y$

We are given values for the spring constant, the mass, and gravity. Using these values will allow use to solve for the displacement.

$mg=\frac{1}{2}k\Delta y$

$8kg* -9.8\frac{m}{s^2}=\frac{1}{2}* 11.1\frac{N}{m}* \Delta y$

$-78.4N=5.55\frac{N}{m}* \Delta y$

$\frac{-78.4N}{5.55\frac{N}{m}}=\Delta y$

$-14.13m=\Delta y$

Note that the displacement will be negative because the spring is stretched in the downward direction due to gravity.

Report an Error

Example Question #2 : Using Spring Equations

How much force is required to compress a spring 0.2m , if it has a spring constant of $200\frac{N}{m}$ ?

Possible Answers:

180N

4000N

200N

100N

40N

Correct answer:

40N

Explanation:

For this problem use Hooke's Law:

$F=k\Delta x$

Plug in our given values:

$F=200\frac{N}{m}*0.2m$

F=40N

Report an Error

Example Question #213 : Motion And Mechanics

How much force is required to compress a spring 9.32m , if it has a spring constant of $181\frac{N}{m}$ ?

Possible Answers:

7861.05N

843.46N

190.32N

1686.92N

168.69N

Correct answer:

1686.92N

Explanation:

For this problem use Hooke's Law:

$F=k\Delta x$

Plug in our given values:

$F=181\frac{N}{m}*9.32m$

F=1686.92N

Report an Error

Example Question #214 : Motion And Mechanics

How much potential energy is created by compressing a spring 0.2m , if it has a spring constant of $200\frac{N}{m}$ ?

Possible Answers:

40J

44J

0.04J

Correct answer:

Explanation:

The formula for spring potential energy is:

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

Plug in our given values and solve:

$PE=\frac{1}{2}*200\frac{N}{m}*(0.2m)^2$

$PE = \frac{1}{2}*200\frac{N}{m}*(0.04m^{2})$

$PE = \frac{1}{2}*8N*m$

J = N*m , so:

$PE=\frac{1}{2}*8J$

PE=4J

Report an Error

Example Question #1 : Using Spring Equations

How much potential energy is generated by compressing a spring 9.32m , if it has a spring constant of $181\frac{N}{m}$ ?

Possible Answers:

4047.79J

1686.92J

3930.52J

7861.05J

8095.58J

Correct answer:

7861.05J

Explanation:

The formula for spring potential energy is

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

Plug in our given values and solve:

$PE=\frac{1}{2}*181\frac{N}{m}*(9.32m)^2$

$PE=90.5\frac{N}{m}*86.8624m^2$

PE = 7861.05N*m

J = N*m , so:

PE=7861.05J

Report an Error

Example Question #11 : Harmonic Motion

A 0.56kg mass is placed at the end of a spring. The spring is compressed 0.2m . What is the maximum velocity of the mass if the spring has a spring constant of $200\frac{N}{m}$ ?

Possible Answers:

$14.29\frac{m}{s}$

$7.56\frac{m}{s}$

$4\frac{m}{s}$

$8\frac{m}{s}$

$3.78\frac{m}{s}$

Correct answer:

$3.78\frac{m}{s}$

Explanation:

If we're looking for the maximum velocity, that will happen when all the energy in the system is kinetic energy.

We can use the law of conservation of energy to see PE_i=KE_f . So, if we can find the initial potential energy, we can find the final kinetic energy, and use that to find the mass's final velocity.

The formula for spring potential energy is:

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

Plug in our given values and solve:

$PE=\frac{1}{2}*200\frac{N}{m}*(0.2m)^2$

$PE=\frac{1}{2}*200\frac{N}{m}*(0.04m^{2})$

$PE=\frac{1}{2}*8N*m$

J = N*m , so:

$PE=\frac{1}{2}*8J$

PE=4J

The formula for kinetic energy is:

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

Since $PE_{i}=KE_{f}$ , that means that KE=4J .

We can plug in that information to the formula for kinetic energy to solve for the maximum velocity:

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

$4J=\frac{1}{2}*0.56kg*v^2$

8J=0.56kg*v^2

$\frac{8J}{0.56kg}=v^2$

$14.29\frac{J}{kg} = v^{2}$

Since $J = \frac{kg*m^{2}}{s^{2}}$ , that means that $\frac{J}{kg}=\frac{m^{2}}{s^{2}}$

$14.29\frac{m^2}{s^2}=v^2$

$\sqrt{14.29\frac{m^2}{s^2}}=\sqrt{v^2}$

$3.78\frac{m}{s}=v$

Report an Error

Example Question #3 : Using Spring Equations

A 0.23kg mass is placed at the end of a spring. If the spring is compressed 9.32m , what will be the mass's final velocity if the spring has a spring constant of $181\frac{N}{m}$ ?

Possible Answers:

$130.73\frac{m}{s}$

$261.45\frac{m}{s}$

$15722.1\frac{m}{s}$

$7861.05\frac{m}{s}$

$68356.96\frac{m}{s}$

Correct answer:

$261.45\frac{m}{s}$

Explanation:

If we're looking for the maximum velocity, that will happen when all the energy in the system is kinetic energy.

We can use the law of conservation of energy to see PE_i=KE_f . So, if we can find the initial potential energy, we can find the final kinetic energy, and use that to find the mass's final velocity.

The formula for spring potential energy is:

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

Plug in our given values and solve:

$PE=\frac{1}{2}*181\frac{N}{m}*(9.32m)^2$

$PE=90.5\frac{N}{m}*86.8624m^2$

PE=7861.05N*m

J = N*m , so:

PE=7861.05J

The formula for kinetic energy is:

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

Since $PE_{i}=KE_{f}$ , we know that KE=7861.05J .

We can plug this information into the formula for kinetic energy and use it to solve for the maximum velocity:

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

$7861.05J=\frac{1}{2}*0.23kg*v^2$

$2(7861.05J)=2(\frac{1}{2}*0.23kg*v^{2})$

15722.1J=0.23kg*v^2

$\frac{15722.1J}{0.23kg}{}=v^2$

$68356.96\frac{J}{kg}=v^{2}$

Since $J = \frac{kg*m^{2}}{s^{2}}$ , that means that $\frac{J}{kg}=\frac{m^{2}}{s^{2}}$ .

$68356.96\frac{m^2}{s^2}=v^2$

$\sqrt{68356.96\frac{m^2}{s^2}}=\sqrt{v^2}$

$261.45\frac{m}{s}=v$

Report an Error

Example Question #11 : Using Spring Equations

A 41N force is used to stretch a spring 0.11m . What is the spring constant?

Possible Answers:

$4.51\frac{N}{m}$

$372.73\frac{N}{m}$

$0.223\frac{N}{m}$

$3388.43\frac{N}m{}$

$0.496\frac{N}{m}$

Correct answer:

$372.73\frac{N}{m}$

Explanation:

The formula for the force required to stretch or compress a spring is:

$F=k\Delta x$

We are given the force and the distance, allowing us to solve for the spring constant.

41N=k*0.11m

$\frac{41N}{0.11m}=k$

$372.73\frac{N}{m}=k$

Report an Error

Example Question #222 : Motion And Mechanics

What is the potential energy stored in a spring that is stretched 0.67m and has a spring constant of $412\frac{N}{m}$ ?

Possible Answers:

184.95J

92.47J

568.64J

276.04J

138.02J

Correct answer:

92.47J

Explanation:

Spring potential energy is given by the equation:

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

We are given the spring constant and the distance that the spring is stretched. Using these values, we can find the energy stored in the spring.

$PE=(\frac{1}{2})(412\frac{N}{m})(0.67m)^2$

$PE=(206\frac{N}{m})(0.4489m^2)$

PE=92.47J

Report an Error

View Tutors

Pushpendra
Certified Tutor

university of allahabad, Bachelor of Science, Mathematics.

View Tutors

Shravanthi
Certified Tutor

Sri Venkata Sai medical college, Bachelor of Science, Community Health and Preventive Medicine.

View Tutors

Kristine
Certified Tutor

St. Dominic Savio, Masters in Education, Special Education.

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ISEE Tutors in Chicago, Statistics Tutors in San Francisco-Bay Area, Reading Tutors in Philadelphia, English Tutors in Seattle, French Tutors in Dallas Fort Worth, Reading Tutors in Seattle, French Tutors in Phoenix, Chemistry Tutors in Phoenix, SAT Tutors in San Francisco-Bay Area, SAT Tutors in Philadelphia

Popular Courses & Classes

ACT Courses & Classes in Denver, GRE Courses & Classes in Chicago, MCAT Courses & Classes in Chicago, ISEE Courses & Classes in Phoenix, SAT Courses & Classes in Houston, GRE Courses & Classes in Atlanta, ISEE Courses & Classes in Boston, Spanish Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Denver, ACT Courses & Classes in Washington DC

Popular Test Prep

MCAT Test Prep in Phoenix, ACT Test Prep in Denver, LSAT Test Prep in Philadelphia, ISEE Test Prep in Washington DC, LSAT Test Prep in Los Angeles, LSAT Test Prep in New York City, ISEE Test Prep in San Francisco-Bay Area, GMAT Test Prep in Denver, ACT Test Prep in Washington DC, MCAT 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

High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #11 : Harmonic Motion

Example Question #212 : Motion And Mechanics

Example Question #2 : Using Spring Equations

Example Question #213 : Motion And Mechanics

Example Question #214 : Motion And Mechanics

Example Question #1 : Using Spring Equations

Example Question #11 : Harmonic Motion

Example Question #3 : Using Spring Equations

Example Question #11 : Using Spring Equations

Example Question #222 : Motion And Mechanics

All High School Physics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: