Call Now to Set Up Tutoring:

(888) 888-0446

High School Physics : Forces

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 #71 : Forces

A 2500kg satellite orbits 3.8*10^8m above the Earth. The satellite runs into another stationary satellite of equal mass and the two stick together. What is their resulting velocity?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Possible Answers:

$2.7*10^{-3}\frac{m}{s}$

$1.33*10^{11}\frac{m}{s}$

$5.33*10^7\frac{m}{s}$

$1.07*10^8\frac{m}{s}$

$2.67*10^7\frac{m}{s}$

Correct answer:

$2.67*10^7\frac{m}{s}$

Explanation:

We can use the conservation of momentum to solve. Since the satellites stick together, there is only one final velocity term.

m_1v_1+m_2v_2=(m_1+m_2)v_3

We know the masses for both satellites are equal, and the second satellite is initially stationary.

$(2500kg*v_1)+(2500kg*0\frac{m}{s})=(2500kg+2500kg)v_3$

Now we need to find the velocity of the first satellite. Since the satellite is in orbit (circular motion), we need to find the tangential velocity. We can do this by finding the centripetal acceleration from the centripetal force.

Recognize that the force due to gravity of the Earth on the satellite is the same as the centripetal force acting on the satellite. That means F_c=F_G .

Solve for F_G for the satellite. To do this, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})$

F_G=6.87N

Now that we know the force, we can find the acceleration. Remember that centripetal force is F_c=m*a_c . Set our two forces equal and solve for the centripetal acceleration.

F_G=F_c=m_s*a_c

6.87N=m_s*a_c

6.87N=2500kg*a_c

$2.7*10^{-3}\frac{m}{s^2}=a_c$

Now we can find the tangential velocity, using the equation for centripetal acceleration. Again, remember that the radius is equal to the sum of the radius of the Earth and the height of the satellite!

$a_c=\frac{v^2}{r}$

$2.7*10^{-3}\frac{m}{s^2}=\frac{v^2}{1.45*10^{17}m}$

$(2.7*10^{-3}\frac{m}{s^2})(1.45*10^{17}m)={v^2}$

$2.85*10^{15}\frac{m^2}{s^2}=v^2$

$\sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}$

$5.33*10^7\frac{m}{s}=v$

This value is the tangential velocity, or the initial velocity of the first satellite. We can plug this into the equation for conversation of momentum to solve for the final velocity of the two satellites.

m_1v_1+m_2v_2=(m_1+m_2)v_3

$(2500kg*5.33*10^{7}\frac{m}{s})+(2500kg*0\frac{m}{s})=(2500kg+2500kg)v_3$

$1.3325*10^{11}\frac{kg*m}{s}=(5000kg)v_3$

$\frac{1.3325*10^{11}\frac{kg*m}{s}}{5000kg}=v_3$

$2.67*10^7\frac{m}{s}=v_3$

Report an Error

Example Question #72 : Forces

A 2500kg satellite orbits 3.8*10^8m above the Earth. What is the period of the satellite's orbit?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Possible Answers:

2.84*10^8s

$2.7*10^{-3}s$

$1.71*10^{11}s$

5420s

5.33*10^7s

Correct answer:

$1.71*10^{11}s$

Explanation:

The period describes how long it takes the satellite to make one full orbit. If you go back to the definition of velocity, $v=\frac{\Delta x}{\Delta t}$ , we can apply that to our new circular orbit, in which the distance is equal to the circumference of the circle and the time is equal to the period: $v=\frac{2\pi r}{T}$ . The circumference divided by the period will give us the average velocity.

The problem gives us the radius, but we need to find the tangential velocity. We can do this by first solving for the centripetal acceleration from the centripetal force.

Recognize that the force due to gravity of the Earth on the satellite is the same as the centripetal force acting on the satellite. That means F_c=F_G .

Solve for F_G for the satellite. To do this, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})$

F_G=6.87N

Now that we know the force, we can find the acceleration. Remember that centripetal force is F_c=m*a_c . Set our two forces equal and solve for the centripetal acceleration.

F_G=F_c=m_s*a_c

6.87N=m_s*a_c

6.87N=2500kg*a_c

$2.7*10^{-3}\frac{m}{s^2}=a_c$

$a_c=\frac{v^2}{r}$

$2.7*10^{-3}\frac{m}{s^2}=\frac{v^2}{1.45*10^{17}m}$

$(2.7*10^{-3}\frac{m}{s^2})(1.45*10^{17}m)={v^2}$

$2.85*10^{15}\frac{m^2}{s^2}=v^2$

$\sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}$

$5.33*10^7\frac{m}{s}=v$

We now have a value for the tangential velocity, which we can use in the equation for velocity from the beginning to find the period.

$v=\frac{\Delta x}{\Delta t}=\frac{2\pi r}{T}$

$5.33*10^7\frac{m}{s}=\frac{2\pi(1.45*10^{17}m)}{T}$

$T=\frac{2\pi(1.45*10^{17}m)}{5.33*10^7\frac{m}{s}}$

$T=1.71*10^{11}s$

Report an Error

Example Question #73 : Forces

Two planets are 3*10^8m apart. If the first planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the gravitational force between them?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$1.86*10^{30}N$

$4.7*10^{-1}N$

$1.24*10^{20}N$

$1.24*10^{19}N$

$2.23*10^{18}N$

Correct answer:

$1.24*10^{20}N$

Explanation:

To solve, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the values for the mass of each planet, as well as the distance (radius) between them. Using these values and the gravitational constant, we can solve for the force of gravity.

$F_G=G\frac{m_1m_2}{r^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

Report an Error

Example Question #74 : Forces

Two planets are 3*10^8m apart. If the first planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the acceleration on the smaller planet?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$51048\frac{m}{s^2}$

$1.96*10^{-7}\frac{m}{s^2}$

$0.47\frac{m}{s^2}$

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

$2.1\frac{m}{s^2}$

Correct answer:

$0.47\frac{m}{s^2}$

Explanation:

Remember that Newton's second law states that F=ma . The force acting upon the planet in question will be the force due to gravity. Once we find that, we can find the acceleration.

To solve for the force, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the values for the mass of each planet, as well as the distance (radius) between them. Using these values and the gravitational constant, we can solve for the force of gravity.

$F_G=G\frac{m_1m_2}{r^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

Now that we know the force of gravity, we can use Newton's second law and the mass of the smaller planet to solve for the acceleration of gravity.

F=ma

$1.24*10^{20}N=(2.65*10^{20}kg)*a$

$\frac{1.24*10^{20}N}{2.65*10^{20}kg}=a$

$0.47\frac{m}{s^2}=a$

Report an Error

Example Question #75 : Forces

Two planets are 3*10^8m apart. If the first planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the acceleration on the larger planet?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$1.96*10^{-7}\frac{m}{s^2}$

$0.47\frac{m}{s^2}$

$5.1*10^6\frac{m}{s^2}$

$2.14\frac{m}{s^2}$

$5.1*10^3\frac{m}{s^2}$

Correct answer:

$1.96*10^{-7}\frac{m}{s^2}$

Explanation:

Remember that Newton's second law states that F=ma . The force acting upon the planet in question will be the force due to gravity. Once we find that, we can find the acceleration.

To solve for the force, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the values for the mass of each planet, as well as the distance (radius) between them. Using these values and the gravitational constant, we can solve for the force of gravity.

$F_G=G\frac{m_1m_2}{r^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

Now that we know the force of gravity, we can use Newton's second law and the mass of the larger planet to solve for the acceleration of gravity.

F=ma

$1.24*10^{20}N=(6.33*10^{26}kg)*a$

$\frac{1.24*10^{20}N}{6.33*10^{26}kg}=a$

$1.96*10^{-7}\frac{m}{s^2}=a$

Report an Error

Example Question #642 : Newtonian Mechanics

An astronaut has a mass of 71kg . He travels to a new planet and observes his weight is 510N on this planet's surface. If the radius of the planet is 6*10^7m , what is the mass of the planet?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Possible Answers:

$3.87*10^{26}kg$

$8.34*10^{31}kg$

$6.46*10^{18}kg$

$3.26*10^{22}kg$

$2.87*10^{-2}kg$

Correct answer:

$3.87*10^{26}kg$

Explanation:

To solve, use Newton's law of universal gravitation:

$F=G\frac{m_1*m_2}{r^2}$

Remember that the weight of the astronaut is the same as the gravitational force acting between the planet and the astronaut.

$F_g=mg=G\frac{m_1*m_2}{r^2}$

We are given the gravitational constant, the radius of the planet, the mass of the astronaut, and the magnitude of the force generated. Using these values in the universal gravitation equation, we can solve for the mass of the planet.

$510N=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(71kg)m_2}{(6*10^7m)^2}$

$\frac{510N}{6.67*10^{-11}\frac{m^3}{kg*s^2}}=\frac{(71kg)m_2}{(6*10^7m)^2}$

$7.65*10^{12}\frac{kg^2}{m^2}=\frac{(71kg)m_2}{3.6*10^{15}m^2}$

$(7.65*10^{12}\frac{kg^2}{m^2})(3.6*10^{15}m^2)=(71kg)m_2$

$2.75*10^{28}kg^2=(71kg)m_2$

$3.87*10^{26}=m_2$

Report an Error

Example Question #3 : Universal Gravitation

Two satellites in space, each with equal mass, are 80m apart from each other. If the force of gravity between them is $2.88*10^{-10}N$ , what is the mass of each satellite?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

Possible Answers:

1602.98kg

1.989kg

2022.44kg

2405.33kg

166.28kg

Correct answer:

166.28kg

Explanation:

To solve this problem, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the value of the force, the distance (radius), and the gravitational constant. We are also told that the masses of the two satellites are equal. Since the masses are equal, we can reduce the numerator of the law of gravitation to a single variable.

$F_G=G\frac{M^2}{r^2}$

Now we can use our give values to solve for the mass.

$2.88*10^{-10}N=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{M^2}{(80m)^2})$

$\frac{2.88*10^{-10}N}{(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})}=\frac{M^2}{6400m^2}$

$4.32\frac{kg^2}{m^2}=\frac{M^2}{6400m^2}$

$6400m^2* 4.32\frac{kg^2}{m^2}=M^2$

27648kg^2=M^2

$\sqrt{27648kg^2}=\sqrt{M^2}$

166.28kg=M

Report an Error

Example Question #1 : Net Force

Jerry pushes a 3.2kg box with 1.5N of force. What is the resultant acceleration?

Possible Answers:

$1.2\frac{m}{s^2}$

$0.06\frac{m}{s^2}$

$4.78\frac{m}{s^2}$

$0.66\frac{m}{s^2}$

$0.47\frac{m}{s^2}$

Correct answer:

$0.47\frac{m}{s^2}$

Explanation:

Newton's second law states that $\vec{F}=m\vec{a}$ .

In this problem, $\vec{F}=1.5N$ and m=3.2kg .

Plug these into the equation to solve for acceleration.

$\vec{F}=m\vec{a}$

1.5N=3.2kg(a)

$\frac{1.5N}{3.2kg}=a$

$0.47\frac{m}{s^2}=a$

Report an Error

Example Question #1 : Calculating Force

A 1.12kg ball rolls forward with a net acceleration of $1.11\frac{m}{s^2}$ . What is the net force on the ball?

Possible Answers:

1.21N

0.10N

1.8N

1.10N

1.24N

Correct answer:

1.24N

Explanation:

Newton's second law states that $\vec{F}=m\vec{a}$ .

Plug in the values given to us and solve for the force.

$\vec{F}=(1.12kg)(1.11\frac{m}{s^2})$

$\vec{F}=1.24N$

Report an Error

Example Question #1 : Net Force

Louisa rolls a ball with 31N of force. She observes that it has a constant linear acceleration of $2.1\frac{m}{s^2}$ . What is the mass of the ball?

Possible Answers:

15.12kg

12.1kg

1.41kg

14.76kg

33.22kg

Correct answer:

14.76kg

Explanation:

Newton's second law states that $\vec{F}=m\vec{a}$ .

Plug in the given values to solve for the mass.

$\vec{F}=m\vec{a}$

$31N=m(2.1\frac{m}{s^2})$

$\frac{31N}{2.1\frac{m}{s^2}}=m$

14.76kg=m

Report an Error

View Tutors

Alex
Certified Tutor

Vanderbilt University, Bachelor of Science, Psychology. Boston University, Master of Science, Speech-Language Pathology.

View Tutors

Chinette
Certified Tutor

Univesity of Cagayan Valley, Bachelor in Arts, Arts, General. St Joseph College of Que zon City, Masters in Education, Specia...

View Tutors

Shweta
Certified Tutor

All India Management Association , Master's/Graduate, Human Resource Management . Indian Law Institute , Master's/Graduate, C...

All High School Physics Resources

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

Popular Subjects

Calculus Tutors in Seattle, Spanish Tutors in Chicago, GRE Tutors in Chicago, Math Tutors in New York City, Reading Tutors in San Diego, Algebra Tutors in Phoenix, ACT Tutors in Chicago, Statistics Tutors in Denver, Biology Tutors in Houston, GRE Tutors in Phoenix

Popular Courses & Classes

SAT Courses & Classes in Boston, GMAT Courses & Classes in Boston, ISEE Courses & Classes in Seattle, GMAT Courses & Classes in Miami, ACT Courses & Classes in New York City, ISEE Courses & Classes in Boston, GRE Courses & Classes in Houston, SSAT Courses & Classes in Philadelphia, GRE Courses & Classes in San Diego, GMAT Courses & Classes in Seattle

Popular Test Prep

MCAT Test Prep in Denver, GMAT Test Prep in San Diego, ACT Test Prep in New York City, GRE Test Prep in Houston, GMAT Test Prep in Boston, ACT Test Prep in Seattle, ISEE Test Prep in Denver, GRE Test Prep in New York City, GMAT Test Prep in Atlanta, MCAT Test Prep in Miami

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 : Forces

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #71 : Forces

Example Question #72 : Forces

Example Question #73 : Forces

Example Question #74 : Forces

Example Question #75 : Forces

Example Question #642 : Newtonian Mechanics

Example Question #3 : Universal Gravitation

Example Question #1 : Net Force

Example Question #1 : Calculating Force

Example Question #1 : Net Force

All High School Physics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: