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High School Physics : Forces

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Example Questions

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Example Question #31 : Universal Gravitation

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:

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

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

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

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

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

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

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

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

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

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

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Example Question #71 : Specific 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:

$2.7*10^{-3}s$

5.33*10^7s

5420s

2.84*10^8s

$1.71*10^{11}s$

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.

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

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

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

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

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Example Question #31 : Universal Gravitation

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:

$2.23*10^{18}N$

$1.24*10^{20}N$

$1.24*10^{19}N$

$4.7*10^{-1}N$

$1.86*10^{30}N$

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.

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

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

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

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

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

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Example Question #34 : Universal Gravitation

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:

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

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

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

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

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

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

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

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

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

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Example Question #681 : Ap Physics 1

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:

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

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

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

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

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

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

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

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

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

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Example Question #682 : Ap Physics 1

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:

$8.34*10^{31}kg$

$2.87*10^{-2}kg$

$6.46*10^{18}kg$

$3.26*10^{22}kg$

$3.87*10^{26}kg$

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.

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

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

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

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

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

$3.87*10^{26}=m_2$

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Example Question #5 : 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:

2022.44kg

1602.98kg

1.989kg

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

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Example Question #1 : Net Force

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

Possible Answers:

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

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

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

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

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

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)

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

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

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Example Question #2 : 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:

0.10N

1.24N

1.21N

1.10N

1.8N

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.

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

$\vec{F}=1.24N$

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Example Question #1 : Calculating 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:

1.41kg

12.1kg

14.76kg

15.12kg

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

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High School Physics : Forces

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #31 : Universal Gravitation

Example Question #71 : Specific Forces

Example Question #31 : Universal Gravitation

Example Question #34 : Universal Gravitation

Example Question #681 : Ap Physics 1

Example Question #682 : Ap Physics 1

Example Question #5 : Universal Gravitation

Example Question #1 : Net Force

Example Question #2 : Calculating Force

Example Question #1 : Calculating Force

All High School Physics Resources

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