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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

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

Example Questions

Example Question #45 : Forces

A 0.05kg ball falls off a cliff. What is the force of gravity on the ball?

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

Possible Answers:

49N

We need to know the time the ball is in the air in order to solve

We need to know the height of the cliff in order to solve

0.49N

-0.49N

Correct answer:

-0.49N

Explanation:

Newton's second law states:

F=ma

In this case the acceleration will be the constant acceleration due to gravity on Earth.

F=mg

Use the acceleration of gravity and the mass of the ball to solve for the force on the ball.

$F=(0.05kg)(-9.8\frac{m}{s^2})$

F=-0.49N

The answer is negative because the force is directed downward. Since gravity is always acting downward, a force due to gravity will always be negative.

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Example Question #45 : Forces

An astronaut weighs 150N on Earth. On a distant moon, she weighs 130N . What is the acceleration due to gravity on this moon?

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

Possible Answers:

$15.3\frac{m}{s^2}$

$8.5\frac{m}{s^2}$

$2.04\frac{m}{s^2}$

$13.3\frac{m}{s^2}$

$0.12\frac{m}{s^2}$

Correct answer:

$8.5\frac{m}{s^2}$

Explanation:

First we need to find the mass of the astronaut using Newton's second law.

F=ma

We know the total weight of the astronaut and the acceleration due to gravity on Earth, allowing us to solve for her mass.

F=mg

$150N=m*9.8\frac{m}{s^2}$

$\frac{150N}{9.8\frac{m}{s^2}}=m$

15.3kg=m

Now that we know her mass, we can look at her weight on the distant moon. We know her weight and mass, allowing us to solve for the acceleration due to gravity in this new environment.

F=ma

130N=15.3kg*a

$\frac{130N}{15.3kg}=a$

$8.5\frac{m}{s^2}=a$

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Example Question #12 : Understanding Gravity And Weight

The mass of the moon is less than that of Earth, causing it to have a gravitational acceleration less than $9.8\frac{m}{s^2}$ . Which of the following could be the weight of an object on the moon, if the object weighs 10N on Earth?

Possible Answers:

10N

15N

10.5N

Correct answer:

Explanation:

Newton's second law states that:

F=ma

We know from the problem that the acceleration due to gravity on the moon is less than the acceleration due to gravity on Earth. The mass of the object, however, will remain constant. The result is that the force of gravity on the object while on the moon will be less than the force on the object while on Earth.

$F_{g\ moon}<F_{g\ Earth}$

This means that the weight of the object while on the moon must be less than 10N . Since the object has a weight on Earth, however, we know that its weight on the moon cannot be zero. This would imply that either the acceleration due to gravity on the moon is zero, or that the mass is zero, neither of which is possible. This allows us to eliminate from the answers.

The only other option that is less than 10N is .

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Example Question #51 : Forces

An astronaut has a mass of 65kg and Mars has an acceleration due to gravity of $3.8\frac{m}{s^2}$ . What is her weight on Mars?

Possible Answers:

65N

247N

637N

17.1N

884N

Correct answer:

247N

Explanation:

Weight is a very specific force, determined by the acceleration due to gravity acting on a given mass. Using Newton's second law, we can see that weight will be equal to the equation:

F_g=mg

We are given the mass of the astronaut and the acceleration due to gravity on Mars. Using these values, we can calculate her weight on Mars.

$F_g=(65kg)(3.8\frac{m}{s^2})$

F_g=247N

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Example Question #1 : Universal Gravitation

Two satellites in space, each with a mass of 2000kg , are 1500m apart from each other. What is the force of gravity between them?

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

Possible Answers:

$3.12*10^{-1}N$

$1.19*10^{-10}N$

$2.66*10^{-11}N$

$6.22*10^{-8}N$

$8.87*10^{-10}N$

Correct answer:

$1.19*10^{-10}N$

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 constant, as well as the satellite masses and distance (radius). Using these values we can solve for the force.

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(2000kg) (2000kg)}{(1500m)^2})$

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{4000000kg^2}{2250000m^2})$

$F_G=1.19*10^{-10}N$

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Example Question #604 : Newtonian Mechanics

Two satellites in space, each with a mass of 1723kg , are 890m apart from each other. What is the force of gravity between them?

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

Possible Answers:

$2.5*10^{-20}N$

$2.2*10^{8}N$

$1.1*10^{3}N$

$1.25*10^{-19}N$

$2.5*10^{-10}N$

Correct answer:

$2.5*10^{-10}N$

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 constant, as well as the satellite masses and distance (radius). Using these values we can solve for the force.

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(1723kg )(1723kg)}{(890m)^2})$

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{2968729kg^2}{792100m^2})$

$F_G=2.5*10^{-10}N$

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Example Question #1 : Universal Gravitation

Two asteroids in space are in close proximity to each other. Each has a mass of $6.69*10^{15}kg$ . If they are 100,000m apart, what is the gravitational force between them?

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

Possible Answers:

$5.98*10^{12}N$

$3.01*10^{-8}N$

$1.49*10^{-32}N$

$1.11*10^{30}N$

$2.99*10^{11}N$

Correct answer:

$2.99*10^{11}N$

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 constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(6.69*10^{15}kg )(6.69*10^{15}kg )}{(100,000m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$F_G=2.99*10^{11}N$

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Example Question #2 : Universal Gravitation

Two asteroids in space are in close proximity to each other. Each has a mass of $6.69*10^{15}kg$ . If they are 100,000m apart, what is the gravitational acceleration that they experience?

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

Possible Answers:

$4.47*10^{-5}\frac{m}{s^2}$

$6.69*10^{15}\frac{m}{s^2}$

$5.12*10^{4}\frac{m}{s^2}$

$3.89*10^{10}\frac{m}{s^2}$

$2.99*10^{11}\frac{m}{s^2}$

Correct answer:

$4.47*10^{-5}\frac{m}{s^2}$

Explanation:

Given that F=ma , we already know the mass, but we need to find the force in order to solve for the acceleration.

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

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

We are given the constant, as well as the satellite masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{(6.69*10^{15}kg)(6.69*10^{15}kg)}{(100,000m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$F_G=2.99*10^{11}N$

Now we have values for both the mass and the force, allowing us to solve for the acceleration.

F=ma

$2.99*10^{11}N=(6.69*10^{15}kg)a$

$\frac{2.99*10^{11}N}{6.69*10^{15}kg}{}=a$

$4.47*10^{-5}\frac{m}{s^2}=a$

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Example Question #2 : Universal Gravitation

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass 5.33*10^8kg , are $10*10^{10}m$ apart. What is the gravitational force on the LARGER asteroid?

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

Possible Answers:

$3.55*10^{-6}N$

$4.74*10^{-6}N$

$2.53*10^{-5}N$

$4.61*10^{-10}N$

3.79*10^5N

Correct answer:

$2.53*10^{-5}N$

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 constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})$

$F_G=2.53*10^{-5}N$

It actually doesn't matter which asteroid we're looking at; the gravitational force will be the same. This makes sense because Newton's 3rd law states that the force one asteroid exerts on the other is equal in magnitude, but opposite in direction, to the force the other asteroid exerts on it.

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Example Question #3 : Universal Gravitation

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass 5.33*10^8kg are $10*10^{10}m$ apart. What is the gravitational force on the SMALLER asteroid?

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

Possible Answers:

$1.11*10^{12}N$

$4.74*10^{-14}N$

$2.53*10^{-5}N$

3.79*10^5N

$3.55*10^{-24}N$

Correct answer:

$2.53*10^{-5}N$

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 constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})$

$F_G=2.53*10^{-5}N$

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All High School Physics Resources

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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #45 : Forces

Example Question #45 : Forces

Example Question #12 : Understanding Gravity And Weight

Example Question #51 : Forces

Example Question #1 : Universal Gravitation

Example Question #604 : Newtonian Mechanics

Example Question #1 : Universal Gravitation

Example Question #2 : Universal Gravitation

Example Question #2 : Universal Gravitation

Example Question #3 : Universal Gravitation

All High School Physics Resources

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