High School Physics : Forces

Study concepts, example questions & explanations for High School Physics

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Example Questions

Example Question #1 : Understanding Gravity And Weight

A ball that weighs \(\displaystyle 4N\) on Earth weighs \(\displaystyle 19N\)on a recently discovered planet. What is the force of gravity on this new planet?  

Give your answer with the correct number of significant figures.

Possible Answers:

\(\displaystyle 50\frac{m}{s^2}\)

\(\displaystyle 50.0\frac{m}{s^2}\)

\(\displaystyle 48.0\frac{m}{s^2}\)

\(\displaystyle 48\frac{m}{s^2}\)

\(\displaystyle 47.5\frac{m}{s^2}\)

Correct answer:

\(\displaystyle 50\frac{m}{s^2}\)

Explanation:

Weight is defined as the force of gravity on an object. We can use Newton's second law to write an equation for weight.

\(\displaystyle F_g=mg\)

If the ball weighs \(\displaystyle 4N\) on Earth, then its mass can be found using this equation and the acceleration of gravity on Earth.

\(\displaystyle 4N = m(9.8\frac{m}{s^2})\)

\(\displaystyle m = \frac{4N}{(9.8\frac{m}{s^2})} = 0.40816 kg\)

Use this mass and the given weight on the new planet to find the acceleration of gravity on this new planet. Though our initial values (and thus our final values) only allow one significant figure, we will not round until the end of all calculations. This ensures that we preserve accuracy before adjusting for precision.

\(\displaystyle 19N = (0.40816kg)a\)

\(\displaystyle \frac{19N}{0.40816kg} = 46.55\frac{m}{s^2}\)

Adjust this value to one significant figure by rounding up. The zero in the tens place is before the decimal, and is not considered significant.

\(\displaystyle 46.55\frac{m}{s^2}\approx 50\frac{m}{s^2}\)

 

Example Question #41 : Forces

An object is placed in the direct center of the Earth. What would be the perceived weight of the object?

Possible Answers:

The object would have infinite weight downward

The object would have infinite weight upward

The weight of the object would be equal to its weight at the Earth's surface

We must know the mass of the object to draw a conclusion

The object would have no weight

Correct answer:

The object would have no weight

Explanation:

We must use Newton's law of universal gravitation to solve this question.

\(\displaystyle F_G=G\frac{m_1m_2}{r^2}\)

There are three variables that really change the force of gravity: the mass of each object and the distance between the bodies.

The important thing to recognize here, though, is that when an object is in the center of the earth, the mass of the earth is distributed symmetrically all around it. It's like being in the center of a giant bubble. Because the mass is symmetrically distributed, the mass that is trying to pull the object in each direction is equal. Essentially, the mass pulling upward cancels out the mass pulling downward, and the mass pulling right cancels out the mass pulling left.

This happens for the entirety of the circle, leaving you with a net force of zero acting upon the object.

Example Question #23 : Specific Forces

A satellite is in orbit \(\displaystyle 3.7*10^7m\) above the Earth. What is the relationship between the acceleration due to gravity on the satellite, versus the acceleration due to gravity on the Earth's surface?

Possible Answers:

Acceleration due to gravity on the satellite will be \(\displaystyle \small 0\frac{m}{s^2}\)

Acceleration due to gravity on the satellite will be less than \(\displaystyle \small g\)

Acceleration due to gravity on the satellite will be equal to \(\displaystyle \small g\)

Acceleration due to gravity on the satellite will be greater than \(\displaystyle \small g\)

We would need to know the mass of the satellite to solve

Correct answer:

Acceleration due to gravity on the satellite will be less than \(\displaystyle \small g\)

Explanation:

For this problem, we are comparing the force of gravity on the surface, or weight, to the force of gravity on the satellite. We can use Newton's second law to find the weight of the satellite, and the law of universal gravitation to find the gravity on the satellite. These two terms will be equal to one another.

\(\displaystyle W=mg\)

\(\displaystyle F_G=G\frac{m_1m_2}{r^2}\)

\(\displaystyle mg=G\frac{m_1m_2}{r^2}\)

Let's call \(\displaystyle m_1\) the Earth and \(\displaystyle m_2\) the mass of the satellite.

\(\displaystyle m_sg=G\frac{m_Em_s}{r^2}\)

Notice that the masses of the satellite cancel out.

\(\displaystyle g=G\frac{m_E}{r^2}\)

This formula gives us the acceleration of gravity in terms of the mass of the Earth and the distance from the Earth's center. We can write two separate equations, one for the surface and one for the satellite. Since the mass of the Earth doesn't change and \(\displaystyle G\) is a constant, the only variable that can change is \(\displaystyle r\), the distance between the objects.

\(\displaystyle g_{surface}=G\frac{m_E}{r_E^2}\)

\(\displaystyle g_{satellite}=G\frac{m_E}{(r_E+h)^2}\)

On Earth, \(\displaystyle r\) is the radius of the earth. For the satellite, \(\displaystyle r\) is the radius of the Earth PLUS the orbiting distance; therefore \(\displaystyle r_s>r_E\). Because we are dividing by our \(\displaystyle r^2\), a greater \(\displaystyle r\) gives us a smaller \(\displaystyle g\).

The satellite in space will have a smaller acceleration due to gravity. It will not be zero, but it will be smaller than the acceleration on the surface.

Example Question #511 : High School Physics

\(\displaystyle 2500kg\) satellite orbits \(\displaystyle 3.8*10^8m\) above the Earth. What is the satellite's acceleration due to gravity?

\(\displaystyle m_E=5.97*10^{24}kg\)

\(\displaystyle r_E=6.37*10^5m\)

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

Possible Answers:

\(\displaystyle 2.7*10^{-3}\frac{m}{s^2}\)

\(\displaystyle 9.81\frac{m}{s^2}\)

\(\displaystyle 6.87\frac{m}{s^2}\)

\(\displaystyle 1.32*10^{-5}\frac{m}{s^2}\)

\(\displaystyle 6.24*10^8\frac{m}{s^2}\)

Correct answer:

\(\displaystyle 2.7*10^{-3}\frac{m}{s^2}\)

Explanation:

To solve this problem, use the law of universal gravitation.

\(\displaystyle F_G=G\frac{m_1m_2}{r^2}\)

Remember that \(\displaystyle r\) 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.

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

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

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

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

\(\displaystyle F_G=6.87N\)

Now that we know the force, we can find the acceleretion. Remember that weight is equal to the mass times acceleration due to gravity.

\(\displaystyle F_G=W=m*g\)

Set our two forces equal and solve for the acceleration.

\(\displaystyle 6.87N=m_s*g\)

\(\displaystyle 6.87N=2500kg*g\)

\(\displaystyle 2.7*10^{-3}\frac{m}{s^2}=g\)

Example Question #42 : Forces

\(\displaystyle 2500kg\) satellite orbits \(\displaystyle 3.8*10^8m\) above the Earth. What is the gravitational acceleration on the Earth caused by the satellite?

\(\displaystyle m_E=5.97*10^{24}kg\)

\(\displaystyle r_E=6.37*10^5m\)

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

Possible Answers:

\(\displaystyle 6.87\frac{m}{s^2}\)

\(\displaystyle 9.81\frac{m}{s^2}\)

\(\displaystyle 0\frac{m}{s^2}\)

\(\displaystyle 2.7*10^{-3}\frac{m}{s^2}\)

\(\displaystyle 1.15*10^{-24}\frac{m}{s^2}\)

Correct answer:

\(\displaystyle 1.15*10^{-24}\frac{m}{s^2}\)

Explanation:

To solve this problem, use the law of universal gravitation.

\(\displaystyle F_G=G\frac{m_1m_2}{r^2}\)

Remember that \(\displaystyle r\) 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.

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

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

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

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

\(\displaystyle F_G=6.87N\)

Now that we know the force, we can find the acceleretion. Remember that weight is equal to the mass times acceleration due to gravity.

\(\displaystyle W=m*g\).

Set our two forces equal and solve for the acceleration.

\(\displaystyle 6.87N=m_s*g\)

\(\displaystyle 6.87N=5.97*10^{24}kg*g\)

\(\displaystyle \frac{6.87N}{5.97*10^{24}kg}=g\)

\(\displaystyle 1.15*10^{-24}\frac{m}{s^2}=g\)

Example Question #21 : Specific Forces

A woman stands on the edge of a cliff and drops two rocks, one of mass \(\displaystyle m\) and one of \(\displaystyle 3m\), from the same height. Which one experiences the greater force?

Possible Answers:

The rock with mass \(\displaystyle m\)

We need to know the height of the cliff to determine the answer

They both experience the same force

We need to know the volume of the rocks to determine the answer

The rock with mass \(\displaystyle 3m\)

Correct answer:

The rock with mass \(\displaystyle 3m\)

Explanation:

The formula for force is given by Newton's second law:

\(\displaystyle F=ma\)

Both rocks will experience the same acceleration, \(\displaystyle g\), or the acceleration due to gravity.

\(\displaystyle F_g=mg\)

Use the mass of each rock in this equation to find which rock experiences a greater force.

\(\displaystyle F_{g,m}=(m)g\)

\(\displaystyle F_{g,3m}=(3m)g=3(mg)\)

We can see that the force on the rock with mass of \(\displaystyle 3m\) is equal to three times for the force on the rock with mass of \(\displaystyle m\). The heavier rock experiences the greater force.

\(\displaystyle F_{g,3m}=3(F_{g,m})\)

Example Question #46 : Forces

A woman stands on the edge of a cliff and drops two rocks, one of mass \(\displaystyle m\) and one of \(\displaystyle 3m\), from the same height. Which one experiences the greater acceleration?

Possible Answers:

We need to know the density of the rocks in order to solve

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

The rock with mass \(\displaystyle 3m\)

The rock with mass \(\displaystyle m\)

They experience the same acceleration

Correct answer:

They experience the same acceleration

Explanation:

Even though the rocks have different masses, the acceleration on both will be \(\displaystyle g\), the acceleration due to gravity. We can look at Newton's second law to see the force experienced by the rocks:

\(\displaystyle F=ma\)

When objects are in free-fall, the acceleration will be equal to the acceleration from gravity, regardless of the mass of the object.

\(\displaystyle F_g=mg\)

Example Question #11 : Understanding Gravity And Weight

\(\displaystyle 0.05kg\) ball falls off a cliff. What is the force of gravity on the ball?

\(\displaystyle g=-9.8\frac{m}{s^2}\)

Possible Answers:

\(\displaystyle 0.49N\)

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

\(\displaystyle 49N\)

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

\(\displaystyle -0.49N\)

Correct answer:

\(\displaystyle -0.49N\)

Explanation:

Newton's second law states:

\(\displaystyle F=ma\)

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

\(\displaystyle F=mg\)

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

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

\(\displaystyle F=-0.49N\)

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.

Example Question #41 : Forces

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

\(\displaystyle g=-9.8\frac{m}{s^2}\)

Possible Answers:

\(\displaystyle 15.3\frac{m}{s^2}\)

\(\displaystyle 8.5\frac{m}{s^2}\)

\(\displaystyle 0.12\frac{m}{s^2}\)

\(\displaystyle 2.04\frac{m}{s^2}\)

\(\displaystyle 13.3\frac{m}{s^2}\)

Correct answer:

\(\displaystyle 8.5\frac{m}{s^2}\)

Explanation:

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

\(\displaystyle F=ma\)

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

\(\displaystyle F=mg\)

\(\displaystyle 150N=m*9.8\frac{m}{s^2}\)

\(\displaystyle \frac{150N}{9.8\frac{m}{s^2}}=m\)

\(\displaystyle 15.3kg=m\)

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.

\(\displaystyle F=ma\)

\(\displaystyle 130N=15.3kg*a\)

\(\displaystyle \frac{130N}{15.3kg}=a\)

\(\displaystyle 8.5\frac{m}{s^2}=a\)

Example Question #21 : Specific Forces

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

Possible Answers:

\(\displaystyle 10N\)

\(\displaystyle 10.5N\)

\(\displaystyle 3N\)

\(\displaystyle 0N\)

\(\displaystyle 15N\)

Correct answer:

\(\displaystyle 3N\)

Explanation:

Newton's second law states that:

\(\displaystyle F=ma\)

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.

\(\displaystyle F_{g\ moon}< F_{g\ Earth}\)

This means that the weight of the object while on the moon must be less than \(\displaystyle 10N\). 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 \(\displaystyle 0N\) from the answers.

The only other option that is less than \(\displaystyle 10N\) is \(\displaystyle 3N\).

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