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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #31 : Specific Forces

A girl is standing on two feet on the ground. She then lifts her right foot in the air. How does this change the normal force affecting her?

Possible Answers:

None of these

It will be doubled

It will be quadrupled

No change

It will be cut in half

Correct answer:

No change

Explanation:

Normal force is independent of surface area, and only dependent on the opposite force being applied, in this case gravity. While the pressure on her foot may change, the force of gravity pushing her down stays the same.

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Example Question #32 : Forces

A 68 kg person stands on solid ground without moving. How much force is exerted on the person by the ground? Assume gravity is $9.81\frac{m}{s^2}$ .

Possible Answers:

68 N

.667 N

667 N

24 N

34 N

Correct answer:

667 N

Explanation:

The ground exerts an equal force as the person exerts on it, which is the normal force. This is simply the mass of the person times the acceleration due to gravity which is $9.81\frac{m}{s^2}$ . This ends with the normal force and the force exerted by the ground on the person being 667 N .

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Example Question #33 : Forces

A 35kg block remains stationary on the ground. Assuming that the acceleration due to gravity, , is $9.8 \frac{m}{s^2}$ , and that the only forces acting on the block are the force of gravity and normal force, what is the value of the normal force exerted by the ground on the block?

Possible Answers:

35J

35N

73.6N

343N

343J

Correct answer:

343N

Explanation:

This question tests your understanding of the concept of normal force. In a block that remains stationary, in which the only two forces acting upon it are the force of gravity, and normal force, you therefore know that the normal force must be equal to the force of gravity, since the block is not accelerating or decelerating (i.e. the sum of the forces is equal to zero).

Thus, we can set the normal force equal to the force of gravity, and solve as follows:

$F_{N} = F_{g}$

$F_{N} = m * g$

$F_{N} = 35 kg * 9.8 \frac{m}{s^{2}}$

$F_{N} = 343 N$

Therefore, the normal force exerted on the block by the ground is 343N .

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Example Question #1 : Universal Gravitation

A certain planet has three times the radius of Earth and nine times the mass. How does the acceleration of gravity at the surface of this planet (a_g) compare to the acceleration at the surface of Earth (g)?

Possible Answers:

$\frac{g}{9}$

$\frac{g}{3}$

Correct answer:

Explanation:

The acceleration of gravity is given by the equation $a_{g} = \frac{GM}{r^{2}}$ , where G is constant.

For Earth, $a_{g} = \frac{GM_{earth}}{r_{earth}^{2}} = g$ .

For the new planet,

$a_{g} = \frac{G(9M_{earth})}{\left ( 3r_{earth} \right )^2} = \frac{G(9M_{earth})}{9r_{earth{}}^2} = \frac{GM_{earth}}{r_{earth}^{2}}} = g$ .

So, the acceleration is the same in both cases.

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Example Question #31 : Forces

A new planet is discovered with mass $3.21\cdot 10^{23}kg$ and with a diameter of $8.53\cdot 10^{3}km$ . What is the lowest escape velocity required to escape this planet's gravitational pull?

$G=6.67\cdot10^{-11}\frac{N\cdot m^2}{kg^2}$

Possible Answers:

$7862\frac{m}{s}$

$942\frac{m}{s}$

$1286\frac{m}{s}$

$3169\frac{m}{s}$

$4753\frac{m}{s}$

Correct answer:

$3169\frac{m}{s}$

Explanation:

The equation to calculate the escape velocity from a planet is

$v_{e}=\sqrt{\frac{2GM_{p}}{R_{p}}}$

The diameter of the planet is given and can be divided by two to determine the radius of the planet. By plugging in the given values, the escape velocity threshold can be determined:

$v_{e}=\sqrt{\frac{2(6.67\cdot10^{-11}\frac{N\cdot m^2}{kg^2})(3.21\cdot10^{23}kg)}{\frac{(8.53\cdot10^{6}m)}{}2}}=3169\frac{m}{s}$

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Example Question #33 : Forces

There are two isolated stars orbiting each other. The first star has a mass of $2\times10^{12}kg$ and the second star has a mass of $4.5\times10^{14}kg$ . If the stars are 2,000km away, what is the gravitational force felt by the first star?

$G=6.67\times10^{-11}\frac{m^3}{kg\cdot s}$

Possible Answers:

$1.50\mathrm{x}10^9N$

$1.50\mathrm{x}10^4N$

$1.50\mathrm{x}10^{14}N$

None of the other answers

Correct answer:

$1.50\mathrm{x}10^4N$

Explanation:

We need to know Newton's law of universal gravitation to solve this problem:

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

It is important to note that both suns will feel the same gravitational force. However, since they have different masses, they will accelerate at different rates.

Plugging in the variables we have, we get:

$F = 6.67\mathrm{x}10^{-11} \frac{2\mathrm{x}10^{12}kg \cdot 4.5\mathrm{x}10^{14}kg}{(2\mathrm{x}10^6m)^2}$

$F = 1.50\mathrm{x}10^4N$

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Example Question #2 : 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:

$6.22*10^{-8}N$

$2.66*10^{-11}N$

$1.19*10^{-10}N$

$8.87*10^{-10}N$

$3.12*10^{-1}N$

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 #41 : Specific Forces

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:

$2.2*10^{8}N$

$2.5*10^{-10}N$

$1.1*10^{3}N$

$2.5*10^{-20}N$

$1.25*10^{-19}N$

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 #4 : 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:

1.989kg

2022.44kg

1602.98kg

166.28kg

2405.33kg

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 : 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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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #31 : Specific Forces

Example Question #32 : Forces

Example Question #33 : Forces

Example Question #1 : Universal Gravitation

Example Question #31 : Forces

Example Question #33 : Forces

Example Question #2 : Universal Gravitation

Example Question #41 : Specific Forces

Example Question #4 : Universal Gravitation

Example Question #1 : Universal Gravitation

All AP Physics 1 Resources

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