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AP Physics 1 : Specific Forces

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #642 : Newtonian Mechanics

The center of the moon is 3.84*10^8m from the earth.

The earth has a mass of $6.0*10^{24}kg$ .

Estimate lunar velocity.

Possible Answers:

None of these

$2.14*10^6\frac{m}{s}$

$.84*10^6\frac{m}{s}$

$1.04*10^6\frac{m}{s}$

$1.51*10^6\frac{m}{s}$

Correct answer:

$1.04*10^6\frac{m}{s}$

Explanation:

In an orbit, the centripital force will be equal to the gravitational force:

$m_m\frac{v^2}{r}=G\frac{m_e*m_m}{r^2}$

Where:

is the distance to the center of the earth from the center of the moon.

is the linear velocity of the moon

m_m is the mass of the moon

m_e is the mass of the earth

is the universal gravity constant, $6.674*10^{-11}N*\frac{m^2}{kg^2}$

Simplifying and plugging in values:

${v^2}=6.67*10^{-11}\frac{7.35*10^{22}}{1.75*10^6}$

Solving for :

$v=1.04*10^6\frac{m}{s}$

*note how the mass of the moon was irrelevant

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Example Question #643 : Newtonian Mechanics

Distance from earth to the moon: 384000 km

Mass of moon: $7.35*10^{22}kg$

Universal gravitation constant: $G=6.674*10^{-11}\frac{N*m^2}{kg^2}$

Estimate the gravitational force of the moon on a person of mass 101kg on the surface of the earth.

Possible Answers:

$2.45*10^{-3}N$

$8.54*10^{-3}N$

$6.18*10^{-3}N$

$3.36*10^{-3}N$

$8.81*10^{-3}N$

Correct answer:

$3.36*10^{-3}N$

Explanation:

Use the law of universal gravitation:

$F_{gravity}=G\frac{m_{1}m_{2}}{r^2}$

Convert to and plug in values:

$F_{gravity}=6.674*10^{-11}\frac{7.35*10^{22}*101}{(3.84*10^8)^2}$

$F_{gravity}=3.36*10^{-3}N$

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Example Question #681 : Ap Physics 1

Mass of moon: $7.35*10^{22}kg$

Moon radius: 1.7*10^6m

$G=6.674*10^{-11}N*\frac{m^2}{kg^2}$

A spacecraft is orbiting 50km above the surface of the moon. Determine the velocity of the spacecraft.

Possible Answers:

$1700\frac{m}{s}$

$1200\frac{m}{s}$

$1300\frac{m}{s}$

$1900\frac{m}{s}$

$1500\frac{m}{s}$

Correct answer:

$1700\frac{m}{s}$

Explanation:

In an orbit, the centripetal force will be equal to the gravitational force:

$m_s\frac{v^2}{r}=G\frac{m_s*m_m}{r^2}$

Where:

is the distance to the moon's center. This will be the sum of the radius of the moon and the distance above the moon's center

is the linear velocity

m_s is the mass of the ship

m_m is the mass of the moon

is the universal gravity constant, $6.674*10^{-11}N*\frac{m^2}{kg^2}$

Simplify and plug in values:

${v^2}=6.67*10^{-11}\frac{7.35*10^{22}}{1.75*10^6}$

Solve for :

$v=1700\frac{m}{s}$

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Example Question #681 : Ap Physics 1

Mass of Earth: $5.97*10^{24}kg$

Mass of Moon: $7.25*10^{22}kg$

Distance between Earth and Moon: 3.84*10^8m

If a spaceship were to travel in a straight line towards the moon, at approximately what distance from the Earth would the forces of gravity due to the earth and moon be equal in magnitude?

Possible Answers:

2.41*10^8m

None of these

2.12*10^8m

1.11*10^8m

3.41*10^8m

Correct answer:

3.41*10^8m

Explanation:

Use the following equations to set up our calculation:

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

F_E=F_M

$R_{ship-earth}+R_{ship-moon}=R_{moon-earth}$

$G\frac{m_s m_E}{(R_{ship-earth})^2}=G\frac{m_s m_M}{(R_{ship-moon})^2}$

Combine equations and simplify:

$\frac{m_E}{(R_{ship-earth})^2}=\frac{m_M}{(R_{earth-moon}-R_{ship-earth})^2}$

Plug in values:

$\frac{5.97*10^{24}}{(R_{ship-earth})^2}=\frac{7.25*10^{22}}{(3.84*10^8-R_{ship-earth})^2}$

Solve for $R_{ship-earth}$ by converting to a quadratic equation:

$0=((\frac{7.25*10^{22}}{5.97*10^{24}})-1)R_{ship-earth}^2+2*3.84*10^8*R_{ship-earth}-(3.84*10^8)^2$

Use the quadratic formula to solve:

$R_{ship-earth}=3.41*10^8m$

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Example Question #41 : Universal Gravitation

Pluto radius: 1.186*10^6 m

Pluto mass: $1.3*10^{22} kg$

Determine the gravity constant, $g_{pluto}$ on the surface of Pluto.

Possible Answers:

$g_{pluto}=-2.4\frac{m}{s^2}$

$g_{pluto}=-.616\frac{m}{s^2}$

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

$g_{pluto}=-3.1\frac{m}{s^2}$

$g_{pluto}=-14.8\frac{m}{s^2}$

Correct answer:

$g_{pluto}=-.616\frac{m}{s^2}$

Explanation:

Set both forms of gravitational force equal to each other.

$m_{object}g_{pluto}=-G\frac{m_{pluto} m_{object}}{r^2}$

Simplify:

$g_{pluto}=-G\frac{m_{pluto}}{r^2}$

Plug in values:

$g_{pluto}=-6.674*10^{-11}\frac{1.3*10^{22}}{(1.186*10^6)^2}$

$g_{pluto}=-.616\frac{m}{s^2}$

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Example Question #41 : Universal Gravitation

How would the linear velocity of the Moon be different if it's mass was doubled? Assume that the distance to the Earth stayed the same.

Possible Answers:

Doubled

Unchanged

Halved

Quadrupled

Quartered

Correct answer:

Unchanged

Explanation:

Set centripetal force equal to gravitational force:

$m_M\frac{v^2}{r}=G\frac{m_M m_E}{r^2}$

$\frac{v^2}{r}=G\frac{m_E}{r^2}$

The mass of the Moon, m_M cancels out, thus, there is no effect.

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Example Question #43 : Universal Gravitation

If on earth you have a weight, , what would your new weight be if you were standing on a planet with the same mass as earth, but with half the radius?

Possible Answers:

$\frac{1}{2}$ your earth weight.

The same as your earth weight.

times your earth weight

$\frac{1}{2}$ your earth weight.

times your earth weight.

Correct answer:

times your earth weight.

Explanation:

Your weight is a function of how much force is pulling you down towards the planet, not just your mass. To find force when your standing on a different planet, you would need to use Newton's Law of Universal Gravitation.

$F = - \frac{G M}{r^{2}}$

To compare the forces from each planet, you would set this equation equal to itself.

$- \frac{G M_{earth}}{r_{earth}^{2}} = - \frac{G M_{planet}}{r_{planet}^{2}}$

We then cancel the common terms, in this formula that's the negative sign, (the gravitational constant), and the mass of the earth/planet (because they're the same). After that we can substitute the radius of the new planet for half of the earth radius.

$r_{planet}=\frac{1}{2}{}r_{earth}$

We need to remember that when we square the radius in the Law of Universal Gravitation, we also need to square the $\frac{1}{2}$ , making it $\frac{1}{4}$ . Because the $\frac{1}{4}$ is in the denominator, we take the inverse of it, so you would feel times more force standing on the new planet. Therefore you would have times as much weight.

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Example Question #51 : Universal Gravitation

An electronic scale is used to find the mass of a lead cube at sea level. The scale and lead cube are then transported to the top of a mountain, over $20000\textup{ ft}$ above sea level. How does the weight reading compare to the weight given at sea level?

Possible Answers:

It is impossible to determine

It will be larger

None of these

It will be smaller

It will be the same

Correct answer:

It will be smaller

Explanation:

The electronic scale measures based on the normal force the scale provides to the object. This in turn is based on the force of gravity on the object by the earth.

$F_G=-G\frac{m_1 m_2}{r^2}$

As height above sea level increases, as does , the distance to the center of the Earth. As increases, F_G decreases. This would decrease the normal force which would decrease the reading on the scale.

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Example Question #83 : Specific Forces

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

Mass of moon: $7.3*10^{22}kg$

Radius of moon: 1737km

A spring of rest length 15cm is placed upright on the moon. A mass of 15kg is gently placed on top and the spring contracts by 1cm . Determine the spring constant.

Possible Answers:

None of these

$\frac{2415N}{m}$

$\frac{3388N}{m}$

$\frac{4449N}{m}$

$\frac{3751N}{m}$

Correct answer:

$\frac{2415N}{m}$

Explanation:

First, estimate the acceleration due to gravity close to the moon's surface:

F=ma

$F=-G\frac{m_1m_2}{r^2}$

Combining equations and solving for the acceleration:

$a_{mass}=-G\frac{m_{moon}}{r_{moon}^2}$

Converting to and pugging in values:

$a_{mass}=-6.67*10^{-11}\frac{7.3*10^{22}}{(1737*10^3)^2}$

$a_{mass}=-1.61\frac{m}{s^2}$

Using

$F_{total}=F_1+F_2+...$

$F_{gravity}=ma_{gravity}$

$F_{total}=kx+ma_{gravity}$

Solving for

$k=\frac{F_{total}-ma_{gravity}}{x}$

Converting to and plugging in values:

$k=\frac{0-15*-1.61}{.01}$

(Since the mass is at rest, the acceleration and thus the net force is zero)

$k=\frac{2415N}{m}$

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Example Question #53 : Universal Gravitation

A rope is used to accelerate a 5kg mass upwards at $2\frac{m}{s^2}$ . Determine the tension in the rope.

Possible Answers:

202N

90N

None of these

59N

18N

Correct answer:

59N

Explanation:

Using

F=ma

$F_{total}=F_1+F_2+...$

Combining equations:

$ma=F_{tension}+mg$

Solving for $F_{tension}$ and plugging in values:

$F_{tension}=5*2-5*(-9.8)$

$F_{tension}=59N$

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AP Physics 1 : Specific Forces

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #642 : Newtonian Mechanics

Example Question #643 : Newtonian Mechanics

Example Question #681 : Ap Physics 1

Example Question #681 : Ap Physics 1

Example Question #41 : Universal Gravitation

Example Question #41 : Universal Gravitation

Example Question #43 : Universal Gravitation

Example Question #51 : Universal Gravitation

Example Question #83 : Specific Forces

Example Question #53 : Universal Gravitation

All AP Physics 1 Resources

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