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AP Physics C: Mechanics : Forces

Study concepts, example questions & explanations for AP Physics C: Mechanics

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All AP Physics C: Mechanics Resources

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Example Questions

Example Question #1 : Calculating Weight

A large planet exerts a gravitational force five times stronger than that experienced on the surface of Earth. What is the weight of a 50kg object on this planet?

Possible Answers:

50N

2450N

2675N

250N

Correct answer:

2450N

Explanation:

The weight of the object on Earth's surface is:

$F_E=ma=(50kg)(9.8\frac{m}{s^2})=490N$

The force on the new planet is five times that on Earth, so we can simply multiply:

F=5(F_E)=5(490N)=2450N

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Example Question #2 : Calculating Weight

A space woman finds herself in an unkown planet with gravity $5\frac{m}{s^2}$ . If her weight on Earth is 500N, what is her weight on the unkown planet?

$g_{Earth}=10\frac{m}{s^2}$

Possible Answers:

5 N

250 N

500 N

225 N

25 N

Correct answer:

250 N

Explanation:

We know that the weight of an object is given by:

w = mg

is the mass of the object and is the gravitational acceleration of whatever planet the object happens to be on.

We know the gravity on the unkown planet, so the weight of the woman is given by:

$w_{u}=mg_{u}$

We need only to find the mass of the woman to solve the problem. Since the mass of the woman is constant, we can use the information about her weight on Earth to figure out her mass.

$m=\frac{w_{e}}{g_{e}}=\frac{500N}{10\frac{m}s^2{}}=50kg$

Use this mass to solve for her weight on the new planet.

$w_{u}=mg_{u}=(50kg)(5\frac{m}{s^2})=250N$

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Example Question #1 : Understanding Orbits

If $v_{escape}$ is the escape velocity from the surface of a planet of mass and radius . What is the velocity necessary for an object launched from the surface of a planet of the same mass, but with a radius that is .25R , to escape the gravitational pull of this smaller, denser planet?

Possible Answers:

$\frac{1}{4}v_{escape}$

$2v_{escape}$

$4v_{escape}$

Not enough information

$v_{escape}$ (the same velocity for both planets)

Correct answer:

$2v_{escape}$

Explanation:

Escape velocity is the velocity at which the kinetic energy of an object in motion is equal in magnitude to the gravitational potential energy. This allows us to set two equations equal to each other: the equation for kinetic energy of an object in motion and the equation for gravitational potential energy. Therefore:

$\frac{1}{2}mv^{2}=\frac{GMm}{r}$

$mv^{2}=\frac{2GMm}{r}$

$v^{2}=\frac{2GM}{r}$

$v=\sqrt{\frac{2GM}{r}}$

In this case, $v=v_{escape}$ . Substitute.

$v_{escape}=\sqrt{\frac{2GM}{r}}$

Since the radius of the new planet is 0.25 or $\frac{1}{4}$ the radius of the original planet, and they have the same masses, we can equate:

$v^{*}=\sqrt{\frac{2GM}{\frac{R}{4}}}=\sqrt{4}\sqrt{\frac{2GM}{R}}=2\sqrt{\frac{2GM}{R}}=2v_{escape}$

Where $v_{escape}$ is the escape velocity of the larger planet and $v^{*}$ is the escape velocity of the smaller planet.

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Example Question #31 : Forces

A comet is in an elliptical orbit about the Sun, as diagrammed below:

Eliptical orbit

The mass of the comet is very small compared to the mass of the Sun.

In the diagram above, how does the net torque on the comet due to the Sun's gravitational force compare at the two marked points?

Possible Answers:

It cannot be determined from the information given

The torque is zero at both a and b

The torque is the same at points a and b, but it is not zero

The torque is greater at point b

The torque is greater at point a

Correct answer:

The torque is zero at both a and b

Explanation:

Since the direction of the gravitational force is directly towards the center of the Sun, it lies in the plane of the comet's orbit. Since torque is $\small r\cdot F = Fr\sin \theta$ , the torque is always zero. That is why the comet's angular momentum is conserved in orbit.

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Example Question #111 : Ap Physics C

A satellite of mass is in an elliptical orbit about the Earth. It's velocity at perigee is $\small v_{p}$ and its orbital radius at perigee is $\small r_{p}$ . If the radius at apogee is $\small 2r_{p}$ , what is its velocity at apogee?

Possible Answers:

$\small 4v_p$

$\frac{v_{p}}{4}$

$\small \frac{v_{p}}{2}$

$\small v_{p}$

$\small 2v_p$

Correct answer:

$\small \frac{v_{p}}{2}$

Explanation:

Since the gravitational force cannot exert torque on the satellite about the Earth's center, angular momentum is conserved in this orbit:

$\small L_p=L_a$

$\small mv_pr_p=mv_ar_a$

$\small v_pr_p=v_a(2r_p)$

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Example Question #112 : Ap Physics C

A satellite is in an elliptical orbit about the Earth. If the satellite needs to enter a circular orbit at the apogee of the ellipse, in what direction will it need to accelerate?

Possible Answers:

This maneuver cannot be accomplished

Opposite the direction of travel

Away from Earth

Towards Earth

In the direction of travel

Correct answer:

In the direction of travel

Explanation:

Since the speed of the satellite at apogee is too low for a circular orbit at that orbital radius, the satellite needs to speed up to circularize the orbit.

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Example Question #1 : Newton's Laws And Force Diagrams

A $\small 3kg$ paint bucket hangs from the right end of a meter stick, oriented horizontally. The left end of the meter stick rests on a fulcrum so that it may rotate about that point. A rope is to be attached $\small 60cm$ from the left end, so that the system does not rotate. What is the minimum force this rope must support, assuming the meter stick is massless?

Possible Answers:

44N

20N

74N

29N

49N

Correct answer:

49N

Explanation:

Relevant equations:

$\tau = r \times F$

$\tau_{net} = I \alpha$

$\tau_{net}= \tau_{ccw}- \tau_{cw}$

Determine the clockwise torque caused by the bucket.

$\tau_{cw}=(1m)F_g = (1m)(3kg)(9.8\frac{m}{s^2})=29.4 N*m$

Write an expression for the counterclockwise torque caused by the rope.

$\tau_{ccw}=(0.60m)F_{rope}$

Combine the torque of the rope and the torque of the bucket into the net torque equation.

$\tau_{net}=(0.60m)F_{rope}-29.4N*m$

Since the system has no angular acceleration, net torque must be zero, allowing us to solve for the force of the rope.

$0 = (0.60m)F_{rope}-29.4N*m$

$F_{rope}=\frac{29.4N*m}{0.60m}$

$F_{rope}=49N$

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Example Question #32 : Forces

An object at rest will remain at rest unless acted on by a(n) __________.

Possible Answers:

internal force

external force

positive force

negative force

Correct answer:

external force

Explanation:

The correct answer is external force. External forces applied to an object will result in non-zero acceleration, causing the object to move. In contrast, internal forces contribute to the properties of the object and do not result in acceleration of the object.

Either positive or negative forces can result in the acceleration of an object. The sign on the force simply conveys information about its relative direction.

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Example Question #2 : Newton's Laws And Force Diagrams

A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?

$g=10\frac{m}{s^2}$

Rocket2

Possible Answers:

$20 \frac{m}{s^2}$

$0\frac{m}{s^2}$

$29 \frac{m}{s^2}$

$-20 \frac{m}{s^2}$

$-29 \frac{m}{s^2}$

Correct answer:

$20 \frac{m}{s^2}$

Explanation:

When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.

In addition, the rocket experiences the downward force of its own weight. This is given by:

$W = mg = 1000 kg * 10 \frac{m}{s^2} = 10000 N$

Rocket

We know that $F_{net} = ma$ .

We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.

We know that $F_{net} = F - W$ (since is directed upwards and is directed downwards).

$F_{net} = 30000 N - 10000 N = 20000 N$

Finally we solve for acceleration:

$a = \frac{F_{net}}{m}=\frac{20000N}{1000kg}=20\frac{m}{s^2}$

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Example Question #3 : Newton's Laws And Force Diagrams

3masses

Three boxes tied by two ropes move across a frictionless surface pulled by a force as shown in the diagram. Which of the following is an expression of the acceleration of the system?

Possible Answers:

$a = \frac{F}{m}$

$a = \frac{F}{4m}$

None of these

$a = \frac{F}{6m}$

$a = \frac{F}{3m}$

Correct answer:

$a = \frac{F}{6m}$

Explanation:

Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.

A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between and as $T_{1}$ . Similarly, we will label the tension experienced by the rope between and as $T_{2}$ .

For the box on the right (of mass ) we know that the force pulls to the right and the rope pulls it to the left with a tension of $T_{1}$ .

We get the equation: $F - T_{1} = 2ma$ . is the mass of the box and is its acceleration (which is the same for the other boxes).

For the middle box (of mass ) we have a rope pulling on it on the right with tension $T_{1}$ and a rope pulling on the left with tension $T_{2}$ .

So we get the equation: $T_{1}- T_{2}= ma$ . is the mass of the box and is its acceleration (which is the same for the other boxes).

Finally, for the box on the left (of mass ) we have only one rope pulling it to the right with a tension $T_{2}$ .

So we get the equation: $T_{2}=3ma$ . is the mass of the box and is its acceleration (which is the same for the other boxes).

Now it is just a matter of simple substitution. We have three equations:

$T_{2}=3ma$

$T_{1}- T_{2}= ma$

$F - T_{1} = 2ma$

From them we can get an expression for the force. Isolate $T_{1}$ in the second equation.

$T_{1}= ma + T_{2}$

Substitute the expression of $T_{2}$ from the first equation and simplify.

T_1= ma + 3ma = 4ma

Use this value of $T_{1}$ in the third equation to get an expression for the force.

$F = 2ma + T_{1} = 2ma + 4ma = 6ma$

So we have:

F = 6ma

Solve for acceleration.

$a = \frac{F}{6m}$

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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Forces

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #1 : Calculating Weight

Example Question #2 : Calculating Weight

Example Question #1 : Understanding Orbits

Example Question #31 : Forces

Example Question #111 : Ap Physics C

Example Question #112 : Ap Physics C

Example Question #1 : Newton's Laws And Force Diagrams

Example Question #32 : Forces

Example Question #2 : Newton's Laws And Force Diagrams

Example Question #3 : Newton's Laws And Force Diagrams

All AP Physics C: Mechanics Resources

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