AP Physics C: Mechanics : Forces

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

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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:

Correct answer:

Explanation:

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

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

Example Question #31 : Forces

A space woman finds herself in an unkown planet with gravity . If her weight on Earth is 500N, what is her weight on the unkown planet?

Possible Answers:

Correct answer:

Explanation:

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

 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:

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.

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

Example Question #101 : Mechanics Exam

If  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 , to escape the gravitational pull of this smaller, denser planet?

Possible Answers:

Not enough information

 (the same velocity for both planets)

Correct answer:

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:

In this case,  . Substitute.

Since the radius of the new planet is  or  the radius of the original planet, and they have the same masses, we can equate:

Where  is the escape velocity of the larger planet and  is the escape velocity of the smaller planet.

Example Question #11 : Gravity

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 greater at point a

The torque is greater at point b

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

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 , the torque is always zero. That is why the comet's angular momentum is conserved in orbit.

Example Question #11 : Gravity

A satellite of mass  is in an elliptical orbit about the Earth. It's velocity at perigee is  and its orbital radius at perigee is . If the radius at apogee is , what is its velocity at apogee?

Possible Answers:

Correct answer:

Explanation:

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

Example Question #1 : Understanding Orbits

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:

Opposite the direction of travel

In the direction of travel

Towards Earth

This maneuver cannot be accomplished

Away from Earth

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.

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

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  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:

Correct answer:

Explanation:

Relevant equations:

Determine the clockwise torque caused by the bucket.

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

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

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

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

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

Possible Answers:

positive force

negative force

internal force

external 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.

Example Question #1 : 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?

 

Rocket2

Possible Answers:

Correct answer:

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:

Rocket

We know that .

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  (since  is directed upwards and  is directed downwards).

 

Finally we solve for acceleration:

Example Question #111 : Mechanics Exam

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:

None of these

Correct answer:

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 . Similarly, we will label the tension experienced by the rope between  and  as .

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 .

2m

We get the equation:  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  and a rope pulling on the left with tension .

M

So we get the equation:  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 .

3m

So we get the equation:  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:

From them we can get an expression for the force. Isolate  in the second equation.

Substitute the expression of  from the first equation and simplify.

Use this value of  in the third equation to get an expression for the force.

So we have:

Solve for acceleration.

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