### All AP Physics 1 Resources

## Example Questions

### Example Question #101 : Fundamentals Of Force And Newton's Laws

Which answer choice below includes only scalar quantities?

**Possible Answers:**

Distance, speed, time

Acceleration, energy, displacement

Force, time, velocity

Velocity, displacement, force

Displacement, time, acceleration

**Correct answer:**

Distance, speed, time

Scalar quantities are those that can be described with magnitude only, as opposed to vectors, which include both magnitude and direction components. Distance, speed, and time are all scalars. Displacement is not a scalar, as it involves both the distance and the direction moved from a starting point. Velocity also includes a direction component, and is therefore a vector quantity.

### Example Question #331 : Forces

A person who weighs 700 N is standing on a scale that is on the floor of an elevator lift. If the elevator accelerates downward at , what will the scale read?

**Possible Answers:**

**Correct answer:**

According to Newton's second law,

There are two forces acting on the person standing on the scale in the elevator. Those two forces are the person's weight due to gravity (which acts in a downward direction), and the normal force from standing on the scale (which acts in an upward direction). So, the net force equation can be written as:

represents the person's weight, which was given in the problem to be 700 N. is the normal force due to standing on the scale; this is what we are trying to find, because it is what the scale would read. We subtract the two forces because they point in opposite directions, and so one must be positive and the other negative.

We also know the acceleration, , is .

We were not given the mass of the person, but we can find it because we know the person's weight is 700 N. The equation that relates a person's mass to their weight near the surface of the Earth is:

Now we can solve for the normal force by using the mass we found in the original equation above:

### Example Question #331 : Forces

What units make up a Newton?

**Possible Answers:**

**Correct answer:**

The answer is exemplified by the centripetal force equation:

.

When we combine the units of each of the equation's components, we arrive at

which, when simplified, becomes

### Example Question #331 : Forces

Two toy cars are moving on a frictionless track. Car 1 is moving at when it collides into Car 2, which was at rest. After the collision, Car 2 moves at and Car 1 moves at . If Car 1 weighs 0.05 kg, how much does Car 2 weigh?

**Possible Answers:**

**Correct answer:**

First, lets identify the given information:

Since this is a collision problem, we can apply the law of conservation of momentum:

Because Car 2 is initially at rest, this equation can be simplified:

The equation can then be rearranged to solve for the mass of Car 2:

By plugging in the known values, we can determine the mass of Car 2:

### Example Question #941 : Ap Physics 1

Consider a cube with a mass of and sides of . Calculate the pressure, , on the floor from such a cube.

**Possible Answers:**

**Correct answer:**

To calculate the pressure on the floor, we need to know the force acting on the floor, as well as the area over which it acts.

The force acting is simply the weight of the cube,

The area is the area of one side of the cube as it rests on the floor,

Therefore, the pressure is given as:

Plug in known values and solve.

### Example Question #1 : Other Force Fundamentals

During a mission to the Moon, astronauts were traveling towards the moon at , however, for most of the flight to and from the Moon, they were not slammed into their seats, and could move freely about the aircraft, why is that?

**Possible Answers:**

Because they were being pulled towards the Moon

None of these

Because there is no gravity in space

Because they were not accelerating

The astronauts had very strong muscles and could push themselves up

**Correct answer:**

Because they were not accelerating

Once the spacecraft had reached a constant speed, so had it's occupants. Thus, the astronauts were also moving at and they were experiencing no forces from their seats.

Thus, the astronauts were experiencing zero net force.

### Example Question #337 : Forces

Three forces act on an object that is in equilibrium and suspended in air.* * is ** **pulling on the object from the positive horizontal axis. * *is pulling on the object from the negative horizontal axis.

**is acting on the object straight down on the negative vertical axis. What is the magnitude of**

*?*

**Possible Answers:**

**Correct answer:**

and** **cancel each other out in the horizontal direction due to the fact that they are equal and opposite about the vertical axis. Now that the resultant force in the x-direction is zero, we must find what

**must be in order to zero out the other two forces' y-components, which are:**

These forces are pointing in the positive y-direction (pulling the object up). Therefore, in order to have a resultant force of zero in the vertical direction, there must be a downward force of

Since we are asked for the magnitude, we are not particularly concerned with the sign of this value.

### Example Question #1 : Other Force Fundamentals

A cowboy is lassoing a cow at the rodeo. Identify the force that is causing the object to travel in a circular path.

**Possible Answers:**

Electrical

Tension

Spring

Gravity

Normal

**Correct answer:**

Tension

In these conceptual problems, it is always best to draw an accurate force diagram. In string problems, there is always a tension force, and here, it is causing the circular motion. The tension force causes the lasso to swing in a circle and pulls it toward the center of the circle.

### Example Question #1 : Other Force Fundamentals

Which of the following is not one of Newton's fundamental laws?

**Possible Answers:**

All of these correctly represent Newton's laws

An object in motion will remain in that state of motion unless an outside force acts on that object

The acceleration experienced by an object is directly proportional to the force applied to that object, and inversely proportional to the mass of that object

In order for an object to be in motion, a force needs to act on it

For every action, there will always be an opposite reaction equal in magnitude

**Correct answer:**

In order for an object to be in motion, a force needs to act on it

To answer this question, we'll need to be familiar with Isaac Newton's fundamental laws of physics, which forms the basis of classical mechanics.

Newton's first law states that an object in motion will remain in that state of motion indefinitely, unless an outside force acts on this object.

Newton's second law can be stated as follows:

This expression tells us that the acceleration experienced by an object is directly proportional to the applied force and inversely proportional to the object's mass.

Newton's third law states that for every action, there is always an equal and opposite reaction. Another way of saying this is that if one object were to exert a force on a second object, then the second object would exert the same force against the first object.

One of the answer choices states the following

"In order for an object to be in motion, a force needs to act on it."

While this may seem intuitive based on our everyday experience, and it may also show some resemblance to Newton's first law, this statement is actually incorrect. **Force is not responsible for motion - force is responsible for changes in motion.**

### Example Question #1 : Other Force Fundamentals

You and your friend are pushing on a box with a force parallel to the ground of 40 N. The box weighs 5 kg, and the coefficient of kinetic friction between the box and the ground is 0.75. What is the net force on the box in the horizontal direction?

**Possible Answers:**

**Correct answer:**

The force due to kinetic friction is calculated by the equation

where is the normal force on the box from the ground. The frictional force is in the opposite direction of the force applied by you and your friend. Therefore, the net force can be calculated:

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