Force & Motion

Question 1

Define:

force

Answer 1

It is the change in velocity per unit time that is experienced by a given mass. Force can also be thought of as a pull, push, or other action exerted upon an object.

Note that force can be positive or negative, depending on the direction of the velocity.

Question 2

What quantity is measured in newtons?

Answer 2

• Newton (N) is the SI unit for force.
• 1 Newton = 1 kg*m/s².

Question 3

How is force related to change in momentum?

Answer 3

Force and change in momentum are directly proportional.

Change in momentum, or Δp, is equal to the product of force and time.

Question 4

Describe:

Newton’s first law of motion

Answer 4

Also called the Law of Inertia, it states that an object in motion will continue with constant velocity unless acted on by a net force.

Similarly, an object at rest will continue to remain at rest until acted on by a net force.

Question 5

What must be true about the acceleration of an object, if all forces acting on it cancel?

Answer 5

The object has zero acceleration.

Since all forces cancel, there is no net force and velocity will not change. No change in velocity means no acceleration.

Question 6

In Newton’s second law of motion, what relationship exists between force, mass, and acceleration in?

Answer 6

F_net = ma

Net force and acceleration are both vectors, and will be pointing in the same direction.

Question 7

What change in force is required to make an object move with twice its original acceleration?

Answer 7

Twice the original force must be applied.

According to Newton’s second law, F=ma, net force and acceleration are directly proportional.

Question 8

How does Newton’s third law of motion describe the forces between two objects?

Answer 8

F_1on2 = -F_2on1

For every force exerted by one object on a second, an equal and opposite force is exerted by the second back on the first.

Question 9

An orange exerts a 5N force on an apple in free space. What magnitude of force must be exerted by the apple on the orange?

Answer 9

5N

From Newton’s third law, every force exerted must have an equal and opposite force. The negative sign is already factored in, since the question specified direction.

Question 10

What formula is used to calculate gravitational force?

Answer 10

F_g = Gm₁m₂ / r²

Where:

G = gravitational constant (N*m²/kg²)
m₁ and m₂ = masses (kg)
r = distance between masses (m)

Question 11

What change in gravitational force between two objects must take place for the distance between them to double?

Answer 11

Force must be decreased to one-quarter of its previous value.

Since F is proportional to 1/r², doubling r will reduce the force by a factor of 4.

Question 12

What change in gravitational force will occur between two objects if both the distance between them and the mass of each object halves?

Answer 12

No change in force will occur.

Since each mass is directly proportional to force, halving each mass will result in 1/4 the original force. But, since force is proportional to 1/r², halving r will result in a force that is 4 times greater. The net change in F is the product of both factors: 4 (1/4) = 1, or no change.

Question 13

On Earth, what convenient relationship can be demonstrated for the force on a mass due to gravity?

Answer 13

F = mg

Where:

F = force (N)
m = mass (kg)
g = acceleration due to gravity (9.8 m/s²)

Question 14

What is the magnitude of the force acting downward on an apple with a mass of 0.2 kg on Earth?

Answer 14

2N

F = mg= (0.2)(9.8) ≈ 2N

Note that if the apple is resting on the ground, it will also be subject to an equivalent normal force pointing upwards.

Question 15

Define:

weight

Answer 15

It is the term for the force on an object due to gravity.

W = mg

Weight is often confused with mass; an object with one weight on Earth will have a different weight on the moon, but its mass will remain constant.

Question 16

What will the proportional weight of an object be on the moon, if the moon has 1/6 of Earth’s gravity?

Answer 16

The object will have 1/6 the weight it had on Earth.

Since W is proportional to the acceleration due to gravity, moving to an area with 1/6 g will produce 1/6 W.

Question 17

How could an object’s percieved (or instantaneous) weight seemingly change?

Answer 17

It is the force pushing back on an object against gravity. This is a specific type of normal force.

A person might experience a decreased instantaneous weight if the ground falls away beneath them, such as in an elevator accelerating downwards.

Question 18

How will the instantaneous weight compare for a person in an elevator that is accelerating upwards vs. at rest?

Answer 18

Instantaneous weight while accelerating upwards will be greater than at rest.

This happens because the upward acceleration adds to the force that the person experiences pushing up on him.

If the elevator were accelerating downwards, the person would feel less weight. In fact, if the elevator could accelerate downwards at 9.8 m/s², the person would experience weightlessness.

Question 19

Define:

normal force

Answer 19

When two objects are touching, an opposing force exists between the two objects and perpendicular to the surface in contact.

Often, on the MCAT, normal force opposes the force due to gravity.

Question 20

What is the normal force for a book with a mass of 0.5kg sitting stationary on a table?

Answer 20

5N directed from the table on the book

The force pulling down on the book due to gravity is F = mg = (0.5)(10) = 5 N downwards on the table. Since the book is stationary, we know that normal force is equal and opposite; hence it must also be 5 N in magnitude.

Question 21

What relationship exists between normal force and friction for a static object?

Answer 21

It is a force that opposes any two surfaces in contact from sliding over each other. The relationship involved is:

f_s ≤ μ_sF_N

Where:

f_s = force of static friction (N)
μ_s = coefficient of static friction (unitless)
F_N = normal force (N)

Question 22

Describe what happens to a box if a person pushes on it horizontally with a force of 1/2 μ_sF_N, μ_sF_N, and 2μ_sF_N, respectively.

Answer 22

• In the first case, no motion occurs. 1/2 μ_sF_N will be exerted back on the box by static friction.
• In the second case, no motion occurs. μ_sF_N will be exerted back on the box by static friction.
• In the last case, the box will accelerate forward and begin sliding. 2μ_sF_N is greater than the maximum force with which static friction can resist motion.

Question 23

What relationship exists between normal force and friction for a sliding object?

Answer 23

Kinetic friction is a force that opposes the motion of any two surfaces in contact that are already sliding relative to each other. The relationship involved is:

f_k = μ_kF_N

Where:

f_k = force of kinetic friction (N)
μ_k = coefficient of kinetic friction (unitless)
F_N = normal force (N)

Question 24

Describe what would happen to an already-sliding box, if a person continues to push horizontally on it with a force of 1/2 μ_kF_N, μ_kF_N, and 2μ_kF_N, respectively.

Answer 24

• In the first case, the box will eventually come to rest. 1/2 μ_kF_N is less than the force of kinetic friction acting against the box.
• In the second case, the box will continue its sliding motion. μ_kF_N will exactly cancel the force of kinetic friction against the box and no net force will exist.
• In the last case, the box will accelerate forward. 2μ_kF_N is greater than the force of kinetic friction acting against the box.

Question 25

Why is there an equal sign for the force of kinetic friction, but an inequality sign for the force of static friction?

Answer 25

* **Kinetic friction** only depends on the composition of the two moving surfaces in contact and hence has only one value. * **Static friction** opposes any amount of force applied while the object remains immobile; its value can vary anywhere from zero to the maximum for those surfaces.

Question 26

In which of the following scenarios is static friction between the surfaces **higher** than kinetic friction? * A brick on ice * A brick on asphalt * An ice block on glass * An ice block on rubber

Answer 26

All of them. ## Footnote Static friction is always higher than kinetic friction for any pair of surfaces in contact. Another way to think of this question is that it always requires more force to start an object sliding than it does to keep it sliding.

Question 27

What formulas give the component forces for the force due to gravity on an inclined plane? ## Footnote Note that, by convention, the "x" axis is along the slope of the plane and the "y" axis is perpendicular to the plane.

Answer 27

F_x=mgsinθ F_y=mgcosθ ## Footnote Since these formulas are the opposite of the way we normally compute components, a way to remember the x component is "sine is for slope."

Question 28

What calculation will give you the normal force on a box of mass m, on an inclined plane with angle θ?

Answer 28

F_N = -mgcosθ ## Footnote Since the box is not sinking into the slope, nor launching up off of the slope, the net force must be zero in the perpendicular direction of the slope. Hence, F_y = -F_N.

Question 29

What is the force of static friction if the box does not start to slide down the plane?

Answer 29

F_s = -mg sin(θ) ## Footnote Two forces are parallel to the slope of the plane: the component of gravity parallel to the plane, with magnitude mg sin(θ), and the force of static friction, pointing in the opposite direction. These forces must be equal and opposite to prevent motion.

Question 30

In the pulley system demonstrated below, m₂ is twice the mass of m₁ and no net movement is occurring. What relationship exists between the tension in the string at points 1 and 2?

Answer 30

The tensions are **equal**. ## Footnote A key to all pulley questions is to remember that the tension in the string running through the system is always constant, since it is all one string.

Question 31

What is the force F in the pulley apparatus below, with respect to the mass m₁, assuming that all masses are stationary?

Answer 31

F = m₁g ## Footnote Two forces are acting on m₁: gravity and tension. If the box isn't moving, they must be equal and opposite, so T = -m₁g. Since tension is constant through the string, the force on m₂ from tension must also be F = -T = m₁g.

Question 32

What is the force F in the pulley apparatus below, with respect to the mass m₂, assuming that all masses are stationary and no friction exists?

Answer 32

F = -m₂gsinθ ## Footnote Two forces are acting on the box along the inclined plane: the x component of gravity and the tension in the string. If the box isn't moving, they must be equal and opposite, so F = T = -m₂gsinθ.

Question 33

What is the formula for the **centripetal acceleration** acting on a body following a circular path with a constant speed v?

Answer 33

a = v²/r ## Footnote On the MCAT, all circular motion occurs at a constant speed.

Question 34

How does the acceleration experienced by a child on a merry-go-round change if the speed of the merry-go-round doubles?

Answer 34

The child's acceleration **increases** by a factor of 4. ## Footnote Since centripetal acceleration is proportional to the velocity squared, if the velocity doubles, the acceleration increases by the square of that amount, or 4x.

Question 35

What is the **centripetal force F** experienced by a body traveling in a circle of radius r at a constant speed v?

Answer 35

F = mv²/r ## Footnote Where: v = velocity (m/s) a = acceleration (m/s²) m = mass (kg) r = radius of circle (m) This formula simply comes from F = ma, where a is the centripetal acceleration.

Question 36

How do the centripetal forces on Objects 1 and 2 differ, if both are traveling the same circular path at the same speed, but Object 2 is twice as heavy as Object 1?

Answer 36

The centripetal force on Object 2 (the heavier one) is **twice** that on Object 1. ## Footnote Centripetal force is given as F = mv²/r. Since F is proportional to m, with all else equal, the force on the heavier object will be proportionally twice as large.

Question 37

How do the centripetal forces on Objects 1 and 2 differ, if both are the same mass and traveling the same circular path, but Object 2 is moving three times as quickly as Object 1?

Answer 37

The centripetal force on Object 2 is **nine times** that on Object 1. ## Footnote Centripetal force is given as F = mv²/r. Since F is proportional to v squared, with all else equal, the force on the heavier object will be larger by the square of the velocity difference.

Question 38

# Define: center of mass

Answer 38

It is the **weighted average of the location** of all of the mass in the object. ## Footnote On the MCAT, for the purposes of translational motion, an object can be approximated as a point object, of equivalent mass, located at the center of mass.

Question 39

Where is the **center of mass** of a solid disk of constant density located?

Answer 39

The center of mass of the disk is located directly at the **center of the disk**. ## Footnote An object's center of mass is not always located at its geometric center, but this is always the case for symmetric solids of constant density.

Question 40

# Define: vector field

Answer 40

It is a function that assigns a vector to every point in space. ## Footnote The only fields on the MCAT are **force fields**, which can be used to interpret the force an object at any given point in space will experience. Common examples are electric fields and gravitational fields.

Question 41

What is the **direction of the force** experienced by an object placed at point A in the gravitational field depicted below?

Answer 41

The object feels a **force down** and to the **left**. ## Footnote This is the direction the vectors in the vicinity of point A are pointing, and these vectors represet the forces experienced by objects at these locations. On the MCAT, fields are assumed to be of uniform density until told otherwise.

Question 42

Where is the **magnitude of the field** depicted below greater, at point A or at point B?

Answer 42

The magnitude of the field is **greater** at point B. ## Footnote The magnitude of a field can be estimated from the density of the field lines in this location. The greater the density, the larger the magnitude of the field. The field lines are packed more tightly at point B than at point A, so the magnitude of the field must be greater there.