Physics Mc Forces Newtons Laws

AP* PHYSICS B

Forces & Newton’s Laws

Teacher Packet

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Objective To review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions on forces and Newton’s laws. Standards Forces and Newton’s laws are addressed in the topic outline of the College Board AP* Physics Course Description Guide as described below. AP Physics Exam Connections Topics relating to forces and Newton’s laws are tested every year on the multiple choice and in most years on the free response portion of the exam. The list below identifies free response questions that have been previously asked over forces and Newton’s laws. These questions are available from the College Board and can be downloaded free of charge from AP Central. http://apcentral.collegeboard.com. Free Response Questions

2008 Question 2 2008 Form B Question 2 2007 Question 1 2007 Form B Question 1 2006 Question 2 2006 Form B Question 1 2005 Question 1 2005 Form B Question 1 2003 Question 1 2003 Form B Question 1 2000 Question 1 2000 Question 2

I. Newtonian Mechanics B. Newton’s laws of motion (including friction and centripetal force)

1. Static equilibrium (first law) 2. Dynamics of a single particle (second law) 3. Systems of two or more bodies (third law)

F. Oscillations and Gravitation 4. Newton’s law of gravity

http://apcentral.collegeboard.com/




What I Absolutely Have to Know to Survive the AP* Exam

Force is any push or pull. It is a vector. Newton’s Second Law is the workhorse of the AP Physics B exam. It allows you to write down mathematical relationships that are true. Thus, for a single body, if you pick any direction and sum up all the positive and negative forces that act on the body along that line, the sum will equal the product of the body’s mass and its acceleration along that line. A Free Body Diagram allows you to identify all of the forces acting on a single body. Neglect one force or add a fictitious force on your FBD and you are in trouble. • Newton’s 1st Law: in an inertial frame of reference, an object in a state of constant velocity (including

zero velocity) will continue in that state unless impinged upon by a net external force. If ΣF=0, then a=0 and the object is at rest or moving at a constant velocity in a straight line. The converse is true also, if an object is in a state of constant velocity (including zero velocity) then a=0 and ΣF=0.

• Newton’s 2nd Law: A net force acting on a mass causes that mass to accelerate in the direction of the net force. The acceleration (vector) is directly proportional to the net force (vector) acting on the mass and

inversely proportional to the mass of the object being accelerated. Fa= or m

F maΣΣ =

• Newton’s 3rd Law: For every action force, there exists an equal and opposite reaction force. Let’s say you hit a table with your fist. Doing so, applies a force to the table which, if great enough will break the table. Likewise, the table applies a force to your fist which, if great enough, will break your fist. The size and direction of the force you apply must be equal and opposite the force the table applies to you. Hence, the only way forces can be generated are in action/reaction pairs which occur on different objects. If you try to apply 800 Newtons of force to a table that can only provide 600 Newtons of reaction force back on you, you will never succeed. The table will break as soon as you exceed 600 Newtons, which is the maximum force it can apply to you.

Key Formulas and Relationships

max

1 22

( )

( )

net

W

f s s N

f k k N

G

mF mgF F static

F F kinetic

Gm mFr

μ

μ

Σ = ==

≤

=

=

F F a

ΣF = Sum of the forces is the Net Force Newtons (N) = 2

kgms

a = acceleration m = mass FW = weight g = acceleration due to gravity

maxf sF = maximum static frictional force

f kF = kinetic frictional force FN = normal force FG = gravitational force r = distance between the centers of two masses



Basic Kinds of Forces

gF mg= Always directed toward the Earth’s center. Force on a free falling body, if we neglect air friction.

Weight gF , W

1 22G

m mF Gr

=

A force of attraction between any two massive objects. When the Earth is one of the two bodies involved, then the force felt by the second body while positioned on the Earth’s surface will always be directed toward the Earth’s center.

GF Gravitational

A force of support, provided to an object by a surface in which the object is in contact. Always directed perpendicular to and away from the surface providing the support.

Normal NF , N

In the figure above, a box is supported by a table. The figure shows all the forces acting on the box and is called a Free Body Diagram (FBD). If a box, rests on a level table, then the NF W mg= = . Notice that the normal force sometimes equals the weight but not always. If the box is placed on an inclined plane, then the cosNF mg θ= , the component of the weight that is equal and opposite the normal force. For the inclined plane above, the normal force and the weight are not equal and not even in the same direction.

FN

W

θ

FN

y

xmgcosθ

W



Friction fF

riction is produced by the atomic interaction between two bodies as

Fthey either slide over one another (kinetic friction) or sit motionless incontact with one another (static friction).

maxf s s

, f

NF F≤ μ (static) Static Friction opposes the intended direction of relative sliding. The

of

static frictional force will only be as high as it needs to be to keep the system in equilibrium. If successively greater and greater forces are applied, the static frictional force will counter each push with a forceequal and opposite magnitude until the applied force is great enough to shear the bonding between the two surfaces. When you calculate sf or

f sF using the equation above, you are finding the maximum static frictional force, one of an infinite number of possible frictional forcthat could be exerted between the two bodies.

es sμ is a proportionality

constant called the coefficient of static friction. t is the ratio of the stfrictional force between the surfaces divided by the normal force acting on the surface.

I atic

(kinetic)f k k NF Fμ= Kinetic Friction or Dynamic Friction or Sliding Friction is always

is a opposite the direction of motion. The statement that kinetic frictionfunction of the normal force only (surface area is independent) is true only when dealing with rigid bodies that are sliding relative to each other. When you calculate kf or fF using the equation above, you arfinding the single, constant

e kinetic frictional force that exists between t

two bodies sliding relative to one another. No matter their velocity (assuming heating does not alter the coefficient of kinetic friction) thkinetic frictional force will always be the same. k

he

e μ is a proportionality

constant called the coefficient of kinetic friction. It is the ratio of the kinetic frictional force between the surfaces divided by the normal foracting on the surface

ce

WARNING! The two quantities sf and kf may look the same, but they c ty

ee body

al

tell us different things. Kinetic fri tion is pically less than static frictionfor the same two surfaces in contact. Note that the normal force sometimes equals the weight but not always. When you draw a frdiagram of forces acting on an object or system of objects, be sure to include the frictional force as opposing the relative motion (or potentifor relative motion) of the two surfaces in contact.



Tension T

is a force that is applied to a body by a rope, string, or cable. TF

TF is applied along the line of the string and away from the body in question.

, T F

Applied SubscriptF

ush me, pull you force that does not fall into one of the above he force

Pcategories, for example, a friend shoves you. The magnitude of tis characterized by an F, with any subscript that makes sense to solve the problem. Later in the year, you will encounter additional forces, like the electric and magnetic forces.

Strategy on Force Problems

1. Take one body in the system a ram (FBD) for it. the direction of the

nd draw a Free Body Diag2. Choose x and y axes and place them beside your FBD. One axis must be in

acceleration you are trying to find. If there is no acceleration, then 0FΣ = . If there are forces on the FBD that are not along the x and y directio d 3. ns, fin their respective x

4. the forces in the x direction and set them equal to and y components. Using Newton’s 2nd Law, sum xma . If a

ma5. e nu er of

second equation is needed, sum the forces in the y direction and set them equal to y . Repeat the above process for all the bodies in the system or until you have the sam mb

equations as unknowns and solve the problem.

Effective Problem Solving Strategies

Free Body Diagram

A Free B as a box showing all the forces be

d

hen drawing Free Body Diagrams show only the force(s) that act on the body

.

(FBD)

ody Diagram is normally depicted acting on the body. These forces are depicted as arrows. They don’t have to drawn to scale, but they should have a length that is appropriate for their magnitude. Also, the force vectors do need to be directionally accurate anlabeled. Do not include components of the force vectors on your FBD. Win question and do not show forces that the body applies to other bodies. Also, do not include velocity or acceleration vectors on your free body force diagrams, since you will lose points for extraneous vectors on your FBD



Example 1

A rope supports an empty bucket of mass 3.0 kg. Determine the tension in the rope when the bucket is (a) at rest and (b) the bucket is accelerated upward at 2.0 m/s2. Solution Step 1 Draw a Free Body Diagram for each body in the system.

Step 2 Choose x and y axes. In this case we could choose the y axes to be in the same dimension as the tension and the weight. Step 3 If there are off axes forces, then find the x and y components. Step 4 Using Newton’s 2nd law, sum the forces in the y dimension and set equal to may. (a) When the bucket is at rest, the net force is zero, so that the tension in the rope equals the weight of the bucket.

( ) 2

0

3.0 10 30

y y

T

T

F ma

F mgmF mg kg Ns

Σ =

− =

⎛ ⎞= = =⎜ ⎟⎝ ⎠

(b) When the bucket accelerates upward, the net force is may.

FT

W



( )

2 23.0 10 2.0 36 upward

y y

T

T

T

F ma

F mg maF mg ma m g a

m mF kg Ns s

Σ =

− =

= + = +

⎛ ⎞= + =⎜ ⎟⎝ ⎠

Example 2

In the diagram below, two bodies of different masses (M1 and M2) are connected by a string which passes over a pulley of negligible mass and friction. What is the magnitude of the acceleration of the system in terms of the given quantities and fundamental constants?

Solution

Step 1: Draw a FBD for each body in the system.

There are two forces acting on each of the bodies: weight downward and the tension in the string upward. The tension is distributed throughout the string. The pulley (negligible mass and friction) changes only the direction of motion, not the tension, so the tension is the same on each side of the pulley. Our FBDs should look like this:

M2 M1



Step 2 Choose x and y axes. In this case we could choose the y axes to be in the same dimension as the tension and the weight for both masses. Step 3 If there are off axes forces, then find the x and y components. Step 4 Using Newton’s 2nd law, for each body sum the forces in the y dimension and set equal to may and/or the x dimension. For m1 the tension is positive and the weight is negative since the acceleration is upward.

1

1 1

y y

T

F m a

F W m a

Σ =

− =

For m2 the tension is negative and the weight is positive since the acceleration is downward. 2

2 2

y y

T

F m a

F W m a

Σ =

− + =

T T

W1

W2



( )

( ) ( )( )( )

1 1

2 2

2 1 1 2

2 12 1 2 1

1 2 1 2 1 2

T

T

F W m aF W m a

W W m m a

m m gW W m g m gam m m m m m

− =− + =

− = +

−− −= = =

+ + +

Example 3

A block of mass m rests on a horizontal table. A string is tied to the block, passed over a pulley, and another block of mass M is hung on the other end of the string, as shown in the figure below. The coefficient of kinetic friction between block m and the table is μk. Find the magnitude of the acceleration of the system in terms of the given quantities and fundamental constants.

m

M

μ k



g

Solution Step 1: Draw a FBD for each body. Note that according to the FBD, the vertical acceleration of the block on the table is zero, since the normal force is directed upward and the weight force is directed downward. There is a horizontal acceleration for the block since the tension is greater than the frictional force.

Step 2 Choose x and y axes. In this case we could choose the y axes to be in the same dimension as the tension and the weight for both masses. We could choose the x axis to be in the same dimension as the frictional force and the tension pulling the mass to the right. Step 3 If there are off axes forces, then find the x and y components. Step 4 Using Newton’s 2nd law, for each body sum the forces in the y dimension and set equal to may and/or the x dimension.

For block m the net force vertically is 0 since the block is accelerating to the right and not upward nor downward. Summing the forces in the y dimension we find that the normal force equals the weight of the block.

1

1

0y y

N

N

F ma

F WF W m

Σ =

− == =

Summing the forces in the horizontal dimension, the direction of the acceleration for block m, we find that

T

T

W2W1

FN

fk



max x

k

FT f maΣ =− =

For mass two, summing the forces in the vertical dimension, the vertical acceleration for block M, we find that

2

ay yF M

T W Ma

Σ =

− + =

Adding the two equations together we determine the acceleration of the system

( )

( ) ( )( )( )

2

2

2

Since

k

k

k k

k

k k N k

T f maT W Ma

W f m M aW f Mg mgam M m M

M m ga

m Mf F mg

μ

μ

μ μ

− =− + =

− = +

− −= =

+ +

−=

+

= =



Example 4

Three blocks of mass m1, m2, and m3 are connected by a string passing over a pulley attached to a plane inclined at an angle θ as shown below.

The coefficient of kinetic friction between blocks m2 , m1 and the table is μk. Assuming that m3 is large enough to descend and cause the system to accelerate, determine the acceleration of the system in terms of the given quantities and fundamental constants. Solution Step 1: Draw a FBD for each body. When drawing a FBD, be sure the orientation of the box representing the body is the same as the actual orientation of the body in the problem. Note the weight force is always directed straight downward and the normal force perpendicular to the surface.

θ

m1

m2

m3` θ



Step 2 Choose x and y axes. Place axes next to FBD. One axis must be in the direction of the acceleration you are trying to determine. Step 3 If there are off axes forces, then find the x and y components. There is one force that is off axis, the weight force. The weight of each block must be broken into two vectors parallel and perpendicular to the inclined plane.

1

1 1

y1 1

2

2 2

y2 2

the parallel and perpendicular components of the weight force aresin

w cos

m the parallel and perpendicular components of the weight force aresin

w cos

x

x

mw F m g

F m g

w F m gF m g

θθ

θθ

⊥

⊥

= =

= =

= =

= =

Step 4 Using Newton’s 2nd law, for each body, sum the forces in the y dimension and set equal to may. Sum the forces in the x dimension and set equal to max.

T2

W3

FN2

T1

f

T2 FN1

T1

W2 f

W1



1

y

1

1

x 1

1 1

1 1 1 1

vertical motion perpendicular to the incline planeF 0

cos 0cos

horizontal motion parallel to the incline planeF

cos sin

N

N

k

k

m

F m gF m g

m aT f F m aT m g m g m a

θθ

μ θ θ

Σ =

− ==

Σ =− − =

− − =

2

y

2

2

x 2

2 1 2

2 1 2 2 2

vertical motion perpendicular to the incline planeF 0

cos 0cos

horizontal motion parallel to the incline planeF

cos sin

N

N

k

k

m

F m gF m g

m aT T f F m aT T m g m g m a

θθ

μ θ θ

Σ =

− ==

Σ =− − − =

− − − =

3

y 3

2 3 3

2 3 3

vertical motion F

m

m a

T W m aT m g m a

Σ =

− + =− + =

Looking at the motion along the axis of acceleration, up the incline plane



)

2

2

2 3

a( ) ( )( ) ( )

( ) ( )( ) ( ) (

1 1 1 1

2 1 2 2 2

2 1 2 1 2 1

2 1 2 1

2 3 3

2 3 3

3 3 1 2 1 2

3 1 2 1

cos sincos cos

cos sin

cos sin

cos sin

cos sin

k

k

k

k

k

k

T m g m g m aT T m g m g m aT g m m g m m m a m

T g g m m m a m a

T m g m aT m g m a

m g m a g g m m m a m a

m g g g m m m m m

μ θ θμ θ θ

μ θ θ

μ θ θ

μ θ θ

μ θ θ

− − =− − − =

− + − + =

− + + = +

− + == −

− − + + = +

− + + = + +

( ) ( )( )

+

3 1 2

1 2 3

cos sink

a

m g g g m ma

m m mμ θ θ− + +

=+ +



Free Response Question 1 (15 pts) Two blocks, both of mass 0.5 kg, are connected to each other by a thin string which is passed over a pulley as shown in the diagram. Block 1 sits on a rough horizontal part where the coefficient of static friction is 0.6 and the coefficient of kinetic friction is 0.4. Block 2 sits on a frictionless incline which forms an angle θ with the horizontal.

θ

1

2

A. On the diagram below, draw and label vectors to represent the forces acting on block

2.

1 point for each correctly drawn and labeled vector (tension, normal, weight)

(3 points max) T N

W

1 point deducted for each extraneous vector with a minimum score being 0



B. Determine the maximum angle θ the incline can make without the blocks sliding

down.

(6 points max) 1 point for a correct statement that when the blocks are not sliding the ΣF=0

When block 2 is not sliding, block 1 is not sliding and the ΣF=0, so the maximum angle means maximal frictional force.

( )

max

max

1

Block 20

0sin

Block 10

0

Thussinsin

sin 0.6

sin 0.637

s

s s N

s N

s

s

FT FT mg

FT f

T f F

F mgmg mg

θ

μ

μ θμ θ

θ μ

θθ

−

Σ =

− =

=

Σ =

− =

= =

=== =

=

= °

1 point for a correct application of NSL to block 2 such that the tension equals the parallel component of the weight 1 point for a correct application of NSL to block 1 such that the tension equals the maximum frictional force 1 point for a correct statement that the maximum angle occurs when the frictional force is maximum on block 1 using 0.6 for the coefficient of friction 1 point for a correct statement that the maximum frictional force on block 1 equals the parallel component of the weight on block 2 1 point for the correct answer including correct units and reasonable number of significant digits



C. Assume θ is 30º, find the magnitude of the acceleration of the system

(6 points max)

( ) ( )2

2

Block 1F

Block 2F

sin

sinsin 2

sin2

10 sin 30 0.4 10

2

0.5

k

k

k

k

k

maT f maT mg ma

maT F maT mg ma

T mg maT mg ma

mg mg mag g a

m msa

mas

μ

θ

μθ

θ μθ μ

°

Σ =

− =− =

Σ =

− + =

− + =

− =− + =

− =−

=

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠=

=

1 point for any indication that there are two forces acting on Block 1 parallel to the surface 1 point for a correct application of NSL to block 1 that includes the two forces (tension and friction) and a non zero acceleration

2s

he g

1 point for a correct statement that tfrictional force is kinetic and usin0.4 for the coefficient of friction 1 point for any indication that there are two forces acting on Block 2 parallel to the surface 1 point for a correct application of NSL to block 2 that includes the two forces (tension and parallel component of the weight) and a non zero acceleration 1 point for the correct answer including correct units and reasonable number of significant digits



Question 2 (10 pts) A block of mass m rests on an air table (no friction), and is pulled with a force probe, producing the Force vs. Acceleration graph shown below.

A. Determine the mass of the block.

Newton’s 2nd law states that a

Fm net= . This

ratio is the slope.

2 2

9 N 3 1.5 kg6m/s 2m/s

F Nm slopea

Δ −= = = =

Δ −

1 point for using the slope of the F vs. a graph to determine the mass or the correct use of Newton’s second law 1 point for the correct answer including correct units and reasonable number of significant digits

The block is now placed on a rough horizontal board having a coefficient of static friction μs = 0.2, and a coefficient of kinetic (sliding) friction μk = 0.1.



B. What is the minimum value of the force F which will cause the block to just begin to move?

The block will just begin to move when the force F overcomes the maximum static frictional force:

( ) ( ) ( )min

2min 0.2 1.5kg 10m/s 3N

s s N sF f F mg

F

μ μ= = =

= =

1 point for the correct equation for the maximum static frictional force 1 point for the correct answer or an answer consistent with part (A), including correct units and reasonable number of significant digits

The block rests on the rough horizontal board. One end of the board is slowly lifted until the block just begins to slide down the board. At the instant the block begins to slide, the angle of inclination for the board is θ.

C. Determine the relationship between the angle θ and the coefficient of static friction, μs.

(6 points max) 1 point for a correct statement of net force parallel to the plane is zero or FBD indicating the same

At the instant the block is just about to move, the maximum frictional force directed up the incline is equal and opposite to the parallel component of the weight down the incline, and the normal force is equal and opposite to the perpendicular component of the weight.

1 point for a correct statement of net force perpendicular to the plane is zero or FBD indicating the same

max

max

max

00

sin

0

0cos

Thussin tancos

x

s

s

y

N

N

ss

N

Ff F

f mg

F

F FF mg

f mgF mg

θ

θ

θμ θθ

⊥

Σ =− =

=

Σ =

− ==

= = =

1 point for a correct statement that the maximum static frictional force equals the parallel component of the weight or FBD indicating the same 1 point for a correct statement that the normal force equals the perpendicular component of the weight or FBD indicating the same 1 point for a correct statement that μs is the ratio between the parallel and perpendicular components of the weight

1 point for the correct relationship between θ and μs



Multiple Choice Questions 1 – 2

A block of mass 2m is suspended by a string, the other end of which is passed over a pulley of negligible mass and friction and tied to a block of mass m.

2m

m

1. The acceleration of the system is

a) 13

g

b) 12

g

c) 23

g

d) g

e) 43

g

ΣF = ma Find net force on object

The forces acting on each block are shown above. From this, we write two net force equations, one for each block (the smaller block is block 1) and apply ΣF = ma.

1

2 2 2

F T mg ma

F mg T ma

= −

= −

=

=∑∑

These equations can be added together, eliminating T.

2 2

2 2

ma T mg

ma mg T

ma ma mg mg

= −

+ = −

+ = −

Simplifying and solving for a gives

3

ga =

A

mg

2mg

T

T



2. The tension in the string is

a) 13

mg

b) 12

mg

c) 23

mg

d) mg

e) 43

mg

ΣF = ma Use either net force equation from above and substitute for

a to solve for T Find net force on object

3

34

3

ma T mg

gm T

mgT mg

mgT

= −

= −

= +

=

⎛ ⎞⎜ ⎟⎝ ⎠

mg

E

3. An object of mass m is pushed to the right by a force F that forms an angle θ with the horizontal as shown. The object moves with constant velocity. The coefficient of kinetic friction between the object and the surface is a) cosF θ b) sinF θ

c) cossin

θθ

d) cosFmg

θ

e) cossinF

F m

F θ

m

gθ

θ +



Resolve the force vector F into components and draw a FBD. We know the net force in both directions is zero, since the velocity is constant.

Resolve for vector into perpendicular and parallel components ΣF = ma Ff = μFN

Object moving with constant velocity has zero net force.

Analyzing the forces in the vertical direction, we can write

sinN yF mg F mg F θ= + = + and cosf xF F F θ= = We now apply the equation for kinetic friction:

f NF Fμ= Substituting from our other two equations:

( )cos sinF mg Fθ μ θ= + And solving for μ gives

( )cos

sin

F

mg F

θμ

θ=

+

E mg Fy

Fx

FN

4. A 75 kg man and a 25 kg girl stand facing each other on a frictionless sheet of ice. The girl exerts a force of 30 N on the man, causing him to accelerate to the north. The magnitude and direction of the reaction force on the girl is Magnitude Direction a) 10 N north b) 10 N south c) 30 N north d) 30 N south e) 90 N south Reaction forces are equal and opposite

The force exerted by the girl on the man is 30 N to the north, therefore the reaction force must be equal (30 N) and in the opposite direction (south)

D



Questions 5 – 6

A block of mass 1 kg is sitting on top of a block of mass 2 kg. The 1 kg block is tied to a thin thread which extends horizontally to a wall where the other end is attached. The 2 kg block is pulled by an external horizontal force F with constant velocity to the right. The coefficient of kinetic friction between the two blocks is .5 and the coefficient of kinetic friction between the 2kg block and the surface is .8.

1 kg

2 kg F

T

5. As the bottom block is pulled to the right, the tension in the thread T is most nearly a) .2 N b) 1 N c) 5 N d) 8 N e) 15 N

We know the net force on the top block is zero (it remains at rest). Drawing the forces on the top block:

Ff = μFN Object moving with constant velocity has zero net force.

The force of kinetic friction opposes the relative motion of the two blocks. Since the bottom block moves to the right, friction is attempting to pull the top block to the right so the two don’t move relative to each other. From the above diagram, we can see that T = Ff,k1. Substituting and solving for T gives (1 designates the top block, 2 the bottom)

( ) ( ) ( )1, ,1 ,1 ,1 1

2.5 1 10

5

f k k N kT F F m g

mT kgs

T N

μ μ= = =

=

=

C

FN

m1g

Ff,k1 T



6. The force F required to pull the bottom block with constant velocity is most nearly a) 15 N b) 21 N c) 24 N d) 29 N e) 34 N Ff = μFN Object moving with constant velocity has zero net force.

The net force on the bottom block is zero as well (it moves with constant speed). The forces on the bottom block are:

In the y-direction: 1 2NF m g m g= +

In the x-direction: , 1 , 2f k f kF F F= + Ff,k1 is already known to be 5 N, so we find Ff,k2 and substitute to find F.

( )( ) ( )

5 5, 2 , 2 1 2

5 .8 10 2 12

5 24

29

F N F N m g mk N k

mF N kg kgs

F N N

g

F N

μ μ= + = + +

= + +

= +

=

⎡ ⎤⎢ ⎥⎣ ⎦

D

FN

m1g

Ff,k1

F

m2g

Ff,k2

7. An object moves with constant velocity while three forces act on it. Which of the following must be true? I. the three forces have equal magnitude II. the vector sum of the three forces is zero III. the forces must be perpendicular to the direction the object is traveling a) I only b) II only c) I and II d) I and III e) II and III




Since the object is moving with zero acceleration, it must experience zero net force: the vector sum of the three forces must equal zero. No other requirement is needed; the forces can be of any magnitude and direction so long as their vector sum is zero.

B

8. A 4 kg mass is hung vertically from an ideal spring, which causes it to stretch 8 cm. If the 4 kg mass is replaced by a mass of 8 kg, how far will the spring stretch? a) 2 cm b) 4 cm c) 8 cm d) 16 cm e) 64 cm Hooke’s Law: Fs = -kx Hooke’s Law shows that the stretch of a spring is directly

proportional to the force it exerts: hence double the force on a spring and the distance it is stretched will double as well. Thus, the 8 kg mass with stretch the spring 16 cm.

D

9. As a 40 N ball falls through Earth’s atmosphere, it experiences a drag force, which is given by the expression , where b is known to be 5 kg/s. The terminal velocity of the ball close to Earth’s surface is

DF b= − v

a) 5 m/s b) 8 m/s c) 25 m/s d) 40 m/s e) 45 m/s Object moving with constant velocity has zero net force. Terminal velocity occurs when FD = Fg

The force of drag increases with the velocity, reducing the acceleration until the force of drag equals the weight. At this point we can write

40

408

5

D g

T

T

F F

bv N

N mv skgs

=

=

= =

B



Questions 10 – 11

An object of mass m is at rest on an inclined plane as shown above. The inclined plane forms an angle θ with the horizontal. The coefficient of static friction between the object and the plane is μs.

m

θ

10. Which of the following diagrams correctly shows the forces acting on the object?

a) b) c)

d) e)

Resolve forces on an incline Direction of forces

D correctly shows the forces. Weight always pulls an object down, a normal force is always perpendicular to the surface, and friction is always parallel to the surface. In this case, friction points up the ramp because it opposes the motion of the object sliding down the ramp.

D

11. The force of static friction acting on the object is a) smgμ b) cossmgμ θ c) sinsmgμ θ d) cosmg θ e) sinmg θ



Resolve forces on an incline

In this case, the relationship between static friction and the coefficient of static friction cannot be used because it is an inequality: ,0 f s s

Static friction is variable NF Fμ≤ ≤ . We must determine the force

of friction by force analysis. We know that the net force is zero (it is at rest), so parallel to the ramp we have:

, ,f s g xF F= Resolving the weight vector and substituting:

E

, sinf sF mg θ=

Fg

θ

Fg,x = Fgsinθ

Fg,y = Fgcosθ

12. Three forces act on an object that is moving at constant velocity. Two of the forces are shown on the diagram above. Which of the following correctly shows the third force vector?

a) b) c)

d) e)


In order for the object to move with constant velocity, the vector sum must be zero. Adding the vector shown in answer choice A would accomplish this.

A



3. A 20 N bucket is at the bottom of a 9 m deep well and is initially at rest. The bucket

ewton’s Second Law & Use kinematics to find the acceleration of the bucket:

1is then pulled to the top of the well by a string that exerts a constant tension force, such that it reaches the top in 3.0 s. During this interval, the tension in the string is a) 4 N b) 6 N c) 20 Nd) 24 N e) 40 N NKinematics

( )( )

21x x v t at0 0

2

222

21

22 2 9

23

x at

x m mast s

=

= = =

The forces on the bucket are:

And the net fo ce equation is

= + +

rF T mg= −∑

Substituting Newton’s Second Law and solving for T:

( )22 2 20

24

ma T mg= −

mT ma mg kg Ns

T N

= + = +

=

D



4. Which of the following graphs shows an object whose net force is zero?

terpreting Motion Zero net force means the velocity must be constant. Graphs

C

1

a) I only

t

v III.

t

x II.

t

x I.

b) II onlyc) I and II d) I and IIIe) II and III

InGraphs Object moving with

I and II show objects moving with constant velocity: graph III shows an object with decreasing velocity. Thus, the answer is C constant velocity has

zero net force.

15. The diagram above shows the path of a pr

P

Q

R

ojectile as it moves from left to right n of

) P

nd R ration is the same at all points

y le: we know that the only

E

without friction or drag. At which point along the projectile’s path is the acceleratiothe projectile the greatest? ab) Q c) R d) P ae) the accele

A projectile is free The diagram shows a projectifalling due to gravit force acting on a projectile is gravity, which is constant.

Thus, the acceleration is the same at all points. The projectile moves horizontally due to its initial velocity andinertia.

Physics Mc Forces Newtons Laws

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