Chapter 7: Forces, Acceleration and Newton’s Laws of Motion
In this chapter we study the branch of physics known as dynamics. Dynamics is the relationship between an
object’s acceleration and the net force acting on the object. So we are again visiting our constant acceleration
model except now looking at the causal force model necessary for this condition. In the last chapter we studied at
great length that when the net force is zero, the object does not accelerate. Now we will study in greater detail the
behavior of an object when the net force is not zero.
Part 1: Newton’s 3 Laws
First Law
An object at rest, or in motion at a constant speed in a straight line (in other words at a constant velocity), will
remain in that state unless acted on by a net unbalanced force. In other words, if the forces on an object add up to
zero, the object will not accelerate. It may not move at all, or it may move at a constant velocity. Both of these are
states of equilibrium. This means that if an object is moving at a constant velocity, there are either no forces acting
on the object, or the forces that are acting on it must add up to zero.
Matter has a property called inertia, which is defined as the resistance of an object to a change in motion. It is a
property that depends only on how much mass the object has. If a puck is given a push on an air table, it will keep
moving after it leaves contact with the hand. When asked what force causes the puck keeps going, never say the
force of inertia. Inertia is not a force; it is just a resistance to a change in motion. The puck keeps going because
there is no force causing it to slow down.
Second Law
If a net unbalanced force acts on an object the object will accelerate in the direction of that force. The acceleration
will be directly proportional to the unbalanced force and inversely proportional to the object’s mass. The law tells us
that if the forces on an object don’t add up to zero the object will accelerate. We will look at the applications of the
second law in more detail in part 2.
Third Law
Whenever object A exerts a force on object B, object B exerts an equal and opposite force back on A. Forces
always occur in pairs, and the two forces always have the same magnitudes. The force of A on B is called the
action force, and the force of B back on A is called the reaction force.
I think the best way to begin to develop an understanding of the laws is to pose some questions and use
Newton’s Laws to answer the questions.
Sample problem:
1) Crystal Line is pushing the box across the floor with a constant force and the box moves at a speed of 1.0 m/s.
Answer the following questions about the forces acting on the box.
a) What forces are acting on the box? How do you know?
1. Crystal pushing the box to the right. I was told in the problem.
2. The floor pushes the box to the left. If there was no force pushing to the left, Crystal’s
force would be the only horizontal force and the box would not move at a constant speed
because the net force could not be zero.
3. The Earth is pulling the box down. The earth pulls everything on it (and close to it) down.
4. The floor pushing the box up. The box is not accelerating downward so there must be a
force balancing the Earth’s pull downward.
b) Draw a complete force diagram for the forces acting on Crystal and the box. Indicate which forces, if any are
Newton’s 3rd law pairs.
FNfC
FNbC
Ffs fC
FgEC
These two are a 3rd law pair.
Third Law pairs always have the
same objects but in the opposite
order. Crystal is pushing on the
box to the right and the box is
pushing her equally to the left.
There is a third law pair for every
force but they are not acting on
Crystal or the box.
FNfb
FNCb
FfK fb
FgEb
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c) What is the net horizontal force on the box? How do you know?
The net horizontal force is zero because the box is moving horizontally at a constant velocity.
d) What is the net vertical force on the box? How do you know?
The net vertical force is zero because the box is at rest vertically.
e) What is the reaction force to each force of the forces listed in a)?
1) The box is pushing Crystal to the left.
2) The box is pushing the floor to the right. Tip: When you do action/reaction questions, take the action
statement, switch the position of the two objects around
3) The box is pulling the Earth up.
and change the direction to the opposite.
4) The box is pushing the floor down.
switch words around
Action: The Earth is pulling the box down.
Reaction: The box is pulling the Earth up.
change direction
2) The hand is causing the ball to accelerate upward at 2.0 m/s2.
a) What forces are acting on the ball?
The hand is pushing the ball up and the Earth is pulling the ball down.
FNhb
b) Which of the two forces is greater? How do you know?
The hand force is greater. I know because the ball is accelerating upward. If the weight
force were larger, the ball would be accelerating downward.
c) Draw a force diagram for the ball. (Done to the right.)
d) What is the net force on the ball?
The net force is the force of the hand minus the weight of the ball.
FgEb
e) What is the reaction force to the weight of the ball?
Since the weight is caused by the Earth pulling the ball down, the reaction force is the ball pulling the Earth
up. (If you are ever tempted to make the reaction force another force on the same object, don’t do it.
Action and reaction forces are never acting on the same object.)
f) Which is greater, the force of the Earth pulling the ball down or the force of the ball pulling the Earth up?
The forces are exactly the same as stated in Newton’s third law.
g) If the two forces are equal, why doesn’t the Earth move upward when the ball pulls it?
The earth has so much more mass that the force causes very little acceleration. (Newton’s 2nd law)
Part 2: Solving problems using Newton’s 2nd law
Newton’s second law when written as an equation is F = ma, which states that the sum of the
kg  m
forces is equal to mass times acceleration. We learned that the sum of the forces can also be called 1N  1
s2
the “net force”, allowing us to write the equation as Fnet = ma. The unit of force is the newton (N).
The amount of force called 1 newton is the force necessary to accelerate one kilogram of mass at a
rate of 1 meter per second squared. When using Newton’s 2nd law and the unit the Newton, mass
must be in kilograms and acceleration must be in m/s2.
The process for solving force problems involving motion is very similar to equilibrium force problems. The
difference is that the forces do not add up to zero unless the velocity is constant. If the sum is not equal to zero,
then the net (unbalanced) is set equal to mass times acceleration.
Problem solving strategy for dynamics problems:
1. Make a drawing of the situation. Identify your knowns and unknowns.
2. Construct a force diagram for each necessary object in your drawing. Draw arrows to represent each
known and unknown force on the object.
3. Impose a coordinate system on your diagram. On problems involving inclines, you will want to tilt the axis
until the x-axis is parallel to the plane. On problems where objects are connected by strings, you will want
to make an axis follow the path of the string.
4. Write the equation for each object in component form and set them equal to ma.
5. Solve for the unknowns.
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Often you will have to use a kinematics equation along with Newton’s 2nd law. If so, the process will be one of
these two types:
F
1:  F or net = a to solve for a, then kinematics to solve for any of x, y, t, or v.
m
m
2: Any of x, y, t, or v and kinematics to solve for a, then ma = F to solve for net force.
Sample Problems:
3) A 0.100 kg rock thrown vertically upward by Ruth Liss at 10.m/s. If her hand moved 50.cm while pushing on
the rock, what was her average force on the rock?
Fhand Knowns:
Solution:
m = .100 kg
vi = 0
vf = 10 m/s
y = .500 m
Unknowns:
a, F
W
Fh - W = ma, which becomes Fh - .1 kg(9.8m/s2) = (.100kg)a
a is not given so we must use kinematics to solve for a.
a
m 2
v 2f  v i2 (10 s )  0

 100
2y
2(.50m)
m
s2
 Fh = .100kg(100m/s2) +.98N
Fh = 10.N + .98N = 11 N
4) Jerry Atric pulls his 200.N sister in a 50.N sled with a rope at 30.0 above horizontal. If the rope has a tension
of 20.0 N and the sled has no friction; a) what is the normal force? b) at what rate will the sled accelerate?
20N
N
30o
250N
Knowns:
W sis = 200 N
W sled = 50 N
T = 20 N
 = 30o
Unknowns:
N, Fnet, a
Solution:
a) Use vertical forces to solve for normal. The net force is set equal to zero
because sled is not moving upward.
Upward forces are the Earth pushing the sled up (normal force) and the
vertical component of the rope force. Downward is the weight of the sled
and sister. We know they all add to zero because the sled has zero upward
velocity.
vertical: Normal + vertical component of rope – W = 0
N + sin30(20N) - 250 N = 0  N = 240 N
b) The only horizontal force in this problem is the horizontal component of the
rope force. Since there is only one horizontal force it can’t be balanced.
Set the horizontal force equal to ma.
Horizontal: cos30(20N) = ma  17.3 N = ma
Don’t forget to calculate the mass of the objects if you are given their weights.
250N
17.3N
m  Wg 
 25.5kg a =
.679 sm2
25.5kg
9.8 sm2
5) A box of mass 10.0 kg is placed on a frictionless incline 2.0 meters long. How hard is the plane pushing on the
box? After the box is released, how much time will it take to get to the bottom?
N
F
30o
F W
Knowns:
m = 10.0 kg
vi = 0
d = 2.0 m
 = 30o
Unknowns:
N, F, a, t
Solution:
Set the y-axis perpendicular to the plane and the x-axis parallel.
Perpendicular the normal force is pushing up and the
perpendicular component of the weight force pulling down. The
box is not moving perpendicular so the forces add to zero.
perpendicular: N - cos30 (10.0kg)(9.8 m/s2) = 0 or N = 84.9 N.
Parallel: the only force is the component of the weight force.
parallel: sinmg = ma so a = sing = sin30(9.8 m/s2) = 4.9m/s2
Use the calculated acceleration and kinematics to solve for time
t
2 x
a

2(2m)
4.9m/s2
 .90s (Notice mass doesn’t matter!)
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Two body problems:
On two body problems you will get the objects are hooked together so they move as one. Therefore both objects
will accelerate at the same rate. In these problems, make your coordinate system follow the string connecting
the objects. In other words, up will be positive for one of the masses and negative for the other.
6) In the system below, block A has a mass of 4.0 kg, and block B has a mass of 1.0 kg. If the pulley and table
are frictionless, what is the acceleration of the two blocks, and how much tension is in the string?
T
Solution:
N
A
T
A
coordinate
system: x-axis
follows string, +
to right
B
B
WA
WB
vertical:
horizontal:
N-W=0
T = mAa
W B-T = mBa
W B = (mA + mB)a
There were two unknowns in each equation,
so the equations were added together.
m
Knowns:
mA = 4.0 kg
mB = 1.0 kg
g = 9.80 m/s2
1.0kg(9.8 s 2 )
WB
9.8N
m
a


 1.96 s 2
m A  mB 4.0kg  1.0kg 5.0kg
Unknowns:
a
T
m
T  m A a  4.0kg(1.96 s 2 )  7.8N
This problem can also be done without the addition of equations if you are able to look at the whole system
and identify the weight of block B as the net force (weight A and N cancel out, and the two Ts cancel out). By
setting that net force equal to the total mass time acceleration, you have WB = (mA + mB)a. Then write the
equation for one of the blocks and solve for the tension.
7) On the Atwood’s machine below, the pulley is frictionless and massless. What is the acceleration of the two
blocks, and how much tension is in the string?
Choosing your coordinate system to
follow the path of the string
simplifies the problem!
10.N
5.0N
Knowns:
W 1 = 10 N
W 2 = 5.0 N
g = 9.80 m/s2
Unknowns:
A and T
T
T
10.N
5.0N
Solution:
First write the equation for each object:
Left weight:
W 1 - T = m1a
Right weight: T - W 2 = m2a
(There are two unknowns in each equation, so the
equations need to be added together.)
W 1 - T = m1a
T - W 2 = m2a
W 1 - W 2 = (m1 + m2)a
Solve for a then take that and solve for T
a
W1
W1 - W2 10.0N  5.0N
5.0N


 3.27 m2
10N 5N
s
m1  m2
1.53kg
2
9.8m/s
W2
T  W1  m1a  10N 
(
10N
9.8m/ s 2
(3.27
)  6.67N
m
)
s2
Part 3: The friction force
Whenever two objects are in contact and at least one object moves, or tries to move, there is a force called
friction opposing that motion. Even smooth surfaces have tiny peaks and valleys that the surfaces are moved
across. If there were such a thing as a perfectly smooth surface, there would still be electrostatic attractions
causing a force. The smallest possible force needed to get an object started in motion is called static friction (f s).
The force which must be overcome to keep an object in motion over a surface is called kinetic friction (fk). The
kinetic friction force is smaller than the static friction force. In our problems, we will only be concerned with the
kinetic friction.
The amount of the friction force depends on two factors. One is the amount of force with which the two surfaces
are pressed together. This force is the normal force. The other factor is the roughness of the two surfaces. The
variable that covers this is called the coefficient of friction. The symbol is , pronounced “mew”. The equation for
the coefficient of friction is:

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f
N
coefficient of friction =
friction force
normal force
Sample problem:
8) A box with a mass of 20.kg is pushed across a surface with a constant horizontal force of 40.0 N. What is the
coefficient of friction?
Knowns:
Solution: (first write 2nd Law equations)
N
m = 20 kg
hor: F - f = 0 (We know this because the object is moving at a
F = 40 N
constant speed.)
g = 9.80 m/s2
ver: N - W = 0
Unknowns:
f = 40 N and N = 20kg(9.8m/s2) = 196 N.
f
F
N, f, 
f
40N
 
 .20 (Notice that  is a ratio, and has no units.)
N 196N
W
Using an inclined plane to determine 
A clever method of determining the coefficient of friction for two surfaces is to make use of an inclined plane.
Let’s say you want to know the coefficient of friction between rubber tires and concrete. You would place a piece
of rubber tire on a piece of concrete, then tilt the concrete until a small tap on the rubber would cause it to slide
down the concrete at a constant speed. Let’s say that the angle for this to occur is 35.
N
2nd Law equations:
F - f = 0 which becomes f = sin35W
N - F = 0 which becomes N = cos35W
f
F

35o
F
f
sin35 o W

 tan35 o .70 (Notice that the weight does not matter.)
o
N cos35 W
This shows that whenever gravity causes an object to slide down an
incline at a constant speed, the coefficient of friction between the two
surfaces is the tangent of the angle of the incline.   tan
W
Sample force problems involving the friction equation:
9) Billy Yunair is driving his 1000.kg car at 25 m/s and has to make an emergency stop. If he brakes with the
wheels locked on a surface where  = .50, how far will the car travel before stopping?
N
Knowns:
Solution:
m = 1000 kg
vi = 25 m/s
vf = 0
= .50
Unknowns:
W, N, f, a, d
fk
W
vert: N - W = 0 which becomes N = 1000kg(9.8 m/s2) = 9800 N
hor: -f = ma since f = N, we can write -N = ma
Putting in values is -.50(9800 N) = (1000 kg)a, a = -4.9 m/s2
d
v 2f  v i2 0  (25 ms ) 2

 64m
2a
2(4.9 sm2 )
10) In the system below, block A has a mass of 5.0 kg, and block B has a mass of 2.0 kg. If the coefficient of
friction between the table and the block is .30, what is the acceleration of the two blocks?
A
T
N
B
f
A
Solution:
vert: N - W 1 = 0, N = 5kg(9.8m/s2) = 49N
T
B
W1
W2
Knowns:
mA = 5.0 kg
mB = 2.0 kg
 = .30
Unknown:
a
Hor: T -f = m1a
W 2-T = m2a
W 2 -f = (m1 + m2)a
f = N = .30(49 N) = 14.7 N,
2.0kg(9.8 sm2 )  14.7N 4.9N
W2  f
a


.70 sm2
m1  m 2
5.0kg  2.0kg
7.0kg
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11) A box of weighing 150 N will slide with a constant speed down a ramp inclined at 15. At what rate will the
box accelerate down a ramp made of the same material inclined at 30?
N
Solution:
Knowns:
First, to find , we make use of the fact that if the block slides at a
W = 150 N
constant speed,  = tan, so  = tan 15 = .27.
1 = 15o
2 = 30o
perp: N - cos30(150 N) = 0 or N = 130. N
Unknowns:
F
par: sinW - f = ma, or sinW - N = ma

30o
f
75N - 34.8N
sin30o (150N)  .27(130N)  150N 2 a, a =
 2.6 m2
a
9.8m/s
s
F W
15.3kg
12) A 20.kg table is being pushed across a floor where  = .20. The force on the table is 40.0 N at 30 above
horizontal. A) What is the friction force? B) What is the acceleration of the table?
40 N
N
30o
f
W
Knowns:
m = 20 kg
F = 40 N
 = 30
= .20
Unknowns:
N
f
a
Solution:
A) to find the friction force we must first find the
normal force.
N + sin30(40N) - W = 0
N = 20 kg(9.8 m/s2) - 20 N = 78 N
f = N = .20(78 N) = 15.6 N
B) cos30(40 N) - f = ma, 34.6 N - 15.6 N = (20kg)a
a = .95 m/s2
Problems:
1) a) Suppose Itsok Steubiyue is in the act of vertically jumping upward from the ground. His legs are flexed and
pushing on the ground so the center of his body is accelerating upward. Draw separate force diagrams for
Itsok and the earth. Show the relative sizes of the forces by using longer arrows for larger forces and equal
length arrows for forces of the same magnitude. For part a, write out each force in words (In the form of object
pushing object direction.) and somehow identify which of the forces are Newton’s 3rd law pairs.
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b) Repeat the instructions in part a), but now Itsok has left the ground and is moving upward. (You do not
need to write forces in words.)
c) Repeat the instructions in part a), but now Itsok is in contact with the floor again and slowing down. (You do
not need to write forces in words.)
2) For each of the situations below, draw a force diagram for each block at the moment they hit each other on a
table without any friction. Show the relative sizes of the forces by using longer arrows for larger forces and
equal length arrows for forces of the same magnitude. For each section tell which block exerts the largest
force on the other and which block has the greatest acceleration in the collision.
a) B moving and A is at rest
B 2 kg
A 1 kg
b) A moving and B is at rest
B 2 kg
A 1 kg
c) both A and B are moving
B 2 kg
A 1 kg
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3) For each of the situations below, draw a force diagram for each block as it moves across the table with some
friction. Show the relative sizes of the forces by using longer arrows for larger forces and equal length arrows
for forces of the same magnitude. For each section tell which block exerts the largest force on the other.
a) constant velocity
A 1 kg
B 2 kg

A
B
b) constantly gaining speed
A 1 kg
B 2 kg

A
B
c) constantly gaining speed at the same rate as blocks in part b)
A 1 kg
B 1 kg

A
B
d) constantly losing speed
B 2 kg
B
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A 1 kg

A
4) Harry Back tosses a ball into the air, but it is no longer in contact with his hand. The
top position is the highest point in the trip where it stops going up.
a) On the each position of the ball, draw and label a vector for each force acting on it.
b) Now put into words the object causing each force and the direction it is acting.
Lower position
Top position
Forward:
Backward:
Upward:
Downward:
c) Do the horizontal forces add to zero? How do you know?
d) Do the vertical forces add to zero? How do you know?
e) What are the reaction forces to each force listed in b)?
f) What is the horizontal equation for the ball’s net force?
g) What is the vertical equation for the ball’s net force?
h) What is the ball’s acceleration at the very top?
5) The ball rolling to the right across the horizontal table at a constant speed.
a) On the ball outline, draw and label a vector for each force acting on it.
b) Now put into words the object causing each force and the direction it is acting.
Forward:
Backward:
Upward:
Downward:
c) Do the horizontal forces add to zero? How do you know?
d) Do the vertical forces add to zero? How do you know?
e) What are the reaction forces to each force listed in b)?
f) What is the horizontal equation for the ball’s net force?
g) What is the vertical equation for the ball’s net force?
h) Draw a force diagram for a ball that is slowing
down as it goes across the table.
i) What direction is the net force on the ball in h)? How do you know? What is the equation for
the net force?
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6) The ball has rolled off the table and is falling toward the floor?
a) On the ball, draw and label a vector for each force acting on it.
b) Now put into words the object causing each force and the direction it is acting.
Forward:
Backward:
Upward:
Downward:
c) Do the horizontal forces add to zero? How do you know?
d) Do the vertical forces add to zero? How do you know?
e) What are the reaction forces to each force listed in b)?
f) What is the horizontal equation for the ball’s net force?
g) What is the vertical equation for the ball’s net force?
h) If a ball with 5 times the mass were to roll off the table at the same speed, how would the net force on it
compare to the net force on the lighter ball? How do you know?
i) How would the distance the heavy ball travels before hitting the floor compare to the distance the light ball
travels? How do you know?
7) The block is being pulled across the rough table and is gaining speed.
The string is at a constant angle .
a) On the block, draw and label a vector for each force acting on the block.
b) Which is exerting more force; the earth downward or the table upward? How do you know?
c) Do the vertical forces add to zero? How do you know?
d) Do the horizontal forces add to zero? How do you know?
e) What is the reaction force to the string pulling the block left?
f) What horizontal force is pulling the block to the left?
g) What is the equation for the force in f)?
h) What is the equation for the net horizontal force?
After the block is moving rapidly the string then breaks loose from the block.
i) On the block to the right, draw and label a vector for each force acting on the block.
j) Describe the motion of the block after the string breaks. (While block is on table.)
k) What force causes the block to keep moving forward after the string breaks?
l) Are the forces balanced after the string breaks? If not, what forces are not balanced?
m) If the table had been frictionless, describe the motion after the string breaks. (While block is on table.)
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8) Minnie Sota is riding in an elevator. On each small diagram of Minnie, draw and label a vector for each force
acting on Minnie for the situation described. If the magnitudes of the forces are not equal, show this by
making the vectors different length.
a) the elevator is at rest
b) the elevator is decreasing
c) the elevator is decreasing
its speed upward
its speed downward
d) the elevator is increasing
its speed upward
e) the elevator is moving
downward at 1.5 m/s.
f) The elevator cable has
broken
g) In which situations are the forces on Minnie balanced? How do you know?
h) In which situations is/are the upward forces greater than the downward? How do you know?
i) Put the downward force in words. What is the reaction force for it?
j) What is the equation for the net force on Minnie in a)? What is the equation in b)? In e)?
9) Ann Tique is skiing down a hill that has no friction.
a) On Ann, draw and label a vector for each of the two forces acting on her.
b) Now put into words the object causing each force and the direction it is acting.
c) What are the reaction forces to each force listed in b)?
d) Which is greater, the normal force on Ann or the weight force?
e) What is the direction of the net force on Ann? How do you know?
f) Since you (hopefully) put no forces parallel to the hill put in words the force making her go down the hill.
g) What is the equation for the force in e)?
h) What is the equation for the net force on Ann?
Ann got careless and flew off the hill. She is now in the air.
i) On Ann, draw and label a vector for each of the forces acting on her.
j) Ann smashes into the ground while traveling at 80 miles/hour. Which force is greater, the force
of Ann on the Earth, or the Earth on Ann? How do you know?
k) When Ann hits the ground, which has the greatest acceleration; Ann or the Earth? How do you know?
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10) A force F causes a mass m to accelerate at a rate of a m/s2.
a) What will the rate be with twice the force on the same mass?
b) What will the rate be with twice the force on half the mass?
c) What will the rate be with half the force on four times the mass?
d) What will the rate be with 10 times the force on 1/10th the mass?
11) A car engine has the ability to produce a maximum force of 15,000 N.
a) If placed in an 820 kg car, what is the cars
b) If the force opposing the car’s motion is 2,400N, at
maximum rate of acceleration?
what rate will the car accelerate?
12) Amanda Hugg (mass of 50.0 kg) is water skiing. She is pulled from rest to 15.0 m/s in 2.00 seconds by a
horizontal towrope.
a) What is the magnitude of the net force acting on Amanda?
b) After she reaches 15m/s she stays at that
speed for 10.0 s. What is the net force on
her?
c) If the tension in the rope is 60.N, how large is the
force resisting her motion? (In part b.)
13) Sara N. Dippity’s 1000.kg car is accelerated at 1.25 m/s2.
b) If there is a friction force of 140.N opposing the
a) What is the magnitude of the net force on her
cars motion, how hard is the ground pushing the
car?
car forward?
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14) As Sue Perlative rode her bicycle she was observed to undergo the motion represented by the graph below.
Draw a quantitative graph for net force versus time if the bicycle and rider had a mass of 80.0 kg.
15
F (N)
V (m/s)
10
5
0
0
2
4
6
t (s)
0
2
4
6
t (s)
15) A 12,000 kg Mack truck smashes into a 660 kg Honda Civic that was at rest. The Civic flies off at 15 m/s. If
they are in contact for .50 s;
a) How much force did the truck exert on the Civic?
b) How much force did the Honda exert on the truck?
c) What was the rate of acceleration for the truck?
16) A grocery bag can withstand 200.N of force against the bottom before it breaks. If there are 15.0 kg of
groceries in the bag, what is the maximum acceleration upward the bag can be lifted at?
17) Dan Sinkween, a 60.0 kg enterprising physics student is riding an elevator while standing on a bathroom
scale that gives the value in newtons. What does the scale read in the following situations?
a) moving upward at 3.0 m/s?
b) accelerating upward at 1.0 m/s2.
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c) accelerating downward at 1.3 m/s2
d) If the scale reads 720 N, at what rate is the elevator accelerating?
e) His friend Kit Chentabel has a reading of 820 N while accelerating at 2.40 m/s2 upwards. What is Kit’s
mass?
18) What is the acceleration of the blocks on the inclines below if there is no friction?
a)
b)
20kg
100N
30o
25
19) The surface below is frictionless. a) What is the normal force exerted by the table on the block?
b) What is the acceleration of the block?
30 N
37o
20 N
20) On the frictionless table below, what is the acceleration of the blocks and the tension in the strings for the
following?
a) block A is 3.0 kg and B is .50 kg
b) block A is 19.6 N and B is 4.9 N
A
B
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21) On the Atwood’s machine below, each of the blocks is 1.0 m from the floor and the pulley and strings are
massless. For the following, what is the acceleration, the time for a block to hit the floor, and the tension in
the string?
a) block A is 2.0 kg and B is 2.5 kg
b) block A is 10. N and B is 15 N
A
B
22) Ann Phibian (mass 30 kg) is being pulled at a 2.0 m/s on a sled mass (5.0 kg). The rope has a tension of 50.0
N. What is the force of friction and the coefficient of friction if the rope is horizontal?
23) A box with a weight of 300.N is pushed with a horizontal force of 150.N. If the coefficient of friction is .30,
what is the acceleration of the box?
15kg
24) The box at right is sliding down the incline at a constant speed.
a) What is the friction force?
b) What is the normal force?
c) What is ?
20o
25) Clara Nett is speeding down the highway at 30.0 m/s in her 1987 Corvette when she sees her boyfriend hitch
hiking. The coefficient of friction between the tires and the road is .30.
A) If she and the car have a mass of 800.kg, what is the force of friction between her tires and the road?
B) How much time does it take the car to stop?
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26) On the table below, if  = .15, what is the acceleration of the blocks and the tension in the strings if block A is
8.0 kg and B is 1.5 kg?
A
B
27) Paul Lution is skiing down a slope inclined at 25. Paul weighs 750 N and  = .10 for the slope and his skis.
If the slope is 300.m long, and he starts from rest, how long will it take him to get to the bottom?
28) The box below has a mass of 30.0 kg and  between the box and the floor is .25. If the following forces push
the box, calculate the friction force and the acceleration of the box for each case.
a) A 120 N force at 20 below horizontal
20o
30.0 kg
b) A 120 N force at 20 above horizontal
20o
29) Ima Fendid and Ona Roll are both pulling on Bobby Pinn. Ima is pulling South with a force of 90.N, and Ona
is pulling with a force 70.N at 45 N of W. If Bobby weighs 850N, how fast will he accelerate and in what
direction?
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Extra problems:
1) To accelerate a 50.N weight on a horizontal frictionless surface at 4.0 m/s2 requires how much force?
b) To accelerate the weight upward at 4.0 m/s2 requires how much force?
c) If the object is being pulled upward with a rope that can withstand a maximum tension of 120 N, what is the
maximum acceleration of the weight?
d) To accelerate the weight at 4.0 m/s2 on a surface where  = .30 requires how much force?
2) A 1300.kg car moving at 20.0 m/s stops in 80.0 m. What is the value of the force stopping the car?
3) Do number 16 with these blocks: block A is 200.g and B is 40.g
4) A 5.0 kg box is pushed horizontally on the floor to a velocity of 3.0 m/s and released. If  = .40, how far will the
box travel before stopping?
5) Do number 18 with the rope is 40. above horizontal?
6) Do problem 22 with block A is 15 N and B is 65 N.
7) Claire Voyant is on a sliding trip. She starts 200 m from the bottom of a 15 hill. If Claire and her sled have a
mass of 80.0 kg, and the coefficient of friction is .10, how much time will it take for her to go down the hill?
8) The force on the block below causes it to accelerate at 2.4 m/s2. What is the  between the surfaces?
65N
20
15 kg
9) What is the acceleration of the blocks on the incline below if there is no friction?
100N
50N
25o
10) On the table to the right,  = .40 . Calculate a, T1 and T2 .
10 kg
4.0kg
T1
T2
6.0kg
11) On the inclined plane below, the coefficient of friction is .30. If A is 7.0 kg and B is 6.0 kg, what is the
acceleration and the tension in the string? (Block A goes to the right.)
A
B
30o
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