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Physics 11 Assignment KEY
Dynamics – Chapters 4 & 5
Note: for all dynamics problem-solving questions, draw appropriate free body diagrams and
use the aforementioned problem-solving method.
1. Define the following terms:
Inertia - the natural tendency of an object to
stay at rest or continue its motion in a straight
line at constant speed in the absence of
outside forces; objects with greater mass have
greater inertia
Dynamics - the study of the motions of bodies
while considering their masses and the
responsible forces
Mechanics - the branch of physics comprising
kinematics and dynamics; simply, the how and
the why of simple motion
Newton’s Laws of Motion - three fundamental
laws of motion which are the basis of
Newtonian mechanics are: 1) an object will
remain at rest or in straight-line motion unless
acted on by an outside force; 2) the
acceleration of an object is proportional to the
force acting on it and
inversely proportional to its mass; 3) for every
action force on an object, the object exerts and
equal and opposite reaction force
Force - an action, like a push or a pull, that
causes a change in motion of an object
Inertial frame of reference - a frame of
reference in which the law of inertia is valid; it
is a non-accelerating frame of reference
Non-inertial frame of reference - an
accelerating frame of reference
Mass - the quantity of matter an object
contains; determined through the inertial
properties of an object or its gravitational
influence on other objects
Inertial mass - the property of matter that
resists a change in motion
Gravitational mass - the property of matter
that determines the strength of the gravitational
force
Contact force - the force exerted by an object
in direct contact with another object
Non-contact force - the force that acts even
though objects are separated by a distance,
such as attraction or repulsion between
magnets
Weight - the force that gravity exerts on an
object because of its mass
Static frictional force - a frictional force that
acts to keep an object at rest; measured as the
force required to move an object from rest
Kinetic frictional force - a frictional force that
acts to slow the motion of an object; measured
as the force required to just keep an object
sliding over another object
Coefficient of friction - the ratio of frictional
force to the normal force between two object
surfaces
Normal force - a force that acts in a direction
perpendicular to the common contact surface
between two objects
Net force - the vector sum of all forces acting
on an object
2. Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull?
The child tends to remain at rest (Newton’s 1st Law), unless a force acts on her. The force
is applied to the wagon, not the child, and so the wagon accelerates out from under the
child, making it look like the child falls backwards relative to the wagon. If the child is
standing in the wagon, the force of friction between the child and the bottom of the wagon
will produce an acceleration of the feet, pulling the feet out from under the child, also
making the child fall backwards.
3. If the acceleration of a body is zero, are no forces acting on it?
If the acceleration of an object is zero, then by Newton’s second law, the net force must be
zero. There can be forces acting on the object as long as the vector sum of the forces is
zero.
4. Why do you push harder on the pedals of a bicycle when first starting out than when moving at a
constant speed?
You push harder on the pedals of a bicycle when first starting out than when moving at
constant velocity because your applied force must overcome static friction. Static friction
is greater than kinetic friction. Once the bicycle is moving a constant velocity, less applied
force is required to equal kinetic friction.
5. Only one force acts on an object. Can the object have zero acceleration? Can it have zero
velocity?
If only one force act on an object, then the net force is greater than zero and, according to
the 2nd law of motion, the acceleration cannot be zero and it cannot have zero velocity.
6. The force of gravity on a 2-kg rock is twice as great as that on a 1-kg rock. Why then doesn’t the
heavier rock fall faster?
The acceleration of both rocks is found by dividing their weight (the force of gravity on
them) by their mass. The 2-kg rock has a force of gravity on it that is twice as great as the
force of gravity on the 1-kg rock, but also twice as great a mass as the 1-kg rock, so the
acceleration is the same for both.
7. A person exerts an upward force of 40 N to hold a bag of groceries. Describe the ‘reaction’ force
by stating (a) its magnitude, (b) its direction, (c) on what body it is exerted, and (d) by what body it
is exerted.
(a) The magnitude is 40 N.
(b) The direction is downward.
(c) It is exerted on the person.
(d) It is exerted by the bag of groceries.
8. Why is the stopping distance of a truck much shorter than for a train going the same speed?
The stopping distance for a truck is much shorter than that of a train going the same speed
because, even though the rate of slowing down, i.e. acceleration, is the same, the truck’s
mass is significantly less than the train’s mass. Therefore, according to the Second law of
motion, a greater force will be required to slow down the train and thus a longer period of
time and greater stopping distance.
9. You can hold a heavy box against a rough wall and prevent it from slipping down by pressing only
horizontally. How can the application of a horizontal force keep an object from moving vertically?
By pressing the block against the rough wall, you increase the normal force of the wall on
the block. As the normal force increases, the amount of static friction between the block
and the wall increases. This static friction force is vertical and opposes gravity or the
weight of the block. Provided that the static friction force equals the force of weight, the
block will not slide down the wall.
10. a) A box sits at rest on a rough 30o inclined plane. Draw free body diagram, showing all the forces
acting on the box. b) How does the diagram change if the box were sliding down the plane? c)
How does the diagram change if the box were sliding up the plane?
(a)
(b)
(c)
In (a) the friction is static and opposes the impending motion down the plane.
In (b) the friction is kinetic and opposes the motion down the plane.
In (c) the friction is kinetic and opposes the motion up the plane.
11. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s 2?
The net force, Fnet, is due to some applied force, Fa, and
friction, Ff. Since we do not know any of these individual
forces, solve for Fnet.
Fnet  m  a  (60.0 kg) (1.15 m/s 2 )  69.0 N
12. A net force of 255 N accelerates a bike and rider at 2.20 m/s 2. What is the mass of the bike and
rider?
The net force, Fnet, is due to some applied force, Fa, and
friction, Ff and is 255 N.
Fnet  m  a
m 
Fnet
255 N

 116 kg
a
2.20 m/s 2
13. How much tension must a rope withstand if it is used to accelerate a 1050-kg car horizontally at
1.20 m/s2? Ignore friction.
The net force, Fnet, is due to tension in the rope, FT.
Fnet  FT  m  a  (1050 kg) (1.20 m/s 2 )  1260 N
14. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the Moon (g = 1.7 m/s2), (c) on Mars
(g = 3.7 m/s2), and (d) in outer space traveling with constant velocity?
The mass remains the same in all
four situations, 66 kg, but the
weight is different:
(a)
FG  m  g  (66 kg)(-9.81 m/s 2 )  647 N
(b)
FG  m  g  (66 kg)(-1.7 m/s 2 )  112 N
(c)
FG  m  g  (66 kg)(-3.7 m/s 2 )  244 N
(d)
FG  m  g  (66 kg)(0 m/s 2 )  0 N
15. A 20.0-kg box rest on a table. (a) What is the weight of the box and the normal force acting on it?
(b) A 10.0-kg box is placed on top of the 20.0-kg box. Determine the normal force that the table
exerts on the 20.0-kg box and the normal force that the 20.0-kg box exerts on the 10.0-kg box.
(a) The weight of the box is:
FG  m  g  (20.0 kg)(-9.81m/s2 )  -196 N
The normal force acting on it is:
FN  FG  m  g  (20.0 kg)(-9.81m/s2 )  196 N
(b) We select both objects and apply the Second Law of Motion:
FN  FG  (m20  m10 )  g  (10.0 kg + 20.0 kg)(-9.81m/s 2 )  294 N
If we select the top block as the object, then:
FN  FG  m  g  (10.0 kg)(-9.81m/s2 )  98.0 N
16. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?
The average force required to stop the car would be a
combination of the braking force and the frictional force (call it
FA). The acceleration can be found from the car’s 1-D motion:
v f  vi  at
v f  vi
0 m/s - 25 m/s
 3.13 m/s2
t
8.0 s
We apply the Second Law of Motion:
a 

FA = m  a = (1100 kg)(  3.13 m/s2 )  -3440 N
17. What average force is needed to accelerate a 7.00-g pellet from rest to 175 m/s over a distance of
0.700 m along the barrel of a rifle?
The average force required to accelerate Bullet Bill comes from
the explosion in the gun (call it FA). The acceleration can be
found:
18. A 10.0-kg bucket is lowered by a rope in which there is 63.0 N of tension. What is the
acceleration of the bucket? Is it up or down?
Fnet = FG  FT = m g  FT = m  a
(10 kg)(9.81 m/s2 )  (63 N) = (10 kg)(a)
 a  3.51 m/s2 (down )
19. A person stands on a bathroom scale in a motionless elevator. When the elevator begins to
move, the scale briefly reads only 0.75 of the person’s weight. Calculate the acceleration of the
elevator and find the direction of acceleration.
The scale reads the force the person exerts on the scale. From the Third
Law of Motion, this is also the magnitude of the normal force acting on
the person.
FG  FN = m  a
m g  FN = m  a
mg  0.75mg  ma
 a  (1-0.75)(9.81 m/s 2 )  2.45 m/s 2 ( down)
20. If the coefficient of kinetic friction between a 35-kg crate and the floor is 0.30, what horizontal force
is required to move the crate at a steady speed across the floor? What horizontal force is required
if the coefficient of kinetic friction is zero?
Because the crate is moving at constant speed, the acceleration
is zero. Using the Second Law of Motion for each bucket:
Fnet = FA - F f = 0
 FA  F f  k FN  (0.30)(35 kg)(9.81 m/s2 )=103 N
If µk is reduced to zero then the friction is eliminated and no FA
is required to keep the crate moving at constant speed (i.e. First
Law of Motion)
21. A force of 40.0 N is required to start a 5.0-kg box moving across a horizontal concrete floor. (a)
What is the coefficient of static friction between the box and the floor? (b) If the 40.0-N force
continues, the box accelerates at 0.70 m/s2. What is the coefficient of kinetic friction?
At the moment the box begins to move, the applied force equals
the frictional force and acceleration is instantaneously zero:
(a)
Fnet = FA - F f = 0
 FA  F f  s FN  40.0 N
s 
Ff
FN

40.0 N
 0.815
(5.0 kg)(9.81 m/s 2 )
(b)
Fnet = FA - F f = m  a = (5.0 kg)(0.70 m/s 2 ) = 3.5 N
 F f  FA  3.5 N = 40.0 N - 3.5 N = 36.5 N
s 
Ff
FN

36.5 N
 0.744
(5.0 kg)(9.81 m/s 2 )
22. An Atwood machine consists of masses 3.8 kg and 4.2 kg. What is the acceleration of the
masses? What is the tension in the rope?
 m  m1 
 4.2 kg  3.8 kg 
  (9.81 m/s 2 )
  0.491 m/s 2
a  g  2
m

m
4
.
2
kg

3
.
8
kg


1 
 2
23. The smaller mass on an Atwood machine is 5.2 kg. If the masses accelerate at 4.6 m/s 2, what is
the mass of the second object? What is the tension in the rope?
 m  m1 

a  g  2
 m2  m1 
am2  m1   g m2  m1 
am2  am1  gm2  gm1
am2  gm2  am1  gm1
m2 a  g   m1  a  g 
  4.6 m/s 2  9.81 m/s 2
ag
  5.2 kg
m2  m1 
2
2
 ag 
 4.6 m/s  9.81 m/s
m2  14.4 kg



24. Stacie, who has a mass of 45 kg, starts down a slide that is inclined at an angle of 45° with the
horizontal. If the coefficient of kinetic friction between Stacie’s shorts and the slide is 0.25, what is
her acceleration?
25. You are shadowing a nurse in the emergency room of a local hospital. An orderly wheels in a
patient who has been in a very serious accident and has had severe bleeding. The nurse quickly
explains to you that in a case like this, the patient’s bed will be tilted with the head downward to
make sure the brain gets enough blood. She tells you that, for most patients, the largest angle that
the bed can be tilted without the patient beginning to slide off is 32.0° from the horizontal.
a. On what factor or factors does this angle of tilting depend?
The coefficient of static friction between the patient and the bed’s sheets.
b. Find the coefficient of static friction between a typical patient and the bed’s sheets.