The Glenelg School of Abu Dhabi
Answer Key Practice Sheets #4-Term 2
Subject:
Unit/Topic:
Pre-AP Physics
Chapter 4: Dynamics;
Newton’s Laws of Motion
Student’s
Name:
Grade
Level:
11
Date:
Term 2
NOTE
The questions in the Practice Sheets are intended to help you
practice solve problems and build up Problem-Solving strategies.
The questions are not necessarily selected to be given in your
tests or final exams.
Ch 4: Dynamics; Newton’s Laws of Motion
Summary of Equations
1- 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑
2- 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
=
=
𝑑
𝑎 = acceleration
𝑥 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡 = time
𝑣 = velocity or speed
𝑡
×−×0
𝑡
=
∆𝑥
𝑡
3- When the velocity is constant (𝑎 = 0)

𝑥 = 𝑥0 + 𝑣𝑡
4- The kinematic equations (When the acceleration 𝑎 is constant)
𝑣 = 𝑣0 + 𝑎𝑡
𝑥 = 𝑥0 + 𝑣0 𝑡 + ½𝑎𝑡 2
𝑣 2 = 𝑣02 + 2𝑎(𝑥 − 𝑥0 )
𝑣+𝑣
𝑥 = ( 2 0 ) 𝑡 + 𝑥0
∆𝑦 = (
or
or
or
∆𝑥 = 𝑣0 𝑡 + ½𝑎𝑡 2
𝑣 2 = 𝑣02 + 2𝑎∆𝑥
𝑣+𝑣
∆𝑥 = ( 2 0 ) 𝑡
𝑣 + 𝑣0
)𝑡
2
Answer Key Practice Sheets #4-Ch 4
The Glenelg School of Abu Dhabi: Page 1 of 12
The Glenelg School of Abu Dhabi
Summary of Equations
𝑎 = acceleration
𝑦 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡 = time
𝑣 = velocity or speed
5- Free Fall
𝑣 = 𝑣0 + 𝑔𝑡
∆𝑦 = 𝑣0 𝑡 + ½𝑔𝑡 2
𝑣 2 = 𝑣02 + 2𝑔∆𝑦
𝑎 = acceleration
𝐹 = force
𝑚 = mass
𝑓𝑓𝑟𝑖𝑐 = friction force
6-Net Force
∑ 𝐹 = 𝐹𝑛𝑒𝑡 = 𝑚𝑎
GEOMETRY AND TRIGONOMETRY
Rectangle: 𝐴 = 𝑏 × ℎ
1
Triangle: 𝐴 = 2 𝑏 × ℎ
Trapezoid: 𝐴 =
1
2
(𝑏1 + 𝑏2 ) × ℎ
Right Triangle
𝑎2 + 𝑏 2 = 𝑐 2 ; 𝐬𝐢𝐧 𝜽 =
𝒂
𝒄
; 𝐜𝐨𝐬 𝜽 =
CONSTANTS
Acceleration due to
gravity
at Earth’s surface
g=9.80 m/s2
𝒃
𝒄
; 𝐭𝐚𝐧 𝜽 =
𝒂
𝒃
UNITS
PREFIXES
Name
Symbol
meter
m
kilogram
kg
second
s
newton
N
Answer Key Practice Sheets #4-Ch 4
Factor
109
106
103
10-2
10-3
10-6
10-9
10-12
Prefix
giga
mega
kilo
centi
milli
micro
nano
pico
Symbol
G
M
k
c
m
𝜇
n
p
The Glenelg School of Abu Dhabi: Page 2 of 12
The Glenelg School of Abu Dhabi
VALUES OF
TRIGONOMETRIC
FUNCTIONS FOR COMMON
ANGLES
𝜃
sin 𝜃
cos 𝜃
tan 𝜃
0°
0
1
0
1
2
or 0.50
37°
3
5
√3
2
or 0.87
√3
3
or 0.60
4
5
or 0.80
3
4
45°
√2
2
or 0.71
√2
2
or 0.71
53°
4
5
or 0.80
3
5
or 0.60
1
2
or 0.50
60°
√3
2
30°
90°
Answer Key Practice Sheets #4-Ch 4
or 0.87
1
0
or 0.58
or 0.75
1
4
3
or 1.33
√3 or 1.73
∞
The Glenelg School of Abu Dhabi: Page 3 of 12
The Glenelg School of Abu Dhabi
Multiple Choice Questions
1) When you sit on a chair, the resultant force on you is
A) zero.
B) up.
C) down.
D) depending on your weight.
Answer: A
2) The acceleration of an object is inversely proportional to
A) the net force acting on it.
B) its position.
C) its velocity.
D) its mass.
Answer: D
3) A net force F accelerates a mass m with an acceleration a. If the same net force is applied to
mass 2m, then the acceleration will be
A) 4a.
B) 2a.
C) a/2.
D) a/4.
Answer: C
4) Action-reaction forces
A) sometimes act on the same object.
B) always act on the same object.
C) may be at right angles.
D) always act on different objects.
Answer: D
5) Action-reaction forces are
A) equal in magnitude and point in the same direction.
B) equal in magnitude but point in opposite directions.
C) unequal in magnitude but point in the same direction.
D) unequal in magnitude and point in opposite directions
Answer: B
6) An object of mass m sits on a flat table. The Earth pulls on this object with force mg, which
we will call the action force. What is the reaction force?
A) The table pushing up on the object with force mg.
B) The object pushing down on the table with force mg.
C) The table pushing down on the floor with force mg.
D) The object pulling upward on the Earth with force mg.
Answer: D
Answer Key Practice Sheets #4-Ch 4
The Glenelg School of Abu Dhabi: Page 4 of 12
The Glenelg School of Abu Dhabi
7) An example of a force which acts at a distance is
A) tension.
B) weight.
C) static friction.
D) kinetic friction.
Answer: B
8) A stone is thrown straight up. At the top of its path, the net force acting on it is
A) greater than its weight.
B) greater than zero, but less than its weight.
C) instantaneously equal to zero.
D) equal to its weight.
Answer: D
9) An object sits on a frictionless surface. A 16-N force is applied to the object, and it accelerates
at 2.0 m/s2. What is the mass of the object?
A) 4.0 kg
B) 8.0 kg
C) 32 kg
D) 78 N
Answer: B
10) Starting from rest, a 4.0-kg body reaches a speed of 8.0 m/s in 2.0 s. What is the net force
acting on the body?
A) 4.0 N
B) 8.0 N
C) 16 N
D) 32 N
Answer: C
11) A person on a scale rides in an elevator. If the mass of the person is 60.0 kg and the elevator
2
accelerates downward with an acceleration of 4.90 m/s , what is the reading on the scale?
A) 147 N
B) 294 N
C) 588 N
D) 882 N
Answer: B
Answer Key Practice Sheets #4-Ch 4
The Glenelg School of Abu Dhabi: Page 5 of 12
The Glenelg School of Abu Dhabi
12) Two horizontal forces act on a 5.0-kg mass. One force has a magnitude of 8.0 N and is directed
due north. The second force toward the east has a magnitude of 6.0 N. What is the acceleration of
the mass?
A) 1.6 m/s2 due north
B) 1.2 m/s2 due east
C) 2.0 m/s2 at 53° N of E
D) 2.0 m/s2 at 53 m E of N
Answer: C
13) In the Atwood machine shown in the adjacent figure , if M = 0.60 kg and m = 0.40 kg, what is
the magnitude of the acceleration of the system? (Ignore friction and the mass of the pulley.)
A) 5.3 m/s2
B) 3.9 m/s2
C) 2.0 m/s2
D) 0.98 m/s2
Answer: C
14) Two boxes of masses m and 2m are in contact with each other on a frictionless surface.
(See Fig. above) What is the acceleration of the more massive box?
A) F/m
B) F/(2m)
C) F/(3m)
D) F/(4m)
Answer: C
15) Two boxes of masses m and 2m are in contact with each other on a frictionless surface.
(See Fig. above) What is the net force on the more massive box?
A) 2/3 F
B) F
C) 3/2 F
D) 2F
Answer: A
Answer Key Practice Sheets #4-Ch 4
The Glenelg School of Abu Dhabi: Page 6 of 12
The Glenelg School of Abu Dhabi
Free Response Questions
1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.25 m/s2?
Answer
Use Newton’s second law to calculate the force.
 F  ma   60.0 kg  1.25 m s  
2
75.0 N
2. How much tension must a rope withstand if it is used to accelerate a 960-kg car horizontally
along a frictionless surface at 1.20 m s 2 ?
Answer
Use Newton’s second law to calculate the tension.
F  F
T


 ma   960 kg  1.20 m s 2  1.15 103 N
3. What average force is needed to accelerate a 7.00-gram pellet from rest to 125 m s over a
distance of 0.800 m along the barrel of a rifle?
Answer
The average force on the pellet is its mass times its average acceleration. The average acceleration is
found from the Equation below For the pellet, v0  0 , v  125m s , and x  x0  0.800 m .
aavg
125 m s   0


 9770
2  x  x0 
2  0.800 m 
2
v 2  v02


m s2

Favg  maavg  7.00  103 kg 9770 m s 2  68.4 N
4. How much tension must a rope withstand if it is used to accelerate a 1200-kg car vertically
upward at 0.80 m/s2
Answer
Choose up to be the positive direction. Write Newton’s 2nd law for the vertical
direction, and solve for the tension force.
F  F
T
 mg  ma  FT  m  g  a 


FT
mg
FT  1200 kg  9.80 m s 2  0.80 m s 2  1.3 104 N
Answer Key Practice Sheets #4-Ch 4
The Glenelg School of Abu Dhabi: Page 7 of 12
The Glenelg School of Abu Dhabi
5. 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 regular weight. Calculate the acceleration of the
elevator, and find the direction of acceleration.
Answer
There will be two forces on the person – their weight, and the normal force of the
scales pushing up on the person. A free-body diagram for the person is shown.
Choose up to be the positive direction, and use Newton’s 2nd law to find the
acceleration.
F  F
N
mg
 mg  ma  0.75mg  mg  ma 

FN

a  0.25 g  0.25 9.8 m s 2  2.5 m s 2
Due to the sign of the result, the direction of the acceleration is down. Thus the elevator must have
started to move down since it had been motionless.
6. A 650-N force acts in a northwesterly direction. A second 650-N force must be exerted in what
direction so that the resultant of the two forces points westward? Illustrate your answer with a
vector diagram.
Answer
The two forces must be oriented so that the northerly component of the first
force is exactly equal to the southerly component of the second force. Thus the second force must act
southwesterly. See the diagram.
F1
F2
 
F1  F2
Answer Key Practice Sheets #4-Ch 4
The Glenelg School of Abu Dhabi: Page 8 of 12
The Glenelg School of Abu Dhabi


7. The two forces F1 and F2 in the figures (a) and (b) shown below act on a 27.0-kg object on a
frictionless tabletop. If F1=10.2 N and F2 =16.0 N find the net force on the object and its acceleration
for (a) and (b).
Answer
F1
The net force in each case is found by vector addition with components.
(a)
FNet 
FNet x   F1  10.2 N
 10.2 2   16.0 2

FNet y   F2  16.0 N
 19.0 N
16.0
  tan 1
F2
Fnet
10.2
The actual angle from the x-axis is then 237.5 .
F
19.0 N
a  Net 
 0.703m s 2 at 237 o
m
27.0 kg
 57.5o
o
F2
(b) FNet x  F1 cos30o  8.83 N
FNet y  F2  F1 sin 30o  10.9 N
Fnet

30o
F1
Answer Key Practice Sheets #4-Ch 4
The Glenelg School of Abu Dhabi: Page 9 of 12
The Glenelg School of Abu Dhabi
8. The figure below shows a block mass m1  on a smooth horizontal surface, connected by a thin
cord that passes over a pulley to a second block m2 , which hangs vertically.
(a) Draw a free-body diagram for each block, showing the force of gravity on each, the force
(tension) exerted by the cord, and any normal force.
(b) Apply Newton’s second law to find formulas for the acceleration of the system and for the
tension in the cord. Ignore friction and the masses of the pulley and cord.
Answer
(a) See the free-body diagrams included.
y
(b) For block 1, since there is no motion in the vertical direction,
we have FN1  m1 g . We write Newton’s 2nd law for the x
direction:
F
1x
 FT  m1a1x . For block 2, we only need to
x
FT
FN1
FT
consider vertical forces:  F2 y  m2 g  FT  m2 a2 y . Since the two
m2 g
m1g
blocks are connected, the magnitudes of their accelerations will
be
the same, and so let a1 x  a2 y  a . Combine the two force equations from above, and solve for a by
substitution.
FT  m1a
m2 g  FT  m2 a
m1a  m2 a  m2 g 
ag
Answer Key Practice Sheets #4-Ch 4

m2 g  m1a  m2 a 
m2
m1  m2
FT  m1a  g
m1m2
m1  m2
The Glenelg School of Abu Dhabi: Page 10 of 12
The Glenelg School of Abu Dhabi
9. A student attaches a rope to a 196 N box of books. He pulls with a force of 200.0 N at an
angle of 37.0o with the horizontal. Assume the magnitude of the force of friction is 15.0 N.
(cos 53.0o≈0.600 & sin 53.0o≈0.800)
+y
F =200.0 N
53.0o
+x
(a) Draw the free body diagram of the box.
Fx
Fn
Ff
Fy
Fg
(b) Find the x and y components of F.
𝑭𝒙 = 𝑭 𝐜𝐨𝐬 𝜽
𝑭𝒙 = 𝟐𝟎𝟎. 𝟎 𝐜𝐨𝐬 𝟓𝟑. 𝟎𝒐
𝑭𝒙 = 𝟐𝟎𝟎. 𝟎(𝟎. 𝟔𝟎𝟎) = 𝟏𝟐𝟎. 𝑵
𝑭𝒚 = 𝑭 𝐬𝐢𝐧 𝜽
𝑭𝒚 = 𝟐𝟎𝟎. 𝟎 𝐬𝐢𝐧 𝟓𝟑. 𝟎𝒐
𝑭𝒚 = 𝟐𝟎𝟎. 𝟎(𝟎. 𝟖𝟎𝟎) = 𝟏𝟔𝟎. 𝑵
(c) Find the normal force (Fn).
𝑭𝒏 + 𝑭𝟏𝒚 = 𝑭𝒈
𝑭𝒏 + 𝟏𝟔𝟎. = 𝟏𝟗𝟔
𝑭𝒏 = 𝟏𝟗𝟔 − 𝟏𝟔𝟎. 𝟎 = 𝟑𝟔 𝑵 , in the +y direction
Answer Key Practice Sheets #4-Ch 4
The Glenelg School of Abu Dhabi: Page 11 of 12
The Glenelg School of Abu Dhabi
(d) Find the net force along the x-axis (ΣFx).
∑ 𝑭𝒙 = 𝑭𝒙 − 𝑭𝒇𝒓𝒊𝒄𝒕𝒊𝒐𝒏
∑ 𝑭𝒙 = 𝟏𝟐𝟎. −𝟏𝟓. 𝟎
∑ 𝑭𝒙 = 𝟏𝟎𝟓 𝑵, in the + x direction
(e) Find the mass of the crate (m).
𝒎=
𝑭𝒈
𝒈
𝒎=
𝟏𝟗𝟔
= 𝟐𝟎. 𝟎 𝒌𝒈
𝟗. 𝟖𝟎
(f) Find the acceleration of the crate.
𝒂=
∑ 𝑭𝒙
𝟏𝟎𝟓
=
= 𝟓. 𝟐𝟓 𝒎/𝒔𝟐 , 𝒊𝒏 𝒕𝒉𝒆 + 𝒙 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏
𝒎
𝟐𝟎. 𝟎
Practice Makes Perfect
Giancoli 4th Edition
Physics for Scientists & Engineers
End of Practice Sheets #4
Answer Key Practice Sheets #4-Ch 4
The Glenelg School of Abu Dhabi: Page 12 of 12