The Glenelg School of Abu Dhabi
Answer Key Practice Sheets #5-Term 2
Subject:
Unit/Topic:
Student’s
Name:
Pre-AP Physics
Grade
Level:
Chapter 5
Date:
11
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 5: Using Newton’ Laws: Friction, Circular Motion, Drag
Forces
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 #5-Ch 5
The Glenelg School of Abu Dhabi: Page 1 of 7
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
𝜇 = coefficient of friction
6-Net Force
∑ 𝐹 = 𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝑓𝑓𝑟𝑖𝑐 ≤ 𝜇𝑁
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 =
Answer Key Practice Sheets #5-Ch 5
𝒃
𝒄
; 𝐭𝐚𝐧 𝜽 =
𝒂
𝒃
The Glenelg School of Abu Dhabi: Page 2 of 7
The Glenelg School of Abu Dhabi
Table of information
CONSTANTS
UNITS
PREFIXES
Acceleration due to gravity
at Earth’s surface
g=9.80 m/s2
Name
Symbol
Factor
Prefix
Symbol
meter
m
Coulomb’s Law Constant
𝑘 = 9.0 × 109 N.m2/C2
kilogram
kg
109
106
103
10-2
10-3
10-6
10-9
10-12
giga
mega
kilo
centi
milli
micro
nano
pico
G
M
k
c
m
𝜇
n
p
second
s
newton
N
coulomb
C
* Remark: use √2 = 1.41 & √3 = 1.73
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
45°
√3
2
or 0.87
√3
3
or 0.60
4
5
or 0.80
3
4
√2
2
or 0.71
√2
2
or 0.71
53°
4
5
or 0.80
3
5
or 0.60
or 0.87
1
2
or 0.50
60°
√3
2
30°
90°
Answer Key Practice Sheets #5-Ch 5
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 7
The Glenelg School of Abu Dhabi
Multiple Choice Questions
16) The coefficients of static and kinetic frictions for plastic on wood are 0.50 and 0.40,
respectively. How much horizontal force would you need to apply to a 3.0 N plastic calculator to
start it moving from rest?
A) 0.15 N
B) 1.2 N
C) 1.5 N
D) 2.7 N
17) An object slides on a level surface in the +x direction. It slows and comes to a stop with a
2
constant acceleration of -2.45 m/s . What is the coefficient of kinetic friction between the object
and the floor?
A) 0.25
B) 0.50
C) 4.9
D) Impossible to determine without knowing the mass of the object.
18) A 10-kg box sitting on a horizontal surface is pulled by a 5.0-N force. A 3.0-N friction force
retards the motion. What is the acceleration of the object?
A) 0.20 m/s2
B) 0.30 m/s2
C) 0.50 m/s2
D) 5.0 m/s2
19) A horizontal force of 5.0 N accelerates a 4.0-kg mass, from rest, at a rate of 0.50 m/s2 in the
positive direction. What friction force acts on the mass?
A) 2.0 N
B) 3.0 N
C) 4.0 N
D) 5.0 N
20) A wooden block slides directly down an inclined plane, at a constant velocity of 6.0 m/s. What
Is the coefficient of kinetic friction, if the plane makes an angle of 25° with the horizontal?
A) 0.47
B) 0.42
C) 0.37
D) 0.91
21) An object with a mass m slides down a rough 37° inclined plane where the coefficient of kinetic
friction is 0.20. What is the acceleration of the object?
2
A) 4.3 m/s
B) 5.9 m/s2
2
C) 6.6 m/s
D) 7.8 m/s2
Answer: A
Answer Key Practice Sheets #5-Ch 5
The Glenelg School of Abu Dhabi: Page 4 of 7
The Glenelg School of Abu Dhabi
22) An object slides down a smooth inclined plane with a certain acceleration. The acceleration of
the object is proportional to
A) cosθ
B) θ.
C) sinθ.
D) the mass of the object
E) none of the above.
Free Response Questions
10. 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  k is zero?
Answer
A free-body diagram for the crate is shown. The crate does not
accelerate vertically, and so FN  mg . The crate does not accelerate
horizontally, and so FP  Ffr . Putting this together, we have


FP  Ffr  k FN  k mg   0.30  35 kg  9.8 m s 2  103  1.0  102 N
FN
Ffr
FP
mg
If the coefficient of kinetic friction is zero, then the horizontal force required is 0 N , since there is
no friction to counteract. Of course, it would take a force to START the crate moving, but once it
was moving, no further horizontal force would be necessary to maintain the motion.
11. What is the maximum acceleration a car can undergo if the coefficient of static friction between
the tires and the ground is 0.80?
Answer
FN
Ffr
mg
A free-body diagram for the accelerating car is shown. The car does not accelerate vertically, and so
FN  mg . The static frictional force is the accelerating force, and so Ffr  ma . If we assume the maximum
acceleration, then we need the maximum force, and so the static frictional force would be its maximum value
of s FN . Thus we have
Ffr  ma   s FN  ma   s mg  ma 


a   s g  0.80 9.8 m s 2  7.8 m s 2
Answer Key Practice Sheets #5-Ch 5
The Glenelg School of Abu Dhabi: Page 5 of 7
The Glenelg School of Abu Dhabi
12. A 15.0-kg box is released on a 32º incline and accelerates down the incline at 0.30 m s 2 .
(a) Find the friction force impeding its motion.
(b) What is the coefficient of kinetic friction?
FN
Ffr
y
x


mg
Answer
Start with a free-body diagram. Write Newton’s 2nd law for each direction.
F
F
x
 mg sin   Ffr  max
y
 FN  mg cos   ma y  0
Notice that the sum in the y direction is 0, since there is no motion (and hence no acceleration) in the y
direction. Solve for the force of friction.
mg sin   Ffr  max 

Ffr  mg sin   max  15.0 kg   9.80 m s 2
 sin 32   0.30 m s
o
2
  73.40 N  73 N
Now solve for the coefficient of kinetic friction. Note that the expression for the normal force comes from the
y direction force equation above.
Ffr
73.40 N
Ffr  k FN  k mg cos   k 

 0.59
mg cos  15.0 kg  9.80 m s 2 cos 32o



13. A box sits at rest on a rough 30º inclined plane.
(a) Draw the free-body diagram, showing all the forces acting on the box.
(b) How would the diagram change if the box were sliding down the plane?
(c) How would it change if the box were sliding up the plane after an initial shove?
Answer
FN
Ffr
y
x
 mg

(a) Here is a free-body diagram for the box at rest on the plane. The force of friction is a STATIC
frictional force, since the box is at rest.
(b) If the box were sliding down the plane, the only change is that the force of friction would be a
KINETIC frictional force.
(c) If the box were sliding up the plane, the force of friction would be a KINETIC frictional force,
and it would point down the plane, in the opposite direction to that shown in the diagram.
Answer Key Practice Sheets #5-Ch 5
The Glenelg School of Abu Dhabi: Page 6 of 7
The Glenelg School of Abu Dhabi
14. A 50.0 kg person on skis is going down a hill sloped at 37.0o. The force of friction between the
skis and the snow is 14.0 N.
(a) Draw the free-body diagram, showing all the forces acting on the skies.
Fn
Ff
37O
Fgx
Fgy
Fg
(b) Find the components of the weight that are parallel and perpendicular to the plane.
𝐹𝑔𝑥 = 𝐹𝑔 sin 𝜃
𝐹𝑔𝑥 = (50.0 × 9.8 )sin 37.0𝑜
𝐹𝑔𝑥 ≈ 295 𝑁
𝐹𝑔𝑦 = 𝐹𝑔 cos 𝜃
𝐹𝑔𝑦 = (50.0 × 9.8 )cos 37.0𝑜
𝐹𝑔𝑦 ≈ 391 𝑁
(c) Find the net force acting on the skis in the x-direction.
𝑅𝑥 = 𝐹𝑔𝑥 − 𝐹𝑓
𝑅𝑥 = 295 𝑁 − 14.0 𝑁 = 281 𝑁
(c) Find the acceleration of the skis in the x-direction.
𝑎=
𝑅𝑥 281
=
= 5.62 𝑚/𝑠 2
𝑚 50.0
Practice Makes Perfect
Giancoli 4th Edition
Physics for Scientists & Engineers
End of Practice Sheets #5
Answer Key Practice Sheets #5-Ch 5
The Glenelg School of Abu Dhabi: Page 7 of 7