PHYS 165 Module 5 Examination
Physics: Portage Learning | 2026/2027 Academic Year
25 Questions | Actual Exam Answer Key
Abstract
This document presents the complete 25-question actual examination for the PHYS 165 Module 5
assessment within the 2026/2027 Portage Learning university-level physics curriculum. Module 5
addresses the foundational principles of Newtonian mechanics, encompassing Newton's three laws of
motion, free-body diagram construction and analysis, static and kinetic friction, circular motion and
centripetal acceleration, and applied dynamics involving multi-force systems and connected bodies.
The examination is structured across four principal content domains that collectively develop the
analytical and mathematical skills necessary for solving complex dynamic problems. Each domain
requires the application of vector decomposition, force superposition, and the systematic use of
Newton's second law to predict the motion of objects under the influence of multiple forces. The
questions emphasize both conceptual mastery of force laws and quantitative problem-solving
proficiency consistent with proven methodologies in university-level physics instruction.
Content Area Overview
Content Area Questions Key Topics Weight
Newton's 3 Laws, FBD analysis,
Newton's Laws & Free Body Diagrams 8 normal force, tension, inclined planes, 32%
Atwood machine
Static friction, kinetic friction,
Static & Kinetic Friction 5 coefficients, inclined surfaces, 20%
equilibrium
Centripetal force, centripetal
Circular Motion & Centripetal Acceleration 6 acceleration, banked curves, vertical 24%
circles, period
Connected bodies, pulley systems,
Applied Dynamics & Multi-Force Systems 6 friction on walls, multi-block systems, 24%
elevators
TOTAL 25 100%
Examination Questions
Domain: Newton's Laws & Free Body Diagrams
1. A 5.0 kg object experiences a net force of 30 N to the right. What is the acceleration of
the object?
A. 6.0 m/s^2 to the right
B. 150 m/s^2 to the right
C. 0.17 m/s^2 to the right
D. 35 m/s^2 to the right
Correct Answer: A
Rationale: By Newton's second law, F_net = ma, so a = F_net / m = 30 N / 5.0 kg = 6.0 m/s^2. The
acceleration is in the same direction as the net force (to the right). This is the most fundamental
equation in dynamics, linking force, mass, and acceleration directly. The inverse proportionality
between mass and acceleration means heavier objects require greater forces to achieve the same
acceleration.
Why Wrong: B multiplies F and m instead of dividing. C divides m by F (inverted). D adds m to F.
1
, Reference: PHYS 165 Module 5, 2026/2027 Portage Learning Curriculum; Halliday, Resnick &
Walker, Fundamentals of Physics, 12th Ed., Ch. 5-6
2. A 12 kg box rests on a horizontal surface. What is the normal force acting on the box?
A. 0 N
B. 12 N
C. 118 N
D. 235 N
Correct Answer: C
Rationale: On a horizontal surface with no vertical acceleration, the normal force equals the
gravitational force: N = mg = (12 kg)(9.8 m/s^2) = 117.6 N, which rounds to 118 N. This follows
from applying Newton's second law in the vertical direction: N - mg = 0, since a_y = 0. The normal
force is a contact force perpendicular to the surface.
Why Wrong: A assumes no forces act on the box. B uses N = m, confusing mass with force. D uses N
= 2mg.
Reference: PHYS 165 Module 5, 2026/2027 Portage Learning Curriculum; Halliday, Resnick &
Walker, Fundamentals of Physics, 12th Ed., Ch. 5-6
3. A 60 kg person stands in an elevator that is accelerating upward at 2.0 m/s^2. What
is the apparent weight (normal force) exerted by the floor on the person?
A. 588 N
B. 468 N
C. 708 N
D. 120 N
Correct Answer: C
Rationale: Applying Newton's second law in the vertical direction: N - mg = ma, so N = m(g + a) =
60(9.8 + 2.0) = 60(11.8) = 708 N. The apparent weight exceeds the actual weight because the floor
must provide enough upward force to both support against gravity and accelerate the person
upward. This is a classic application of free-body diagram analysis in non-inertial reference frames.
Why Wrong: A uses N = mg, ignoring the acceleration. B uses N = m(g - a), which applies for
downward acceleration. D uses N = ma only.
Reference: PHYS 165 Module 5, 2026/2027 Portage Learning Curriculum; Halliday, Resnick &
Walker, Fundamentals of Physics, 12th Ed., Ch. 5-6
4. A 3.0 kg block is pulled across a frictionless horizontal surface by a force of 15 N
applied at an angle of 30 degrees above the horizontal. What is the horizontal
acceleration of the block?
A. 4.3 m/s^2
B. 5.0 m/s^2
C. 7.5 m/s^2
D. 8.7 m/s^2
Correct Answer: A
Rationale: Only the horizontal component of the applied force contributes to horizontal acceleration.
F_horizontal = F * cos(30) = 15 * 0.866 = 13.0 N. Using Newton's second law: a = F_horizontal / m
= 13..0 = 4.33 m/s^2, which rounds to 4.3 m/s^2. The vertical component of the force affects the
normal force but not the horizontal motion on a frictionless surface. Free-body diagram analysis is
essential here to resolve force components correctly.
Why Wrong: B uses F/m = 15/3 = 5.0, ignoring the angle. C uses F*sin(30)/m. D uses
F/(m*cos(30)).
Reference: PHYS 165 Module 5, 2026/2027 Portage Learning Curriculum; Halliday, Resnick &
Walker, Fundamentals of Physics, 12th Ed., Ch. 5-6
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Physics: Portage Learning | 2026/2027 Academic Year
25 Questions | Actual Exam Answer Key
Abstract
This document presents the complete 25-question actual examination for the PHYS 165 Module 5
assessment within the 2026/2027 Portage Learning university-level physics curriculum. Module 5
addresses the foundational principles of Newtonian mechanics, encompassing Newton's three laws of
motion, free-body diagram construction and analysis, static and kinetic friction, circular motion and
centripetal acceleration, and applied dynamics involving multi-force systems and connected bodies.
The examination is structured across four principal content domains that collectively develop the
analytical and mathematical skills necessary for solving complex dynamic problems. Each domain
requires the application of vector decomposition, force superposition, and the systematic use of
Newton's second law to predict the motion of objects under the influence of multiple forces. The
questions emphasize both conceptual mastery of force laws and quantitative problem-solving
proficiency consistent with proven methodologies in university-level physics instruction.
Content Area Overview
Content Area Questions Key Topics Weight
Newton's 3 Laws, FBD analysis,
Newton's Laws & Free Body Diagrams 8 normal force, tension, inclined planes, 32%
Atwood machine
Static friction, kinetic friction,
Static & Kinetic Friction 5 coefficients, inclined surfaces, 20%
equilibrium
Centripetal force, centripetal
Circular Motion & Centripetal Acceleration 6 acceleration, banked curves, vertical 24%
circles, period
Connected bodies, pulley systems,
Applied Dynamics & Multi-Force Systems 6 friction on walls, multi-block systems, 24%
elevators
TOTAL 25 100%
Examination Questions
Domain: Newton's Laws & Free Body Diagrams
1. A 5.0 kg object experiences a net force of 30 N to the right. What is the acceleration of
the object?
A. 6.0 m/s^2 to the right
B. 150 m/s^2 to the right
C. 0.17 m/s^2 to the right
D. 35 m/s^2 to the right
Correct Answer: A
Rationale: By Newton's second law, F_net = ma, so a = F_net / m = 30 N / 5.0 kg = 6.0 m/s^2. The
acceleration is in the same direction as the net force (to the right). This is the most fundamental
equation in dynamics, linking force, mass, and acceleration directly. The inverse proportionality
between mass and acceleration means heavier objects require greater forces to achieve the same
acceleration.
Why Wrong: B multiplies F and m instead of dividing. C divides m by F (inverted). D adds m to F.
1
, Reference: PHYS 165 Module 5, 2026/2027 Portage Learning Curriculum; Halliday, Resnick &
Walker, Fundamentals of Physics, 12th Ed., Ch. 5-6
2. A 12 kg box rests on a horizontal surface. What is the normal force acting on the box?
A. 0 N
B. 12 N
C. 118 N
D. 235 N
Correct Answer: C
Rationale: On a horizontal surface with no vertical acceleration, the normal force equals the
gravitational force: N = mg = (12 kg)(9.8 m/s^2) = 117.6 N, which rounds to 118 N. This follows
from applying Newton's second law in the vertical direction: N - mg = 0, since a_y = 0. The normal
force is a contact force perpendicular to the surface.
Why Wrong: A assumes no forces act on the box. B uses N = m, confusing mass with force. D uses N
= 2mg.
Reference: PHYS 165 Module 5, 2026/2027 Portage Learning Curriculum; Halliday, Resnick &
Walker, Fundamentals of Physics, 12th Ed., Ch. 5-6
3. A 60 kg person stands in an elevator that is accelerating upward at 2.0 m/s^2. What
is the apparent weight (normal force) exerted by the floor on the person?
A. 588 N
B. 468 N
C. 708 N
D. 120 N
Correct Answer: C
Rationale: Applying Newton's second law in the vertical direction: N - mg = ma, so N = m(g + a) =
60(9.8 + 2.0) = 60(11.8) = 708 N. The apparent weight exceeds the actual weight because the floor
must provide enough upward force to both support against gravity and accelerate the person
upward. This is a classic application of free-body diagram analysis in non-inertial reference frames.
Why Wrong: A uses N = mg, ignoring the acceleration. B uses N = m(g - a), which applies for
downward acceleration. D uses N = ma only.
Reference: PHYS 165 Module 5, 2026/2027 Portage Learning Curriculum; Halliday, Resnick &
Walker, Fundamentals of Physics, 12th Ed., Ch. 5-6
4. A 3.0 kg block is pulled across a frictionless horizontal surface by a force of 15 N
applied at an angle of 30 degrees above the horizontal. What is the horizontal
acceleration of the block?
A. 4.3 m/s^2
B. 5.0 m/s^2
C. 7.5 m/s^2
D. 8.7 m/s^2
Correct Answer: A
Rationale: Only the horizontal component of the applied force contributes to horizontal acceleration.
F_horizontal = F * cos(30) = 15 * 0.866 = 13.0 N. Using Newton's second law: a = F_horizontal / m
= 13..0 = 4.33 m/s^2, which rounds to 4.3 m/s^2. The vertical component of the force affects the
normal force but not the horizontal motion on a frictionless surface. Free-body diagram analysis is
essential here to resolve force components correctly.
Why Wrong: B uses F/m = 15/3 = 5.0, ignoring the angle. C uses F*sin(30)/m. D uses
F/(m*cos(30)).
Reference: PHYS 165 Module 5, 2026/2027 Portage Learning Curriculum; Halliday, Resnick &
Walker, Fundamentals of Physics, 12th Ed., Ch. 5-6
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