PHYS 165 Module 5 Exam Physics Portage
Learning Forces Newton's Laws Applications
Official Practice Exam Actual Exam 2026/2027 with
Detailed Rationales | Complete Exam-Style
Questions | Pass Guaranteed – A+ Graded
══════════════════════════════════════
SECTION 1: NEWTON'S LAWS — CONCEPTUAL Q1 – Q5
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Question 1 of 25
A hockey puck slides across a frictionless frozen pond and gradually slows due to air
resistance. A student claims this violates Newton's First Law because the puck slows
without any visible push. What is the correct analysis of this situation?
A. The puck slows because an unbalanced force (air resistance) acts on it, which is entirely
consistent with Newton's First Law. ✓ CORRECT
B. The puck slows because its inertia naturally decreases as it moves across the ice.
C. The puck should maintain constant velocity forever since the pond surface is frictionless.
D. The puck slows because Newton's First Law only applies to objects that are initially at
rest.
Correct Answer: A
Rationale: Newton's First Law states that an object maintains its velocity unless acted upon
by a net external force, and the puck slows precisely because air resistance provides an
unbalanced force opposing its motion. Choice B incorrectly suggests inertia decreases with
motion, but inertia is an intrinsic property of mass and never diminishes. Remember that the
First Law applies to both moving and stationary objects whenever the net force is zero.
Question 2 of 25
A 1200 kg car accelerates from rest to 25 m/s in 10 seconds along a straight highway. What
net force must be acting on the car during this interval?
A. 300 N
B. 3000 N ✓ CORRECT
C. 12000 N
D. 250 N
, Correct Answer: B
Rationale: Newton's Second Law gives F = ma, and the car's acceleration is a = Δv/Δt = 25/10
= 2.5 m/s², so the required net force is F = (1200 kg)(2.5 m/s²) = 3000 N. Choice A results
from forgetting to multiply by mass after finding the acceleration, a common error when
students conflate acceleration with force. Always verify that your final units are newtons, not
m/s².
Question 3 of 25
A physics textbook rests on a laboratory table. The table exerts an upward normal force of 25
N on the book. According to Newton's Third Law, what is the true reaction force to this
normal force?
A. The weight of the book pulling downward on the Earth.
B. The gravitational force of the Earth pulling downward on the book.
C. The book pushing downward on the table with a force of 25 N. ✓ CORRECT
D. The table pushing upward on the Earth with a force of 25 N.
Correct Answer: C
Rationale: Newton's Third Law requires that the reaction force to the table pushing up on the
book is the book pushing down on the table with equal magnitude and opposite direction.
Choice A describes the gravitational interaction between the book and Earth, which is a
different force pair altogether, not the reaction to the normal force. When identifying Third
Law pairs, ensure the two forces act on different objects and arise from the same interaction.
Question 4 of 25
A warehouse worker pushes a heavy crate across a level concrete floor at a steady, constant
velocity. Which statement correctly describes the relationship between the worker's applied
horizontal force and the friction force?
A. The applied force is greater than the friction force because the crate is moving.
B. The applied force is less than the friction force because the crate is heavy.
C. The applied force is zero since the crate moves at constant velocity without accelerating.
D. The applied force equals the kinetic friction force because the acceleration is zero. ✓
CORRECT
Correct Answer: D
Rationale: Because the crate moves at constant velocity, its acceleration is zero, so the net
force must be zero by Newton's Second Law; therefore the applied force exactly balances the
kinetic friction force. Choice A incorrectly assumes a greater force is needed to maintain
motion, confusing the static friction needed to start motion with the usually smaller kinetic
friction that acts during sliding. At constant velocity, always set the sum of forces equal to
zero, not ma.
Learning Forces Newton's Laws Applications
Official Practice Exam Actual Exam 2026/2027 with
Detailed Rationales | Complete Exam-Style
Questions | Pass Guaranteed – A+ Graded
══════════════════════════════════════
SECTION 1: NEWTON'S LAWS — CONCEPTUAL Q1 – Q5
══════════════════════════════════════
Question 1 of 25
A hockey puck slides across a frictionless frozen pond and gradually slows due to air
resistance. A student claims this violates Newton's First Law because the puck slows
without any visible push. What is the correct analysis of this situation?
A. The puck slows because an unbalanced force (air resistance) acts on it, which is entirely
consistent with Newton's First Law. ✓ CORRECT
B. The puck slows because its inertia naturally decreases as it moves across the ice.
C. The puck should maintain constant velocity forever since the pond surface is frictionless.
D. The puck slows because Newton's First Law only applies to objects that are initially at
rest.
Correct Answer: A
Rationale: Newton's First Law states that an object maintains its velocity unless acted upon
by a net external force, and the puck slows precisely because air resistance provides an
unbalanced force opposing its motion. Choice B incorrectly suggests inertia decreases with
motion, but inertia is an intrinsic property of mass and never diminishes. Remember that the
First Law applies to both moving and stationary objects whenever the net force is zero.
Question 2 of 25
A 1200 kg car accelerates from rest to 25 m/s in 10 seconds along a straight highway. What
net force must be acting on the car during this interval?
A. 300 N
B. 3000 N ✓ CORRECT
C. 12000 N
D. 250 N
, Correct Answer: B
Rationale: Newton's Second Law gives F = ma, and the car's acceleration is a = Δv/Δt = 25/10
= 2.5 m/s², so the required net force is F = (1200 kg)(2.5 m/s²) = 3000 N. Choice A results
from forgetting to multiply by mass after finding the acceleration, a common error when
students conflate acceleration with force. Always verify that your final units are newtons, not
m/s².
Question 3 of 25
A physics textbook rests on a laboratory table. The table exerts an upward normal force of 25
N on the book. According to Newton's Third Law, what is the true reaction force to this
normal force?
A. The weight of the book pulling downward on the Earth.
B. The gravitational force of the Earth pulling downward on the book.
C. The book pushing downward on the table with a force of 25 N. ✓ CORRECT
D. The table pushing upward on the Earth with a force of 25 N.
Correct Answer: C
Rationale: Newton's Third Law requires that the reaction force to the table pushing up on the
book is the book pushing down on the table with equal magnitude and opposite direction.
Choice A describes the gravitational interaction between the book and Earth, which is a
different force pair altogether, not the reaction to the normal force. When identifying Third
Law pairs, ensure the two forces act on different objects and arise from the same interaction.
Question 4 of 25
A warehouse worker pushes a heavy crate across a level concrete floor at a steady, constant
velocity. Which statement correctly describes the relationship between the worker's applied
horizontal force and the friction force?
A. The applied force is greater than the friction force because the crate is moving.
B. The applied force is less than the friction force because the crate is heavy.
C. The applied force is zero since the crate moves at constant velocity without accelerating.
D. The applied force equals the kinetic friction force because the acceleration is zero. ✓
CORRECT
Correct Answer: D
Rationale: Because the crate moves at constant velocity, its acceleration is zero, so the net
force must be zero by Newton's Second Law; therefore the applied force exactly balances the
kinetic friction force. Choice A incorrectly assumes a greater force is needed to maintain
motion, confusing the static friction needed to start motion with the usually smaller kinetic
friction that acts during sliding. At constant velocity, always set the sum of forces equal to
zero, not ma.
