PHYS 165 Module 6 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 6
assessment within the 2026/2027 Portage Learning university-level physics curriculum. Module 6
addresses the foundational principles of work, energy, linear momentum, and collisions, which
constitute the core analytical framework for understanding classical mechanics phenomena involving
force interactions, energy transformations, and momentum transfer. The examination is structured
across four principal content domains: Work, Kinetic Energy, and the Work-Energy Theorem;
Potential Energy, Conservative Forces, and Conservation of Energy; Linear Momentum, Impulse, and
Collisions; and Center of Mass and Systems of Particles. Each domain requires students to apply
mathematical reasoning, physical laws, and proven methodologies to solve quantitative problems
involving scalar and vector calculations, energy conservation principles, and collision dynamics. The
questions emphasize both conceptual understanding and computational proficiency, reflecting the
rigorous analytical standards expected in university-level physics education.
Content Area Overview
Content Area Questions Key Topics Weight
Work by constant/variable forces,
Work, Kinetic Energy & the Work-Energy
7 kinetic energy, work-energy theorem, 28%
Theorem
power
Gravitational PE, elastic PE,
Potential Energy, Conservative Forces &
6 conservative vs. non-conservative 24%
Conservation of Energy
forces, energy conservation
Momentum, impulse-momentum
Linear Momentum, Impulse & Collisions 6 theorem, elastic/inelastic collisions, 24%
recoil
Center of mass calculation, system
Center of Mass & Systems of Particles 6 momentum, internal/external forces, 24%
explosions
TOTAL 25 100%
Examination Questions
Domain: Work, Kinetic Energy & the Work-Energy Theorem
1. A 5.0 kg object is subjected to a horizontal force of 20 N over a displacement of 3.0 m
on a frictionless surface. What is the work done by the force?
A. 15 J
B. 60 J
C. 45 J
D. 100 J
Correct Answer: B
Rationale: Work is calculated as W = Fd cos(theta). Since the force is horizontal and the
displacement is horizontal, theta = 0 degrees, so cos(0) = 1. Therefore, W = (20 N)(3.0 m)(1) = 60 J.
The work-energy theorem directly relates this work to the change in kinetic energy of the object.
Why Wrong: A results from using W = F + d instead of F multiplied by d. C arises from
miscalculating cos(0) as 1.5. D reflects using W = Fd^2 or an incorrect force value.
1
, Reference: PHYS 165 Module 6, Lesson 1; Halliday, Resnick & Walker, Fundamentals of Physics,
12th Ed., Ch. 7
2. A 2.0 kg particle accelerates from rest to a speed of 8.0 m/s. What is the net work
done on the particle?
A. 8.0 J
B. 16 J
C. 32 J
D. 64 J
Correct Answer: D
Rationale: By the work-energy theorem, the net work done equals the change in kinetic energy:
W_net = Delta(K) = (1/2)mv_f^2 - (1/2)mv_i^2. Since the particle starts from rest, v_i = 0, giving
W_net = (1/2)(2.0 kg)(8.0 m/s)^2 = 64 J. This is a direct application of the theorem.
Why Wrong: A results from using W = mv instead of (1/2)mv^2. B uses W = mv_f. C uses W =
(1/2)mv_f without squaring the velocity.
Reference: PHYS 165 Module 6, Lesson 1; Young & Freedman, University Physics, 15th Ed., Ch. 6
3. A force F = (3x^2) N acts on a particle moving along the x-axis from x = 0 to x = 2.0 m.
How much work does this force do?
A. 6.0 J
B. 8.0 J
C. 12 J
D. 24 J
Correct Answer: B
Rationale: For a variable force, work is computed by integration: W = integral from 0 to 2.0 of
(3x^2) dx = [x^3] from 0 to 2.0 = (2.0)^3 - 0 = 8.0 J. This demonstrates the generalization of work
for position-dependent forces using calculus.
Why Wrong: A results from evaluating x^2 at x = 2 instead of integrating. C arises from multiplying
3 by 2^2 = 12 without integration. D uses 3 times 2^3 = 24, confusing the integral result.
Reference: PHYS 165 Module 6, Lesson 1; Halliday, Resnick & Walker, Ch. 7
4. A 10 kg box is pushed across a rough horizontal surface with a coefficient of kinetic
friction of 0.30. The box moves 4.0 m at constant velocity. How much work is done by
friction?
A. -120 J
B. -40 J
C. 40 J
D. 120 J
Correct Answer: A
Rationale: The friction force is f_k = mu_k * N = mu_k * mg = (0.30)(10 kg)(9.8 m/s^2) = 29.4 N.
Since friction opposes displacement, the work done by friction is W_f = -f_k * d = -(29.4 N)(4.0 m) =
-117.6 J, which rounds to approximately -120 J. The negative sign indicates that friction removes
energy from the system.
Why Wrong: B uses only mg without the coefficient. C has the wrong sign, ignoring that friction
opposes motion. D has the wrong sign and may use g = 10 m/s^2 incorrectly.
Reference: PHYS 165 Module 6, Lesson 2; Young & Freedman, University Physics, 15th Ed., Ch. 6
5. The kinetic energy of an object is doubled. By what factor does its speed change?
A. It doubles
B. It increases by a factor of the square root of 2
C. It quadruples
2
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 6
assessment within the 2026/2027 Portage Learning university-level physics curriculum. Module 6
addresses the foundational principles of work, energy, linear momentum, and collisions, which
constitute the core analytical framework for understanding classical mechanics phenomena involving
force interactions, energy transformations, and momentum transfer. The examination is structured
across four principal content domains: Work, Kinetic Energy, and the Work-Energy Theorem;
Potential Energy, Conservative Forces, and Conservation of Energy; Linear Momentum, Impulse, and
Collisions; and Center of Mass and Systems of Particles. Each domain requires students to apply
mathematical reasoning, physical laws, and proven methodologies to solve quantitative problems
involving scalar and vector calculations, energy conservation principles, and collision dynamics. The
questions emphasize both conceptual understanding and computational proficiency, reflecting the
rigorous analytical standards expected in university-level physics education.
Content Area Overview
Content Area Questions Key Topics Weight
Work by constant/variable forces,
Work, Kinetic Energy & the Work-Energy
7 kinetic energy, work-energy theorem, 28%
Theorem
power
Gravitational PE, elastic PE,
Potential Energy, Conservative Forces &
6 conservative vs. non-conservative 24%
Conservation of Energy
forces, energy conservation
Momentum, impulse-momentum
Linear Momentum, Impulse & Collisions 6 theorem, elastic/inelastic collisions, 24%
recoil
Center of mass calculation, system
Center of Mass & Systems of Particles 6 momentum, internal/external forces, 24%
explosions
TOTAL 25 100%
Examination Questions
Domain: Work, Kinetic Energy & the Work-Energy Theorem
1. A 5.0 kg object is subjected to a horizontal force of 20 N over a displacement of 3.0 m
on a frictionless surface. What is the work done by the force?
A. 15 J
B. 60 J
C. 45 J
D. 100 J
Correct Answer: B
Rationale: Work is calculated as W = Fd cos(theta). Since the force is horizontal and the
displacement is horizontal, theta = 0 degrees, so cos(0) = 1. Therefore, W = (20 N)(3.0 m)(1) = 60 J.
The work-energy theorem directly relates this work to the change in kinetic energy of the object.
Why Wrong: A results from using W = F + d instead of F multiplied by d. C arises from
miscalculating cos(0) as 1.5. D reflects using W = Fd^2 or an incorrect force value.
1
, Reference: PHYS 165 Module 6, Lesson 1; Halliday, Resnick & Walker, Fundamentals of Physics,
12th Ed., Ch. 7
2. A 2.0 kg particle accelerates from rest to a speed of 8.0 m/s. What is the net work
done on the particle?
A. 8.0 J
B. 16 J
C. 32 J
D. 64 J
Correct Answer: D
Rationale: By the work-energy theorem, the net work done equals the change in kinetic energy:
W_net = Delta(K) = (1/2)mv_f^2 - (1/2)mv_i^2. Since the particle starts from rest, v_i = 0, giving
W_net = (1/2)(2.0 kg)(8.0 m/s)^2 = 64 J. This is a direct application of the theorem.
Why Wrong: A results from using W = mv instead of (1/2)mv^2. B uses W = mv_f. C uses W =
(1/2)mv_f without squaring the velocity.
Reference: PHYS 165 Module 6, Lesson 1; Young & Freedman, University Physics, 15th Ed., Ch. 6
3. A force F = (3x^2) N acts on a particle moving along the x-axis from x = 0 to x = 2.0 m.
How much work does this force do?
A. 6.0 J
B. 8.0 J
C. 12 J
D. 24 J
Correct Answer: B
Rationale: For a variable force, work is computed by integration: W = integral from 0 to 2.0 of
(3x^2) dx = [x^3] from 0 to 2.0 = (2.0)^3 - 0 = 8.0 J. This demonstrates the generalization of work
for position-dependent forces using calculus.
Why Wrong: A results from evaluating x^2 at x = 2 instead of integrating. C arises from multiplying
3 by 2^2 = 12 without integration. D uses 3 times 2^3 = 24, confusing the integral result.
Reference: PHYS 165 Module 6, Lesson 1; Halliday, Resnick & Walker, Ch. 7
4. A 10 kg box is pushed across a rough horizontal surface with a coefficient of kinetic
friction of 0.30. The box moves 4.0 m at constant velocity. How much work is done by
friction?
A. -120 J
B. -40 J
C. 40 J
D. 120 J
Correct Answer: A
Rationale: The friction force is f_k = mu_k * N = mu_k * mg = (0.30)(10 kg)(9.8 m/s^2) = 29.4 N.
Since friction opposes displacement, the work done by friction is W_f = -f_k * d = -(29.4 N)(4.0 m) =
-117.6 J, which rounds to approximately -120 J. The negative sign indicates that friction removes
energy from the system.
Why Wrong: B uses only mg without the coefficient. C has the wrong sign, ignoring that friction
opposes motion. D has the wrong sign and may use g = 10 m/s^2 incorrectly.
Reference: PHYS 165 Module 6, Lesson 2; Young & Freedman, University Physics, 15th Ed., Ch. 6
5. The kinetic energy of an object is doubled. By what factor does its speed change?
A. It doubles
B. It increases by a factor of the square root of 2
C. It quadruples
2
