PHYS 165 Module 6 Exam Work Energy
Power Official Practice Exam Actual Exam
2026/2027 with Detailed Rationales |
Complete Exam-Style Questions | Pass
Guaranteed – A+ Graded
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SECTION 1: WORK & KINETIC ENERGY Q1 – Q5
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Question 1 of 25
A warehouse worker pushes a 25 kg crate across a level floor by applying a constant force of
45 N directed 30° below the horizontal. The crate moves 8.0 m in a straight line. How much
work does the worker do on the crate?
A. 360 J
B. 312 J ✓ CORRECT
C. 180 J
D. 225 J
Correct Answer: B
Rationale: The work done by a constant force is calculated using W = Fd cosθ, where θ is the
angle between the force and displacement vectors; here, W = (45 N)(8.0 m)cos(30°) = 312 J.
Choice A incorrectly uses W = Fd and ignores the horizontal component, which is a common
error when the angle is given but not applied. When you see an angle in a work problem,
always verify whether the force is parallel to the displacement or if a trigonometric factor is
needed.
Question 2 of 25
A delivery truck traveling along a straight highway increases its speed from 15 m/s to 25 m/s.
If the net work done on the 1200 kg truck during this acceleration is 480 kJ, what is the
truck's mass?
A. 1200 kg
B. 960 kg
, C. 2133 kg
D. 2400 kg ✓ CORRECT
Correct Answer: D
Rationale: The work-energy theorem states that the net work done on an object equals the
change in its kinetic energy, so W_net = ½m(v² − v₀²); solving for mass yields m = 2W_net /
(v² − v₀²) = 2(480,000 J) / (625 m²/s² − 225 m²/s²) = 2400 kg. Choice A results from
mistakenly dividing the work by the change in speed (400 m²/s²) rather than using the
difference of squared velocities. Always square the velocities first when applying the
work-energy theorem to avoid this algebraic trap.
Question 3 of 25
A physics student notices that the kinetic energy of a moving cart has doubled while its mass
remains unchanged. By what factor has the speed of the cart increased?
A. √2 ✓ CORRECT
B. 2
C. 4
D. ½
Correct Answer: A
Rationale: Since kinetic energy is proportional to the square of the speed (KE = ½mv²),
doubling the kinetic energy requires the speed to increase by a factor of √2. Choice B
confuses the linear relationship between energy and speed with the actual quadratic
dependence, which is a frequent error on energy exams. Remember that kinetic energy scales
with v², so any change in KE corresponds to the square root of that change in speed.
Question 4 of 25
A 5.0 kg box slides from rest down a frictionless ramp that is 3.0 m high at its top end. What
is the total work done by gravity on the box as it reaches the bottom of the ramp?
A. −147 J
B. 245 J
C. 147 J ✓ CORRECT
D. 73.5 J
Correct Answer: C
Rationale: The work done by gravity depends only on the vertical displacement of the object,
so W_gravity = mgΔh = (5.0 kg)(9.8 m/s²)(3.0 m) = 147 J. Choice A incorrectly assigns a
negative sign, which would only be appropriate if the object were being lifted upward against
gravity rather than falling. On energy problems involving ramps, the work done by gravity is
always mgΔh regardless of the path length or ramp angle.
Power Official Practice Exam Actual Exam
2026/2027 with Detailed Rationales |
Complete Exam-Style Questions | Pass
Guaranteed – A+ Graded
══════════════════════════════════════
SECTION 1: WORK & KINETIC ENERGY Q1 – Q5
══════════════════════════════════════
Question 1 of 25
A warehouse worker pushes a 25 kg crate across a level floor by applying a constant force of
45 N directed 30° below the horizontal. The crate moves 8.0 m in a straight line. How much
work does the worker do on the crate?
A. 360 J
B. 312 J ✓ CORRECT
C. 180 J
D. 225 J
Correct Answer: B
Rationale: The work done by a constant force is calculated using W = Fd cosθ, where θ is the
angle between the force and displacement vectors; here, W = (45 N)(8.0 m)cos(30°) = 312 J.
Choice A incorrectly uses W = Fd and ignores the horizontal component, which is a common
error when the angle is given but not applied. When you see an angle in a work problem,
always verify whether the force is parallel to the displacement or if a trigonometric factor is
needed.
Question 2 of 25
A delivery truck traveling along a straight highway increases its speed from 15 m/s to 25 m/s.
If the net work done on the 1200 kg truck during this acceleration is 480 kJ, what is the
truck's mass?
A. 1200 kg
B. 960 kg
, C. 2133 kg
D. 2400 kg ✓ CORRECT
Correct Answer: D
Rationale: The work-energy theorem states that the net work done on an object equals the
change in its kinetic energy, so W_net = ½m(v² − v₀²); solving for mass yields m = 2W_net /
(v² − v₀²) = 2(480,000 J) / (625 m²/s² − 225 m²/s²) = 2400 kg. Choice A results from
mistakenly dividing the work by the change in speed (400 m²/s²) rather than using the
difference of squared velocities. Always square the velocities first when applying the
work-energy theorem to avoid this algebraic trap.
Question 3 of 25
A physics student notices that the kinetic energy of a moving cart has doubled while its mass
remains unchanged. By what factor has the speed of the cart increased?
A. √2 ✓ CORRECT
B. 2
C. 4
D. ½
Correct Answer: A
Rationale: Since kinetic energy is proportional to the square of the speed (KE = ½mv²),
doubling the kinetic energy requires the speed to increase by a factor of √2. Choice B
confuses the linear relationship between energy and speed with the actual quadratic
dependence, which is a frequent error on energy exams. Remember that kinetic energy scales
with v², so any change in KE corresponds to the square root of that change in speed.
Question 4 of 25
A 5.0 kg box slides from rest down a frictionless ramp that is 3.0 m high at its top end. What
is the total work done by gravity on the box as it reaches the bottom of the ramp?
A. −147 J
B. 245 J
C. 147 J ✓ CORRECT
D. 73.5 J
Correct Answer: C
Rationale: The work done by gravity depends only on the vertical displacement of the object,
so W_gravity = mgΔh = (5.0 kg)(9.8 m/s²)(3.0 m) = 147 J. Choice A incorrectly assigns a
negative sign, which would only be appropriate if the object were being lifted upward against
gravity rather than falling. On energy problems involving ramps, the work done by gravity is
always mgΔh regardless of the path length or ramp angle.
