Physics for Scientists Engineers Knight 5th Ed. All Chapters
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Section 1: Newton's Laws - Kinematics & Dynamics (20 Questions)
Q1: A car accelerates uniformly from rest to 25 m/s in 10 seconds. What is its
acceleration?
A. 0.25 m/s²
B. 1.5 m/s²
C. 2.5 m/s² [CORRECT]
D. 5.0 m/s²
Correct Answer: C
Rationale: Using the kinematic equation a = Δv/Δt = (25 m/s – 0 m/s) / 10 s = 2.5 m/s²;
uniform acceleration means constant acceleration throughout the interval (Knight, 5th
ed., Ch. 2).
Q2: A ball is thrown vertically upward with an initial velocity of 20 m/s. How long does it
take to reach its maximum height? (g = 9.8 m/s²)
A. 1.0 s
B. 2.0 s [CORRECT]
C. 4.0 s
D. 10.0 s
Correct Answer: B
,Rationale: At maximum height, v = 0; using v = v₀ – gt, t = v₀/g = 20 m/s / 9.8 m/s² ≈
2.04 s ≈ 2.0 s; the symmetry of free-fall motion gives equal time up and down (Knight,
5th ed., Ch. 2).
Q3: Two vectors A and B have magnitudes 5.0 m and 3.0 m, respectively. If they point in
opposite directions, what is the magnitude of A + B?
A. 8.0 m
B. 5.0 m
C. 2.0 m [CORRECT]
D. 15.0 m
Correct Answer: C
Rationale: When vectors point in opposite directions, subtract their magnitudes: |A + B|
= |5.0 – 3.0| = 2.0 m; vector addition requires consideration of both magnitude and
direction (Knight, 5th ed., Ch. 3).
Q4: A 5.0 kg object experiences a net force of 15 N. What is its acceleration?
A. 0.33 m/s²
B. 3.0 m/s² [CORRECT]
C. 20 m/s²
D. 75 m/s²
Correct Answer: B
Rationale: Newton's second law: F = ma, so a = F/m = 15 N / 5.0 kg = 3.0 m/s²;
acceleration is directly proportional to net force and inversely proportional to mass
(Knight, 5th ed., Ch. 4).
,Q5: A block slides down a frictionless incline at 30° to the horizontal. What is its
acceleration?
A. 4.9 m/s² [CORRECT]
B. 9.8 m/s²
C. 8.5 m/s²
D. 0 m/s²
Correct Answer: A
Rationale: The component of gravity parallel to the incline is mg sinθ; a = g sinθ = 9.8
m/s² × sin(30°) = 9.8 × 0.5 = 4.9 m/s²; the normal force cancels the perpendicular
component (Knight, 5th ed., Ch. 5).
Q6: In the absence of air resistance, a projectile launched horizontally from a height h
will strike the ground with a vertical velocity component of:
A. Zero
B. √(2gh) [CORRECT]
C. √(gh)
D. 2gh
Correct Answer: B
Rationale: The vertical motion is free fall from rest; using v² = v₀² + 2gΔy with v₀y = 0 and
Δy = h, vy = √(2gh); horizontal and vertical motions are independent (Knight, 5th ed., Ch.
3).
Q7: A 10 kg box rests on a horizontal surface with μs = 0.40 and μk = 0.30. What
minimum horizontal force is required to start the box moving?
A. 29 N
B. 39 N [CORRECT]
, C. 49 N
D. 98 N
Correct Answer: B
Rationale: To start motion, F must overcome maximum static friction: fs,max = μsN = μs
mg = 0.40 × 10 kg × 9.8 m/s² = 39.2 N ≈ 39 N; once moving, kinetic friction (29 N) acts
(Knight, 5th ed., Ch. 5).
Q8: An object moving in a circle at constant speed is accelerating because:
A. Its speed is changing
B. Its direction is changing [CORRECT]
C. Its velocity is constant
D. No force acts on it
Correct Answer: B
Rationale: Centripetal acceleration occurs because velocity is a vector; even at constant
speed, changing direction means changing velocity; a = v²/r directed toward the center
(Knight, 5th ed., Ch. 6).
Q9: A 2.0 kg object on a frictionless surface is pulled by two forces: 10 N east and 6 N
west. What is its acceleration?
A. 2.0 m/s² east [CORRECT]
B. 8.0 m/s² east
C. 2.0 m/s² west
D. 8.0 m/s² west
Correct Answer: A
Test Bank Actual Exam 2026/2027 – Comprehensive Exam
with Detailed Rationales | 100% Verified | Pass Guaranteed –
A+ Graded
Section 1: Newton's Laws - Kinematics & Dynamics (20 Questions)
Q1: A car accelerates uniformly from rest to 25 m/s in 10 seconds. What is its
acceleration?
A. 0.25 m/s²
B. 1.5 m/s²
C. 2.5 m/s² [CORRECT]
D. 5.0 m/s²
Correct Answer: C
Rationale: Using the kinematic equation a = Δv/Δt = (25 m/s – 0 m/s) / 10 s = 2.5 m/s²;
uniform acceleration means constant acceleration throughout the interval (Knight, 5th
ed., Ch. 2).
Q2: A ball is thrown vertically upward with an initial velocity of 20 m/s. How long does it
take to reach its maximum height? (g = 9.8 m/s²)
A. 1.0 s
B. 2.0 s [CORRECT]
C. 4.0 s
D. 10.0 s
Correct Answer: B
,Rationale: At maximum height, v = 0; using v = v₀ – gt, t = v₀/g = 20 m/s / 9.8 m/s² ≈
2.04 s ≈ 2.0 s; the symmetry of free-fall motion gives equal time up and down (Knight,
5th ed., Ch. 2).
Q3: Two vectors A and B have magnitudes 5.0 m and 3.0 m, respectively. If they point in
opposite directions, what is the magnitude of A + B?
A. 8.0 m
B. 5.0 m
C. 2.0 m [CORRECT]
D. 15.0 m
Correct Answer: C
Rationale: When vectors point in opposite directions, subtract their magnitudes: |A + B|
= |5.0 – 3.0| = 2.0 m; vector addition requires consideration of both magnitude and
direction (Knight, 5th ed., Ch. 3).
Q4: A 5.0 kg object experiences a net force of 15 N. What is its acceleration?
A. 0.33 m/s²
B. 3.0 m/s² [CORRECT]
C. 20 m/s²
D. 75 m/s²
Correct Answer: B
Rationale: Newton's second law: F = ma, so a = F/m = 15 N / 5.0 kg = 3.0 m/s²;
acceleration is directly proportional to net force and inversely proportional to mass
(Knight, 5th ed., Ch. 4).
,Q5: A block slides down a frictionless incline at 30° to the horizontal. What is its
acceleration?
A. 4.9 m/s² [CORRECT]
B. 9.8 m/s²
C. 8.5 m/s²
D. 0 m/s²
Correct Answer: A
Rationale: The component of gravity parallel to the incline is mg sinθ; a = g sinθ = 9.8
m/s² × sin(30°) = 9.8 × 0.5 = 4.9 m/s²; the normal force cancels the perpendicular
component (Knight, 5th ed., Ch. 5).
Q6: In the absence of air resistance, a projectile launched horizontally from a height h
will strike the ground with a vertical velocity component of:
A. Zero
B. √(2gh) [CORRECT]
C. √(gh)
D. 2gh
Correct Answer: B
Rationale: The vertical motion is free fall from rest; using v² = v₀² + 2gΔy with v₀y = 0 and
Δy = h, vy = √(2gh); horizontal and vertical motions are independent (Knight, 5th ed., Ch.
3).
Q7: A 10 kg box rests on a horizontal surface with μs = 0.40 and μk = 0.30. What
minimum horizontal force is required to start the box moving?
A. 29 N
B. 39 N [CORRECT]
, C. 49 N
D. 98 N
Correct Answer: B
Rationale: To start motion, F must overcome maximum static friction: fs,max = μsN = μs
mg = 0.40 × 10 kg × 9.8 m/s² = 39.2 N ≈ 39 N; once moving, kinetic friction (29 N) acts
(Knight, 5th ed., Ch. 5).
Q8: An object moving in a circle at constant speed is accelerating because:
A. Its speed is changing
B. Its direction is changing [CORRECT]
C. Its velocity is constant
D. No force acts on it
Correct Answer: B
Rationale: Centripetal acceleration occurs because velocity is a vector; even at constant
speed, changing direction means changing velocity; a = v²/r directed toward the center
(Knight, 5th ed., Ch. 6).
Q9: A 2.0 kg object on a frictionless surface is pulled by two forces: 10 N east and 6 N
west. What is its acceleration?
A. 2.0 m/s² east [CORRECT]
B. 8.0 m/s² east
C. 2.0 m/s² west
D. 8.0 m/s² west
Correct Answer: A
