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2026 RIO SALADO PHY 111 FINAL EXAM 2026-2027 BANK
QUESTIONS WITH DETAILED VERIFIED ANSWERS EXAM
QUESTIONS WILL COME FROM HERE (100% CORRECT
ANSWERS A+ GRADED
1. A car travels 100 km due west in 2 hours, then turns and travels 50
km due east in 1 hour. What is the magnitude of the average velocity
for the entire trip?
A) 50 km/h
B) 33.3 km/h
C) 16.7 km/h
D) 0 km/h
Answer: C) 16.7 km/h
Explanation: Average velocity is total displacement divided by total
time. The displacement is 100 km west minus 50 km east, which is 50
km west. Total time is 3 hours. The magnitude is (50 km) / (3 h) = 16.7
km/h.
2. An object moves with a constant speed of 10 m/s in a circular path of
radius 2 m. What is the magnitude of its centripetal acceleration?
A) 5 m/s²
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B) 20 m/s²
C) 50 m/s²
D) 0 m/s²
Answer: C) 50 m/s²
Explanation: Centripetal acceleration a_c is given by v²/r. Substituting
the given values, (10 m/s)² / (2 m) = = 50 m/s².
3. Which of Newton's laws states that for every action, there is an equal
and opposite reaction?
A) Newton's First Law
B) Newton's Second Law
C) Newton's Third Law
D) Law of Universal Gravitation
Answer: C) Newton's Third Law
Explanation: Newton's Third Law formally states that if body A exerts a
force on body B, then body B exerts a force on body A that is equal in
magnitude and opposite in direction.
4. A net force of 20 N is applied to an object of mass 4 kg, initially at
rest on a frictionless surface. What is its speed after 3 seconds?
A) 5 m/s
B) 10 m/s
C) 15 m/s
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D) 20 m/s
Answer: C) 15 m/s
Explanation: First, find acceleration using F = ma, so a = F/m = 20 N / 4
kg = 5 m/s². Then, for constant acceleration from rest, v = at = (5
m/s²)(3 s) = 15 m/s.
5. A 2 kg book is held stationary against a vertical wall by a horizontal
force of 50 N. If the coefficient of static friction between the book and
wall is 0.4, what is the magnitude of the frictional force?
A) 20 N
B) 19.6 N
C) 50 N
D) 4 N
Answer: B) 19.6 N
Explanation: The book is stationary, so the forces are in equilibrium.
The vertical gravitational force mg = (2 kg)(9.8 m/s²) = 19.6 N
downward must be balanced by the upward static friction force. Thus,
friction is 19.6 N, not necessarily equal to the maximum possible static
friction.
6. The work-energy theorem states that the net work done on an object
is equal to its change in:
A) Potential energy
B) Kinetic energy
C) Total mechanical energy
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D) Velocity
Answer: B) Kinetic energy
Explanation: The work-energy theorem is W_net = ΔKE. It directly
relates the net work to the change in an object's kinetic energy.
7. A crane lifts a 200 kg crate vertically upward a distance of 10 m at
constant speed. How much work does the crane do on the crate?
A) 0 J
B) 2,000 J
C) 19,600 J
D) 1,960 J
Answer: C) 19,600 J
Explanation: The crane must apply a force equal to the weight mg =
(200 kg)(9.8 m/s²) = 1960 N. Work done is force times distance in the
direction of the force, so W = (1960 N)(10 m) = 19,600 J.
8. What is the SI unit of power?
A) Joule
B) Watt
C) Newton
D) Horsepower
Answer: B) Watt
2026 RIO SALADO PHY 111 FINAL EXAM 2026-2027 BANK
QUESTIONS WITH DETAILED VERIFIED ANSWERS EXAM
QUESTIONS WILL COME FROM HERE (100% CORRECT
ANSWERS A+ GRADED
1. A car travels 100 km due west in 2 hours, then turns and travels 50
km due east in 1 hour. What is the magnitude of the average velocity
for the entire trip?
A) 50 km/h
B) 33.3 km/h
C) 16.7 km/h
D) 0 km/h
Answer: C) 16.7 km/h
Explanation: Average velocity is total displacement divided by total
time. The displacement is 100 km west minus 50 km east, which is 50
km west. Total time is 3 hours. The magnitude is (50 km) / (3 h) = 16.7
km/h.
2. An object moves with a constant speed of 10 m/s in a circular path of
radius 2 m. What is the magnitude of its centripetal acceleration?
A) 5 m/s²
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B) 20 m/s²
C) 50 m/s²
D) 0 m/s²
Answer: C) 50 m/s²
Explanation: Centripetal acceleration a_c is given by v²/r. Substituting
the given values, (10 m/s)² / (2 m) = = 50 m/s².
3. Which of Newton's laws states that for every action, there is an equal
and opposite reaction?
A) Newton's First Law
B) Newton's Second Law
C) Newton's Third Law
D) Law of Universal Gravitation
Answer: C) Newton's Third Law
Explanation: Newton's Third Law formally states that if body A exerts a
force on body B, then body B exerts a force on body A that is equal in
magnitude and opposite in direction.
4. A net force of 20 N is applied to an object of mass 4 kg, initially at
rest on a frictionless surface. What is its speed after 3 seconds?
A) 5 m/s
B) 10 m/s
C) 15 m/s
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D) 20 m/s
Answer: C) 15 m/s
Explanation: First, find acceleration using F = ma, so a = F/m = 20 N / 4
kg = 5 m/s². Then, for constant acceleration from rest, v = at = (5
m/s²)(3 s) = 15 m/s.
5. A 2 kg book is held stationary against a vertical wall by a horizontal
force of 50 N. If the coefficient of static friction between the book and
wall is 0.4, what is the magnitude of the frictional force?
A) 20 N
B) 19.6 N
C) 50 N
D) 4 N
Answer: B) 19.6 N
Explanation: The book is stationary, so the forces are in equilibrium.
The vertical gravitational force mg = (2 kg)(9.8 m/s²) = 19.6 N
downward must be balanced by the upward static friction force. Thus,
friction is 19.6 N, not necessarily equal to the maximum possible static
friction.
6. The work-energy theorem states that the net work done on an object
is equal to its change in:
A) Potential energy
B) Kinetic energy
C) Total mechanical energy
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D) Velocity
Answer: B) Kinetic energy
Explanation: The work-energy theorem is W_net = ΔKE. It directly
relates the net work to the change in an object's kinetic energy.
7. A crane lifts a 200 kg crate vertically upward a distance of 10 m at
constant speed. How much work does the crane do on the crate?
A) 0 J
B) 2,000 J
C) 19,600 J
D) 1,960 J
Answer: C) 19,600 J
Explanation: The crane must apply a force equal to the weight mg =
(200 kg)(9.8 m/s²) = 1960 N. Work done is force times distance in the
direction of the force, so W = (1960 N)(10 m) = 19,600 J.
8. What is the SI unit of power?
A) Joule
B) Watt
C) Newton
D) Horsepower
Answer: B) Watt
