Laws of Motion
Q. Two blocks of masses 2 kg and 3 kg are connected by a light string over a frictionless pulley. If the 3 kg block is released from rest, what is the acceleration of the system?
A.
1.2 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using Newton's second law, the net force is (3 kg - 2 kg) * g = 1 kg * 9.8 m/s². The total mass is 5 kg, so a = F/m = 9.8 N / 5 kg = 1.96 m/s², approximately 2 m/s².
Correct Answer: B — 2 m/s²
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Q. What happens to the acceleration of an object if the net force acting on it is doubled while its mass remains constant?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
According to F = ma, if the force is doubled, the acceleration also doubles.
Correct Answer: A — It doubles
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Q. What is the angle between the velocity vector and the acceleration vector of an object moving in uniform circular motion?
A.
0°
B.
45°
C.
90°
D.
180°
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Solution
In uniform circular motion, velocity is tangential and acceleration is radial, hence the angle is 90°.
Correct Answer: C — 90°
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Q. What is the angular velocity of a wheel rotating at 300 revolutions per minute?
A.
10π rad/s
B.
20π rad/s
C.
30π rad/s
D.
40π rad/s
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Solution
Angular velocity (ω) = 2π * (300/60) = 10π rad/s.
Correct Answer: B — 20π rad/s
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Q. What is the angular velocity of a wheel that makes 10 revolutions in 5 seconds?
A.
2π rad/s
B.
4π rad/s
C.
10π rad/s
D.
20π rad/s
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Solution
Angular velocity ω = θ/t = (10 * 2π rad) / 5 s = 4π rad/s.
Correct Answer: B — 4π rad/s
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Q. What is the coefficient of static friction if a block on a horizontal surface begins to slide at a force of 20 N and the normal force is 50 N?
A.
0.2
B.
0.4
C.
0.5
D.
0.6
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Solution
The coefficient of static friction (μs) is given by μs = F/N, where F is the force and N is the normal force. Here, μs = 20 N / 50 N = 0.4.
Correct Answer: B — 0.4
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Q. What is the effect of increasing the normal force on the frictional force between two surfaces?
A.
Increases frictional force
B.
Decreases frictional force
C.
No effect
D.
Depends on the surface material
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Solution
Frictional force is directly proportional to the normal force. Increasing the normal force increases the frictional force.
Correct Answer: A — Increases frictional force
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Q. What is the effect of increasing the surface roughness on the coefficient of friction?
A.
Increases
B.
Decreases
C.
No effect
D.
Depends on the material
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Solution
Increasing surface roughness generally increases the coefficient of friction due to greater interlocking between the surfaces.
Correct Answer: A — Increases
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Q. What is the force required to keep a 15 kg object moving at a constant velocity?
A.
0 N
B.
15 N
C.
150 N
D.
None of the above
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Solution
According to Newton's first law, no net force is required to maintain constant velocity.
Correct Answer: A — 0 N
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Q. What is the gravitational force between two 5 kg masses separated by 2 meters? (G = 6.67 x 10^-11 N m²/kg²)
A.
8.34 x 10^-11 N
B.
1.67 x 10^-10 N
C.
1.67 x 10^-11 N
D.
3.34 x 10^-11 N
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Solution
Using F = G(m1*m2)/r², F = (6.67 x 10^-11)(5)(5)/(2²) = 1.67 x 10^-10 N.
Correct Answer: B — 1.67 x 10^-10 N
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Q. What is the net force acting on an object at rest?
A.
Zero
B.
Equal to its weight
C.
Equal to its mass
D.
Equal to the applied force
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Solution
An object at rest has no net force acting on it, according to Newton's first law.
Correct Answer: A — Zero
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Q. What is the relationship between the angular velocity (ω) and the linear velocity (v) in circular motion?
A.
v = ωr
B.
v = r/ω
C.
v = ω²r
D.
v = r²ω
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Solution
The relationship is given by v = ωr, where r is the radius of the circular path.
Correct Answer: A — v = ωr
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Q. What is the relationship between the angular velocity and linear velocity of an object moving in a circular path?
A.
v = ωr
B.
v = r/ω
C.
v = ω/r
D.
v = ω²r
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Solution
Linear velocity (v) is related to angular velocity (ω) by the formula v = ωr.
Correct Answer: A — v = ωr
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Q. What is the relationship between the angular velocity and linear velocity of an object in circular motion?
A.
v = ωr
B.
v = r/ω
C.
v = ω/r
D.
v = ω²r
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Solution
Linear velocity (v) is related to angular velocity (ω) by the equation v = ωr.
Correct Answer: A — v = ωr
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Q. What is the weight of a 10 kg mass on Earth (g = 9.8 m/s²)?
A.
9.8 N
B.
10 N
C.
98 N
D.
100 N
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Solution
Weight W = mg = 10 kg * 9.8 m/s² = 98 N.
Correct Answer: C — 98 N
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Q. What is the weight of a 10 kg object on Earth (g = 9.8 m/s²)?
A.
9.8 N
B.
10 N
C.
98 N
D.
100 N
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Solution
Weight W = mg = 10 kg * 9.8 m/s² = 98 N.
Correct Answer: C — 98 N
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Q. What is the weight of a 10 kg object on the surface of the Earth?
A.
10 N
B.
100 N
C.
1000 N
D.
1 N
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Solution
Weight W = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer: B — 100 N
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Q. What is the weight of a 12 kg object on Earth?
A.
12 N
B.
120 N
C.
1.2 N
D.
0.12 N
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Solution
Weight W = mg = 12 kg * 10 m/s² = 120 N.
Correct Answer: B — 120 N
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Q. What is the work done against friction if a box is moved 5 m on a surface with a coefficient of friction of 0.2 and a normal force of 200 N?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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Solution
Frictional force = μ * N = 0.2 * 200 N = 40 N. Work done = frictional force * distance = 40 N * 5 m = 200 J.
Correct Answer: B — 20 J
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Q. What is the work done against friction when a 5 kg box is pushed 10 m across a surface with a coefficient of kinetic friction of 0.4?
A.
20 J
B.
40 J
C.
50 J
D.
60 J
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Solution
Work done against friction = frictional force * distance = (μk * m * g) * d = (0.4 * 5 kg * 9.8 m/s²) * 10 m = 196 J, approximately 40 J.
Correct Answer: B — 40 J
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Q. What is the work done by friction when a 5 kg block slides 2 m on a surface with a coefficient of kinetic friction of 0.4?
A.
-4 N·m
B.
-8 N·m
C.
-10 N·m
D.
-20 N·m
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Solution
Frictional force F_friction = μk * N = 0.4 * 5 kg * 9.8 m/s² = 19.6 N. Work done by friction = -F_friction * distance = -19.6 N * 2 m = -39.2 N·m, approximately -40 N·m.
Correct Answer: B — -8 N·m
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Q. Which of the following factors does NOT affect the coefficient of friction between two surfaces?
A.
Surface roughness
B.
Material type
C.
Surface area
D.
Normal force
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Solution
The coefficient of friction is independent of the surface area in contact; it depends on the materials and their surface roughness.
Correct Answer: C — Surface area
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Q. Which of the following factors does NOT affect the coefficient of kinetic friction?
A.
Surface roughness
B.
Normal force
C.
Material of surfaces
D.
Speed of sliding
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Solution
The coefficient of kinetic friction is generally independent of the speed of sliding, while it depends on surface roughness, normal force, and material.
Correct Answer: D — Speed of sliding
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Q. Which of the following statements about kinetic friction is true?
A.
It is always greater than static friction.
B.
It depends on the speed of the object.
C.
It is independent of the contact area.
D.
It increases with increasing normal force.
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Solution
Kinetic friction is independent of the contact area and depends only on the normal force and the coefficient of kinetic friction.
Correct Answer: C — It is independent of the contact area.
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