Laws of Motion
Q. A 3 kg object is in free fall. What is the force acting on it due to gravity?
A.
3 N
B.
9 N
C.
30 N
D.
0 N
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Solution
The force due to gravity is F = mg = 3 kg * 9.8 m/s² = 29.4 N, approximately 30 N.
Correct Answer: B — 9 N
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Q. A 3 kg object is moving in a circular path of radius 2 m with a speed of 4 m/s. What is the centripetal force acting on it?
A.
6 N
B.
12 N
C.
24 N
D.
36 N
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Solution
Centripetal force F = mv²/r = 3 kg * (4 m/s)² / 2 m = 24 N.
Correct Answer: B — 12 N
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Q. A 3 kg object is moving in a straight line with a constant velocity of 5 m/s. What is the net force acting on the object?
A.
0 N
B.
3 N
C.
15 N
D.
5 N
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Solution
Since the object is moving with constant velocity, the net force acting on it is 0 N.
Correct Answer: A — 0 N
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Q. A 3 kg object is moving with a velocity of 4 m/s. If a force of 12 N is applied in the opposite direction, what will be its velocity after 2 seconds?
A.
0 m/s
B.
2 m/s
C.
4 m/s
D.
6 m/s
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Solution
Acceleration = F/m = -12 N / 3 kg = -4 m/s². Change in velocity = a * t = -4 m/s² * 2 s = -8 m/s. Final velocity = 4 m/s - 8 m/s = -4 m/s (object stops).
Correct Answer: B — 2 m/s
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Q. A 3 kg object is subjected to a net force of 12 N. What is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Using F = ma, a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer: C — 4 m/s²
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Q. A 4 kg object is at rest on a table. What is the force of static friction acting on it?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
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Solution
If the object is at rest and no external force is applied, the static friction force is 0 N.
Correct Answer: A — 0 N
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Q. A 4 kg object is at rest on a table. What is the force of static friction if no external force is applied?
A.
0 N
B.
4 N
C.
40 N
D.
None of the above
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Solution
If no external force is applied, the force of static friction is 0 N.
Correct Answer: A — 0 N
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Q. A 4 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
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Solution
The force acting on the object is its weight, F = mg = 4 kg * 10 m/s² = 40 N.
Correct Answer: C — 40 N
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Q. A 4 kg object is hanging from a rope. What is the tension in the rope if the object is at rest?
A.
0 N
B.
4 N
C.
40 N
D.
10 N
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Solution
The tension in the rope equals the weight of the object: T = mg = 4 kg * 10 m/s² = 40 N.
Correct Answer: C — 40 N
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Q. A 4 kg object is moving in a circular path of radius 2 m with a speed of 6 m/s. What is the centripetal force acting on the object?
A.
9 N
B.
12 N
C.
18 N
D.
24 N
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Solution
Centripetal force F = mv²/r = 4 kg * (6 m/s)² / 2 m = 4 * 36 / 2 = 72 N / 2 = 36 N.
Correct Answer: B — 12 N
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Q. A 4 kg object is moving in a circular path of radius 2 m with a speed of 6 m/s. What is the centripetal force acting on it?
A.
9 N
B.
12 N
C.
18 N
D.
24 N
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Solution
Centripetal force F = mv²/r = 4 kg * (6 m/s)² / 2 m = 72 N / 2 = 36 N.
Correct Answer: B — 12 N
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Q. A 4 kg object is moving in a straight line with a velocity of 5 m/s. What is the kinetic energy of the object?
A.
50 J
B.
40 J
C.
20 J
D.
10 J
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Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 4 kg * (5 m/s)² = 50 J.
Correct Answer: A — 50 J
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Q. A 4 kg object is moving with a velocity of 2 m/s. What is the change in momentum if the object comes to a stop?
A.
0 kg·m/s
B.
4 kg·m/s
C.
8 kg·m/s
D.
2 kg·m/s
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Solution
Change in momentum = final momentum - initial momentum = 0 - (4 kg * 2 m/s) = -8 kg·m/s.
Correct Answer: C — 8 kg·m/s
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Q. A 4 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
18 J
B.
24 J
C.
36 J
D.
12 J
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Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 4 kg * (3 m/s)² = 18 J.
Correct Answer: B — 24 J
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Q. A 4 kg object is moving with a velocity of 5 m/s. What is its momentum?
A.
10 kg·m/s
B.
20 kg·m/s
C.
15 kg·m/s
D.
25 kg·m/s
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Solution
Momentum p = mv = 4 kg * 5 m/s = 20 kg·m/s.
Correct Answer: B — 20 kg·m/s
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Q. A 4 kg object is pulled with a force of 20 N. If the frictional force is 4 N, what is the net force acting on the object?
A.
16 N
B.
20 N
C.
24 N
D.
4 N
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Solution
Net force = applied force - frictional force = 20 N - 4 N = 16 N.
Correct Answer: A — 16 N
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Q. A 4 kg object is pushed with a force of 16 N. What is the acceleration of the object?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Using F = ma, we have a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer: C — 4 m/s²
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Q. A 4 kg object is pushed with a force of 20 N. If the frictional force is 4 N, what is the acceleration of the object?
A.
4 m/s²
B.
5 m/s²
C.
6 m/s²
D.
7 m/s²
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Solution
Net force = applied force - friction = 20 N - 4 N = 16 N. Acceleration a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer: B — 5 m/s²
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Q. A 4 kg object is subjected to a net force of 12 N. What is the acceleration of the object?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Using F = ma, a = F/m = 12 N / 4 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 4 kg object is subjected to a net force of 16 N. What is its acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Using F = ma, acceleration a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer: C — 4 m/s²
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Q. A 5 kg block is placed on a surface with a coefficient of static friction of 0.6. What is the maximum static friction force before the block starts to move?
A.
30 N
B.
20 N
C.
15 N
D.
25 N
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Solution
Maximum static friction force = μs * N = 0.6 * (5 kg * 9.8 m/s²) = 29.4 N, approximately 30 N.
Correct Answer: A — 30 N
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Q. A 5 kg block is resting on a frictionless surface. A force of 10 N is applied to it. What is the acceleration of the block?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A 5 kg block is resting on a frictionless surface. If a force of 10 N is applied to it, what will be its acceleration?
A.
1 m/s²
B.
2 m/s²
C.
5 m/s²
D.
10 m/s²
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Solution
Using Newton's second law, F = ma, we have a = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A 5 kg block is resting on a horizontal surface. What is the normal force acting on the block?
A.
5 N
B.
10 N
C.
15 N
D.
20 N
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Solution
The normal force equals the weight of the block, which is mass times gravity: 5 kg * 9.8 m/s² = 49 N.
Correct Answer: B — 10 N
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Q. A 5 kg object is at rest on a surface with a coefficient of static friction of 0.4. What is the maximum static friction force?
A.
10 N
B.
20 N
C.
30 N
D.
40 N
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Solution
Maximum static friction force = μs * N = 0.4 * (5 kg * 9.8 m/s²) = 0.4 * 49 N = 19.6 N, approximately 20 N.
Correct Answer: B — 20 N
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Q. A 5 kg object is at rest on a surface. If a horizontal force of 15 N is applied, what is the object's acceleration? (Assume friction is negligible)
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground? (Assume g = 9.8 m/s²)
A.
5 N
B.
9.8 N
C.
49 N
D.
0 N
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Solution
The force acting on the object is its weight, F = mg = 5 kg * 9.8 m/s² = 49 N.
Correct Answer: C — 49 N
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Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
A.
5 N
B.
10 N
C.
15 N
D.
20 N
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Solution
The only force acting on it is its weight, W = mg = 5 kg * 10 m/s² = 50 N.
Correct Answer: B — 10 N
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Q. A 5 kg object is pulled with a force of 25 N. What is the acceleration of the object?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Using F = ma, acceleration a = F/m = 25 N / 5 kg = 5 m/s².
Correct Answer: C — 4 m/s²
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Q. A 5 kg object is pushed with a force of 15 N. If the frictional force is 5 N, what is the acceleration of the object?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
Show solution
Solution
Net force = applied force - friction = 15 N - 5 N = 10 N. Acceleration = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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