Newtons Laws
Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the force exerted by the spring when it is compressed by 0.1 m?
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A.
2 N
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B.
5 N
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C.
10 N
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D.
20 N
Solution
Using Hooke's Law, F = kx = 200 N/m * 0.1 m = 20 N.
Correct Answer: C — 10 N
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Q. A 1 kg object is pushed with a force of 10 N. If the frictional force is 4 N, what is the net force acting on the object?
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A.
6 N
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B.
10 N
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C.
4 N
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D.
0 N
Solution
Net force = applied force - frictional force = 10 N - 4 N = 6 N.
Correct Answer: A — 6 N
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Q. A 10 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
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A.
10 N
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B.
20 N
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C.
30 N
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D.
40 N
Solution
The force acting on the object is its weight, F = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer: B — 20 N
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Q. A 10 kg object is hanging at rest from a rope. What is the tension in the rope?
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A.
0 N
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B.
10 N
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C.
100 N
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D.
50 N
Solution
The tension in the rope must balance the weight of the object. T = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer: C — 100 N
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Q. A 10 kg object is hanging from a rope. What is the tension in the rope when the object is at rest?
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A.
0 N
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B.
10 N
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C.
100 N
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D.
50 N
Solution
At rest, the tension in the rope equals the weight of the object: T = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer: C — 100 N
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Q. A 10 kg object is pushed with a force of 30 N. If the frictional force is 10 N, what is the net force?
-
A.
10 N
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B.
20 N
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C.
30 N
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D.
40 N
Solution
Net force = applied force - friction = 30 N - 10 N = 20 N.
Correct Answer: B — 20 N
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Q. A 12 kg object is at rest on a horizontal surface. What is the normal force acting on it?
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A.
0 N
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B.
12 N
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C.
120 N
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D.
100 N
Solution
The normal force equals the weight of the object: N = mg = 12 kg * 10 m/s² = 120 N.
Correct Answer: C — 120 N
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Q. A 12 kg object is in free fall. What is the force acting on it due to gravity?
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A.
12 N
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B.
24 N
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C.
36 N
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D.
48 N
Solution
The force due to gravity is F = mg = 12 kg * 9.8 m/s² = 117.6 N (approximately 120 N).
Correct Answer: B — 24 N
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Q. A 12 kg object is moving with a velocity of 5 m/s. What is its momentum?
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A.
30 kg·m/s
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B.
60 kg·m/s
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C.
90 kg·m/s
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D.
120 kg·m/s
Solution
Momentum p = mv = 12 kg * 5 m/s = 60 kg·m/s.
Correct Answer: B — 60 kg·m/s
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Q. A 12 kg object is pulled with a force of 24 N. If the object experiences a frictional force of 4 N, what is its acceleration?
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A.
1.67 m/s²
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B.
2 m/s²
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C.
2.5 m/s²
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D.
3 m/s²
Solution
Net force = 24 N - 4 N = 20 N. Acceleration a = F/m = 20 N / 12 kg = 1.67 m/s².
Correct Answer: B — 2 m/s²
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Q. A 15 kg box is pushed with a force of 30 N. If the frictional force opposing the motion is 10 N, what is the acceleration of the box?
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A.
1 m/s²
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B.
2 m/s²
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C.
3 m/s²
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D.
4 m/s²
Solution
Net force = applied force - friction = 30 N - 10 N = 20 N. Acceleration a = F/m = 20 N / 15 kg = 1.33 m/s².
Correct Answer: B — 2 m/s²
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Q. A 15 kg box is pushed with a force of 45 N. What is the acceleration of the box? (Assume no friction)
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A.
1 m/s²
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B.
2 m/s²
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C.
3 m/s²
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D.
4 m/s²
Solution
Using F = ma, we have a = F/m = 45 N / 15 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 15 kg object is at rest on a table. What is the force exerted by the table on the object?
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A.
0 N
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B.
15 N
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C.
150 N
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D.
10 N
Solution
The normal force equals the weight of the object: 15 kg * 10 m/s² = 150 N.
Correct Answer: C — 150 N
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Q. A 15 kg object is in equilibrium. What is the net force acting on it?
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A.
0 N
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B.
15 N
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C.
30 N
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D.
45 N
Solution
In equilibrium, the net force acting on the object is 0 N.
Correct Answer: A — 0 N
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Q. A 15 kg object is pushed with a force of 45 N. What is the acceleration of the object?
-
A.
2 m/s²
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B.
3 m/s²
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C.
4 m/s²
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D.
5 m/s²
Solution
Using F = ma, acceleration a = F/m = 45 N / 15 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 2 kg block is sliding on a frictionless surface. If a force of 6 N is applied, what is its acceleration?
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A.
2 m/s²
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B.
3 m/s²
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C.
4 m/s²
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D.
6 m/s²
Solution
Using F = ma, acceleration a = F/m = 6 N / 2 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 2 kg object is in free fall. What is the net force acting on it?
-
A.
0 N
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B.
2 N
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C.
20 N
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D.
19.6 N
Solution
The net force is equal to the weight of the object: F = mg = 2 kg * 9.8 m/s² = 19.6 N.
Correct Answer: C — 20 N
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Q. A 2 kg object is pulled with a force of 8 N. What is the acceleration of the object?
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A.
2 m/s²
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B.
4 m/s²
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C.
6 m/s²
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D.
8 m/s²
Solution
Using F = ma, we find a = F/m = 8 N / 2 kg = 4 m/s².
Correct Answer: B — 4 m/s²
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Q. A 2 kg object is sliding on a frictionless surface with a velocity of 4 m/s. What is the momentum of the object?
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A.
8 kg·m/s
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B.
2 kg·m/s
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C.
4 kg·m/s
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D.
16 kg·m/s
Solution
Momentum p = mv = 2 kg * 4 m/s = 8 kg·m/s.
Correct Answer: A — 8 kg·m/s
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Q. A 20 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
-
A.
0 N
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B.
20 N
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C.
200 N
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D.
10 N
Solution
The force acting on the object is its weight: F = mg = 20 kg * 10 m/s² = 200 N.
Correct Answer: C — 200 N
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Q. A 25 kg object is pulled with a force of 100 N. What is the acceleration of the object?
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A.
2 m/s²
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B.
3 m/s²
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C.
4 m/s²
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D.
5 m/s²
Solution
Using F = ma, we have a = F/m = 100 N / 25 kg = 4 m/s².
Correct Answer: D — 5 m/s²
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Q. A 25 kg object is subjected to a force of 50 N. What is its acceleration?
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A.
1 m/s²
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B.
2 m/s²
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C.
3 m/s²
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D.
4 m/s²
Solution
Using F = ma, we find a = F/m = 50 N / 25 kg = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A 3 kg block is on a table and a horizontal force of 15 N is applied. If the frictional force is 5 N, what is the acceleration of the block?
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A.
2 m/s²
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B.
3 m/s²
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C.
5 m/s²
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D.
10 m/s²
Solution
Net force = applied force - friction = 15 N - 5 N = 10 N. Acceleration a = F/m = 10 N / 3 kg = 3.33 m/s².
Correct Answer: A — 2 m/s²
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Q. A 3 kg block is pulled with a force of 12 N. If the frictional force is 3 N, what is the acceleration of the block?
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A.
3 m/s²
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B.
4 m/s²
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C.
2 m/s²
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D.
1 m/s²
Solution
Net force = applied force - friction = 12 N - 3 N = 9 N. Acceleration a = F/m = 9 N / 3 kg = 3 m/s².
Correct Answer: B — 4 m/s²
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Q. A 3 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block?
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A.
3.9 m/s²
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B.
4.9 m/s²
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C.
9.8 m/s²
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D.
1.5 m/s²
Solution
The acceleration down the incline is given by a = g * sin(θ) = 9.8 m/s² * sin(30°) = 4.9 m/s².
Correct Answer: A — 3.9 m/s²
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Q. A 3 kg block is sliding on a frictionless surface with a velocity of 4 m/s. What is the momentum of the block?
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A.
12 kg·m/s
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B.
8 kg·m/s
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C.
6 kg·m/s
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D.
4 kg·m/s
Solution
Momentum p = mv = 3 kg * 4 m/s = 12 kg·m/s.
Correct Answer: A — 12 kg·m/s
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Q. A 3 kg object is at rest on a frictionless surface. A force of 9 N is applied. What is the final velocity after 3 seconds?
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A.
1 m/s
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B.
2 m/s
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C.
3 m/s
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D.
4 m/s
Solution
First, find acceleration: a = F/m = 9 N / 3 kg = 3 m/s². Then, v = u + at = 0 + 3 m/s² * 3 s = 9 m/s.
Correct Answer: C — 3 m/s
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Q. A 3 kg object is at rest on a frictionless surface. If a force of 9 N is applied, what will be its velocity after 3 seconds?
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A.
3 m/s
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B.
6 m/s
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C.
9 m/s
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D.
12 m/s
Solution
First, find acceleration: a = F/m = 9 N / 3 kg = 3 m/s². Then, v = u + at = 0 + 3 m/s² * 3 s = 9 m/s.
Correct Answer: B — 6 m/s
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Q. A 3 kg object is at rest on a table. If a horizontal force of 12 N is applied, what is the acceleration of the object?
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A.
4 m/s²
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B.
0 m/s²
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C.
3 m/s²
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D.
12 m/s²
Solution
Using F = ma, acceleration a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer: A — 4 m/s²
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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?
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A.
0 N
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B.
3 N
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C.
15 N
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D.
5 N
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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