Q. A block of mass 5 kg is resting on a frictionless surface. A force of 10 N is applied to it. What will be its acceleration? (2020)
A.
2 m/s²
B.
5 m/s²
C.
10 m/s²
D.
20 m/s²
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Solution
Using Newton's second law, F = ma, acceleration (a) = F/m = 10 N / 5 kg = 2 m/s².
Correct Answer: B — 5 m/s²
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Q. A block of mass 5 kg is resting on a frictionless surface. If a force of 15 N is applied, what will be its velocity after 3 seconds? (2022)
A.
3 m/s
B.
5 m/s
C.
9 m/s
D.
15 m/s
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Solution
Acceleration a = F/m = 15 N / 5 kg = 3 m/s². Velocity after 3 seconds = a × t = 3 m/s² × 3 s = 9 m/s.
Correct Answer: C — 9 m/s
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Q. A block of mass 5 kg is resting on a horizontal surface. If a horizontal force of 15 N is applied, what is the acceleration of the block? (Assume no friction) (2021)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using F = ma, acceleration a = F/m = 15 N / 5 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A block of mass 5 kg is sliding down a frictionless incline of angle 30°. What is the acceleration of the block? (2022)
A.
4.9 m/s²
B.
5 m/s²
C.
9.8 m/s²
D.
3.9 m/s²
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Solution
Acceleration a = g sin(θ) = 9.8 m/s² × sin(30°) = 4.9 m/s².
Correct Answer: A — 4.9 m/s²
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Q. A car accelerates from rest at a constant rate of 2 m/s². How far does it travel in 5 seconds? (2021)
A.
10 m
B.
20 m
C.
25 m
D.
50 m
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Solution
Using the equation of motion: s = ut + (1/2)at². Here, u = 0, a = 2 m/s², t = 5 s. So, s = 0 + (1/2)(2)(5²) = 25 m.
Correct Answer: C — 25 m
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Q. A car accelerates from rest at a rate of 2 m/s². What is its speed after 5 seconds? (2023)
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
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Solution
Using the formula v = u + at, where u = 0, a = 2 m/s², and t = 5 s, we get v = 0 + 2 * 5 = 10 m/s.
Correct Answer: B — 10 m/s
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Q. A car accelerates from rest to a speed of 30 m/s in 10 seconds. What is the acceleration of the car? (2022)
A.
3 m/s²
B.
2 m/s²
C.
1 m/s²
D.
4 m/s²
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Solution
Acceleration (a) = (final velocity - initial velocity) / time = (30 m/s - 0) / 10 s = 3 m/s².
Correct Answer: A — 3 m/s²
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Q. A cyclist accelerates at 3 m/s² for 4 seconds. What is the final velocity if the initial velocity is 2 m/s? (2023)
A.
10 m/s
B.
11 m/s
C.
14 m/s
D.
20 m/s
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Solution
Using v = u + at, we have v = 2 m/s + (3 m/s² * 4 s) = 2 + 12 = 14 m/s.
Correct Answer: B — 11 m/s
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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the acceleration? (2023)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Acceleration = (final velocity - initial velocity) / time = (10 m/s - 0) / 5 s = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the cyclist's acceleration? (2021)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Acceleration a = (final velocity - initial velocity) / time = (10 m/s - 0) / 5 s = 2 m/s².
Correct Answer: B — 2 m/s²
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Q. A cyclist of mass 70 kg accelerates at 2 m/s². What is the force exerted by the cyclist on the ground? (2021)
A.
70 N
B.
140 N
C.
210 N
D.
280 N
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Solution
Using F = ma, F = 70 kg * 2 m/s² = 140 N.
Correct Answer: B — 140 N
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Q. A force of 15 N is applied to a 3 kg mass. What is the resulting acceleration? (2023)
A.
3 m/s²
B.
5 m/s²
C.
7 m/s²
D.
10 m/s²
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Solution
Using F = ma, acceleration a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer: B — 5 m/s²
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Q. A force of 30 N is applied to a 6 kg object. What is the object's acceleration? (2023)
A.
3 m/s²
B.
4 m/s²
C.
5 m/s²
D.
6 m/s²
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Solution
Using F = ma, we find a = F/m = 30 N / 6 kg = 5 m/s².
Correct Answer: B — 4 m/s²
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Q. An object is thrown vertically upwards with a velocity of 20 m/s. How high will it rise? (Assume g = 10 m/s²) (2020)
A.
20 m
B.
30 m
C.
40 m
D.
50 m
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Solution
Using h = v²/(2g) = (20 m/s)² / (2 × 10 m/s²) = 20 m.
Correct Answer: A — 20 m
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Q. An object moves with a constant velocity of 10 m/s. What is the net force acting on it? (2023)
A.
0 N
B.
10 N
C.
20 N
D.
100 N
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Solution
If the object moves with constant velocity, the net force acting on it is 0 N.
Correct Answer: A — 0 N
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Q. If a 10 kg object is dropped from a height of 20 m, what is the velocity just before it hits the ground? (Assume g = 10 m/s²) (2019)
A.
10 m/s
B.
20 m/s
C.
14 m/s
D.
15 m/s
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Solution
Using v² = u² + 2gh, v = √(0 + 2 × 10 m/s² × 20 m) = √400 = 20 m/s.
Correct Answer: B — 20 m/s
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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground (g = 10 m/s²)? (2023)
A.
10 m/s
B.
20 m/s
C.
30 m/s
D.
40 m/s
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Solution
Using v = √(2gh), for h = 10 m, v = √(2 * 10 m/s² * 10 m) = √200 = 20 m/s.
Correct Answer: B — 20 m/s
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Q. If a 2 kg object is dropped from a height, what is its velocity just before hitting the ground? (g = 9.8 m/s², h = 20 m) (2023)
A.
14 m/s
B.
19.6 m/s
C.
20 m/s
D.
28 m/s
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Solution
Using v² = u² + 2gh, where u = 0, we find v = √(2 * 9.8 * 20) = 19.6 m/s.
Correct Answer: B — 19.6 m/s
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Q. If a 3 kg object is subjected to a net force of 12 N, what is its acceleration? (2023)
A.
4 m/s²
B.
3 m/s²
C.
2 m/s²
D.
1 m/s²
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Solution
Acceleration a = F/m = 12 N / 3 kg = 4 m/s².
Correct Answer: A — 4 m/s²
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Q. If a 5 kg object is moving with a velocity of 3 m/s, what is its momentum? (2023)
A.
5 kg·m/s
B.
10 kg·m/s
C.
15 kg·m/s
D.
20 kg·m/s
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Solution
Momentum p = mv = 5 kg * 3 m/s = 15 kg·m/s.
Correct Answer: C — 15 kg·m/s
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Q. If a 5 kg object is subjected to a net force of 20 N, what is its acceleration? (2023)
A.
2 m/s²
B.
4 m/s²
C.
5 m/s²
D.
10 m/s²
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Solution
Using F = ma, we find a = F/m = 20 N / 5 kg = 4 m/s².
Correct Answer: B — 4 m/s²
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Q. If a force of 10 N is applied to a mass of 2 kg, what is the acceleration of the mass? (2020)
A.
2 m/s²
B.
5 m/s²
C.
10 m/s²
D.
20 m/s²
Show solution
Solution
Using F = ma, acceleration a = F/m = 10 N / 2 kg = 5 m/s².
Correct Answer: B — 5 m/s²
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Q. If a force of 15 N is applied to a 3 kg object, what is the net force acting on the object if it is already moving with a velocity of 5 m/s? (2022)
A.
15 N
B.
3 N
C.
0 N
D.
18 N
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Solution
The net force is still 15 N as it does not depend on the velocity of the object.
Correct Answer: A — 15 N
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Q. If a force of 15 N is applied to a mass of 3 kg, what is the acceleration of the mass? (2023)
A.
3 m/s²
B.
4 m/s²
C.
5 m/s²
D.
6 m/s²
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Solution
Using F = ma, a = F/m = 15 N / 3 kg = 5 m/s².
Correct Answer: B — 4 m/s²
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Q. What is the net force acting on a 12 kg object moving with a constant velocity? (2023)
A.
0 N
B.
12 N
C.
24 N
D.
36 N
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Solution
According to Newton's first law, if the object is moving with constant velocity, the net force is 0 N.
Correct Answer: A — 0 N
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Q. What is the net force acting on a 15 kg object moving with a constant velocity? (2021)
A.
0 N
B.
15 N
C.
30 N
D.
45 N
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Solution
If the object is moving with constant velocity, net force is 0 N (Newton's first law).
Correct Answer: A — 0 N
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Q. What is the net force acting on a 15 kg object that is accelerating at 4 m/s²? (2020)
A.
60 N
B.
30 N
C.
45 N
D.
75 N
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Solution
Net force F = m × a = 15 kg × 4 m/s² = 60 N.
Correct Answer: A — 60 N
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