Laws of Motion: Essential Concepts for Competitive Exams

Laws of Motion

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 = 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

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 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

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 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

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

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

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

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

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

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

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

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

Solution

Net force F = m × a = 15 kg × 4 m/s² = 60 N.

Correct Answer: A — 60 N

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Laws of Motion

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