Newton's Laws for Competitive Exam Success

Newtons Laws

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?

A. 12 kg·m/s
B. 8 kg·m/s
C. 6 kg·m/s
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?

A. 1 m/s
B. 2 m/s
C. 3 m/s
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?

A. 3 m/s
B. 6 m/s
C. 9 m/s
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?

A. 4 m/s²
B. 0 m/s²
C. 3 m/s²
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?

A. 0 N
B. 3 N
C. 15 N
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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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

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 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

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

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

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

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 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

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

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

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 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

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

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

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 50 kg crate is at rest on a flat surface. What is the normal force acting on the crate?

A. 0 N
B. 50 N
C. 500 N
D. 1000 N

Solution

The normal force equals the weight of the crate, which is N = mg = 50 kg * 10 m/s² = 500 N.

Correct Answer: C — 500 N

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Q. A 50 kg object is at rest on a surface. What is the normal force acting on it?

A. 0 N
B. 50 N
C. 100 N
D. 500 N

Solution

The normal force equals the weight of the object, which is F = mg = 50 kg * 9.8 m/s² = 490 N.

Correct Answer: C — 100 N

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Q. A 50 kg object is in free fall. What is the net force acting on it?

A. 50 N
B. 100 N
C. 200 N
D. 500 N

Solution

The net force is equal to the weight of the object, F = mg = 50 kg * 9.8 m/s² = 490 N (approximately 500 N).

Correct Answer: B — 100 N

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Q. A 50 kg object is moving with a constant velocity. What can be said about the net force acting on it?

A. It is zero
B. It is equal to its weight
C. It is equal to the applied force
D. It is maximum

Solution

If the object is moving with constant velocity, the net force acting on it is zero.

Correct Answer: A — It is zero

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Q. A 50 kg object is pulled with a force of 200 N. What is the acceleration of the object?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, we have a = F/m = 200 N / 50 kg = 4 m/s².

Correct Answer: D — 5 m/s²

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Q. A 6 kg object is in free fall. What is the force acting on it due to gravity?

A. 6 N
B. 60 N
C. 600 N
D. 0 N

Solution

Force due to gravity F = mg = 6 kg * 10 m/s² = 60 N.

Correct Answer: B — 60 N

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Q. A 6 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?

A. 27 J
B. 36 J
C. 54 J
D. 18 J

Solution

Kinetic energy KE = 0.5 * m * v² = 0.5 * 6 kg * (3 m/s)² = 27 J.

Correct Answer: B — 36 J

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Q. A 6 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

Solution

Using F = ma, acceleration a = F/m = 12 N / 6 kg = 2 m/s².

Correct Answer: B — 3 m/s²

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Q. A car accelerates from rest at a rate of 2 m/s². What is the net force acting on the car if its mass is 1000 kg?

A. 200 N
B. 500 N
C. 1000 N
D. 2000 N

Solution

Using F = ma, F = 1000 kg * 2 m/s² = 2000 N.

Correct Answer: D — 2000 N

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Q. A car of mass 1000 kg accelerates at 2 m/s². What is the net force acting on the car?

A. 2000 N
B. 500 N
C. 1000 N
D. 1500 N

Solution

Using F = ma, the net force is F = 1000 kg * 2 m/s² = 2000 N.

Correct Answer: A — 2000 N

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Q. A car of mass 1000 kg is moving with a velocity of 20 m/s. If the brakes are applied and the car comes to a stop in 5 seconds, what is the average force applied by the brakes?

A. 2000 N
B. 4000 N
C. 5000 N
D. 6000 N

Solution

The change in momentum is 1000 kg * 20 m/s = 20000 kg·m/s. The average force is F = Δp/Δt = 20000 kg·m/s / 5 s = 4000 N.

Correct Answer: B — 4000 N

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Q. A car of mass 1000 kg is moving with a velocity of 20 m/s. What is the net force required to bring it to rest in 5 seconds?

A. 4000 N
B. 2000 N
C. 1000 N
D. 500 N

Solution

First, find the deceleration: a = (final velocity - initial velocity) / time = (0 - 20 m/s) / 5 s = -4 m/s². Then, F = ma = 1000 kg * 4 m/s² = 4000 N.

Correct Answer: A — 4000 N

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Q. A force of 10 N is applied to a 2 kg object. What is the object's acceleration?

A. 5 m/s²
B. 10 m/s²
C. 2 m/s²
D. 20 m/s²

Solution

Using F = ma, acceleration a = F/m = 10 N / 2 kg = 5 m/s².

Correct Answer: A — 5 m/s²

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Q. A force of 10 N is applied to a 2 kg object. What is the resulting acceleration?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, we have a = F/m = 10 N / 2 kg = 5 m/s².

Correct Answer: A — 2 m/s²

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Newtons Laws

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