Laws of Motion Study Guide for Competitive Exams

Laws of Motion

Circular Motion Friction Newtons Laws

Q. A 1 kg ball is thrown upwards with a force of 10 N. What is the net force acting on the ball at the peak of its motion?

A. 0 N
B. 10 N
C. 5 N
D. 20 N

Solution

At the peak, the only force acting on the ball is its weight (10 N down), so the net force is 0 N.

Correct Answer: A — 0 N

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Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (Assume g = 10 m/s²)

A. 20 m
B. 30 m
C. 40 m
D. 50 m

Solution

Using the formula h = v²/(2g), h = (20 m/s)² / (2 * 10 m/s²) = 400 / 20 = 20 m.

Correct Answer: B — 30 m

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Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum extension of the spring when the mass is released from rest?

A. 0.5 m
B. 1 m
C. 2 m
D. 0.25 m

Solution

Using Hooke's law, F = kx, where F = mg = 1 kg * 9.8 m/s² = 9.8 N. Thus, x = F/k = 9.8 N / 200 N/m = 0.049 m.

Correct Answer: A — 0.5 m

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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?

A. 2 N
B. 5 N
C. 10 N
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 mass is attached to a spring with a spring constant of 200 N/m. What is the maximum force exerted by the spring when it is compressed by 0.1 m?

A. 2 N
B. 5 N
C. 20 N
D. 10 N

Solution

Using Hooke's law, F = kx = 200 N/m * 0.1 m = 20 N.

Correct Answer: C — 20 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?

A. 6 N
B. 10 N
C. 4 N
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 block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be the acceleration of the block?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be its acceleration? (Assume no friction)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using F = ma, a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be its acceleration?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

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Q. A 10 kg object is at rest on a horizontal surface. If a force of 30 N is applied, what is the object's acceleration assuming a frictional force of 10 N?

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

Solution

Net force = 30 N - 10 N = 20 N. Using F = ma, a = F/m = 20 N / 10 kg = 2 m/s².

Correct Answer: B — 3 m/s²

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Q. A 10 kg object is at rest on a surface. If a force of 30 N is applied, what is the object's acceleration assuming the friction is negligible?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².

Correct Answer: C — 3 m/s²

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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?

A. 10 N
B. 20 N
C. 30 N
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?

A. 0 N
B. 10 N
C. 100 N
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?

A. 0 N
B. 10 N
C. 100 N
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 moving in a circular path of radius 5 m with a speed of 10 m/s. What is the centripetal force acting on the object?

A. 20 N
B. 10 N
C. 5 N
D. 50 N

Solution

Centripetal force F = mv²/r = 10 kg * (10 m/s)² / 5 m = 200 N.

Correct Answer: A — 20 N

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Q. A 10 kg object is moving in a circular path of radius 5 m with a speed of 4 m/s. What is the centripetal force acting on the object?

A. 8 N
B. 16 N
C. 32 N
D. 40 N

Solution

Centripetal force F = mv²/r = 10 kg * (4 m/s)² / 5 m = 32 N.

Correct Answer: B — 16 N

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Q. A 10 kg object is moving with a velocity of 2 m/s. If it comes to rest, what is the change in momentum?

A. 20 kg·m/s
B. 10 kg·m/s
C. 5 kg·m/s
D. 0 kg·m/s

Solution

Initial momentum = mv = 10 kg * 2 m/s = 20 kg·m/s. Final momentum = 0. Change in momentum = 20 kg·m/s - 0 = 20 kg·m/s.

Correct Answer: A — 20 kg·m/s

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Q. A 10 kg object is moving with a velocity of 3 m/s. What is its momentum?

A. 10 kg·m/s
B. 30 kg·m/s
C. 3 kg·m/s
D. 0.3 kg·m/s

Solution

Momentum p = mv = 10 kg * 3 m/s = 30 kg·m/s.

Correct Answer: B — 30 kg·m/s

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Q. A 10 kg object is moving with a velocity of 5 m/s. What is its momentum?

A. 10 kg·m/s
B. 25 kg·m/s
C. 50 kg·m/s
D. 75 kg·m/s

Solution

Momentum p = mv = 10 kg * 5 m/s = 50 kg·m/s.

Correct Answer: C — 50 kg·m/s

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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
B. 20 N
C. 30 N
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 box is pulled with a force of 60 N at an angle of 30 degrees to the horizontal. What is the horizontal component of the force?

A. 30 N
B. 60 N
C. 51.96 N
D. 20 N

Solution

The horizontal component is F_horizontal = F * cos(θ) = 60 N * cos(30°) = 60 N * (√3/2) = 51.96 N.

Correct Answer: C — 51.96 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?

A. 0 N
B. 12 N
C. 120 N
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?

A. 12 N
B. 24 N
C. 36 N
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 kinetic energy?

A. 150 J
B. 120 J
C. 100 J
D. 75 J

Solution

Kinetic energy KE = 0.5 * m * v² = 0.5 * 12 kg * (5 m/s)² = 150 J.

Correct Answer: B — 120 J

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Q. A 12 kg object is moving with a velocity of 5 m/s. What is its momentum?

A. 30 kg·m/s
B. 60 kg·m/s
C. 90 kg·m/s
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?

A. 1.67 m/s²
B. 2 m/s²
C. 2.5 m/s²
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?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
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)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
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 box is pushed with a force of 60 N on a rough surface with a frictional force of 15 N. What is the box's acceleration?

A. 3 m/s²
B. 4 m/s²
C. 5 m/s²
D. 6 m/s²

Solution

Net force = 60 N - 15 N = 45 N. Acceleration a = F/m = 45 N / 15 kg = 3 m/s².

Correct Answer: B — 4 m/s²

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Q. A 15 kg box is pushed with a force of 60 N. If the frictional force is 20 N, what is the acceleration of the box?

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

Solution

Net force = applied force - friction = 60 N - 20 N = 40 N. Acceleration a = F/m = 40 N / 15 kg = 2.67 m/s².

Correct Answer: B — 3 m/s²

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Laws of Motion

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