Laws of Motion Study Guide for Competitive Exams

Laws of Motion

Circular Motion Friction Newtons Laws

Q. A 5 kg object is subjected to a net force of 15 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, a = F/m = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 5 kg object is subjected to a net force of 15 N. What is the object's acceleration?

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

Solution

Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 5 kg object is subjected to a net force of 25 N. What is its acceleration?

A. 2 m/s²
B. 5 m/s²
C. 10 m/s²
D. 15 m/s²

Solution

Using F = ma, a = F/m = 25 N / 5 kg = 5 m/s².

Correct Answer: C — 10 m/s²

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Q. A 5 kg object is thrown vertically upward with a speed of 20 m/s. What is the maximum height reached by the object? (g = 10 m/s²)

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

Solution

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

Correct Answer: B — 30 m

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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 50 kg person jumps off a diving board with a speed of 5 m/s. What is the momentum of the person just before hitting the water?

A. 100 kg·m/s
B. 200 kg·m/s
C. 250 kg·m/s
D. 300 kg·m/s

Solution

Momentum p = mv = 50 kg * 5 m/s = 250 kg·m/s.

Correct Answer: A — 100 kg·m/s

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Q. A 6 kg object is at rest on a horizontal surface. If a horizontal force of 12 N is applied, what is the acceleration assuming no friction?

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

Solution

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

Correct Answer: B — 2 m/s²

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Q. A 6 kg object is dropped from a height. What is the force acting on it just before it hits the ground?

A. 6 N
B. 60 N
C. 12 N
D. 0 N

Solution

The force acting on it is its weight, F = mg = 6 kg * 10 m/s² = 60 N.

Correct Answer: B — 60 N

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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 pulled with a force of 24 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, a = F/m = 24 N / 6 kg = 4 m/s².

Correct Answer: B — 3 m/s²

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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 ball is swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to just maintain its circular motion?

A. Weight must be greater than tension
B. Tension must be zero
C. Centripetal force must be zero
D. Weight must be less than tension

Solution

At the highest point, the centripetal force is provided by the weight of the ball. For just maintaining motion, tension can be zero.

Correct Answer: B — Tension must be zero

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Q. A ball is thrown vertically upward with a speed of 20 m/s. How high will it rise before coming to rest? (g = 10 m/s²)

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

Solution

Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s², we get 0 = (20)² + 2(-10)s, solving gives s = 20 m.

Correct Answer: B — 30 m

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Q. A ball is thrown vertically upward with a speed of 20 m/s. How high will it rise before coming to a stop? (g = 10 m/s²)

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

Solution

Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s², we find 0 = (20)² + 2(-10)s, giving s = 20 m.

Correct Answer: C — 40 m

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Q. A ball is thrown vertically upward with a speed of 20 m/s. How high will it rise before coming to rest?

A. 20 m
B. 40 m
C. 10 m
D. 80 m

Solution

Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, a = -9.8 m/s². 0 = (20)² + 2(-9.8)s, solving gives s = 20.4 m, approximately 40 m.

Correct Answer: B — 40 m

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Q. A ball is thrown vertically upward with a speed of 20 m/s. How long will it take to reach the maximum height?

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Solution

Using v = u - gt, where v = 0, u = 20 m/s, and g = 9.8 m/s², we find t = u/g = 20/9.8 ≈ 2.04 s.

Correct Answer: B — 2 s

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

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

Solution

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

Correct Answer: B — 30 m

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Q. A ball is thrown vertically upward with an initial velocity of 20 m/s. How high will it rise before coming to a momentary stop? (g = 10 m/s²)

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

Solution

Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s², we get 0 = (20)² + 2(-10)s, solving gives s = 20 m.

Correct Answer: B — 30 m

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Q. A ball is thrown vertically upwards with a speed of 20 m/s. How high will it rise before coming to rest momentarily? (g = 10 m/s²)

A. 20 m
B. 40 m
C. 10 m
D. 30 m

Solution

Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s². 0 = (20)² + 2(-10)h, solving gives h = 20 m.

Correct Answer: B — 40 m

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

A. 20 m
B. 40 m
C. 30 m
D. 10 m

Solution

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

Correct Answer: B — 40 m

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Q. A ball is tied to a string and swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to remain in circular motion?

A. Tension must be zero
B. Tension must be maximum
C. Weight must be zero
D. Centripetal force must be zero

Solution

At the highest point, the tension can be zero if the centripetal force is provided entirely by the weight.

Correct Answer: A — Tension must be zero

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Q. A ball is tied to a string and swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to just maintain circular motion?

A. Tension = 0
B. Tension = mg
C. Tension > mg
D. Tension < mg

Solution

At the highest point, the centripetal force is provided by the weight, so T + mg = mv²/r, T = 0.

Correct Answer: A — Tension = 0

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Q. A ball is tied to a string and swung in a vertical circle. At the highest point, the tension in the string is 2 N and the weight of the ball is 3 N. What is the speed of the ball at the highest point if the radius of the circle is 1 m?

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

Solution

At the highest point, T + mg = mv²/r. 2 N + 3 N = mv²/1. v² = 5, v = √5 ≈ 2.24 m/s.

Correct Answer: D — 4 m/s

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Q. A block is at rest on a horizontal surface. If the applied force is gradually increased and reaches the maximum static frictional force, what will happen next?

A. The block will remain at rest
B. The block will start moving
C. The block will accelerate
D. The block will slide back

Solution

Once the applied force exceeds the maximum static frictional force, the block will start moving.

Correct Answer: B — The block will start moving

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Q. A block is sliding down a frictionless incline of angle 30 degrees. If the incline has a coefficient of static friction of 0.5, what is the maximum angle at which the block can remain at rest?

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 90 degrees

Solution

The maximum angle for static friction is given by tan(θ) = μs. Here, θ = tan⁻¹(0.5) which is approximately 26.57 degrees, so the block can remain at rest at angles less than this.

Correct Answer: B — 45 degrees

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Laws of Motion

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