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Q. A 3 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, a = F/m = 12 N / 3 kg = 4 m/s².

Correct Answer: C — 4 m/s²

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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 lifted to a height of 3 m. What is the change in gravitational potential energy?

A. 12 J
B. 24 J
C. 36 J
D. 48 J

Solution

Change in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.

Correct Answer: B — 24 J

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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy? (g = 9.8 m/s²)

A. 117.6 J
B. 117 J
C. 120 J
D. 150 J

Solution

Potential energy increase = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.

Correct Answer: A — 117.6 J

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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy?

A. 12 J
B. 24 J
C. 36 J
D. 48 J

Solution

Increase in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.

Correct Answer: B — 24 J

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Q. A 4 kg object is moving in a circular path of radius 2 m with a speed of 6 m/s. What is the centripetal force acting on it?

A. 9 N
B. 12 N
C. 18 N
D. 24 N

Solution

Centripetal force F = mv²/r = 4 kg * (6 m/s)² / 2 m = 72 N / 2 = 36 N.

Correct Answer: B — 12 N

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Q. A 4 kg object is moving in a circular path of radius 2 m with a speed of 6 m/s. What is the centripetal force acting on the object?

A. 9 N
B. 12 N
C. 18 N
D. 24 N

Solution

Centripetal force F = mv²/r = 4 kg * (6 m/s)² / 2 m = 4 * 36 / 2 = 72 N / 2 = 36 N.

Correct Answer: B — 12 N

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Q. A 4 kg object is moving in a straight line with a velocity of 5 m/s. What is the kinetic energy of the object?

A. 50 J
B. 40 J
C. 20 J
D. 10 J

Solution

Kinetic energy KE = 0.5 * m * v² = 0.5 * 4 kg * (5 m/s)² = 50 J.

Correct Answer: A — 50 J

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Q. A 4 kg object is moving with a speed of 5 m/s. If it comes to rest, what is the work done by friction?

A. 50 J
B. 75 J
C. 100 J
D. 125 J

Solution

Work done = change in kinetic energy = 0 - 0.5 × 4 kg × (5 m/s)² = -50 J.

Correct Answer: C — 100 J

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Q. A 4 kg object is moving with a speed of 5 m/s. What is its kinetic energy?

A. 10 J
B. 20 J
C. 50 J
D. 100 J

Solution

Kinetic energy = 0.5 × m × v² = 0.5 × 4 kg × (5 m/s)² = 50 J.

Correct Answer: C — 50 J

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Q. A 4 kg object is moving with a speed of 5 m/s. What is the total mechanical energy if it is at a height of 2 m?

A. 50 J
B. 60 J
C. 70 J
D. 80 J

Solution

Total mechanical energy = Kinetic energy + Potential energy = 0.5 × 4 kg × (5 m/s)² + 4 kg × 9.8 m/s² × 2 m = 50 J + 78.4 J = 128.4 J.

Correct Answer: C — 70 J

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Q. A 4 kg object is moving with a velocity of 2 m/s. What is its kinetic energy?

A. 8 J
B. 4 J
C. 16 J
D. 2 J

Solution

Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 4 * (2^2) = 8 J

Correct Answer: A — 8 J

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Q. A 4 kg object is moving with a velocity of 2 m/s. What is the change in momentum if the object comes to a stop?

A. 0 kg·m/s
B. 4 kg·m/s
C. 8 kg·m/s
D. 2 kg·m/s

Solution

Change in momentum = final momentum - initial momentum = 0 - (4 kg * 2 m/s) = -8 kg·m/s.

Correct Answer: C — 8 kg·m/s

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Q. A 4 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?

A. 18 J
B. 24 J
C. 36 J
D. 12 J

Solution

Kinetic energy KE = 0.5 * m * v² = 0.5 * 4 kg * (3 m/s)² = 18 J.

Correct Answer: B — 24 J

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Q. A 4 kg object is moving with a velocity of 5 m/s. What is its momentum?

A. 10 kg·m/s
B. 20 kg·m/s
C. 15 kg·m/s
D. 25 kg·m/s

Solution

Momentum p = mv = 4 kg * 5 m/s = 20 kg·m/s.

Correct Answer: B — 20 kg·m/s

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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 over a distance of 3 m. If the object starts from rest, what is its final speed?

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

Solution

Work done = Force × Distance = 20 N × 3 m = 60 J. K.E = 0.5 × m × v², 60 J = 0.5 × 4 kg × v², v = 3.87 m/s.

Correct Answer: C — 4 m/s

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Q. A 4 kg object is pushed with a force of 20 N over a distance of 3 m. If the object starts from rest, what is its final speed? (Assume no friction)

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

Solution

Work done = Force × Distance = 20 N × 3 m = 60 J. Kinetic energy = 0.5 × m × v². 60 J = 0.5 × 4 kg × v². v² = 30, v = √30 ≈ 5.48 m/s.

Correct Answer: C — 4 m/s

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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 4 kg object is subjected to a net force of 12 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 = 12 N / 4 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 4 kg object is subjected to a net force of 16 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 = 16 N / 4 kg = 4 m/s².

Correct Answer: C — 4 m/s²

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Q. A 5 kg block is placed on a surface with a coefficient of static friction of 0.6. What is the maximum static friction force before the block starts to move?

A. 30 N
B. 20 N
C. 15 N
D. 25 N

Solution

Maximum static friction force = μs * N = 0.6 * (5 kg * 9.8 m/s²) = 29.4 N, approximately 30 N.

Correct Answer: A — 30 N

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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 5 kg object is at rest on a surface with a coefficient of static friction of 0.4. What is the maximum static friction force?

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

Solution

Maximum static friction force = μs * N = 0.4 * (5 kg * 9.8 m/s²) = 0.4 * 49 N = 19.6 N, approximately 20 N.

Correct Answer: B — 20 N

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Engineering & Architecture Admissions

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