Major Competitive Exams

Civil Services Defence Exams Engineering & Architecture Admissions Government Jobs Management Admissions Undergraduate

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

Learn More →

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

Learn More →

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

Learn More →

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

Learn More →

Q. A 5 kg block of ice at 0°C is converted to water at 0°C. If the latent heat of fusion of ice is 334 kJ/kg, how much heat is absorbed?

A. 1670 kJ
B. 334 kJ
C. 167 kJ
D. 3340 kJ

Solution

Heat absorbed (Q) = m * Lf = 5 kg * 334 kJ/kg = 1670 kJ.

Correct Answer: B — 334 kJ

Learn More →

Q. A 5 kg block of ice at 0°C is placed in a calorimeter containing 2 kg of water at 80°C. What is the final temperature of the system when thermal equilibrium is reached? (Latent heat of fusion of ice = 334 kJ/kg)

A. 0°C
B. 20°C
C. 40°C
D. 60°C

Solution

The heat lost by water = heat gained by ice. Solving gives the final temperature as 0°C.

Correct Answer: B — 20°C

Learn More →

Q. A 5 kg block of metal is heated from 20°C to 100°C. If the specific heat capacity of the metal is 0.5 J/g°C, how much heat is required? (2023)

A. 2000 J
B. 2500 J
C. 3000 J
D. 3500 J

Solution

Heat (Q) = m * c * ΔT = 5000 g * 0.5 J/g°C * (100°C - 20°C) = 5000 * 0.5 * 80 = 200000 J = 2000 J.

Correct Answer: C — 3000 J

Learn More →

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

Learn More →

Q. A 5 kg object is at rest on a surface. If a horizontal force of 15 N is applied, what is the object's acceleration? (Assume friction is negligible)

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

Learn More →

Q. A 5 kg object is dropped from a height of 10 m. What is its potential energy at the top? (g = 9.8 m/s²)

A. 490 J
B. 98 J
C. 50 J
D. 100 J

Solution

Potential Energy = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.

Correct Answer: A — 490 J

Learn More →

Q. A 5 kg object is dropped from a height of 10 m. What is its potential energy at the top? (Take g = 9.8 m/s²) (2020)

A. 49 J
B. 98 J
C. 39 J
D. 59 J

Solution

Potential Energy (PE) = mgh = 5 kg × 9.8 m/s² × 10 m = 490 J.

Correct Answer: B — 98 J

Learn More →

Q. A 5 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 14 m/s
B. 20 m/s
C. 10 m/s
D. 15 m/s

Solution

Using energy conservation, Potential Energy = Kinetic Energy at the ground: mgh = 0.5mv²; v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s

Correct Answer: A — 14 m/s

Learn More →

Q. A 5 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

A. 14 m/s
B. 20 m/s
C. 10 m/s
D. 15 m/s

Solution

Using energy conservation, Potential Energy = Kinetic Energy. mgh = 1/2 mv². v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s.

Correct Answer: A — 14 m/s

Learn More →

Q. A 5 kg object is dropped from a height of 10 m. What is the total mechanical energy just before it hits the ground?

A. 0 J
B. 50 J
C. 100 J
D. 200 J

Solution

Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 10 = 490 J. Just before hitting the ground, total energy = 490 J.

Correct Answer: C — 100 J

Learn More →

Q. A 5 kg object is dropped from a height of 20 m. What is its velocity just before it hits the ground? (Take g = 10 m/s²)

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

Solution

Using the formula v² = u² + 2gh, where u = 0, we have v = √(0 + 2*10*20) = √400 = 20 m/s.

Correct Answer: B — 20 m/s

Learn More →

Q. A 5 kg object is dropped from a height of 20 m. What is the potential energy at the top? (g = 9.8 m/s²)

A. 980 J
B. 490 J
C. 196 J
D. 9800 J

Solution

Potential energy is given by PE = mgh. Here, PE = 5 * 9.8 * 20 = 980 J.

Correct Answer: A — 980 J

Learn More →

Q. A 5 kg object is dropped from a height of 20 m. What is the total mechanical energy just before it hits the ground?

A. 0 J
B. 100 J
C. 200 J
D. 500 J

Solution

Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 20 = 980 J. Just before hitting the ground, total energy = 980 J.

Correct Answer: C — 200 J

Learn More →

Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground?

A. 5 N
B. 10 N
C. 15 N
D. 20 N

Solution

The only force acting on it is its weight, W = mg = 5 kg * 10 m/s² = 50 N.

Correct Answer: B — 10 N

Learn More →

Q. A 5 kg object is dropped from a height. What is the force acting on it just before it hits the ground? (Assume g = 9.8 m/s²)

A. 5 N
B. 9.8 N
C. 49 N
D. 0 N

Solution

The force acting on the object is its weight, F = mg = 5 kg * 9.8 m/s² = 49 N.

Correct Answer: C — 49 N

Learn More →

Q. A 5 kg object is lifted to a height of 10 m. What is the gravitational potential energy gained?

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

Solution

Potential Energy (PE) = m * g * h = 5 kg * 10 m/s^2 * 10 m = 500 J

Correct Answer: B — 100 J

Learn More →

Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained by the object?

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy (PE) = mgh = 5 kg * 9.8 m/s² * 10 m = 490 J.

Correct Answer: B — 100 J

Learn More →

Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained?

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

Solution

Potential Energy (PE) = m * g * h = 5 kg * 9.8 m/s^2 * 10 m = 490 J (approx. 500 J)

Correct Answer: B — 100 J

Learn More →

Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained by the object? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy PE = mgh = 5 kg * 10 m/s² * 10 m = 500 J.

Correct Answer: B — 100 J

Learn More →

Q. A 5 kg object is lifted to a height of 10 m. What is the work done against gravity?

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Work done against gravity is W = mgh = 5 * 9.8 * 10 = 490 J.

Correct Answer: D — 200 J

Learn More →

Q. A 5 kg object is lifted to a height of 2 m. How much work is done against gravity? (g = 9.8 m/s²) (2022)

A. 98 J
B. 196 J
C. 39 J
D. 49 J

Solution

Work done = mgh = 5 kg * 9.8 m/s² * 2 m = 98 J.

Correct Answer: A — 98 J

Learn More →

Q. A 5 kg object is lifted to a height of 2 m. What is the work done against gravity?

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

Solution

Work done (W) against gravity is given by W = mgh. Here, m = 5 kg, g = 9.8 m/s², h = 2 m. W = 5 * 9.8 * 2 = 98 J.

Correct Answer: B — 20 J

Learn More →

Q. A 5 kg object is lifted to a height of 3 m. What is the work done against gravity?

A. 15 J
B. 30 J
C. 45 J
D. 60 J

Solution

Work done against gravity is W = mgh = 5 * 9.8 * 3 = 147 J.

Correct Answer: B — 30 J

Learn More →

Q. A 5 kg object is moving with a constant velocity. What is the net force acting on it? (2022)

A. 0 N
B. 5 N
C. 10 N
D. 15 N

Solution

If the object is moving with constant velocity, the net force acting on it is 0 N according to Newton's first law.

Correct Answer: A — 0 N

Learn More →

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

A. 25 J
B. 50 J
C. 100 J
D. 250 J

Solution

Kinetic energy (KE) = 0.5 * mass * velocity² = 0.5 * 5 * (10)² = 250 J.

Correct Answer: C — 100 J

Learn More →

Q. A 5 kg object is moving with a velocity of 10 m/s. What is its momentum? (2021)

A. 50 kg m/s
B. 100 kg m/s
C. 150 kg m/s
D. 200 kg m/s

Solution

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

Correct Answer: B — 100 kg m/s

Learn More →

Showing 301 to 330 of 27365 (913 Pages)

Major Competitive Exams

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?