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Q. A block of mass 10 kg is resting on a horizontal surface. If the coefficient of static friction is 0.5, what is the maximum static frictional force acting on the block?

A. 25 N
B. 50 N
C. 75 N
D. 100 N

Solution

Maximum static frictional force (Fs) = μs * N = μs * mg = 0.5 * 10 kg * 9.8 m/s² = 49 N, approximately 50 N.

Correct Answer: B — 50 N

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Q. A block of mass 2 kg is lifted to a height of 5 m. What is the work done against gravity? (g = 9.8 m/s²)

A. 98 J
B. 196 J
C. 49 J
D. 20 J

Solution

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

Correct Answer: B — 196 J

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Q. A block of mass 2 kg is placed on a frictionless surface and is pushed with a force of 10 N. What is its acceleration?

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

Solution

Using Newton's second law, F = ma. Here, F = 10 N and m = 2 kg. Therefore, a = F/m = 10/2 = 5 m/s².

Correct Answer: B — 5 m/s²

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Q. A block of mass 2 kg is placed on a frictionless surface and is pushed with a force of 10 N. What is the acceleration of the block?

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

Solution

Using Newton's second law, F = ma, we have a = F/m = 10 N / 2 kg = 5 m/s².

Correct Answer: B — 5 m/s²

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Q. A block of mass 2 kg is pushed along a frictionless surface by a constant force of 10 N. What is the work done by the force when the block moves 5 m?

A. 10 J
B. 20 J
C. 30 J
D. 50 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J.

Correct Answer: B — 20 J

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Q. A block of mass 2 kg is pushed along a frictionless surface by a force of 10 N. What is the work done by the force when the block moves 5 m?

A. 10 J
B. 20 J
C. 30 J
D. 50 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J.

Correct Answer: B — 20 J

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Q. A block of mass 2 kg is pushed along a horizontal surface with a constant force of 10 N. What is the work done by the force after moving the block 5 m?

A. 10 J
B. 20 J
C. 30 J
D. 50 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J.

Correct Answer: B — 20 J

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Q. A block of mass 2 kg is pushed along a horizontal surface with a force of 10 N. If the block moves a distance of 5 m, what is the work done by the force?

A. 10 J
B. 20 J
C. 30 J
D. 50 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J.

Correct Answer: B — 20 J

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Q. A block of mass 2 kg is pushed along a horizontal surface with a force of 10 N. If the block moves a distance of 5 m, what is the work done on the block?

A. 10 J
B. 20 J
C. 30 J
D. 50 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J.

Correct Answer: B — 20 J

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Q. A block of mass 2 kg is pushed along a horizontal surface with a force of 10 N. If the block moves 5 m, what is the work done by the force?

A. 10 J
B. 20 J
C. 30 J
D. 50 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J.

Correct Answer: B — 20 J

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Q. A block of mass 2 kg is pushed along a horizontal surface with a force of 10 N. If the block moves 5 m, what is the work done on the block?

A. 10 J
B. 20 J
C. 30 J
D. 50 J

Solution

Work done = Force × Distance = 10 N × 5 m = 50 J.

Correct Answer: B — 20 J

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Q. A block of mass 2 kg is released 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. 5 m/s

Solution

Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 m/s.

Correct Answer: A — 14 m/s

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Q. A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground?

A. 0 m/s
B. 10 m/s
C. 14 m/s
D. 20 m/s

Solution

Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s.

Correct Answer: C — 14 m/s

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Q. A block of mass 2 kg is released from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.

Correct Answer: B — 10 m/s

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Q. A block of mass 2 kg is sliding down a frictionless incline of angle 30 degrees. What is the acceleration of the block?

A. 4.9 m/s²
B. 9.8 m/s²
C. 3.9 m/s²
D. 1.5 m/s²

Solution

The acceleration a = g * sin(θ) = 9.8 * sin(30°) = 9.8 * 0.5 = 4.9 m/s².

Correct Answer: A — 4.9 m/s²

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Q. A block of mass 2 kg is sliding down a frictionless incline of height 10 m. What is its speed at the bottom?

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 25 m/s

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 m/s.

Correct Answer: B — 14 m/s

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Q. A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom of the incline?

A. 10 m/s
B. 5 m/s
C. 20 m/s
D. 15 m/s

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.81 * 5) = 10 m/s.

Correct Answer: A — 10 m/s

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Q. A block of mass 2 kg is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy at the bottom (1/2 mv²). Thus, v = sqrt(2gh) = sqrt(2 * 9.8 * 5) ≈ 10 m/s.

Correct Answer: B — 10 m/s

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Q. A block of mass 5 kg is placed on a frictionless surface and is pushed with a force of 20 N. What is the acceleration of the block?

A. 2 m/s²
B. 4 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².

Correct Answer: B — 4 m/s²

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Q. A block of mass 5 kg is placed 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 block of mass 5 kg is placed on a frictionless surface. A force of 20 N is applied to it. What is the acceleration of the block?

A. 2 m/s²
B. 4 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².

Correct Answer: B — 4 m/s²

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Q. A block of mass 5 kg is placed 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. 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 block of mass 5 kg is pulled along a horizontal surface with a force of 20 N. If the frictional force acting on the block is 5 N, what is the acceleration of the block? (2021)

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

Solution

Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. 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 block of mass 5 kg is pulled on a horizontal surface with a force of 20 N. If the frictional force is 5 N, what is the acceleration of the block? (2020)

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

Solution

Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. 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 block of mass 5 kg is pulled on a horizontal surface with a force of 20 N. If the frictional force acting on the block is 5 N, what is the acceleration of the block? (2021)

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

Solution

Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Acceleration = Net force / Mass = 15 N / 5 kg = 3 m/s².

Correct Answer: B — 2 m/s²

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Q. A block of mass 5 kg is resting on a frictionless surface. A force of 10 N is applied to it. What will be its acceleration? (2020)

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

Solution

Using Newton's second law, F = ma, acceleration (a) = F/m = 10 N / 5 kg = 2 m/s².

Correct Answer: B — 5 m/s²

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Q. A block of mass 5 kg is resting on a frictionless surface. If a force of 15 N is applied, what will be its velocity after 3 seconds? (2022)

A. 3 m/s
B. 5 m/s
C. 9 m/s
D. 15 m/s

Solution

Acceleration a = F/m = 15 N / 5 kg = 3 m/s². Velocity after 3 seconds = a × t = 3 m/s² × 3 s = 9 m/s.

Correct Answer: C — 9 m/s

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Q. A block of mass 5 kg is resting on a horizontal surface. If a horizontal force of 20 N is applied, what is the acceleration of the block? (Assume no friction)

A. 2 m/s²
B. 4 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².

Correct Answer: A — 2 m/s²

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Q. A block of mass 5 kg is resting on a horizontal surface. If a horizontal force of 15 N is applied, what is the acceleration of the block? (Assume no friction) (2021)

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

Solution

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

Correct Answer: C — 3 m/s²

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Q. A block of mass 5 kg is resting on a horizontal surface. If the coefficient of kinetic friction between the block and the surface is 0.3, what is the frictional force acting on the block when it is sliding?

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

Solution

Frictional force (Ff) = μk * N = μk * mg = 0.3 * (5 kg * 10 m/s²) = 15 N.

Correct Answer: B — 10 N

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Major Competitive Exams

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