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Q. A 20 kg cart is pulled with a force of 80 N. What is the acceleration of the cart? (2023)

A. 2 m/s²
B. 4 m/s²
C. 6 m/s²
D. 8 m/s²

Solution

Using F = ma, we find a = F/m = 80 N / 20 kg = 4 m/s².

Correct Answer: B — 4 m/s²

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Q. A 20 kg object is acted upon by a net force of 80 N. What is the acceleration of the object? (2023)

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

Solution

Acceleration = Net force / Mass = 80 N / 20 kg = 4 m/s².

Correct Answer: C — 4 m/s²

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Q. A 20 kg object is dropped from a height. What is the force acting on it just before it hits the ground?

A. 0 N
B. 20 N
C. 200 N
D. 10 N

Solution

The force acting on the object is its weight: F = mg = 20 kg * 10 m/s² = 200 N.

Correct Answer: C — 200 N

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Q. A 20 kg object is moving with a velocity of 5 m/s. What is its kinetic energy? (2021)

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

Solution

Kinetic energy KE = 0.5 * m * v² = 0.5 * 20 kg * (5 m/s)² = 0.5 * 20 * 25 = 250 J.

Correct Answer: C — 200 J

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Q. A 20 kg object is pulled with a force of 100 N. If the frictional force is 40 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

Net force = applied force - friction = 100 N - 40 N = 60 N. Acceleration a = F/m = 60 N / 20 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 20 kg object is thrown vertically upward with a force of 100 N. What is the net force acting on the object? (2023)

A. 0 N
B. 20 N
C. 80 N
D. 100 N

Solution

Net force = Applied force - Weight = 100 N - (20 kg * 10 m/s²) = 100 N - 200 N = -100 N (downward).

Correct Answer: C — 80 N

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Q. A 200 g piece of metal at 100°C is placed in 300 g of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.18 J/g°C, specific heat of metal = 0.9 J/g°C) (2022)

A. 22.5°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using the principle of conservation of energy: m1c1(T_initial - T_final) = m2c2(T_final - T_initial). Solving gives T_final = 35°C.

Correct Answer: C — 35°C

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Q. A 2000 W heater operates for 3 hours. How much energy does it consume in kilowatt-hours?

A. 6 kWh
B. 5 kWh
C. 4 kWh
D. 3 kWh

Solution

Energy consumed in kilowatt-hours is calculated as E = P * t. Here, P = 2000 W = 2 kW and t = 3 hours. Thus, E = 2 kW * 3 h = 6 kWh.

Correct Answer: A — 6 kWh

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Q. A 25 kg object is acted upon by a force of 100 N. What is its acceleration? (2023)

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

Solution

Using F = ma, we find a = F/m = 100 N / 25 kg = 4 m/s².

Correct Answer: D — 5 m/s²

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Q. A 25 kg object is pulled with a force of 100 N. If the frictional force is 25 N, what is the acceleration of the object? (2023)

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

Solution

Net force = 100 N - 25 N = 75 N. Acceleration = Net force / Mass = 75 N / 25 kg = 3 m/s².

Correct Answer: C — 4 m/s²

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Q. A 25 kg object is pulled with a force of 100 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 = 100 N / 25 kg = 4 m/s².

Correct Answer: D — 5 m/s²

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Q. A 25 kg object is pushed with a force of 100 N. If the frictional force is 25 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

Net force = 100 N - 25 N = 75 N. Acceleration a = F/m = 75 N / 25 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 25 kg object is subjected to a force of 50 N. What is its acceleration?

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

Solution

Using F = ma, we find a = F/m = 50 N / 25 kg = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A 3 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A. 5.1 m
B. 10.2 m
C. 15.3 m
D. 20.0 m

Solution

Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (10)²/(2*9.8) = 5.1 m.

Correct Answer: A — 5.1 m

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Q. A 3 kg block is at rest on a horizontal surface. If a horizontal force of 12 N is applied, what is the block's acceleration assuming no friction?

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

Solution

Acceleration a = F/m = 12 N / 3 kg = 4 m/s².

Correct Answer: A — 2 m/s²

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Q. A 3 kg block is on a table and a horizontal force of 15 N is applied. If the frictional force is 5 N, what is the acceleration of the block?

A. 2 m/s²
B. 3 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Net force = applied force - friction = 15 N - 5 N = 10 N. Acceleration a = F/m = 10 N / 3 kg = 3.33 m/s².

Correct Answer: A — 2 m/s²

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Q. A 3 kg block is pulled with a force of 12 N. If the frictional force is 3 N, what is the acceleration of the block?

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

Solution

Net force = applied force - friction = 12 N - 3 N = 9 N. Acceleration a = F/m = 9 N / 3 kg = 3 m/s².

Correct Answer: B — 4 m/s²

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Q. A 3 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block?

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

Solution

The acceleration down the incline is given by a = g * sin(θ) = 9.8 m/s² * sin(30°) = 4.9 m/s².

Correct Answer: A — 3.9 m/s²

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Q. A 3 kg block is sliding on a frictionless surface with a velocity of 4 m/s. What is the momentum of the block?

A. 12 kg·m/s
B. 8 kg·m/s
C. 6 kg·m/s
D. 4 kg·m/s

Solution

Momentum p = mv = 3 kg * 4 m/s = 12 kg·m/s.

Correct Answer: A — 12 kg·m/s

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Q. A 3 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 stretched?

A. 100 N
B. 200 N
C. 300 N
D. 600 N

Solution

The maximum force exerted by the spring is F = kx. Assuming maximum stretch x = 1 m, F = 200 N/m * 1 m = 200 N.

Correct Answer: B — 200 N

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Q. A 3 kg object is at rest on a frictionless surface. A force of 9 N is applied. What is the final velocity after 3 seconds?

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

Solution

First, find acceleration: a = F/m = 9 N / 3 kg = 3 m/s². Then, v = u + at = 0 + 3 m/s² * 3 s = 9 m/s.

Correct Answer: C — 3 m/s

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Q. A 3 kg object is at rest on a frictionless surface. If a force of 9 N is applied, what will be its velocity after 3 seconds?

A. 3 m/s
B. 6 m/s
C. 9 m/s
D. 12 m/s

Solution

First, find acceleration: a = F/m = 9 N / 3 kg = 3 m/s². Then, v = u + at = 0 + 3 m/s² * 3 s = 9 m/s.

Correct Answer: B — 6 m/s

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Q. A 3 kg object is at rest on a horizontal surface. If a force of 12 N is applied, what is the object's acceleration? (Assume no friction)

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 3 kg object is at rest on a horizontal surface. If a horizontal force of 12 N is applied, 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 = 12 N / 3 kg = 4 m/s².

Correct Answer: B — 3 m/s²

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Q. A 3 kg object is at rest on a table. If a force of 15 N is applied horizontally, what is the object's acceleration?

A. 5 m/s²
B. 10 m/s²
C. 15 m/s²
D. 20 m/s²

Solution

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

Correct Answer: A — 5 m/s²

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Q. A 3 kg object is at rest on a table. If a horizontal force of 12 N is applied, what is the acceleration of the object?

A. 4 m/s²
B. 0 m/s²
C. 3 m/s²
D. 12 m/s²

Solution

Using F = ma, acceleration a = F/m = 12 N / 3 kg = 4 m/s².

Correct Answer: A — 4 m/s²

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Q. A 3 kg object is dropped from a height of 10 m. What is its potential energy at the top? (g = 9.8 m/s²)

A. 294 J
B. 30 J
C. 39.2 J
D. 19.6 J

Solution

Potential Energy (PE) = m × g × h = 3 kg × 9.8 m/s² × 10 m = 294 J.

Correct Answer: A — 294 J

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Q. A 3 kg object is dropped from a height of 10 m. What is the potential energy at the top? (2022)

A. 30 J
B. 60 J
C. 90 J
D. 120 J

Solution

Potential Energy = mass × g × height = 3 kg × 10 m/s² × 10 m = 300 J

Correct Answer: C — 90 J

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Q. A 3 kg object is dropped from a height of 12 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)

A. 15 m/s
B. 10 m/s
C. 12 m/s
D. 20 m/s

Solution

Using conservation of energy, PE = KE: mgh = 1/2 mv²; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 12 m) = 15.49 m/s

Correct Answer: A — 15 m/s

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Q. A 3 kg object is dropped from a height of 12 m. What is the potential energy at the top?

A. 30 J
B. 36 J
C. 60 J
D. 120 J

Solution

Potential energy = mgh = 3 kg × 9.8 m/s² × 12 m = 352.8 J.

Correct Answer: D — 120 J

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Major Competitive Exams

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