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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 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 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 5 m. What is the potential energy at the top? (g = 9.8 m/s²)

A. 14.7 J
B. 29.4 J
C. 39.2 J
D. 49.05 J

Solution

PE = mgh = 3 kg × 9.8 m/s² × 5 m = 147 J.

Correct Answer: C — 39.2 J

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Q. A 3 kg object is lifted to a height of 4 m. What is the gravitational potential energy gained? (2023)

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

Solution

Potential Energy (PE) = m * g * h = 3 kg * 9.8 m/s^2 * 4 m = 117.6 J (approximately 120 J)

Correct Answer: A — 120 J

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Q. A 3 kg object is lifted to a height of 4 m. What is the work done against gravity? (g = 9.8 m/s²)

A. 117.6 J
B. 29.4 J
C. 39.2 J
D. 78.4 J

Solution

Work done (W) = mgh = 3 kg × 9.8 m/s² × 4 m = 117.6 J.

Correct Answer: A — 117.6 J

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Q. A 3 kg object is moving with a velocity of 2 m/s. What is its kinetic energy? (2021)

A. 6 J
B. 12 J
C. 3 J
D. 9 J

Solution

Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 3 kg * (2 m/s)^2 = 6 J

Correct Answer: B — 12 J

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Q. A 3 kg object is moving with a velocity of 6 m/s. What is its momentum? (2021)

A. 18 kg m/s
B. 12 kg m/s
C. 9 kg m/s
D. 6 kg m/s

Solution

Momentum (p) = mv = 3 kg * 6 m/s = 18 kg m/s

Correct Answer: A — 18 kg m/s

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Q. A 3 kg object is pushed with a force of 12 N over a distance of 4 m. What is the work done on the object?

A. 24 J
B. 36 J
C. 48 J
D. 60 J

Solution

Work done (W) = Force (F) × Distance (d) = 12 N × 4 m = 48 J.

Correct Answer: C — 48 J

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Q. A 3 kg object is subjected to a net force of 12 N. What is the acceleration of the object? (2022)

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 thrown upwards with a velocity of 15 m/s. What is its maximum height? (2020)

A. 11.5 m
B. 17.5 m
C. 22.5 m
D. 10.5 m

Solution

Using the formula v^2 = u^2 - 2gh, we find h = u^2 / (2g) = (15 m/s)^2 / (2 * 9.8 m/s²) = 11.5 m.

Correct Answer: B — 17.5 m

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Q. A 4 kg block is pulled with a force of 20 N. What is the acceleration of the block? (2020)

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

Solution

Using F = ma, acceleration (a) = F/m = 20 N / 4 kg = 5 m/s².

Correct Answer: A — 5 m/s²

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Q. A 4 kg object is lifted to a height of 2 m. What is the work done against gravity? (2022)

A. 40 J
B. 80 J
C. 20 J
D. 60 J

Solution

Work done = mass × g × height = 4 kg × 10 m/s² × 2 m = 80 J

Correct Answer: B — 80 J

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Q. A 4 kg object is lifted to a height of 5 m. What is the work done against gravity? (g = 9.8 m/s²)

A. 19.6 J
B. 39.2 J
C. 49 J
D. 196 J

Solution

Work done against gravity = mgh = 4 kg × 9.8 m/s² × 5 m = 196 J.

Correct Answer: B — 39.2 J

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Q. A 4 kg object is moving with a velocity of 5 m/s. What is its kinetic energy? (2021)

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

Solution

Kinetic energy KE = 1/2 mv² = 1/2 * 4 kg * (5 m/s)² = 50 J.

Correct Answer: C — 50 J

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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

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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

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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

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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

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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

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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

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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

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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

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Q. A 5 kg object is moving with a velocity of 4 m/s. What is its momentum? (2023)

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

Solution

Momentum = mass × velocity = 5 kg × 4 m/s = 20 kg m/s.

Correct Answer: C — 20 kg m/s

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Q. A 50 kg crate is pushed with a force of 200 N. If the frictional force is 50 N, what is the acceleration? (2022)

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

Solution

Net force = applied force - friction = 200 N - 50 N = 150 N. Acceleration a = F/m = 150 N / 50 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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Q. A 50 kg crate is pushed with a force of 200 N. If the frictional force is 50 N, what is the acceleration of the crate? (2022)

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

Solution

Net force = 200 N - 50 N = 150 N. Acceleration a = F/m = 150 N / 50 kg = 3 m/s².

Correct Answer: B — 3 m/s²

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