Work, Energy & Power: Essential Concepts for Exam Prep

Work, Energy & Power

Conservation of Energy Power Work Energy Theorem

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

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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

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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

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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

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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

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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

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Q. A 5 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?

A. 22.5 J
B. 30 J
C. 45 J
D. 60 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 5 kg × (3 m/s)² = 22.5 J.

Correct Answer: A — 22.5 J

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Q. A 50 kg object is moving with a velocity of 10 m/s. What is its momentum?

A. 500 kg m/s
B. 1000 kg m/s
C. 1500 kg m/s
D. 2000 kg m/s

Solution

Momentum = mass × velocity = 50 kg × 10 m/s = 500 kg m/s.

Correct Answer: B — 1000 kg m/s

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Q. A 50 kg object is pushed with a force of 200 N over a distance of 10 m. What is the work done? (2000)

A. 2000 J
B. 1000 J
C. 500 J
D. 3000 J

Solution

Work Done = Force * Distance = 200 N * 10 m = 2000 J

Correct Answer: A — 2000 J

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Q. A ball is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A. 11.5 m
B. 15.0 m
C. 22.5 m
D. 30.0 m

Solution

Using the formula h = v² / (2g), h = (15 m/s)² / (2 × 9.8 m/s²) = 11.5 m.

Correct Answer: A — 11.5 m

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Q. A ball is thrown vertically upwards with a speed of 30 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A. 45.9 m
B. 46.0 m
C. 46.1 m
D. 46.2 m

Solution

Using conservation of energy, initial kinetic energy = potential energy at maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (30)²/(2 * 9.8) = 45.9 m.

Correct Answer: B — 46.0 m

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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 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 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 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 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 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 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 body of mass 2 kg is moving with a velocity of 3 m/s. What is its kinetic energy?

A. 9 J
B. 6 J
C. 18 J
D. 12 J

Solution

Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 2 * (3^2) = 9 J

Correct Answer: A — 9 J

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Q. A car accelerates from rest to a speed of 20 m/s. If its mass is 1000 kg, what is the work done by the engine?

A. 200,000 J
B. 100,000 J
C. 50,000 J
D. 400,000 J

Solution

Work Done = Change in Kinetic Energy = 1/2 * m * (v^2) = 1/2 * 1000 * (20^2) = 200,000 J

Correct Answer: A — 200,000 J

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Q. A car accelerates from rest to a speed of 20 m/s. If the mass of the car is 1000 kg, what is the work done by the engine?

A. 200,000 J
B. 100,000 J
C. 50,000 J
D. 400,000 J

Solution

Work Done = Change in Kinetic Energy = 1/2 * m * (v^2) = 1/2 * 1000 * (20^2) = 200,000 J

Correct Answer: A — 200,000 J

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Q. A car accelerates from rest to a speed of 30 m/s. If the mass of the car is 800 kg, what is the work done on the car?

A. 360,000 J
B. 480,000 J
C. 600,000 J
D. 720,000 J

Solution

Work done = Change in Kinetic Energy = 0.5 × mass × (final speed² - initial speed²) = 0.5 × 800 kg × (30 m/s)² = 360,000 J.

Correct Answer: B — 480,000 J

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Q. A car engine does 3000 J of work in 5 seconds. What is the average power output of the engine?

A. 600 W
B. 800 W
C. 1000 W
D. 1200 W

Solution

Using the formula P = W/t, we find P = 3000 J / 5 s = 600 W.

Correct Answer: C — 1000 W

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Q. A car engine does 3000 J of work in 5 seconds. What is the power of the engine?

A. 600 W
B. 800 W
C. 1000 W
D. 1200 W

Solution

Using the formula P = W/t, we have P = 3000 J / 5 s = 600 W.

Correct Answer: A — 600 W

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Q. A car of mass 1000 kg accelerates from rest to a speed of 20 m/s. What is the work done by the engine? (2000)

A. 20000 J
B. 40000 J
C. 50000 J
D. 80000 J

Solution

Kinetic Energy = 0.5 × mass × velocity^2 = 0.5 × 1000 kg × (20 m/s)^2 = 200000 J.

Correct Answer: B — 40000 J

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Q. A car of mass 1000 kg accelerates from rest to a speed of 20 m/s. What is the work done on the car?

A. 20000 J
B. 40000 J
C. 80000 J
D. 100000 J

Solution

Kinetic energy = 0.5 × m × v^2 = 0.5 × 1000 kg × (20 m/s)^2 = 200000 J.

Correct Answer: B — 40000 J

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Q. A car of mass 1000 kg is moving at a speed of 20 m/s. What is its kinetic energy?

A. 200 J
B. 400 J
C. 200,000 J
D. 400,000 J

Solution

Kinetic energy is given by KE = 0.5mv². Here, KE = 0.5 * 1000 * (20)² = 200,000 J.

Correct Answer: C — 200,000 J

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Work, Energy & Power

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