Work, Energy & Power: Essential Concepts for Exam Prep

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Work, Energy & Power

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Work, Energy & Power MCQ & Objective Questions

The concepts of Work, Energy, and Power are fundamental in physics and play a crucial role in various examinations. Understanding these topics not only helps in grasping the principles of mechanics but also enhances your problem-solving skills. Practicing MCQs and objective questions related to Work, Energy, and Power can significantly improve your exam preparation and boost your scores. These practice questions are designed to cover important concepts and help you identify key areas for revision.

What You Will Practise Here

Definitions and units of Work, Energy, and Power
Work done by a constant force and variable force
Kinetic and potential energy concepts
Law of conservation of energy
Power calculations and its significance
Work-energy theorem and its applications
Diagrams illustrating energy transformations

Exam Relevance

The topic of Work, Energy, and Power is frequently tested in CBSE, State Boards, NEET, and JEE examinations. Students can expect questions that assess their understanding of the basic principles, calculations involving formulas, and application of concepts in real-world scenarios. Common question patterns include numerical problems, conceptual questions, and application-based queries that require a deep understanding of the subject matter.

Common Mistakes Students Make

Confusing work done with energy transferred
Misunderstanding the relationship between kinetic and potential energy
Neglecting the direction of forces when calculating work
Overlooking units while solving numerical problems
Failing to apply the conservation of energy principle correctly

FAQs

Question: What is the formula for calculating work done?
Answer: Work done is calculated using the formula: Work = Force × Displacement × cos(θ), where θ is the angle between the force and displacement vectors.

Question: How is power defined in physics?
Answer: Power is defined as the rate at which work is done or energy is transferred, calculated as Power = Work / Time.

Now that you have a clear understanding of the importance of Work, Energy, and Power, it's time to put your knowledge to the test! Solve practice MCQs and objective questions to enhance your understanding and prepare effectively for your exams. Every question you tackle brings you one step closer to mastering these essential concepts!

Conservation of Energy Power Work Energy Theorem

Q. A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?

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

Solution

Using energy conservation, mgh = 0.5 mv², h = v²/(2g) = (10 m/s)²/(2 × 9.8 m/s²) = 5.1 m.

Correct Answer: A — 5 m

Q. A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)

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

Solution

Using energy conservation, initial kinetic energy = mgh. 0.5 × 1 kg × (10 m/s)² = 1 kg × 10 m/s² × h. h = 5 m.

Correct Answer: B — 10 m

Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²)

A. 20 m
B. 30 m
C. 40 m
D. 50 m

Solution

Using energy conservation: Initial K.E. = Potential Energy at max height. 0.5mv² = mgh. h = v²/(2g) = (20 m/s)² / (2 × 10 m/s²) = 20 m.

Correct Answer: B — 30 m

Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)

A. 20.4 m
B. 30.4 m
C. 40.8 m
D. 50.0 m

Solution

Using energy conservation, initial kinetic energy = mgh; 0.5 × 1 kg × (20 m/s)² = 9.8 m × h; h = 20.4 m.

Correct Answer: A — 20.4 m

Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?

A. 10 m
B. 20 m
C. 30 m
D. 40 m

Solution

Using energy conservation: KE_initial = PE_max; 0.5 × m × v² = mgh; h = v²/(2g) = (20 m/s)²/(2 × 9.8 m/s²) = 20.4 m.

Correct Answer: B — 20 m

Q. A 1 kg mass is attached to a spring and compressed by 0.2 m. If the spring constant is 100 N/m, what is the potential energy stored in the spring?

A. 2 J
B. 4 J
C. 6 J
D. 8 J

Solution

Potential energy in a spring is given by PE = 0.5kx². Thus, PE = 0.5 * 100 * (0.2)² = 2 J.

Correct Answer: B — 4 J

Q. A 1 kg mass is dropped from a height of 1 m. What is its speed just before it hits the ground?

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

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 1) = 4.43 m/s.

Correct Answer: B — 2 m/s

Q. A 1 kg mass is dropped from a height of 1 m. What is the speed just before it hits the ground?

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

Solution

Using conservation of energy, v = sqrt(2gh) = sqrt(2 * 9.8 * 1) = 4.43 m/s.

Correct Answer: B — 2 m/s

Q. A 1 kg mass is dropped from a height of 10 m. What is its speed just before it hits the ground?

A. 5 m/s
B. 10 m/s
C. 14 m/s
D. 20 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 10) = 14 m/s.

Correct Answer: C — 14 m/s

Q. A 1 kg mass is dropped from a height of 10 m. What is the speed just before it hits the ground?

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 = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s.

Correct Answer: B — 10 m/s

Q. A 1 kg mass is lifted to a height of 5 m. How much work is done against gravity?

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

Solution

Work done against gravity = mgh = 1 * 9.8 * 5 = 49 J.

Correct Answer: B — 10 J

Q. A 1 kg mass is lifted vertically 10 m. How much work is done against gravity? (g = 9.8 m/s²)

A. 9.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 J

Solution

Work done = mgh = 1 kg × 9.8 m/s² × 10 m = 98 J.

Correct Answer: B — 19.6 J

Q. A 1 kg object is pushed with a force of 10 N over a distance of 3 m. What is the work done?

A. 10 J
B. 20 J
C. 30 J
D. 40 J

Solution

Work done = Force × Distance = 10 N × 3 m = 30 J.

Correct Answer: C — 30 J

Q. A 10 kg object falls freely from a height of 20 m. What is its potential energy at the top?

A. 100 J
B. 200 J
C. 300 J
D. 400 J

Solution

Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.

Correct Answer: B — 200 J

Q. A 10 kg object falls freely from a height of 20 m. What is its speed just before hitting the ground? (g = 9.8 m/s²)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 28 m/s.

Correct Answer: D — 28 m/s

Q. A 10 kg object falls from a height of 20 m. What is its potential energy at the top?

A. 100 J
B. 200 J
C. 300 J
D. 400 J

Solution

Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.

Correct Answer: D — 400 J

Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground?

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 20 m/s.

Correct Answer: C — 20 m/s

Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 10 m/s
B. 14 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 19.8 m/s.

Correct Answer: B — 14 m/s

Q. A 10 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 14 m/s
B. 19.8 m/s
C. 20 m/s
D. 28 m/s

Solution

Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.

Correct Answer: B — 19.8 m/s

Q. A 10 kg object is lifted to a height of 10 m. How much work is done against gravity?

A. 0 J
B. 100 J
C. 200 J
D. 1000 J

Solution

Work done against gravity is equal to the change in potential energy, W = mgh = 10 * 9.8 * 10 = 980 J.

Correct Answer: D — 1000 J

Q. A 10 kg object is lifted to a height of 2 m. What is the work done against gravity?

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

Solution

Work done = mass × g × height = 10 kg × 9.8 m/s² × 2 m = 196 J.

Correct Answer: B — 40 J

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

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

Solution

Work done = mgh = 10 * 9.8 * 5 = 490 J.

Correct Answer: B — 100 J

Q. A 10 kg object is lifted to a height of 5 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 = mgh = 10 * 9.8 * 5 = 490 J.

Correct Answer: B — 100 J

Q. A 10 kg object is moving with a speed of 3 m/s. If it comes to a stop, how much work is done by the friction force?

A. 45 J
B. 90 J
C. 135 J
D. 180 J

Solution

Work done = Change in Kinetic Energy = 0 - (0.5 × 10 kg × (3 m/s)²) = -45 J.

Correct Answer: B — 90 J

Q. A 10 kg object is moving with a speed of 3 m/s. What is its kinetic energy?

A. 45 J
B. 30 J
C. 60 J
D. 90 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 10 kg × (3 m/s)² = 45 J.

Correct Answer: A — 45 J

Q. A 10 kg object is moving with a speed of 5 m/s. What is its total mechanical energy?

A. 125 J
B. 250 J
C. 500 J
D. 1000 J

Solution

Total mechanical energy = KE + PE. KE = 0.5 * 10 * (5)² = 125 J. Assuming PE = 0, total energy = 125 J.

Correct Answer: B — 250 J

Q. A 10 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?

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

Solution

Kinetic energy = 0.5 × m × v² = 0.5 × 10 kg × (3 m/s)² = 45 J.

Correct Answer: C — 45 J

Q. A 10 kg object is moving with a velocity of 3 m/s. What is the total mechanical energy if it is at a height of 5 m?

A. 150 J
B. 180 J
C. 300 J
D. 450 J

Solution

Total Mechanical Energy = Kinetic Energy + Potential Energy = 0.5 × 10 kg × (3 m/s)² + 10 kg × 9.8 m/s² × 5 m = 45 J + 490 J = 535 J.

Correct Answer: C — 300 J

Q. A 10 kg object is moving with a velocity of 4 m/s. What is its momentum?

A. 40 kg m/s
B. 20 kg m/s
C. 10 kg m/s
D. 30 kg m/s

Solution

Momentum (p) = m * v = 10 kg * 4 m/s = 40 kg m/s

Correct Answer: A — 40 kg m/s

Q. A 10 kg object is thrown 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. 22.5 m
C. 15.3 m
D. 10.0 m

Solution

Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (15)²/(2*9.8) = 11.5 m.

Correct Answer: A — 11.5 m

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