Conservation of Energy: Essential for Competitive Exams

Conservation of Energy

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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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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Q. A 2 kg ball is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches?

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

Solution

Using conservation of energy, KE at the bottom = PE at the top. 0.5mv^2 = mgh. Solving gives h = v^2/(2g) = (15^2)/(2*9.8) = 11.47 m.

Correct Answer: B — 10 m

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Q. A 2 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 conservation of energy, KE at launch = PE at max height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2*9.8) = 20.41 m.

Correct Answer: B — 20 m

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Q. A 2 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. 40.4 m
D. 50.4 m

Solution

Using conservation of energy, KE at the bottom = PE at the maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2 * 9.8) = 20.4 m.

Correct Answer: B — 30.4 m

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Q. A 2 kg object is dropped from a height of 15 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A. 17.15 m/s
B. 12.25 m/s
C. 14.14 m/s
D. 10.0 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 * 15) = 17.15 m/s.

Correct Answer: A — 17.15 m/s

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Q. A 2 kg object is moving with a speed of 3 m/s. What is its total mechanical energy?

A. 9 J
B. 12 J
C. 15 J
D. 18 J

Solution

Total mechanical energy = kinetic energy + potential energy. KE = 0.5mv² = 0.5 * 2 * (3)² = 9 J.

Correct Answer: B — 12 J

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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 object is lifted to a height of 4 m. What is the work done against gravity?

A. 12 J
B. 24 J
C. 36 J
D. 48 J

Solution

Work done = mgh = 3 * 9.8 * 4 = 117.6 J.

Correct Answer: B — 24 J

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

A. 150 J
B. 300 J
C. 450 J
D. 600 J

Solution

Kinetic energy = 0.5mv² = 0.5 * 3 * (10)² = 150 J.

Correct Answer: B — 300 J

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

A. 150 J
B. 300 J
C. 450 J
D. 600 J

Solution

Total mechanical energy = KE + PE. Assuming PE = 0, KE = 0.5mv² = 0.5 * 3 * (10)² = 150 J.

Correct Answer: B — 300 J

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Q. A 3 kg object is moving with a speed of 4 m/s. What is its kinetic energy?

A. 12 J
B. 24 J
C. 36 J
D. 48 J

Solution

Kinetic energy = 0.5mv² = 0.5 * 3 * (4)² = 24 J.

Correct Answer: B — 24 J

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Q. A 3 kg object is moving with a speed of 4 m/s. What is its total mechanical energy?

A. 24 J
B. 32 J
C. 48 J
D. 60 J

Solution

Total mechanical energy = KE + PE. Assuming it is at ground level, PE = 0. KE = 0.5mv² = 0.5 * 3 * (4)² = 24 J.

Correct Answer: B — 32 J

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

A. 15 J
B. 25 J
C. 35 J
D. 45 J

Solution

Kinetic energy is given by KE = 0.5mv² = 0.5 * 3 * (5)² = 37.5 J, approximately 35 J.

Correct Answer: B — 25 J

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Q. A 5 kg object is dropped from a height of 10 m. What is the total mechanical energy just before it hits the ground?

A. 0 J
B. 50 J
C. 100 J
D. 200 J

Solution

Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 10 = 490 J. Just before hitting the ground, total energy = 490 J.

Correct Answer: C — 100 J

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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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Conservation of Energy MCQ & Objective Questions

The concept of Conservation of Energy is fundamental in physics and plays a crucial role in various examinations. Understanding this principle not only helps in grasping core scientific concepts but also enhances your performance in exams. Practicing MCQs and objective questions related to Conservation of Energy is an effective way to solidify your knowledge and prepare for important questions that may appear in your school and competitive exams.

What You Will Practise Here

Definition and significance of Conservation of Energy
Key formulas related to energy conservation
Types of energy: kinetic, potential, and mechanical energy
Real-life applications of energy conservation principles
Energy transformation and its implications
Diagrams illustrating energy conservation in different systems
Sample problems and solutions for better understanding

Exam Relevance

The topic of Conservation of Energy is frequently tested in various examinations, including CBSE, State Boards, NEET, and JEE. Students can expect questions that assess their understanding of energy types, energy transformations, and the application of conservation principles in problem-solving. Common question patterns include numerical problems, theoretical questions, and conceptual applications that require a clear grasp of the topic.

Common Mistakes Students Make

Confusing kinetic energy with potential energy and their respective formulas
Overlooking the importance of units in energy calculations
Failing to apply the conservation principle correctly in complex systems
Neglecting to consider energy losses due to friction or air resistance

FAQs

Question: What is the principle of Conservation of Energy?
Answer: The principle states that energy cannot be created or destroyed, only transformed from one form to another.

Question: How can I effectively prepare for Conservation of Energy questions?
Answer: Regular practice of MCQs and understanding key concepts will help you grasp the topic better and perform well in exams.

Start solving practice MCQs on Conservation of Energy today to test your understanding and boost your confidence for upcoming exams. Mastering this topic will not only help you score better but also deepen your appreciation for the laws of physics!

Conservation of Energy

Conservation of Energy MCQ & Objective Questions

What You Will Practise Here

Exam Relevance

Common Mistakes Students Make

FAQs

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