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
Show solution
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
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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
Show solution
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
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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
Show solution
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
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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
Show solution
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
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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
Show solution
Solution
Work done = Force × Distance = 10 N × 3 m = 30 J.
Correct Answer:
C
— 30 J
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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
Show solution
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 10 kg × (3 m/s)² = 45 J.
Correct Answer:
C
— 45 J
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Q. A 2 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
Show solution
Solution
Using energy conservation: mgh = 0.5mv², h = v²/(2g) = (10 m/s)²/(2 × 9.8 m/s²) = 5.1 m.
Correct Answer:
A
— 5 m
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Q. A 2 kg object is dropped from a height of 10 m. What is the speed of the object just before it hits the ground? (g = 9.8 m/s²)
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
30 m/s
Show solution
Solution
Using energy conservation, mgh = 0.5mv²; v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 10 m) = 14 m/s.
Correct Answer:
B
— 14 m/s
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Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the top? (g = 9.8 m/s²)
A.
19.6 J
B.
39.2 J
C.
49 J
D.
98 J
Show solution
Solution
Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer:
B
— 39.2 J
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Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the top?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
Show solution
Solution
Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer:
B
— 20 J
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Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the height?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
Show solution
Solution
Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer:
B
— 20 J
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Q. A 2 kg object is dropped from a height of 5 m. What is the work done by gravity on the object just before it hits the ground?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
Show solution
Solution
Work done by gravity = m × g × h = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer:
B
— 20 J
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Q. A 2 kg object is moving with a speed of 3 m/s. What is its kinetic energy?
A.
6 J
B.
9 J
C.
12 J
D.
18 J
Show solution
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 2 kg × (3 m/s)² = 9 J.
Correct Answer:
B
— 9 J
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Q. A 2 kg object is thrown 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
Show solution
Solution
Using energy conservation: KE_initial = PE_max; 0.5 × 2 kg × (10 m/s)² = 2 kg × 9.8 m/s² × h; h = 5.1 m.
Correct Answer:
A
— 5 m
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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
Show solution
Solution
Potential energy = mgh = 3 kg × 9.8 m/s² × 12 m = 352.8 J.
Correct Answer:
D
— 120 J
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Q. A 3 kg object is dropped from a height of 15 m. What is the potential energy at the top?
A.
30 J
B.
45 J
C.
60 J
D.
75 J
Show solution
Solution
Potential energy = mass × g × height = 3 kg × 9.8 m/s² × 15 m = 441 J.
Correct Answer:
D
— 75 J
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Q. A 3 kg object is dropped from a height of 5 m. What is the potential energy at the height?
A.
15 J
B.
30 J
C.
45 J
D.
60 J
Show solution
Solution
Potential energy = mass × g × height = 3 kg × 9.8 m/s² × 5 m = 147 J.
Correct Answer:
B
— 30 J
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Q. A 3 kg object is moving with a speed of 4 m/s. If a net work of 24 J is done on the object, what will be its final speed?
A.
4 m/s
B.
6 m/s
C.
8 m/s
D.
10 m/s
Show solution
Solution
Initial kinetic energy = 0.5 × m × v^2 = 0.5 × 3 kg × (4 m/s)^2 = 24 J. Final kinetic energy = Initial + Work done = 24 J + 24 J = 48 J. Final speed = √(2 × KE/m) = √(2 × 48 J / 3 kg) = 6.93 m/s.
Correct Answer:
C
— 8 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. If the object starts from rest, what is its final speed? (Assume no friction)
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 12 N × 4 m = 48 J. Kinetic energy = 0.5 × mass × v²; 48 J = 0.5 × 3 kg × v²; v² = 32; v = 4 m/s.
Correct Answer:
C
— 4 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 net work done if friction does 8 J of work?
A.
28 J
B.
32 J
C.
36 J
D.
40 J
Show solution
Solution
Net work done = Work done by force - Work done against friction = (12 N × 4 m) - 8 J = 48 J - 8 J = 40 J.
Correct Answer:
B
— 32 J
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Q. A 3 kg object is pushed with a force of 15 N over a distance of 4 m. If the object experiences a frictional force of 3 N, what is the net work done on the object?
A.
48 J
B.
60 J
C.
72 J
D.
84 J
Show solution
Solution
Net force = Applied force - Friction = 15 N - 3 N = 12 N. Work done = Net force × Distance = 12 N × 4 m = 48 J.
Correct Answer:
B
— 60 J
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Q. A 3 kg object is pushed with a force of 15 N over a distance of 4 m. What is the work done on the object?
A.
30 J
B.
45 J
C.
60 J
D.
75 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer:
C
— 60 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the change in gravitational potential energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
Show solution
Solution
Change in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
Show solution
Solution
Increase in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
B
— 24 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy? (g = 9.8 m/s²)
A.
117.6 J
B.
117 J
C.
120 J
D.
150 J
Show solution
Solution
Potential energy increase = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer:
A
— 117.6 J
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Q. A 4 kg object is moving with a speed of 5 m/s. If it comes to rest, what is the work done by friction?
A.
50 J
B.
75 J
C.
100 J
D.
125 J
Show solution
Solution
Work done = change in kinetic energy = 0 - 0.5 × 4 kg × (5 m/s)² = -50 J.
Correct Answer:
C
— 100 J
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Q. A 4 kg object is moving with a speed of 5 m/s. What is its kinetic energy?
A.
10 J
B.
20 J
C.
50 J
D.
100 J
Show solution
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 4 kg × (5 m/s)² = 50 J.
Correct Answer:
C
— 50 J
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Q. A 4 kg object is moving with a speed of 5 m/s. What is the total mechanical energy if it is at a height of 2 m?
A.
50 J
B.
60 J
C.
70 J
D.
80 J
Show solution
Solution
Total mechanical energy = Kinetic energy + Potential energy = 0.5 × 4 kg × (5 m/s)² + 4 kg × 9.8 m/s² × 2 m = 50 J + 78.4 J = 128.4 J.
Correct Answer:
C
— 70 J
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Q. A 4 kg object is pushed with a force of 20 N over a distance of 3 m. If the object starts from rest, what is its final speed?
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 20 N × 3 m = 60 J. K.E = 0.5 × m × v², 60 J = 0.5 × 4 kg × v², v = 3.87 m/s.
Correct Answer:
C
— 4 m/s
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Q. A 4 kg object is pushed with a force of 20 N over a distance of 3 m. If the object starts from rest, what is its final speed? (Assume no friction)
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 20 N × 3 m = 60 J. Kinetic energy = 0.5 × m × v². 60 J = 0.5 × 4 kg × v². v² = 30, v = √30 ≈ 5.48 m/s.
Correct Answer:
C
— 4 m/s
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Showing 1 to 30 of 56 (2 Pages)
Work Energy Theorem MCQ & Objective Questions
The Work Energy Theorem is a fundamental concept in physics that plays a crucial role in various examinations. Understanding this theorem not only enhances your conceptual clarity but also significantly boosts your performance in exams. Practicing MCQs and objective questions related to the Work Energy Theorem helps you identify important questions and solidify your understanding, making it easier to score better in your school and competitive exams.
What You Will Practise Here
Definition and explanation of the Work Energy Theorem
Key formulas related to work and energy
Understanding kinetic and potential energy
Applications of the Work Energy Theorem in real-life scenarios
Diagrams illustrating work done by forces
Problem-solving techniques for objective questions
Common misconceptions and clarifications
Exam Relevance
The Work Energy Theorem is frequently tested in CBSE, State Boards, NEET, and JEE examinations. Students can expect questions that require them to apply the theorem to solve numerical problems or conceptual questions. Common patterns include direct application of formulas, conceptual understanding of energy transformations, and graphical interpretations. Mastering this topic is essential for achieving high marks in physics.
Common Mistakes Students Make
Confusing work done with energy transfer
Neglecting the direction of forces when calculating work
Misunderstanding the relationship between kinetic and potential energy
Overlooking the significance of units in calculations
Failing to apply the theorem in multi-step problems
FAQs
Question: What is the Work Energy Theorem?Answer: The Work Energy Theorem states that the work done on an object is equal to the change in its kinetic energy.
Question: How can I apply the Work Energy Theorem in exams?Answer: You can apply the theorem by identifying the forces acting on an object and calculating the work done to find changes in energy.
Question: Are there any specific formulas I need to remember?Answer: Yes, key formulas include W = ΔKE (Work done equals change in kinetic energy) and the relationships between kinetic and potential energy.
Now is the time to enhance your understanding of the Work Energy Theorem! Dive into our practice MCQs and test your knowledge to ensure you are well-prepared for your upcoming exams. Your success starts with practice!
