Q. A 10 kg object is dropped from a height of 5 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)
A.
10 m/s
B.
15 m/s
C.
20 m/s
D.
25 m/s
Show solution
Solution
Using energy conservation, Potential Energy = Kinetic Energy at the ground. mgh = (1/2)mv². v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 5 m) ≈ 10 m/s.
Correct Answer:
B
— 15 m/s
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Q. A 10 kg object is lifted to a height of 5 m. What is the potential energy gained? (Take g = 9.8 m/s²) (2021)
A.
490 J
B.
980 J
C.
390 J
D.
590 J
Show solution
Solution
Potential Energy (PE) = mgh = 10 kg × 9.8 m/s² × 5 m = 490 J.
Correct Answer:
B
— 980 J
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Q. A 10 kg object is moving with a speed of 2 m/s. What is its momentum?
A.
20 kg·m/s
B.
10 kg·m/s
C.
5 kg·m/s
D.
15 kg·m/s
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Solution
Momentum (p) = mass × velocity = 10 kg × 2 m/s = 20 kg·m/s.
Correct Answer:
A
— 20 kg·m/s
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Q. A 10 kg object is moving with a speed of 5 m/s. What is its momentum? (2022)
A.
50 kg m/s
B.
10 kg m/s
C.
25 kg m/s
D.
15 kg m/s
Show solution
Solution
Momentum = mass × velocity = 10 kg × 5 m/s = 50 kg m/s.
Correct Answer:
A
— 50 kg m/s
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Q. A 2 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)
A.
14 m/s
B.
20 m/s
C.
10 m/s
D.
15 m/s
Show solution
Solution
Using energy conservation, PE = KE: mgh = 0.5mv². Solving gives 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 dropped from a height of 10 m. What is its potential energy at the top? (Take g = 9.8 m/s²) (2020)
A.
49 J
B.
98 J
C.
39 J
D.
59 J
Show solution
Solution
Potential Energy (PE) = mgh = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer:
B
— 98 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? (Use g = 9.8 m/s²) (2021)
A.
14 m/s
B.
20 m/s
C.
10 m/s
D.
15 m/s
Show solution
Solution
Using energy conservation, Potential Energy = Kinetic Energy. mgh = 1/2 mv². v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s.
Correct Answer:
A
— 14 m/s
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Q. A force of 10 N is applied to move a box 5 m across a frictionless surface. What is the work done on the box? (2021)
A.
10 J
B.
20 J
C.
50 J
D.
5 J
Show solution
Solution
Work done (W) = Force (F) × Distance (d) = 10 N × 5 m = 50 J.
Correct Answer:
B
— 20 J
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Q. A force of 10 N is applied to move an object 5 m in the direction of the force. What is the work done on the object?
A.
10 J
B.
20 J
C.
50 J
D.
5 J
Show solution
Solution
Work done (W) = Force (F) × Distance (d) × cos(θ). Here, θ = 0° (force and displacement are in the same direction). W = 10 N × 5 m × cos(0°) = 10 N × 5 m = 50 J.
Correct Answer:
B
— 20 J
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Q. A force of 10 N is applied to move an object 5 m in the direction of the force. What is the work done? (2021)
A.
10 J
B.
20 J
C.
50 J
D.
5 J
Show solution
Solution
Work done = Force × Distance = 10 N × 5 m = 50 J
Correct Answer:
B
— 20 J
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Q. A force of 15 N is applied at an angle of 30° to the horizontal. What is the horizontal component of the force? (2020)
A.
7.5 N
B.
12.99 N
C.
15 N
D.
10 N
Show solution
Solution
Horizontal component = F × cos(θ) = 15 N × cos(30°) = 15 N × √3/2 ≈ 12.99 N.
Correct Answer:
B
— 12.99 N
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Q. A force of 15 N is applied at an angle of 60° to the horizontal while moving an object 4 m. What is the work done by the force?
A.
30 J
B.
60 J
C.
45 J
D.
20 J
Show solution
Solution
Work done (W) = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.
Correct Answer:
C
— 45 J
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Q. A force of 50 N is applied at an angle of 30° to the horizontal. What is the horizontal component of the force? (2020)
A.
25 N
B.
43.3 N
C.
50 N
D.
35 N
Show solution
Solution
Horizontal component = Fcos(θ) = 50 N × cos(30°) = 50 N × (√3/2) ≈ 43.3 N.
Correct Answer:
B
— 43.3 N
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Q. A force of 50 N is applied at an angle of 60° to the horizontal. What is the horizontal component of the force? (2020)
A.
25 N
B.
50 N
C.
43.3 N
D.
30 N
Show solution
Solution
Horizontal component = F × cos(θ) = 50 N × cos(60°) = 50 N × 0.5 = 25 N.
Correct Answer:
C
— 43.3 N
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Q. An object is moving with a velocity of 10 m/s. What is its kinetic energy if its mass is 2 kg? (2022)
A.
10 J
B.
20 J
C.
30 J
D.
40 J
Show solution
Solution
Kinetic Energy = (1/2)mv² = (1/2) × 2 kg × (10 m/s)² = 100 J
Correct Answer:
B
— 20 J
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Q. An object of mass 10 kg is moving with a velocity of 3 m/s. What is its kinetic energy? (2022)
A.
45 J
B.
30 J
C.
60 J
D.
15 J
Show solution
Solution
Kinetic Energy (KE) = 0.5 × m × v² = 0.5 × 10 kg × (3 m/s)² = 45 J.
Correct Answer:
B
— 30 J
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Q. An object of mass 4 kg is moving with a velocity of 3 m/s. What is its kinetic energy? (2022)
A.
18 J
B.
24 J
C.
36 J
D.
12 J
Show solution
Solution
Kinetic Energy (KE) = 1/2 mv² = 1/2 × 4 kg × (3 m/s)² = 18 J.
Correct Answer:
B
— 24 J
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Q. If a 10 kg object is lifted to a height of 5 m, what is the work done against gravity? (Use g = 9.8 m/s²) (2021)
A.
49 J
B.
98 J
C.
490 J
D.
245 J
Show solution
Solution
Work done = mgh = 10 kg × 9.8 m/s² × 5 m = 490 J.
Correct Answer:
C
— 490 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (Take g = 9.8 m/s²) (2020)
A.
58.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
Show solution
Solution
Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.
Correct Answer:
A
— 58.8 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained by the object? (Use g = 9.8 m/s²)
A.
58.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
Show solution
Solution
Potential Energy (PE) = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.
Correct Answer:
A
— 58.8 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the potential energy gained? (Use g = 9.8 m/s²) (2020)
A.
58.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
Show solution
Solution
Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 3 m = 58.8 J
Correct Answer:
A
— 58.8 J
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Q. If a 2 kg object is lifted to a height of 3 m, what is the work done against gravity? (Use g = 9.8 m/s²) (2023)
A.
58.8 J
B.
39.2 J
C.
19.6 J
D.
29.4 J
Show solution
Solution
Work done = mgh = 2 kg × 9.8 m/s² × 3 m = 58.8 J.
Correct Answer:
A
— 58.8 J
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Q. If a 5 kg object is lifted to a height of 2 m, what is the potential energy gained by the object? (Use g = 9.8 m/s²) (2020)
A.
49 J
B.
98 J
C.
19.6 J
D.
39.2 J
Show solution
Solution
Potential Energy = mgh = 5 kg × 9.8 m/s² × 2 m = 98 J
Correct Answer:
B
— 98 J
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Q. If a car does 2000 J of work in moving a distance of 50 m, what is the average force exerted by the car? (2000)
A.
40 N
B.
50 N
C.
60 N
D.
30 N
Show solution
Solution
Average Force (F) = Work done (W) / Distance (d) = 2000 J / 50 m = 40 N.
Correct Answer:
A
— 40 N
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Q. If a car does 2000 J of work in moving a distance of 50 m, what is the average force exerted? (2023)
A.
40 N
B.
50 N
C.
60 N
D.
30 N
Show solution
Solution
Average Force = Work / Distance = 2000 J / 50 m = 40 N
Correct Answer:
A
— 40 N
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Q. If a car does 3000 J of work in moving a distance of 150 m, what is the average force exerted by the car? (2023)
A.
20 N
B.
30 N
C.
40 N
D.
50 N
Show solution
Solution
Average Force = Work / Distance = 3000 J / 150 m = 20 N
Correct Answer:
B
— 30 N
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Q. If a car does 600 J of work in moving a distance of 30 m, what is the average force exerted by the car? (2023)
A.
20 N
B.
15 N
C.
25 N
D.
10 N
Show solution
Solution
Average Force (F) = Work (W) / Distance (d) = 600 J / 30 m = 20 N.
Correct Answer:
A
— 20 N
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Q. If a car does 6000 J of work in moving a distance of 100 m, what is the average force exerted by the car? (2023)
A.
60 N
B.
50 N
C.
70 N
D.
80 N
Show solution
Solution
Average Force = Work / Distance = 6000 J / 100 m = 60 N
Correct Answer:
A
— 60 N
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Q. If a force of 15 N is applied at an angle of 60° to the horizontal and moves an object 4 m, what is the work done? (2021)
A.
30 J
B.
60 J
C.
45 J
D.
20 J
Show solution
Solution
Work done (W) = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.
Correct Answer:
C
— 45 J
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Q. If a machine has an efficiency of 80% and it does 400 J of useful work, how much energy was supplied to the machine? (2023)
A.
500 J
B.
400 J
C.
320 J
D.
600 J
Show solution
Solution
Efficiency = Useful Work / Total Energy Input. Total Energy Input = Useful Work / Efficiency = 400 J / 0.8 = 500 J.
Correct Answer:
A
— 500 J
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Showing 1 to 30 of 35 (2 Pages)
Force & Energy MCQ & Objective Questions
Understanding the concepts of Force and Energy is crucial for students preparing for school exams and competitive tests. These topics not only form a significant part of the curriculum but also appear frequently in objective questions and MCQs. By practicing Force & Energy MCQ questions, students can enhance their grasp of the subject, leading to better scores in exams.
What You Will Practise Here
Fundamental concepts of Force and Energy
Newton's Laws of Motion and their applications
Types of energy: kinetic, potential, and mechanical energy
Work-energy theorem and its implications
Conservation of energy principles
Important formulas related to Force and Energy
Diagrams illustrating force interactions and energy transformations
Exam Relevance
The topics of Force and Energy are integral to various educational boards, including CBSE and State Boards, as well as competitive exams like NEET and JEE. Students can expect questions that test their understanding of fundamental principles, numerical problems, and conceptual applications. Common question patterns include direct MCQs, assertion-reason type questions, and numerical problems that require the application of formulas.
Common Mistakes Students Make
Confusing the concepts of mass and weight
Misapplying the work-energy theorem in numerical problems
Overlooking the direction of forces in vector problems
Neglecting units while calculating energy and work
FAQs
Question: What are the key formulas related to Force and Energy?Answer: Key formulas include F = ma (Force = mass × acceleration), W = F × d (Work = Force × distance), and KE = 1/2 mv² (Kinetic Energy = 1/2 × mass × velocity squared).
Question: How can I improve my performance in Force and Energy MCQs?Answer: Regular practice of objective questions, understanding the underlying concepts, and reviewing mistakes can significantly improve your performance.
Start solving Force & Energy practice questions today to strengthen your understanding and boost your confidence for upcoming exams. Remember, consistent practice is the key to success!
