Force & Energy Study Materials for Competitive Exams

Force & Energy

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

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

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

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

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

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

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

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

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? (2021)

A. 10 J
B. 20 J
C. 50 J
D. 5 J

Solution

Work done = Force × Distance = 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

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 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

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

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

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

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

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

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

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

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? (Use g = 9.8 m/s²) (2020)

A. 58.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 J

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 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

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? (Take g = 9.8 m/s²) (2020)

A. 58.8 J
B. 19.6 J
C. 29.4 J
D. 39.2 J

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 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

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

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? (2023)

A. 40 N
B. 50 N
C. 60 N
D. 30 N

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 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

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 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

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

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

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

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

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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Force & Energy

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