Chapter Exam

Chapter-5 Work Energy and Power — Online MCQ Test

Q: Work done by a constant force is defined as the product of force, displacement, and the cosine of the angle between them. What is the SI unit of work?

Answer: B. Joule — The SI unit of work is Joule (J), which equals Newton-meter (N·m).

Q: If a force of 10 N is applied on an object and it moves 5 m in the direction of the force, what is the work done?

Answer: C. 50 J — Work = Force × Displacement × cos(θ) = 10 × 5 × cos(0°) = 50 J.

Q: What is the kinetic energy of an object with mass 2 kg moving with a velocity of 3 m/s?

Answer: B. 9 J — Kinetic energy KE = (1/2)mv² = (1/2) × 2 × 3² = 9 J.

Q: Which of the following is an example of a conservative force?

Answer: C. Gravitational force — Gravitational force is conservative as it depends only on position and the work done is path-independent.

Q: A ball is thrown vertically upward with initial velocity 20 m/s. Using the work-energy theorem, find the maximum height reached (g = 10 m/s²).

Answer: B. 20 m — Using work-energy theorem: -mgh = 0 - (1/2)mv²; h = v²/(2g) = 400/(2×10) = 20 m.

Q: A block of mass 5 kg is pushed across a frictionless surface for 10 m by a force of 50 N applied at 60° to the horizontal. What is the work done?

Answer: A. 250 J — Work = F·d·cos(θ) = 50 × 10 × cos(60°) = 50 × 10 × 0.5 = 250 J.

Q: Two objects of mass m₁ and m₂ (m₁ > m₂) are moving with the same kinetic energy. Which has greater momentum?

Answer: A. m₁ — Since KE = p²/(2m), for equal KE, the heavier object has greater momentum: p = √(2m·KE).

Q: A variable force F = 2x N acts on an object along the x-axis. Calculate the work done as the object moves from x = 0 to x = 5 m.

Answer: B. 25 J — Work = ∫F·dx = ∫₀⁵ 2x dx = [x²]₀⁵ = 25 J.

Q: A force F = 3x² N acts on an object from x = 1 m to x = 3 m. Calculate the work done by this variable force.

Answer: C. 26 J — Work = ∫₁³ 3x² dx = [x³]₁³ = 27 - 1 = 26 J.

Q: A block of mass 5 kg is pulled up a rough incline (μ = 0.3, θ = 37°) through distance 10 m. The applied force is 50 N parallel to the incline. Calculate the net work done considering all forces.

Answer: A. 82 J — W_net = W_applied - W_gravity - W_friction = 50(10) - 5(10)sin(37°)(10) - 0.3·5(10)cos(37°)(10) = 500 - 300 - 120 = 82 J.

PHYSICS · CLASS 11 FIRST PUC · Karnataka State Board

Practice Chapter-5 Work Energy and Power with a free chapter-wise online MCQ test for Karnataka State Board CLASS 11 FIRST PUC PHYSICS. This chapter covers: Work done by constant and variable forces - Kinetic energy - Work-energy theorem - Potential energy - Conservative and non-conservative forces - Conservation of mechanical energy -.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter-5 Work Energy and Power — Important Questions & Answers (FAQ)

Frequently asked questions from Karnataka State Board CLASS 11 FIRST PUC PHYSICS — Chapter-5 Work Energy and Power, with answers and explanations. These are sample questions; the exam has its own separate question set.

Work done by a constant force is defined as the product of force, displacement, and the cosine of the angle between them. What is the SI unit of work?

A. Newton
B. Joule ✓
C. Watt
D. Pascal

Answer: B. Joule
The SI unit of work is Joule (J), which equals Newton-meter (N·m).

If a force of 10 N is applied on an object and it moves 5 m in the direction of the force, what is the work done?

A. 2 J
B. 15 J
C. 50 J ✓
D. 0.5 J

Answer: C. 50 J
Work = Force × Displacement × cos(θ) = 10 × 5 × cos(0°) = 50 J.

What is the kinetic energy of an object with mass 2 kg moving with a velocity of 3 m/s?

A. 6 J
B. 9 J ✓
C. 18 J
D. 3 J

Answer: B. 9 J
Kinetic energy KE = (1/2)mv² = (1/2) × 2 × 3² = 9 J.

Which of the following is an example of a conservative force?

A. Friction
B. Air resistance
C. Gravitational force ✓
D. Viscous force

Answer: C. Gravitational force
Gravitational force is conservative as it depends only on position and the work done is path-independent.

A ball is thrown vertically upward with initial velocity 20 m/s. Using the work-energy theorem, find the maximum height reached (g = 10 m/s²).

A. 10 m
B. 20 m ✓
C. 30 m
D. 40 m

Answer: B. 20 m
Using work-energy theorem: -mgh = 0 - (1/2)mv²; h = v²/(2g) = 400/(2×10) = 20 m.

A block of mass 5 kg is pushed across a frictionless surface for 10 m by a force of 50 N applied at 60° to the horizontal. What is the work done?

A. 250 J ✓
B. 500 J
C. 433 J
D. 866 J

Answer: A. 250 J
Work = F·d·cos(θ) = 50 × 10 × cos(60°) = 50 × 10 × 0.5 = 250 J.

Two objects of mass m₁ and m₂ (m₁ > m₂) are moving with the same kinetic energy. Which has greater momentum?

A. m₁ ✓
B. m₂
C. Both have equal momentum
D. Cannot be determined

Answer: A. m₁
Since KE = p²/(2m), for equal KE, the heavier object has greater momentum: p = √(2m·KE).

A variable force F = 2x N acts on an object along the x-axis. Calculate the work done as the object moves from x = 0 to x = 5 m.

A. 10 J
B. 25 J ✓
C. 50 J
D. 100 J

Answer: B. 25 J
Work = ∫F·dx = ∫₀⁵ 2x dx = [x²]₀⁵ = 25 J.

A force F = 3x² N acts on an object from x = 1 m to x = 3 m. Calculate the work done by this variable force.

A. 16 J
B. 24 J
C. 26 J ✓
D. 52 J

Answer: C. 26 J
Work = ∫₁³ 3x² dx = [x³]₁³ = 27 - 1 = 26 J.

A block of mass 5 kg is pulled up a rough incline (μ = 0.3, θ = 37°) through distance 10 m. The applied force is 50 N parallel to the incline. Calculate the net work done considering all forces.