1. A 2 kg body moves at 3 m/s. Its kinetic energy is:
- 4.5 J
- 3 J
- 18 J
- 9 J
Answer: D) 9 J
Explanation: K = 12mv^2 = 12(2)(3)^2 = 9 J. SLO P-11-B-29 · Application
MCQs / Physics / Work and Energy
Work and Energy MCQs — Physics
60 multiple-choice questions on Work and Energy from Physics, each with the correct answer and, where available, a teacher's explanation. Read through them to revise the chapter, then practise the same questions one by one to test yourself.
2. The work done by a variable force is best found from:
- its maximum value × total displacement
- the gradient
- the area under its force–displacement graph
- its minimum value × total displacement
Answer: C) the area under its force–displacement graph
Explanation: The force changes, so integrate (find the area). SLO P-11-B-30 · Application
3. The unit of the area under a force–displacement graph is:
- joule
- N/m
- newton
- watt
Answer: A) joule
Explanation: N × m = J. SLO P-11-B-30 · Application
4. The elastic force of an ideal spring is:
- frictional
- dissipative
- non-conservative
- conservative
Answer: D) conservative
Explanation: Energy stored is fully recoverable. SLO P-11-B-31 · Understanding
5. A horizontal force–displacement graph (constant force) gives an area shaped like:
- a parabola
- a circle
- a triangle
- a rectangle
Answer: D) a rectangle
Explanation: W = Fd. SLO P-11-B-30 · Application
6. The derivation of kinetic energy assumes that the force is:
- varying randomly
- perpendicular to the motion
- constant and acts along the motion
- zero
Answer: C) constant and acts along the motion
Explanation: So the acceleration is uniform. SLO P-11-B-29 · Application
7. A 2 kg body falls 10 m from rest against a constant air resistance of 4.6 N. Its speed on reaching the ground is (g = 9.8):
- 12.2 m/s
- 6.12 m/s
- 24.5 m/s
- 14 m/s
Answer: A) 12.2 m/s
Explanation: mgh - fh = 12mv^2 Rightarrow 196 - 46 = 150 J; v = 12.2 m/s. SLO P-11-B-32 · Application
8. A force rises uniformly from 0 to 30 N over 2 m and then stays at 30 N for another 2 m. The work done is:
- 30 J
- 120 J
- 90 J
- 60 J
Answer: C) 90 J
Explanation: Triangle 30 J + rectangle 60 J = 90 J. SLO P-11-B-30 · Application
9. A 2 kg body falls from rest through 10 m and reaches 12 m/s. The work done against air resistance is (g = 9.8):
- 196 J
- 52 J
- 26 J
- 104 J
Answer: B) 52 J
Explanation: Loss of PE = mgh = 196 J; gain of KE = 144 J; work against friction = 52 J. SLO P-11-B-32 · Application
10. A ball is thrown up and returns to the same point (no air resistance). The total work done by gravity is:
- 2mgh
- mgh
- negative
- zero
Answer: D) zero
Explanation: Up: -mgh; down: +mgh. SLO P-11-B-31 · Understanding
11. A car brakes to a stop. Its kinetic energy is converted by a non-conservative force into:
- gravitational potential energy
- heat in the brakes and tyres
- momentum
- elastic energy
Answer: B) heat in the brakes and tyres
Explanation: Friction dissipates the energy. SLO P-11-B-31 · Understanding
12. A force increases uniformly from 0 to 40 N over a displacement of 0.5 m. The work done is:
- 10 J
- 11 J
- 20 J
- 5 J
Answer: A) 10 J
Explanation: Area of the triangle = 12(40)(0.5) = 10 J. SLO P-11-B-30 · Application
13. A force increases uniformly from 0 to 100 N over a displacement of 2 m. The work done is:
- 105 J
- 100 J
- 50 J
- 200 J
Answer: B) 100 J
Explanation: Area of the triangle = 12(100)(2) = 100 J. SLO P-11-B-30 · Application
14. A 4 kg block moving at 5 m/s slides 5 m along a rough floor against a friction force of 6 N. Its final kinetic energy is:
- 50 J
- 30 J
- 80 J
- 20 J
Answer: D) 20 J
Explanation: Work–energy theorem: Kf = Ki - fd = 50 - 30 = 20 J. SLO P-11-B-32 · Application
15. The unit of kinetic energy, kg m^2, s^-2, is the same as:
- the pascal
- the joule
- the watt
- the newton
Answer: B) the joule
Explanation: 1 J = 1 N m = 1 kg m^2 s^-2. SLO P-11-B-29 · Application
16. A 1000 kg body moving at 20 m/s is brought to rest by a constant resistive force in 50 m. The resistive force is:
- 4000 N
- 400 N
- 8000 N
- 2000 N
Answer: A) 4000 N
Explanation: Work done by friction = loss in KE: fd = 12mu^2 Rightarrow f = 4000 N. SLO P-11-B-32 · Application
17. Which statement about non-conservative forces is true?
- they store potential energy
- work done by them depends on the path and mechanical energy is not conserved
- they are path-independent
- they never do work
Answer: B) work done by them depends on the path and mechanical energy is not conserved
Explanation: e.g. friction, drag, tension in a pushing hand. SLO P-11-B-31 · Understanding
18. A constant force of 12 N acts through 2.5 m along its direction. On a force–displacement graph, the work done (area of the rectangle) is:
- 15 J
- 4.8 J
- 60 J
- 30 J
Answer: D) 30 J
Explanation: W = Fd = 30 J. SLO P-11-B-30 · Application
19. For a body falling through air, mgh = 12mv^2 + fh. The term fh is:
- the gain in PE
- the work done against air resistance
- the final KE
- the momentum
Answer: B) the work done against air resistance
Explanation: Energy lost to the medium. SLO P-11-B-32 · Application
20. On a force–displacement graph, the work done is given by:
- the gradient of the graph
- the area under the graph
- the length of the line
- the y-intercept
Answer: B) the area under the graph
Explanation: W = sum FDelta d. SLO P-11-B-30 · Application
21. If a force is applied at 60^circ to the displacement, the force–displacement graph should use:
- the component Fcos 60^circ along the displacement
- the full force F
- Fsin 60^circ
- zero
Answer: A) the component Fcos 60^circ along the displacement
Explanation: Only the component along the motion does work. SLO P-11-B-30 · Application
22. A constant force of 50 N acts through 4 m along its direction. On a force–displacement graph, the work done (area of the rectangle) is:
- 100 J
- 12.5 J
- 200 J
- 400 J
Answer: C) 200 J
Explanation: W = Fd = 200 J. SLO P-11-B-30 · Application
23. In a resistive medium, the total mechanical energy of a moving body:
- decreases
- remains constant
- doubles
- increases
Answer: A) decreases
Explanation: Resistive forces convert it to heat. SLO P-11-B-32 · Application
24. A cyclist freewheels down a hill at constant speed. The loss of gravitational PE is:
- stored in the bicycle
- equal to the work done against resistive forces
- zero
- converted to kinetic energy
Answer: B) equal to the work done against resistive forces
Explanation: KE doesn't change. SLO P-11-B-32 · Application
25. Work done W = Fd = ma v^22a simplifies to:
- 12mv^2
- mav^2
- 12mv
- mv^2
Answer: A) 12mv^2
Explanation: The a cancels. SLO P-11-B-29 · Application
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