MCQs / Physics / Work and Energy
Work and Energy MCQs — Physics (page 3) 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.
51. A 0.5 kg body falls from rest through 20 m and reaches 15 m/s. The work done against air resistance is (g = 9.8): 20.9 J 98 J 41.8 J 83.5 J Answer: C) 41.8 J
Explanation: Loss of PE = mgh = 98 J; gain of KE = 56.2 J; work against friction = 41.8 J. SLO P-11-B-32 · Application
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52. Which of these is a conservative force? gravitational force viscous drag air resistance friction Answer: A) gravitational force
Explanation: Work done by gravity depends only on height change. SLO P-11-B-31 · Understanding
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53. Air resistance on a falling body causes its total mechanical energy to: become infinite increase stay constant decrease Answer: D) decrease
Explanation: Air drag is non-conservative. SLO P-11-B-31 · Understanding
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54. If the speed of a car is doubled, its kinetic energy becomes: 8 times as large half 4 times as large 2 times as large Answer: C) 4 times as large
Explanation: K propto v^2. SLO P-11-B-29 · Application
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55. A constant force of 300 N acts through 0.2 m along its direction. On a force–displacement graph, the work done (area of the rectangle) is: 60 J 30 J 1500 J 120 J Answer: A) 60 J
Explanation: W = Fd = 60 J. SLO P-11-B-30 · Application
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56. A 60 kg body moves at 5 m/s. Its kinetic energy is: 1500 J 375 J 750 J 150 J Answer: C) 750 J
Explanation: K = 12mv^2 = 12(60)(5)^2 = 750 J. SLO P-11-B-29 · Application
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57. A body sliding up a rough slope loses kinetic energy to: gravitational PE only gravitational PE and work against friction friction only nothing Answer: B) gravitational PE and work against friction
Explanation: Both take energy. SLO P-11-B-32 · Application
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58. When only conservative forces act, the total mechanical energy: increases becomes zero decreases is conserved Answer: D) is conserved
Explanation: KE + PE is constant. SLO P-11-B-31 · Understanding
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59. A force of 10 N acts for the first 2 m, then 20 N for the next 3 m. The total work done is: 50 J 30 J 80 J 100 J Answer: C) 80 J
Explanation: 10(2) + 20(3) = 20 + 60 = 80 J. SLO P-11-B-30 · Application
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60. A force of 25 N at 60^circ to the direction of motion moves a box 4 m. The work done is: 86.6 J 50 J 100 J 25 J Answer: B) 50 J
Explanation: W = Fdcostheta = 25 4 0.5 = 50 J. SLO P-11-B-30 · Application
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