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 A 60 HP electric motor lifts an elevator with a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to (Given 1 HP = 746 W, g = 10 m/s2

  1. a. 1.5 m/s
  2. b. 1.7 m/s
  3. c. 2.0 m/s
  4. d. 1.9 m/s
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Correct option is d) 1.9 ms

P = Fv + Mgv

Applied power = 4000 × v + 20000 v

60 × 246 = 4000 v + 20000 v

v = 1.865

v = 1.9 m/s Rounding it off.
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