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A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.
Part A
How much work is done during this part of its motion by the cable that lowering the elevator? Neglect any friction or air resistance.

Respuesta :

  • Mass of the elevator (m) = 570 Kg
  • Acceleration = 1.5 m/s^2
  • Distance (s) = 13 m
  • Let the force be F.
  • We know, F = ma,
  • Therefore, F = (570 Γ— 1.5) N = 855 N
  • Angle between distance and force (ΞΈ) = 0Β°
  • We know, work done = F s Cos ΞΈ
  • Therefore, work done by the cable during this part
  • = (855 Γ— 13 Γ— Cos 0Β°) J
  • = (855 Γ— 13 Γ— 1) J
  • = 11115 J

Answer:

11115 J

Hope you could get an idea from here.

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