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jeka57 [31]
3 years ago
7

A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.

Physics
1 answer:
jeka943 years ago
6 0
  • 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 × 1.5) N = 855 N
  • Angle between distance and force (θ) = 0°
  • We know, work done = F s Cos θ
  • Therefore, work done by the cable during this part
  • = (855 × 13 × Cos 0°) J
  • = (855 × 13 × 1) J
  • = 11115 J

<u>Answer</u><u>:</u>

<u>1</u><u>1</u><u>1</u><u>1</u><u>5</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

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