Question

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.

Answer 1

• 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

Answer:

11115 J

Hope you could get an idea from here.

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