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2 years ago

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

Physics

1 answer:

jeka942 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

Answer:

11115 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

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