An elevator filled with tourists has a mass of 1650 kg. The elevator accelerates

Question

An elevator filled with tourists has a mass of 1650 kg. The elevator accelerates upward (in the positive direction) from rest at a rate of 1.9m/s^2 for 2.25 s. Calculate the tension in the cable supporting the elevator in newtons. The elevator continues upward at constant velocity for 8.6 s. What is the tension in the cable, in Newtons, during this time? The elevator experiences a negative acceleration at a rate of 0.35 m/s^2 for 22s. What is the tension in the cable, in Newtons, during this period of negative acceleration? How far, in meters, has the elevator moved above its original starting point?

Explanation / Answer

a) Total acceleration = (g + 1.9) = 11.7m/sec^2. (9.8m/sec^2 used as g).
Tension = (1,650 x 11.7) = 19305N.

Distance during acceleration = 1/2 (t^2 x a) = 1/2 (2.25^2 x 1.9) = 4.80 metres.
Velocity attained = (at) = 1.9 x 2.25, = 4.75m/sec.

b) Tension = (1650 x g) = 1650 x 9.8 = 16170N.

Distance during 8.6s. = (4.75 x 8.6) = 40.85 metres.

c) Acceleration = (g - 0.35) = 9.45m/sec^2. Tension = (1,650 x 9.45) = 15,592.5N.

d) Change in velocity = (at) = 0.35 x 2.2, = 0.77m/sec^2. So the elevator has stopped.
Distance to stop = 1/2 (t^2 x a) = 1/2 (2.2^2 x 0.35) = 0.847 metres.
Height from start = (0.847 + 40.85 + 4.75) = 46.44 metres.

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