An elevator loaded with people has a mass of 1750 kg. (a) The elevator accelerat

Question

An elevator loaded with people has a mass of 1750 kg. (a) The elevator accelerates upward (in the positive direction) from rest at a rate of 1.95 m/s^2 for 2.3 s. Calculate the tension in the cable supporting the elevator in newtons. T = (b) The elevator continues upward at constant velocity for 0.6 s. What is the tension in the cable, in Newtons. during this time? (c) The elevator experiences a negative acceleration at a rate of 0.65 m/s^2 for 2.3 s. What is the tension in the cable, in Newtons. During this period of negative acceleration? (d) How far, in meters has the elevator moved above its original starting point?

Explanation / Answer

a) Tension is a force and force = mass * acceleration
Let M = mass (1750kg) A = acceleration

In this case, when the elevator is standing still, the cable tension is given by:
A = 9.8 m/s^2 (acceleration due to gravity being counteracted by the cable)
T = 1750*9.8 = 17150 kg*m/s^2 = 17150 newtons

When accelerating upward:
A = 9.8 m/s^2 + 1.95m/s^2 = 11.75m/s^2
T = 1750 * 11.75 = 20562.5 kg*m/s^2 = 20562.5 newtons ANSWER

Distance during acceleration = 1/2 (t^2 x a) = 1/2 (2.3^2 x 1.95) = 5.157 metres.
Velocity attained = (at) = 1.95 x 2.3, = 4.485m/sec.

b) Tension = (1750 x g) = 1750 x 9.8 = 17150 N.

Distance during 8.6s. = (4.485 x 8.5) = 38.12 metres.

c) Acceleration = (g - 0.65) = 9.15m/sec^2.

Tension = (1,750 x 9.15) = 16012N.

d) Change in velocity = (at) = 0.65 x 2.3, = 1.495m/sec^2. So the elevator has stopped.
Distance to stop = 1/2 (t^2 x a) = 1/2 (2.3^2 x 0.65) = 1.719 metres.
Height from start = (5.157 + 38.12 + 1.719) = 44.996 metres.

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