An elevator loaded with tourists has a mass of 1750 kg.a.) The elevator accelerates upward (in the positive direction) from rest at a rate of 1.55 m/s2 for 2.3 s. Calculate the tension in the cable supporting the elevator in newtonsb.) The elevator continues upward at constant velocity for 8.5 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/s2 for 2.6 s. What is the tension in the cable, in Newtons, during this period of negative accleration?d.) How far, in meters, has the elevator moved above its original starting point?

Question
Asked Oct 8, 2019

An elevator loaded with tourists has a mass of 1750 kg.

a.) The elevator accelerates upward (in the positive direction) from rest at a rate of 1.55 m/s2 for 2.3 s. Calculate the tension in the cable supporting the elevator in newtons

b.) The elevator continues upward at constant velocity for 8.5 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/s2 for 2.6 s. What is the tension in the cable, in Newtons, during this period of negative accleration?

d.) How far, in meters, has the elevator moved above its original starting point?

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Step 1

Note- Since we are authorized to answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts you’d like answered.

Step 2

Given,

m 1750 Kg
a 1.55 m/s2
= 2.3 sec
= 8.5 sec
a 0.65 m/s
t = 2.6 sec
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m 1750 Kg a 1.55 m/s2 = 2.3 sec = 8.5 sec a 0.65 m/s t = 2.6 sec

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Step 3

Fig (a)  Upward acceleration so FN > FG

Fig (b)  Constant Velocity so FNet=0 which means FN = FG

Fig...

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