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An elevator packed with people has a mass of 1800 kg.
A) The elevator accelerates upward (in the positive direction) from rest at a rate of 1.5 m/s2 for 2.2 s. Calculate the tension in the cable supporting the elevator in newtons.
B) The elevator continues upward at constant velocity for 9 s. What is the tension in the cable during this time in newtons?
C) The elevator experiences a negative acceleration at a rate of 0.65 m/s2 for 2 s. What is the tension in the cable, in Newtons, during this period of negative accleration?
D) How far has the elevator moved above its original starting point in meters?

Answer :

erolkayacan

Answer:

a) 20.34 N

b)17.64 N

c)16.47 N

d)38.63m

Explanation:

a) F=m*a. Earth gravity is 9.8m/s2. If the elevator goes upwards, tension on the rope will be higher. To find the tension we need to all accelerates. Total accelerates effects the elevator is a=9.8+1.5=11.3 m/s2. Then;

[tex]F=1.8*11.3=20.34[/tex]

the tension is 20.34N

b) There is no accelerate in this situation therefore the tension is:

F=1.8*9.8=17.64 N

c) In this situation elevator goes down. We need to subtract gravity from elevator acceleration. a=9.8-0.65=9.15m/s2

F=1.8*9.15=16.47 N

d) Total distance is:

[tex]x=0.5*a*t^2+v*t+v*t-0.5*a*t^2\\=0.5*1.5*2.2^2+9*(1.5*2.2)+2*(1.5*2.2)-0.5*0.65*2^2\\=38.63 m[/tex]

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