Question

An elevator packed with passengers has a mass of 1950 kg.

(a) The elevator accelerates upward (in the positive direction) from rest at a rate of 2 m/s2 for 2.05 s. Calculate the tension in the cable supporting the elevator in newtons.
(b) The elevator continues upward at constant velocity for 8 s. What is the tension in the cable, in Newtons, during this time?
(c) The elevator experiences a negative acceleration at a rate of 0.45 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?

Answer 1

Step-by-step explanation:

Below is an attachment of the solution.

Answer 2

a) Fa= Tension in the cable= 23010N

b) 19110N

c) 18232.5N

d=41.28m ,final velocity=0

Step-by-step explanation:

a) Newtons 2nd law is given by Fnet=EF=ma

Fa= m(a + g)

Fa= 1950 (2+9.8)

Fa= 1950×11.8= 23010N

b) Fnet=0

Therefore Fb= W= 1950×9.8= 19110N

c) Fc= m(g - a)

Fc= 1950(9.8 - 0.45)

Fc= 18232.5N

d) First distance

ya= vt + 0.5at^2 = 0 + (0.5)(2)(2.05)^2

ya= 4.203m

yb= vt= 4.203×8=33.62m

yc = vt - (0.5at^2)

yc= 4.203×2.6) - (0.5×0.45×8^2)

yc = 10.93-14.4

yc =-3.46m

Dtotal = -3.46+33.62+4.203

Dtotal=41.28m