An elevator packed with passengers has a mass of 1950 kg. (a) The elevator accel

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? 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?

Explanation / Answer

a) T = m*(g+a)

= 1950*(9.8+2)

= 23010 N

b) T = m*g

= 1950*9.8

= 19110 N

c) T = m*(g - a)

= 1950*(9.8 - 0.45)

= 18232.5 N

d) d = d1 + d2 + d3

= 0.5*2*2.05^2 + (2*2.05)*8 + 2*2.05*2.6 - 0.5*0.45*2.6^2

= 46.1 m

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