Question

An elevator filled with passengers has a mass of 1583 kg.

(a) The elevator accelerates upward from rest at a rate of 1.20 m/s² for 2.00 s. Calculate the tension in the cable (in N) supporting the elevator.

(b)The elevator continues upward at constant velocity for 9.22 s. What is the tension in the cable (in N) during this time?

(c)The elevator decelerates at a rate of 0.600 m/s² for 4.00 s. What is the tension in the cable (in N) during deceleration?

(d)How high has the elevator moved above its original starting point, and what is its final velocity? (Enter the height in m and the final velocity in m/s.)

Height _______ m

Final Velocity ______ m/s

Answer 1

(a)

The equation for the tension in the wire ,

T - mg = m*a

T = m(g + a) = 1583(9.8 +1.20)

T = 17413 N

(b)

calculating tension in the wire

T - mg = m*a

T = m(g+a) = 1583(9.8+0)

T = 15513.4 N

(c)

T - mg = m*a

T = m(g+a) = 1583*(9.8 - 0.600)

T = 14563.6 N

(d)

s = u*t + 1/2*a*t²

first for the accelerating part

s1 = 0 + 1/2*1.20*(2)²

s1 = 2.4 m

velocity after acceleration ,

v = u+a*t = 0+1.20*2

v = 2.40 m/s

s2 = 2.40*9.22 + 0 =

s2 = 22.12 m

velocity after deceleration ,

v = u+a*t = 2.40 + (-0.60)*4

v = 0 m/s

distance covered during deceleration,

s3 = 2.4*4 + 1/2*(-0.60)*(4)²

s3 = 4.8 m

Total diatance covered by the elevator,

s = s1+ s2 + s3 = 2.4 + 22.12 + 4.8

s = 29.32 m

Final velocity is v =0 m/s .