An elevator starts from rest with a constant
upward acceleration and moves 1m in the first
1.5 s. A passenger in the elevator is holding a
9.6 kg bundle at the end of a vertical cord.
The acceleration of gravity is 9.8 m/s2 .
What is the tension in the cord as the ele-
Answer in units of N.
From the equation of the motion we can get the acceleration of the elevator:h = x0+ v0t + at2/2; v0=0, x0=0;thusa = 2h/t2 = 2*1/1.52 = 0.9 m/s2. Then we can find the tension in the cord: T = mg + ma = 9.6 (9.8 + 0.9) = 102.7 N Answer: 102.7 N.