Question

An elevator starts from rest with a constant
upward acceleration and moves 1m in the first
1.9 s. A passenger in the elevator is holding a
5.3 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 elevator accelerates?
Answer in units of N.

Answer 1

Use the fact that it goes 1 m in 1.9 s to get the acceleration, a.

(1/20 a t^2 = (1/2)a*(1s)^2 = 1.7 m
a = 3.4 m/s^2

The tension T in the cord minus the weight equals ma.

T = m (g + a)

m = 5.3 kg and g = 9.8 m/s^2

Do the numbers.

Answer 2

69.96