An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.6 s. A passenger in the elevator is holding a 4.2 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates?

Answer: 44.4 N

Explanation:

from the question we are given the following

distance (s) = 1 m

tine (t) = 1.6 s

mass = 4.2 kg

acceleration due to gravity (g) = 9.8 m/s

acceleration of the bundle (a) can be gotten from the formula s = ut + (1/2) at^2

where s = distance, u = initial velocity (which is 0 in this case because the elevator started from rest), a = acceleration and t = time

1 = (0 x 1.6) + (1/2) x a x 1.6^2

1 = 1.28 x a

a = 0.78 m/s^2

to get the tension we apply the formula tension = (m x a) + (m x g)

tension = (4.2 x 0.78) + (4.2 x 9.8)

tension = 44.4 N