(adsbygoogle = window.adsbygoogle || []).push({}); 1. An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.4 s. A passenger in the elevator is holding a 4.8 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates? The acceleration of gravity is 9.8 m/s^2.

2. Relevant equations

F=ma

3. The attempt at a solution

I did a body diagram and get to the conclusion that T=mg-a. to get the acceleration i first had to get the velocity in the first 1.4s. v=d/t= 1m/1.4s=.714m/s and then get the a=v/t=(.714m/s)/1.4s=.510m/s^2

please correct me if i'm wrong

**Physics Forums - The Fusion of Science and Community**

# Vertical cord tension in an elevator

Have something to add?

- Similar discussions for: Vertical cord tension in an elevator

Loading...

**Physics Forums - The Fusion of Science and Community**