Question

an elevator starts from rest with a constant upward acceleration . It moves 2m in the first 0.6s . A passenger in the elevator is holding a 3kg package by a vertical string . What is the tension in the string during acceleration .(9.8 m/s2) NO LINKS THE ANSWER IS GIVEN AS 62.7N

Answer 1

Let the acceleration of the elevator be a.

As it travel 2m in the first 0.6s

=>by the formula s=0×t+(1/2)at²
=>2=(1/2)a×0.6²=a=2×2/0.36=11.11m/s².

Then tension in string T is given by T-mg=ma

=>T=m(g+a)=3(11.11+9.8)
=3×20.98
=62.7N Ans.....

Answer 2

Answer:

62.7N is the answer to this question i.e the tension in the string