Question

An elevator starts from rest with constant upward acceleration. It moves 2m in the first 0.6 second. A passenger in the elevator is holding a 3kg package by a vertical string find the tension in the string during acceleration

Answer 1

Answer:

Answer. Let the acceleration of the elevator be a. =>2=(1/2)a×0.6²=a=2×2/0.36=11.11m/s²

Answer 2

Given:-

u = 0

M = 3 kg

first D = 2m

t = 0.6 s

To Find:-

Tension = ?

Solution:-

W = mg = 3 kg × 9.8 m/s^2 = 29.4 N

3rd equation of motion

$\sf s = ut \frac{1}{2} a {t}^{2} \\ \\ \sf \ \sf 2 = (0)(0.6) + \frac{1}{2} a {(0.6)}^{2} \\ \\ \sf a = 2 \times \frac{2}{0.36} \\ \\ \sf a = 11.1 \frac{m}{ {s}^{2} }$

Now, using T - W = ma

T = ma + mg = 3(11.1 + 9.8) kg m/s^2

T = 62.7 N