Question

An elevator of mass 4000 kg is accelerating upwards. If the tension in the supporting cable is 48000N, find the acceleration of the elevator. Starting from rest, through what distance does it rise in 3s Take g=$10m{s}^{-2}$.

Answer 1

$T=mg+ma$
$T=m(g+a)$
$48000=4000(g+a)$
$12=g+a=10+a$
$\Rightarrow$ a = $2m{s}^{-2}$
$S=vt+\frac{1}{2}{at}^{2}S=0+\frac{1}{2}\times 2\times (3{)}^2=9m$