Question

A body of mass $10kg$ is hung by a spring-balance in a lift. What woul be the reading of the balance when (i) the lift is ascending with an acceleration of $2m/s^2$, (ii) descending with the same acceleration, (iii)descending with a constant velocity of $2m/s$?$(g=10m/s^2)$

Answer 1

Given that,

Mass of body $m=10kg$

Acceleration $a=2m/s^2$

(I). The lift is ascending with an acceleration

According to figure

$T-mg=ma$

$T=ma+mg$

$\frac{T}{g}=m+\frac{ma}{g}$

$\frac{T}{g}=10+\frac{10\times 2}{10}$

$\frac{T}{g}=12kg$

(II). the descending with the same acceleration

$mg-T=ma$

$T=mg-ma$

$\frac{T}{g}=m-\frac{ma}{g}$

$\frac{T}{g}=10-\frac{10\times 2}{10}$

$\frac{T}{g}=10-2$

$\frac{T}{g}=8kg$

(III). the descending with a constant velocity

$T-mg=0$

$T=mg$

$\frac{T}{g}=10kg$

This is the answer