Question

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads $49\text{N}$, when the lift is stationary. If the lift moves downward with an acceleration of $5{\text{m/s}}^2$, the reading of the spring balance ( in $\text{N}$) will be

Answer 1

When the lift is stationary spring force balances weight.
$kx=mg=49N$
Here $k$ is force constant of spring, $x$ is elongation, True weight is $49N$
$\Rightarrow k=\frac{49}{x}\to \left(1\right)$
and $m=\frac{49}{9.8}=5\text{kg}$ (taking $g=9.8{\text{m/s}}^2)$
When lift moves with acceleration $5{\text{m/s}}^2$ downward we have:
$k{x}_2=mg-5\times m$
Because Pseudo force in lift frame is $5m$ upward, $x_2$ = new elongation)
$\Rightarrow k{x}_2=49-5\times 5=24\text{N}$
So new reading in spring balance is $24\text{N}$