A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring…
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring balance reads $49 \mathrm{~N}$, when the lift is stationary. If the lift moves downward with an acceleration of $5 \mathrm{~m} / \mathrm{s}^2$, the reading of the spring balance will be $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right)$
$15 \mathrm{~N}$
$24 \mathrm{~N}$
$49 \mathrm{~N}$
$74 \mathrm{~N}$
Solution
For downward accelration of the lift,
$\mathrm{Mg}\mathrm{T}=\mathrm{ma}$
$\therefore \quad \mathrm{T}=\mathrm{m}(\mathrm{g}-\mathrm{a})$ ....(i)
$\mathrm{W}=\mathrm{mg}$
$\therefore \quad \mathrm{m}=\frac{\mathrm{W}}{\mathrm{g}}=\frac{49}{9.8}$ ....(ii)
Put (ii) into (i)
$\mathrm{T}=\frac{49}{9.8}(9.8-5)$
$=5(4.8)=24 \mathrm{~N}$