Question

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the balance 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 would be

• A. $24\text{N}$
• B. $74\text{N}$
• C. $15\text{N}$
• D. $49\text{N}$

Answer 1

The correct option is A $24\text{N}$

When the lift is stationary, the spring force $F_s$ in spring balance will be equal to the weight of bag.

$\therefore F_s=W=49\text{N}$

or, $mg=49\Rightarrow m\left(9.8\right)=49$

$\therefore m=\frac{49}{9.8}=5\text{kg}$

When lift is accelerating downwards, the acceleration of bag will be same as that of lift.

Applying Newton's $2^{nd}$ law of motion,

$mg-F_s^{\prime }=ma$

$49-F_s^{\prime }=5\times 5$

$\therefore F_s^{\prime }=49-25=24\text{N}$

Thus, new reading of spring balance will be $24\text{N}$.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: Spring balance measures the}\\ \text{weight of a body based on the spring}\\ \text{force recorded by it.}\end{array}$