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 N, when the lift is stationary. If the lift moves downward with an acceleration of $5m/s^2$, the reading of the spring balance will be:

• A. 24 N
• B. 17 N
• C. 15 N
• D. 49 N

Answer 1

Answer: (A)

24 N

At rest,

$F=mg=49N$

$m=\frac{49}{9.8}$

$m=\frac{49}{98}\times {10}^{-1}=5kg$

$\therefore$When elevator is accelerating downwards with accleration $a=5m/s^2$

$\therefore F={ma}^1=m(g-a)=m(9.8-5)$

$\therefore F=5\times 4.8$

$\therefore F=24N$