Question

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads $49~\mathrm{N}$ 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

• A. 49 N
• B. 24 N
• C. 74 N
• D. 15 N

Hint:

In a downward accelerating lift, apparent weight = $m(g - a)$.

Answer 1

The Correct Option is B

To solve this problem, we need to consider the effect of the lift's motion on the apparent weight of the bag as measured by the spring balance.

1. When the lift is stationary, the spring balance reads the true weight of the bag. According to the problem, this is \(49 \, \mathrm{N}\). Thus, the actual weight of the bag, which equals the gravitational force acting on it, is given by: \(W = mg = 49 \, \mathrm{N}\), where \(g = 9.81 \, \mathrm{m/s}^2\) is the acceleration due to gravity.
   • This implies that the mass \(m\) of the bag is \(m = \frac{49}{9.81} \approx 5 \, \mathrm{kg}\).
2. When the lift moves downward with an acceleration of \(a = 5 \, \mathrm{m/s}^2\), the apparent weight (as read by the spring balance) changes due to the lift's acceleration. The apparent weight \(W'\) is given by: \(W' = m(g - a)\).
3. Substituting the known values into the equation:
   • \(W' = 5 (9.81 - 5) = 5 \times 4.81 = 24.05 \, \mathrm{N}\).
4. Thus, the reading of the spring balance when the lift is moving downward with an acceleration of \(5 \, \mathrm{m/s}^2\) is approximately \(24 \, \mathrm{N}\).

The correct answer is therefore 24 N.