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. 49 N
• B. 24 N
• C. 74 N
• D. 15 N

Answer 1

The correct option is B 24 N

As we know that the spring measures the tension in the string attached to it. So

When lift was stationary

T = mg

Reading of balance = 49 N = T

$\Rightarrow mg=49N\Rightarrow m=5kg$

Now lift has an acceleration.
Applying F.B.D
$\Rightarrow$ T = mg - ma
= 49 - 5(5)
= 49 - 25 = 24 N.