Question

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring balance reads $49\text{ N}$, 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 will be ($g = 9.8\text{ m/s}^2$)

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

Hint:

When an elevator moves downward, you feel lighter because gravity is partially offset by the downward acceleration. You can write this ratio directly to avoid finding mass: $\frac{W_{new}}{W_{old}} = \frac{g - a}{g} = \frac{9.8 - 5}{9.8} = \frac{4.8}{9.8} \approx \frac{1}{2}$. Half of $49\text{ N}$ is roughly $24.5\text{ N}$, pointing instantly to $24\text{ N}$.

Answer 1

The Correct Option is C

Step 1: Find the mass.
At rest the balance reads the true weight $W = mg = 49\,\text{N}$, so with $g = 9.8\,\text{m/s}^2$, \[ m = \frac{49}{9.8} = 5\,\text{kg}. \]
Step 2: Draw the forces.
The bag feels the spring tension $T$ upward and weight $mg$ downward.
Step 3: Apply Newton's second law for downward acceleration.
With the lift accelerating down at $a$, the bag also accelerates down: \[ mg - T = ma. \]
Step 4: Solve for the reading $T$.
\[ T = m(g - a). \]
Step 5: Substitute the values.
With $a = 5\,\text{m/s}^2$, \[ T = 5(9.8 - 5) = 5\times4.8. \]
Step 6: Compute.
\[ T = 24\,\text{N}. \] The apparent weight drops because the lift accelerates downward, option (C). \[ \boxed{T = 24\,\text{N}} \]